Solution manual of modern semiconductor devices for integrated circuits Chenming Calvin Hu

SOLUTION MANUAL

Chapter 1 Visualization of the Silicon Crystal 1.1 (a) Please refer to Figure 1-2. The 8 corner atoms are shared by 8 unit cells and therefore contribute 1 atom. Similarly, the 6 face atoms are each shared by 2 unit cells and contribute 3 atoms. And, 4 atoms are located inside the unit cell.

Hence, there are total 8 silicon atoms in each unit cell. (b) The volume of the unit cell is





Vunit cell  5.43 A  5.43  10 8 cm  1.60  10  22 cm3 , 3

3

and one unit cell contains 8 silicon atoms. The atomic density of silicon is N Si 

8 silicon atoms  5.00  10 22 (silicon atoms) cm  3 .

Vunit cell

Hence, there are 5.001022 silicon atoms in one cubic centimeter. (c) In order to find the density of silicon, we need to calculate how heavy an individual silicon atom is

Mass1 Si atom 

28.1  g/mole   4.67  10 23  g/atom  .

23 6.02  10 atoms/mole 

Therefore, the density of silicon ( Si ) in g/cm3 is ρ Si  N Si  Mass1 Si atom  2.33 g / cm 3 .

Fermi Function 1.2 (a) Assume E = E f in Equation (1.7.1), f(E) becomes ½. Hence, the probability is ½.

(b) Set E = E c + kT and E f = E c in Equation (1.7.1):

f(E) 

1

1 e

 Ec  kT  Ec /kT 

1  0.27 .

1  e1

The probability of finding electrons in states at E c + kT is 0.27.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

* For Problem 1.2 Part (b), we cannot use approximations such as Equations (1.7.2) or (1.7.3) since E-E f is neither much larger than kT nor much smaller than -kT. (c) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a state being empty at E. Using Equation (1.7.1), we can rewrite the problem as f(E  E c  kT)  1  f(E  E c  3kT) 1 1 E c  kT  E f /kT  1  Ec  3kT  E f /kT 1 e 1 e

where E  3 kT  E /kT

E  3kT  E /kT

f f 1 e c 1 e c 1 E  3 kT  E f /kT  E  3kT  E f /kT E  3kT  E f /kT  1 e c 1 e c 1 e c 1 .   E c  3 kT  E f /kT 1 e

1

Now, the equation becomes 1

1 e

E c  kT  E f /kT 

1 e



1



 E c  3 kT  E f /kT

.

This is true if and only if

Ec  kT  E f  Ec  3kT  E f  .

Solving the equation above, we find E f  Ec  2kT .

1.3 (a) Assume E = E f and T > 0K in Equation (1.7.1). f(E) becomes ½. Hence, the probability is ½.

(b) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a state being empty at E. Using Equation (1.7.1), we can rewrite the problem as f(E  Ec )  1  f(E  Ev ) 1 1 Ec  E f /kT  1  Ev  E f /kT 1 e 1 e where

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

E  E /kT

E  E /kT

1 e v f 1 e v f 1 1   . Ev  E f /kT Ev  E f /kT  Ec  E f /kT  E v  E f /kT 1 e 1 e 1 e 1 e 1

Now, the equation becomes 1

1 e

Ec  E f /kT 

1 e



1



 E v  E f /kT

.

This is true if and only if Ec  E f  Ev  E f  .

Solving the equation above, we find Ef 

Ec  Ev .

2

(c) The plot of the Fermi-Dirac distribution and the Maxwell-Boltzmann distribution is shown below.

Probability

1 0.9

Fermi-Dirac Distribution

0.8 0.7 0.6

Maxwell-Boltzmann Distribution

0.5 0.4 0.3 0.2 0.1 0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

(E-E f )/kT

The Boltzmann distribution considerably overestimates the Fermi distribution for small (E-E f )/kT. If we set (E-E f )/kT = A in Equations (1.7.1) and (1.7.2), we have

 1  e  A  1.10 .

A 1  e  Solving for A, we find eA 

1  eA 1.10

 e A  10.11 

A  ln 10.11  2.31 .

Therefore, the Boltzmann approximation is accurate to within 10% for (E-E f )/kT  2.31.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.4 (a) Please refer to the example in Sec. 1.7.2. The ratio of the nitrogen concentration at 10 km above sea level to the nitrogen concentration at sea level is given by E /kT N ( N 2 )10 km e 10 km (E E )/kT   ESea Level /kT  e 10 km Sea Level N ( N 2 ) Sea Level e

where

E10 km  E Sea Level  altitude  mass of N 2 molecule  acceleration of gravity  10 6 cm  28  1.66  10 24 g  980 cm  s 2  4.56  10 14 erg.

The ratio is N ( N 2 )10 km N ( N 2 ) Sea Level

 e ( 4 .5610

14

erg)/( 1.3810 16 erg  K 1 273 K)

 e 1.21  0.30 .

Since nitrogen is lighter than oxygen, the potential energy difference for nitrogen is smaller, and consequently the exponential term for nitrogen is larger than 0.25 for oxygen. Therefore, the nitrogen concentration at 10 km is more than 25% of the sea level N 2 concentration. (b) We know that N (O2 )10 km N (O2 ) Sea Level

 0.25 ,

N ( N 2 )10 km N ( N 2 ) Sea Level

 0.30 , and

N ( N 2 ) Sea Level N (O2 ) Sea Level

 4.

Then, N ( N 2 )10 km N (O2 )10 km



N ( N 2 )10 km N ( N 2 ) Sea Level

 0.30  4 



N ( N 2 ) Sea Level N (O2 ) Sea Level



N (O2 ) Sea Level N (O2 )10 km

1  4.8 .

0.25

It is more N 2 -rich than at sea level.

1.5

1  f E f  E   1 

1

E f  E  E f /kT

1 e E  E  E f /kT e f  E  E  E f /kT 1 e f 1   E f  E  E f /kT 1 e

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 

1 e



1



 E f  E  E f /kT

1

E f  E  E f /kT

1 e  f E f  E 

1.6 (a) 150K

1.0

f(E) 0.5

300K 0.0

Ef

E (b) At 0K, the probability of a state below the Fermi level being filled is 1 and a state above the Fermi level being filled is 0. So a total of 7 states are filled which means there are 14 electrons (since 2 electrons can occupy each state) in the system.

Density of States 1.7 Since the semiconductor is assumed to be, We are asked to use Equations (1.7.2) and (1.7.4) to approximate the Fermi distribution. (This means that the doping concentration is low and E f is not within a few kTs from E c or E v . A lightly doped semiconductor is known as a non-degenerate semiconductor.) The carrier distribution as a function of energy in the conduction band is proportional to Distribution (E)  E  Ec  e





1 / 2  E  E f /kT

,

where e-(E-Ef)/kT is from Equation (1.7.2). Taking the derivative with respect to E and setting it to zero, we obtain

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.





d E  Ec 1/ 2 e  E  E f /kT  1 1 e  E  E f /kT  E  Ec   1 e  E  E f /kT  0 2 E  Ec dE  kT 

The exponential terms cancel out. Solving the remaining equation yields 1 E  Ec 1/ 2  1 E  Ec 1/ 2 kT 2

E  Ec   kT



2

 E  Ec 

kT .

2

So, the number of carriers in the conduction band peaks at E c +kT/2.

Similarly, in the valence band, the carrier distribution as a function of energy is proportional to Distribution (E)  Ev  E  e





1 / 2  E f  E /kT

,

where e-(Ef-E)/kT is Equation (1.7.2). Taking the derivative and setting it to zero, we obtain





d Ev  E 1/ 2 e E f  E /kT  1 Ev  E 1/ 2 e  E f  E /kT  Ev  E 1/ 2  1 e E f  E /kT  0 . dE 2  kT  Again, the exponential terms cancel out, and solving the remaining equation yields 1 Ev  E 1/ 2  1 Ev  E 1/ 2 kT 2



Ev  E   kT 2

 E  Ev 

kT .

2

Therefore, the number of carriers in the valence band peaks at E v -kT/2.

1.8 Since it is given that the semiconductor is non-degenerate (not heavily doped), E f is not within a few kTs from E c or E v . We can use Equations (1.7.2) and (1.7.4) to approximate the Fermi-Dirac distribution. (a) The electron concentration in the conduction band is given by

n

C.B.



Dc(E) f(E) dE   A E  Ec e Ec





 E  E f /kT

dE .

In order to simplify the integration, we make the following substitutions:

E  Ec  x  E  kT x  Ec , kT

1 dE  dx  dE  kT dx, and x : from 0 to  . kT

Now the equation becomes

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.



n   A kTx e





 kTx  E c  E f / kT

0

kT dx  A kT 

3/ 2

e





 E c  E f / kT





x e  x dx

0

where





0

3





x e  x dx    x 2 e  x dx  3 / 2   1

2

0

. (Gamma function)

Hence, the electron concentration in the conduction band is

n

 2

A kT 

3/ 2

e





 E c  E f / kT

.

Similarly, the hole concentration is given by

p   Dv(E)1-f(E)dE   B Ev  E e Ev





 E f  E /kT

dE .

-

V.B.

Again, we make the following substitutions to simplify the integration:

Ev  E 1  x  E   kT x  Ev ,  dE  dx  dE  kT dx, and x : from  to 0 . kT kT Now the equation becomes 0

p    B kTx e





 E f  kTx  E v / kT



kT dx  B kT 

3/ 2

e





 E f  E v / kT





0

x e  x dx

where





0

3



x e  x dx    x 2 e  x dx  3 / 2   1

0

 2

. (Gamma function)

Therefore, the hole concentration in the conduction band is

p

 2

B kT 

3/ 2

e





 E f  E v / kT

.

(b) The word “Intrinsic” implies that the electron concentration and the hole concentration are equal. Therefore,

n p

 2

A kT 

3/ 2

e   Ec  Ei  / kT 

 2

B kT 

3/ 2

e   Ei  Ev  / kT .

This simplifies to

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

A e   E c  Ei  / kT  B e   Ei  Ev  / kT .

Solving for E i yields

Ec  Ev kT  1  Ec  Ev  ln    0.009 eV ; k  8.62  10 5 eV K 1 , T  300 K .

2 2  2 2

Ei 

Hence, the intrinsic Fermi level (E i ) is located at 0.009 eV below the mid-bandgap of the semiconductor.

1.9 The unit step functions set the integration limits. D c (E) is zero for E < E c , and D v (E) is zero for E > E v . Since it is given that the semiconductor is non-degenerate (not heavily doped), E f is not within a few kTs from E c or E v . We can use Equations (1.7.2) and (1.7.4) to approximate the Fermi-Dirac distribution. (a) The electron concentration in the conduction band is given by 

n

C.B.

Dc(E) f(E) dE   A E  Ec e





 E  E f /kT

Ec

dE .

In order to simplify the integration, we make the following substitutions:

E  Ec  x  E  kT x  Ec , kT

1 dE  dx  dE  kT dx, and x : from 0 to  . kT

Now the equation becomes 

n   A kTx e





 kTx  E c  E f / kT

0

kT dx  A kT  e 2





 E c  E f / kT



 x e dx x

0

where 

 x e dx  1 . x

0

Hence, the electron concentration in the conduction band is

n  A kT  e 2





 E c  E f / kT

.

Similarly, the hole concentration is given by

p   Dv(E)1-f(E)dE   B  Ev  E e Ev

V.B.

-





 E f  E /kT

dE .

Again, we make the following substitutions to simplify the integration:

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ev  E 1  x  E   kT x  Ev ,  dE  dx  dE  kT dx, and x : from  to 0 . kT kT Now the equation becomes 0

p    B kTx e





 E f  kTx  E v / kT



kT dx  B kT  e 2





 E f  E v / kT



 x e dx x

0

where 

 x e dx  1 . x

0

Therefore, the hole concentration in the conduction band is

p  B kT  e 2





 E f  E v / kT

.

(b) The word “Intrinsic” implies that the electron concentration and the hole concentration are equal. Therefore,

n  p  A kT  e   E c  Ei  / kT  B kT  e   Ei  E v  / kT .

2

2

This simplifies to

A e   Ec  Ei  / kT  B e   Ei  Ev  / kT .

If we solve for E i , we obtain

Ei 

Ec  Ev kT  1  Ec  Ev  ln    0.009 eV ; k  8.62  10 5 eV K 1 , T  300 K .

2 2  2 2

Hence, the intrinsic Fermi level (E i ) is located at 0.009 eV below the mid-bandgap of the semiconductor.

1.10

(a) The carrier distribution as a function of energy in the conduction band is proportional to 1 / 2  E  E f /kT Distributi on (E)  E  Ec  e , where e-(E-Ef)/kT is from Equation (1.7.2). Taking the derivative and setting it to zero, we obtain





d E  Ec 1/ 2 e E  E f /kT  1 1 e E  E f /kT  E  Ec   1 e E  E f /kT  0 .

2 E  Ec dE  kT  The exponential terms cancel out. Solving the remaining equation yields

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1 E  Ec 1/ 2  1 E  Ec 1/ 2 kT 2



E  Ec   kT 2

 E  Ec 

kT .

2

Hence, the number of carriers in the conduction band peaks at E c +kT/2. (b) The electron concentration in the conduction band is given by

n

C.B.

Dc(E) f(E) dE  

Top of the Conduction Band

Ec

Dc(E) f(E) dE .

We assume that the function f(E) falls off rapidly such that

 



Top of the Conduction Band Top of the Conduction Band

Dc(E) f(E) dE

 0.

Dc(E) f(E) dE

Ec

Now we may change the upper limit of integration from the Top of the Conduction Band to ∞: 

n   A E  Ec e





 E  E f /kT

Ec

dE .

Also, in order to simplify the integration, we make the following substitutions:

E  Ec  x  E  kT x  Ec , kT

1 dE  dx  dE  kT dx, and x : from 0 to  . kT

The equation becomes 

n   A kTx e





 kTx  E c  E f / kT

0

kT dx  A kT 

3/ 2

e





 E c  E f / kT





0

x e  x dx

where





0

3



x e  x dx    x 2 e  x dx  3 / 2   1

0

 2

. (Gamma function)

Therefore, the electron concentration in the conduction band is

n

 2

A kT 

3/ 2

e





 E c  E f / kT

.

(b) The ratio of the peak electron concentration at E = E c +(1/2)kT to the electron concentration at E = E c +40kT is

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.



 / kT

f n( Ec  40kT ) AEc  40kT  EC  e c   1 / 2  E c  0.5 kT  E f  / kT 1   A E 0 .

5 kT E e   c c n( Ec  kT ) 2 1 / 2  E  40 kT  E f  E c  0.5 kT  E f  / kT  40kT / 0.5kT  e c  (40 / 0.5)e  39.5  5.60  1016 .

1 / 2  E  40 kT  E

The ratio is very small, and this result justifies our assumption in Part (b). (c) The kinetic energy of an electron at E is equal to E-E C . The average kinetic energy of electrons is

K .E . 

sum of the kinetic energy of all electrons  total number of electrons

 E  E D (E) f(E) dE  D (E) f(E) dE E  E A E  E e   dE   .   A E  E e dE  c

 C.B.

C.B.

c

c



 E  E f /kT

Ec

c

c



 E  E f /kT

c

Ec

In order to simplify the integration, we make the following substitutions:

E  Ec  x  E  kT x  Ec , kT

1 dE  dx  dE  kT dx, and x : from 0 to  . kT

Now the equation becomes







3/ 2

 kTx  E c  E f / kT

0 

1/ 2

 kTx  E c  E f / kT

 A kTx  e   A kTx  e



0

kT dx kT dx

A kT 

5/ 2



A kT 

3/ 2

e e









 E c  E f / kT

 E c  E f / kT



 x  e dx  x  e dx 0 

3/ 2 x

1/ 2

x

0

where

3  3/ 2 x 1  x 0 x  e dx  0 x 2 e dx  5 / 2  



5

4

(Gamma functions)

and

 1  x 1/ 2  x 0 x  e dx  0 x 2 e dx  3 / 2  . (Gamma functions) 



3

2

Hence, the average kinetic energy is (3/2)kT.

Electron and Hole Concentrations 1.11 (a) We use Equation (1.8.11) to calculate the hole concentration:

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 

2

n  p  ni2  p  ni2 / n  1010 / 10 5 cm 3  1015 cm 3 . (b) Please refer to Equations (1.9.3a) and (1.9.3b). Since N d -N a >> n i and all the impurities are ionized, n = N d -N a , and p = (n i )2/(N d -N a ). (c) Since the Fermi level is located 0.26 eV above E i and closer to E c , the sample is n-type.

If we assume that E i is located at the mid-bandgap (~ 0.55 eV), then E c -E f = 0.29 eV.

1

Ec Ef Ei

2

3

1: E c -E i =0.55 eV 2: E c -E f =0.29 eV 3: E f -E i =0.26 eV

Ev

Using Equations (1.8.5) and (1.8.11), we find

n  Nc e





 E c  E f / kT

 4.01  1014 cm  3 and p  ni2 / n  2.49  105 cm  3 .

Therefore, the electron concentration is 4.011014 cm-3, and the hole concentration is 2.49105 cm-3. * There is another way to solve this problem:

n  ni e

E f  Ei  / kT

 2.20  1014 cm  3 and p  ni2 / n  4.55  105 cm  3 .

(d) If T = 800 K, there is enough thermal energy to free more electrons from siliconsilicon bonds. Hence, using Equation (1.8.12), we first calculate the intrinsic carrier density n i at 800 K:

ni  N c 800 K N v 800 K  e

 

 E g /  2 kT 

 2.56  1016 cm  3 .

where

 2 mdn kT  N c (T  800 K )  2  2   h

3/ 2

 T   2.8  10     300 K 

3/ 2

19

cm  3  1.22  1020 cm  3

and

 2 mdp kT  N v (T  800 K )  2  2  h 

3/ 2

 T   1.04  10     300 K  19

3/ 2

cm  3  4.53  1019 cm  3 .

Clearly, n i at 800K is much larger than N d -N a (which is equal to n from the previous part). Hence the electron concentration is nn i , and the hole concentration is p=(n i )2/nn i . The semiconductor is intrinsic at 800K, and E f is located very close to the mid-bandgap.

Nearly Intrinsic Semiconductor

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.12 Applying Equation (1.8.11) to this problem yields

ni2 / p  n  2 p  2 p 2  ni2  p 

1 ni  7.07  1012 cm  3 and n  1.41  1013 cm  3 .

2

1.13 (a) B is a group III element. When added to Si (which belongs to Group IV), it acts as an acceptor producing a large number of holes. Hence, this becomes a P-type Si film. (b) At T = 300 K, since the intrinsic carrier density is negligible compared to the dopant concentration, p=N a =41016 cm-3, and n = (n i =1010cm-3)2/p = 2500 cm-3.

At T = 600 K,

ni  N c 600 K N v 600 K  e

 

 E g /  2 kT 

 1.16  1015 cm  3

where

 2 mdn kT  N c (T  600 K )  2  2   h

3/ 2

 T   2.8  10     300 K 

3/ 2

19

cm  3  7.92  1019 cm  3

and

 2 mdp kT  N v (T  600 K )  2  2  h 

3/ 2

 T   1.04  10     300 K  19

3/ 2

cm  3  2.94  1019 cm  3 .

The intrinsic carrier concentration is no more negligible compared to the dopant concentration. Thus, we have





p  N a  ni  4  1016  1.16  1015 cm 3  4.12  1016 cm 3 , and





2

n  ni2 / p  1.16  1015 cm 3 / 4.12  1016 cm 3  3.27  1013 cm3 .

The electron concentration has increased by many orders of magnitude. (c) At high temperatures, there is enough thermal energy to free more electrons from silicon-silicon bonds, and consequently, the number of intrinsic carriers increases. (d) Using Equation 1.8.8, we calculate the position of the Fermi level with respect to E v .

E f  Ev  kT lnN v T  / pT   0.34 eV , T  600 K .

At 600 K, the Fermi level is located 0.34 eV above the valence band.

Incomplete Ionization of Dopants and Freeze-out 1.14 From Equation (1.9.1), we know that n + N a - = p + N d +. Since N d + is much larger than N a -, all the samples are n-type, and n  N d + - N a - = 31015 /cm3. This value is assumed to be constant. Using the Equations (1.8.10) and (1.9.3b),

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.





p  ni2 / N d  N a  N c N v exp E g / kT   CT 3 exp E g / kT  , where C is a temperature independent constant. Using the sensitivity of p defined by p/T,

p / T  3  E g / kT  CT 2 exp E g / kT  Therefore, the larger the energy gap is the less sensitive to temperature the minority carrier is. For the definition of the sensitivity of p,

p / T  / p  3  Eg / kT / T The temperature sensitivity of the minority carrier is greater for larger E g .

1.15 (a) Let us first consider the case of n-type doping. The dopant atoms are located at energy E d inside the bandgap, near the conduction band edge. The problem states that we are considering the situation in which half the impurity atoms are ionized, i.e. n=N d /2. In other words, the probability of dopant atoms being ionized is ½, or conversely, the probability that a state at the donor energy E D is filled is ½.

From Problem 1.2 part (a), we know that if f (E D )=1/2, then E D =E f . From Equation 1.8.5,

n  Nce





 Ec  E f / kT

.

We also know that E f =E D and E c -E D =0.05eV.

 2 mdn kT  N c (T )  2 2   h

3/ 2

3/ 2

 T  3  2.8  10    cm .

K 300   N 2 N c (T )  E  E  / kT N c (T )e c f  N c (T )e   Ec  E D  / kT  D   e Ec  E D  / kT .

Nd 2 19

This equation can be solved iteratively. Starting with an arbitrary guess of 100K for T, we find T converges to 84.4 K. Similarly, for boron

 2 mdp kT  N v (T )  2  2   h

3/ 2

3/ 2

 T  3  1.04  10    cm .

300 K   N 2 N v (T )  E  E  / kT N v (T )e f v  N v (T )e   E a  Ev  / kT  a   e E a  Ev  / kT .

Na 2 19

Starting from T =100K, we find T converges to 67.7K.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) We want to find T where n i is 10N d . This can be written as

ni  N c T N v T  e

 

 E g /  2 kT 

 T   1.71  10     300 K  19

3/ 2

e

 

 E g /  2 kT 

 10 N d

where

 2 mdn kT  N c (T )  2 2   h

3/ 2

 2 mdp kT  N v (T )  2  2   h

3/ 2

 T   2.8  1019     300 K 

3/ 2

 T   1.04  10     300 K 

cm  3 and 3/ 2

19

cm 3 .

We need to solve the equation iteratively, as in part (a) for n i =10N d =1017cm-3. Starting from T=300K, we get T=777 K for n i =10N d .

For n i =10N a , we simply replace N d in the equation above with N a . Starting from T =300K, we find T=635 K. (c) If we assume full ionization of impurities at T = 300 K, For arsenic: n  N d  1016 cm 3  ni , p 

2

ni  2.1  104 cm  3 Nd 2

n For boron: p  N a  10 cm  ni , n  i  2.1  105 cm  3 Na 15

3

(d) Please refer to the example in Section 2.8. For arsenic,

E f  E v  kT ln

N v 1.04  1019 cm 3   0.88 eV . p 2.1  10 4 cm 3

For boron,

E f  Ev  kT ln

N v 1.04  1019 cm 3   0.24 eV . p 1015 cm  3

(e) In case of arsenic + boron,





2

n2 1010 cm 3 n  N d  N a  9  10 cm , and p  i   1.11  10 4 cm  3 , and n 9  1015 cm  3  1.04  1019 cm 3   Nv    0.90 eV .     E f  Ev  kT ln   0.026 eV ln 4 3   1 .

11 10 cm  p    15

3

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.16 (a) If we assume full ionization of impurities, the electron concentration is n  N d = 1017cm-3. The hole concentration is p=(n i )2/n=(1010 cm-3)2/1017cm-3=103 cm-3.

The Fermi level position, with respect to E c , is





Ec  E f  kT lnN c / n   0.026 ln 2.8  1019 cm 3 / 1017 cm 3  0.15 eV .

E f is located 0.15 eV below E c . (b) In order to check the full ionization assumption with the calculated Fermi level, we need to find the percentage of donors occupied by electrons.

ED  E f  Ec  E f   Ec  ED   0.1eV , and nD  N d

1

1 e

E D  E f  / kT 

1017 cm 3  15 3 0.1 eV / 0.026 eV   2.09  10 cm  2% of N d .

1 e

Since only 2% of dopants are not ionized, it is fine to assume that the impurities are fully ionized. (c) We assume full ionization of impurities, the electron concentration is n  N d = 1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019 cm-3 = 10 cm-3.

The Fermi level position, with respect to E c , is





Ec  E f  kT lnN c / n   0.026  ln 2.8  1019 cm 3 / 1019 cm 3  0.027 eV .

It is located 0.027 eV below E c .

Again, we need to find the percentage of donors occupied by electrons in order to check the full ionization assumption with the calculated Fermi level.

E D  E f  E c  E f   E c  E D   0.023 eV , and nD  N d

1

1 e

E D  E f  / kT 

1019 cm 3   7.08  1018 cm  3  71% of N d .

1  e 0.023 eV / 0.026 eV 

Since 71% of dopants are not ionized, the full ionization assumption is not correct. (d) For T=30 K, we need to use Equation (1.10.2) to find the electron concentration since the temperature is extremely low. First, we calculate N c and N v at T=30K:

 2 m dn kT  N c (T  30 K )  2   h2  

3/ 2

 T   2.8  10     300 K 

3/ 2

19

cm 3  8.85  1017 cm 3

and

 2 mdp kT  N v (T  30 K )  2   h2  

3/ 2

 T   1.04  10     300 K  19

3/ 2

cm 3  3.29  1017 cm 3 .

The electron concentration is

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

n

N c 30 K N d  Ec  ED  / 2 kT  e  8.43  10 8 cm 3 .

2

And, the hole concentration is

p  ni2 / n  0 where

ni 

N c 30 K N v 30 K  E g  / 2 kT  e  2.32  10 75 cm 3 .

2

Since n i is extremely small, we can assume that all the electrons are contributed by ionized dopants. Hence,

1 8.43  10 cm   8 3 n  N d 1  E D  E f  / kT   8.43  10 cm  1017 cm  3  1 e  8

3

 8.43  10  9 .

The full ionization assumption is not correct since only 8.4310-7% of N d is ionized.

To locate the Fermi level, 1   n     1  0.048 eV .

ED  E f  kT ln 1    Nd    

E c -E f = 0.05-0.048 = 0.002 eV. Therefore, the Fermi level is positioned 0.002 eV below E c , between E c and E D .

1.17 (a) We assume full ionization of impurities, the electron concentration is n  N d = 1016cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1016cm-3 = 104 cm-3.

The Fermi level position, with respect to E c , is





Ec  E f  kT lnN c / n   0.026 ln 2.8  1019 cm 3 / 1016 cm 3  0.21eV .

It is located 0.21 eV below E c .

We need to find the percentage of donors occupied by electrons in order to check the full ionization assumption with the calculated Fermi level.

ED  E f  Ec  E f   Ec  ED   0.16 eV , and





1016 cm 3 13 3 1 nD  N d E D  E f  / kT  1  e0.16 eV / 0.026 eV   2.12  10 cm  2.12  10 % of N d .

1 e 1

Since only 0.21% of dopants are not ionized, the full ionization assumption is correct.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) We assume full ionization of impurities, the electron concentration is n  N d = 1018cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1018cm-3 = 102 cm-3.

The Fermi level position with respect to E c is





Ec  E f  kT lnN c / n  0.026 ln 2.8  1019 cm 3 / 1018 cm 3  0.087 eV .

It is located 0.087 eV below E c .

We need to find the percentage of donors occupied by electrons in order to check the full ionization assumption with the calculated Fermi level.

ED  E f  Ec  E f   Ec  ED   0.037 eV , and nD  N d

1

1 e

E D  E f  / kT 





1018 cm 3 17 3 0.037 eV / 0.026 eV   1.94  10 cm  19% of N d .

1 e

Since 19% of dopants are not ionized, the full ionization assumption is not accurate but acceptable. (c) We assume full ionization of impurities, the electron concentration is n  N d = 1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019cm-3 = 10 cm-3.

The Fermi level position, with respect to E c , is





Ec  E f  kT lnN c / n  0.026 ln 2.8  1019 cm 3 / 1019 cm 3  0.027 eV .

It is located 0.027 eV below E c .

Again, we need to find the percentage of donors occupied by electrons in order to check the full ionization assumption with the calculated Fermi level.

ED  E f  Ec  E f   Ec  ED   0.023 eV , and nD  N d

1

1 e

E D  E f  / kT 





1019 cm3  7.08  1018 cm 3  71% of N d .

1  e 0.023 eV / 0.026 eV 

Since 71% of dopants are not ionized, the full ionization assumption is not correct.

Since N d is not fully ionized and N d (ionized) << N d (not-ionized),

E  E  / kT  E  E  / kT n  N d 1  f ED   N d e D f  N ce c f .

Solving the equation above for E f yields

Ef 

ED  Ec   kT ln N d  .

2

2

N   c

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 2 Mobility 2.1 (a) The mean free time between collisions using Equation (2.2.4b) is

n 

q mn mn

  mn 

 n mn q

 2.85  10 13 sec

where  n is given to be 500 cm2/Vsec (= 0.05 m2/Vsec), and m n is assumed to be m0. (b) We need to find the drift velocity first: v d   n ε  50000 cm / sec .

The distance traveled by drift between collisions is d  v d  mn  0.14 nm .

2.2 From the thermal velocity example, we know that the approximate thermal velocity of an electron in silicon is vth 

3kT  2.29  107 cm / sec . m

Consequently, the drift velocity (v d ) is (1/10)v th = 2.29106 cm/sec, and the time it takes for an electron to traverse a region of 1 m in width is t

10 4 cm  4.37  10 11 sec .

6 2.29  10 cm / sec

Next, we need to find the mean free time between collisions using Equation (2.2.4b):

n 

q mn mn

  mn 

 n mn q

 2.10  1013 sec

where  n is 1400 cm2/Vsec (=0.14 m2/Vsec, for lightly doped silicon, given in Table 2-1), and m n is 0.26m 0 (given in Table 1-3). So, the average number of collision is

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

t

 mn

 207.7 collision  207 collisions .

In order to find the voltage applied across the region, we need to calculate the electric field using Equation (2.2.3b): vd   n ε 

ε  vd  2.29  10 2 cm / sec  1635.71 Vcm 1 .

6

n

1400 cm / V sec

Then, the voltage across the region is V  ε  width  1635.71 Vcm 1  10 4 cm  0.16V .

2.3 (a) Log[]

10

4

1

10

2

3

2

10 100

200

300

400

500

600

700

400

500

600

700

T (K)

(b) If we combine  1 and  2 , Log[] 10

3

2

10 100

200

300

T (K)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The total mobility at 300 K is   1 1   TOTAL (300 K )     1 (300 K )  2 (300 K ) 

1

 502.55 cm 2 / V sec .

(c) The applied electric field is

ε  V  1V  10V / cm .

1 mm

l

The current density is J ndrift  q n nε  q n N d ε  80.41A / cm2 .

Drift 2.4 (a) From Figure 2-8 on page 45, we find the resistivity of the N-type sample doped with 11016cm-3 of phosphorous is 0.5 -cm.

(b) The acceptor density (boron) exceeds the donor density (P). Hence, the resulting conductivity is P-type, and the net dopant concentration is N net = |N d -N a | = p = 91016cm-3 of holes. However, the mobilities of electrons and holes depend on the total dopant concentration, N T =1.11017cm-3. So, we have to use Equation (2.2.14) to calculate the resistivity. From Figure 2-5,  p (N T =1.11017cm-3) is 250 cm2/Vsec. The resistivity is



1





1 qN net  p



1  0.28  cm . q  9  10 cm  250 cm 2 / V sec 16

3





(c) For the sample in part (a),  Nc   2.8  1019 cm 3    E c  E f  kT ln   0.026V ln 1016 cm 3   0.21 eV .

N    d 0.21 eV Ec Ef Ei Ev

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the sample in part (b),  N   1.04  1019 cm 3    0.12 eV E f  Ev  kT ln v   0.026V ln 16 3   9  10 cm   N net  Ec Ei 0.12 eV Ef Ev

2.5 (a) Sample 1: N-type Holes are minority carriers. p = n i 2/N d = (1010cm-3)2/1017cm-3 = 102 cm-3

Sample 2: P-type Electrons are minority carriers. n = n i 2/N a = (1010cm-3)2/1015cm-3 = 105 cm-3 Sample 3: N-type Holes are minority carriers. p = n i 2/N net = (1010cm-3)2/(9.91017cm-3)  102 cm-3 (b) Sample 1: N d = 1017cm-3  n (N d = 1017cm-3) = 750 cm2/Vsec (from Figure 2-4)  = qN d  n = 12 -1cm-1 Sample 2: N a = 1015cm-3  p (N a = 1015cm-3) = 480 cm2/Vsec (from Figure 2-4)  = qN a  p = 12 -1cm-1 Sample 3: N T = N d +N a = 1.011017cm-3  n (N T = 1.011017cm-3) = 750 cm2/Vsec (from Figure 2-4) N net = N d -N a = 0.991017cm-3  = qN net  n = 11.88 -1cm-1 (c) For Sample 1, N   2.8  1019 cm 3    0.15 eV .

Ec  E f  kT ln c   0.026V ln 17 3 N 10 cm    d 0.15 eV Ec Ef Ei Ev

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For Sample 2, N   1.04  1019 cm 3    0.24 eV .

E f  Ev  kT ln v   0.026V ln 15 3  10 cm   Na  Ec Ei 0.24 eV

Ef Ev

For Sample 3,    2.8  1019 cm 3  Nc   0.026V ln   0.15 eV .

Ec  E f  kT ln 16 3   9.9  10 cm   N net  N d  N a  0.15 eV

Ec Ef Ei Ev

2.6 (a) From Figure 2-5,  n (N d = 1016cm-3 of As) is 1250 cm2/Vs. Using Equation (2.2.14), we find



1





1  0.5  cm . qn n

(b) The mobility of electrons in the sample depends not on the net dopant concentration but on the total dopant concentration N T : NT  N d  N a  2  1016 cm 3 .

From Figure 2-5,

 n N T   1140 cm 2 / Vs and

 p N T   390 cm 2 / Vs .

N net = N d -N a = 0. Hence, we can assume that there are only intrinsic carriers in the sample. Using Equation (2.2.14),

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.



1





1 1  qni  n  qpi  p qni  n   p 



1 . q  1  10 cm  1140  390cm 2 / V sec  10

3

The resistivity is 4.08105 -cm. (c) Now, the total dopant concentration (N T ) is 0. Using the electron and hole mobilities for lightly doped semiconductors (from Table 2.1), we have

 n  1400 cm 2 / V sec and

 p  470 cm 2 / V sec .

Using Equation (2.2.14),



1





1 1  qni  n  qpi  p qni  n   p 



1 . q  1  10 cm  1400  470cm 2 / V sec  10

3

The resistivity is 3.34105 -cm. The resistivity of the doped sample in part (b) is higher due to ionized impurity scattering.

2.7 It is given that the sample is n-type, and the applied electric field  is1000V/cm. The hole velocity  dp is 2105cm/s.

(a) From the velocity and the applied electric field, we can calculate the mobility of holes:

 dp =  p ,  p =  dp / = 2105/1000 = 200cm2/V·s.

From Figure 2-5, we find N d is equal to 4.51017/cm3. Hence, n = N d = 4.51017/cm3, and p = n i 2/n = n i 2/ N d = 1020 / 4.51017 = 222/cm3.

Clearly, the minority carriers are the holes. (b) The Fermi level with respect to E c is E f = E c - kTln(N d /N c ) = E c - 0.107 eV. (c) R = L/A. Using Equation (2.2.14), we first calculate the resistivity of the sample:  = q( n n +  p p)  q n n = 1.610-19  400  4.51017 = 28.8/-cm, and =  -1 = 0.035 -cm.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore, R = (0.035)  20m / (10m 1.5m) = 467 .

Diffusion 2.8 (a) Using Equation (2.3.2),

J = qn = qD(dn/dx).

Therefore,

 = D(1/n)(dn/dx) = -D/. (constant) (b) J = q n n = qn and  =  n .

Therefore,  = -D/ n  = -(kT/q)/. (c)  = -1000V/cm = -0.026/. Solving for  yields 0.25m.

2.9 (a) ε  

dV 1 dE v 1      . dx q dx q L qL

(b) E c is parallel to E v . Hence, we can calculate the electron concentration in terms of Ec. n( x )  n 0 e   Ec ( x )  Ec ( 0 )  / kT

where E c ( x )  E c (0)   / L x.

Therefore, n( x )  n 0 e  x / LkT . (c) J n qn n ε  qDn

qni e  x / LkT  n

dn 0 dx

     qDn ni e  x / LkT   0 qL  LkT 

Therefore,

n q



Dn kT  Dn  n . kT q

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 3 Terminology and knowledge 3.1

Integrated semiconductor companies Companies that design and fabricate integrated circuits Fabless With no fabrication/processing facility Foundries Companies that specialize in processing wafers to produce silicon devices Wafer fab Fabrication facility where wafers are processed to produce silicon devices Integrated circuits System of transistors manufactured on silicon wafers CPU Central processor unit that is an active part of computer containing the datapath and control DRAM Dynamic random access memory Flat-panel displays Display panels with flat screens MEMS Micro-electro-mechanical-system DNA chips Silicon chips used for DNA screening Dry oxidation Growth of SiO 2 using oxygen gas Wet oxidation Growth of SiO 2 using water vapor Horizontal furnace Horizontally oriented oxidation furnace Vertical furnace Vertically oriented oxidation furnace

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Photolithography/Optical lithography Process in which the resist is optically patterned and selectively removed from designated areas on a wafer Wafer stepper Equipment used in lithography process Photoresist Ultraviolet-light sensitive material Photomask Quartz photo-plate containing a copy of pattern to be transferred to Si or SiO 2 surface Negative resist Photoresist that becomes polymerized and resistant to a developer when exposed to an UV light Positive resist Photoresist whose stabilizer breaks down when exposed to an UV light, leading to the preferential removal of exposed regions in a developer Strip Removal of photoresist Asher System that removes the resist on a wafer by oxidizing it in oxygen plasma or UV ozone system Lithography field Small area exposed to an UV light during the exposure through a photomask and an optical reduction system Stepper Another name for lithography equipment Step-and-repeat action The process of exposing different parts of a wafer until the whole wafer has been exposed Phase-shift photomask Photomask that produces 180 degree phase difference in neighboring clear features so that their diffraction fringes cancel each other

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Optical Proximity Correction (OPC) Printing a slightly different shape on the photomask to correct distortions resulting from an exposure process Overlay Alignment between 2 separate lithography steps Extreme UV lithography (EUVL) Lithography that uses 13nm wavelength and is expected to result in much higher resolution Soft-x-ray lithography Old name of EUVL Electron-beam lithography (EBL) Lithography using a focused stream of electrons Electron projection lithography (EPL) EBL that exposes a complex pattern simultaneously using a mask and a reduction electron lens system Wet etching Removal of SiO 2 using hydrofluoric acid Isotropic Without preference in direction Dry etching (also known as plasma etching or reactive-ion etching (RIE)) Removal of SiO 2 using plasma and reactive ions Anisotropic With preference in direction Selectivity The extent in which an etching process distinguishes between different materials End-point detector Detector that monitors the emission of characteristic light from the etching products so as to signal when etching should end Plasma process induced damage / Wafer charging damage Damage to devices on wafers due to the use of plasma Antenna Effect Sensitivity of the damage to the size of the conductor

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ion implantation Method of doping in which ions of impurity are accelerated and shot into the semiconductor surface Gas-source doping Method of doping in which a gas reacts with silicon and liberates phosphorus so that phosphorous diffuses into the silicon substrate Solid-source diffusion Method of doping in which the dopants from the thin film coated on the silicon surface diffuse into silicon Anneal Heating of wafers for dopant activation and damage removal Dopant activation Making dopants behave as donors and acceptors by heating the wafers Implantation dose Total number of implanted ions/cm2 Depth/ Range The location of peak concentration below the surface of silicon Straddle Spread of dopant concentration profile

3.2

Diffusion The movement of molecules from an area of high concentration to an area of low concentration Junction depth Thickness of diffusion layer Diffusivity Constant that describes how quickly a given impurity diffuses in silicon for a given furnace temperature Predeposition Portion of diffusion process step with the source present Drive-in Portion of diffusion process step without the source

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Furnace annealing Heating of wafers in a furnace for dopant activation and damage removal Rapid thermal annealing (RTA) Annealing process in which a wafer is rapidly heated to high temperature and cooled quickly down to the room temperature Rapid thermal oxidation Oxidation process in which a wafer is heated to the designated temperature quickly, oxidized, and then cooled rapidly down to the room temperature Rapid thermal chemical vapor deposition (CVD) Chemical vapor deposition process in which a wafer is heated to the designated temperature quickly, the material is deposited on the wafer, and the wafer is cooled rapidly down to the room temperature Laser annealing Annealing process in which a silicon wafer is heated with a pulsed laser Transient enhanced diffusion (TED) Diffusion phenomena in which diffusion rate is increased by crystal damage due to ion implantation Interconnect Metal connection between devices in integrated circuits Inter-metal dielectrics Materials used for electrical isolation between metal layers Crystalline Molecular structure with nearly perfect periodicity Polycrystalline Molecular structure composed of densely packed crystallites or grains of singlecrystals Amorphous Structure with no atomic or molecular ordering Grain boundary Interface between crystal grains Thin-film transistors (TFT) Transistors made of amorphous or polycrystalline silicon, widely used in flat-panel displays

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Sputtering target The source material for sputtering Reactive sputtering Sputtering accompanied by chemical reaction of sputtered ions. For example, Ti sputtered in nitrogen gas forms TiN film on the Si wafer Physical vapor deposition (PVD) Another name for sputtering Step coverage problem The inability of sputtering to deposit uniform films in small holes or vertical features on wafer High-temperature oxide (HTO) Very conformal oxide formed by a CVD process at a high temperature Low-pressure chemical vapor deposition (LPCVD) CVD process at low pressure, which yields good thickness uniformity and low gas consumption Plasma-enhanced chemical vapor deposition (PECVD) CVD using plasma; low deposition temperatures In-situ doping Doping process in which dopant species are introduced during the CVD deposition of polycrystalline silicon Spin-on Process in which liquid materials are spun onto the wafer Epitaxy Special type of thin-film deposition technology that produces a crystalline layer on silicon surfaces that is an extension of the underlying semiconductor crystal arrangement Metallization Interconnection of individual devices by metal lines Via/ Plug Electrical connection between adjacent metal layers Electromigration Migration of metal along the crystal-grain boundaries in a quasi-random manner, causing voids to occur in metal interconnects

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Damascene process Process used to form copper interconnect lines Chemical-mechanical polishing (CMP) Process in which a polishing pad and slurry are used to polish away material and leave a very flat surface Back-end process Metallization; last step of IC fabrication Front-end processes Oxidation, diffusion Planarization Process to obtain a flat surface to improve lithography and etching Multi-chip modules Multiple chips put into one package Solder bumps Electrical connection between chip and package Flip-chip bonding Melting pre-formed solder bumps on IC pads to make all connections simultaneously Burn-in Subjecting IC packages to higher than normal voltage and temperatures; identify potential failures Qualification Routine used to verify the quality of manufacturing and reliability Operation life test Part of qualification process; to find out if the devices last over one thousand operating hours

3.3

(a) Lithography field A small area having the best optical resolution (The beam intensity is uniform within a lithography field.) (b) Misalignment Layer to layer mismatch (Each new mask level should be aligned to one of the previous levels) (c) Selectivity

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The ratio of etching rate of the film to be etched to that of the substrate film below (d) End-point detection Detecting the exposed substrate film after the removal of the desired film. (e) Step coverage The ratio of film thickness deposited on the flat surface to that deposited on the non-flat surface. (f) Electromigration The movement of atoms in a metal film due to momentum transfer from the electrons carrying the current.

3.4

(a) Wet. Otherwise, it would take too long to grow the thick oxide. (b) Dry because it provides better control of the gate (channel) length and vertical wall profile. (c) Arsenic because - it is a donor ion (group V), - it reduces R p and R p, and - it reduces diffusivity. (d) PECVD (e) Sputtering, dry etching, plasmas containing chlorine. (f) Nd Na

Metallurgical junction

Oxidation 3.5

Wet oxidation is faster than dry oxidation because the solid solubility of H 2 O steam into SiO 2 is higher than that of O 2 gas into SiO 2 . This creates a very sharp concentration profile that causes H 2 O to diffuse towards the SiO 2 -Si interface much more effectively than O 2 under the same conditions.

3.6

(a)  

0.2um * 0.2um  0.165um * 0.2um  6.239hr 0.0117um 2 / hr

Solving the given equation for t ox and using quadratic formula,  A  A 2  4 B(t   ) Tox   0.256um .

2 B (b) Linear approximation: Tox  (t   ) => 0.655um (156% error).

A 2 Quadratic approximation: Tox  B(t   ) => 0.320um (25% error).

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

In this case A  Tox , neither linear nor quadratic approximations would be valid.

Deposition 3.7

Poly Silicon: SiH 4  Si(s) + 2H 2 Silicon Nitride: 3SiH 2 Cl 2 + 4NH 3  Si 3 N 4 + 6HCl + 6H 2 LTO: SiH 4 + O 2  SiO 2 + 2HCl + 2H 2 HTO: SiH 2 Cl 2 + 2H 2 O  SiO 2 + 2HCl + 2H 2

Diffusion 3.8 (a) x j  Dt  x j  x j  Dt  (Dt )

x 2j  2 x j x j  x j   Dt  ( Dt )  x 2j  ( Dt )

2 x j x j  x j   ( Dt ) Since x j  x j ,

2 x j x j   ( Dt )  x j 

 ( Dt ) .

2x j

(b) For boron, D(500K) = 10.5Exp[-3.69/(8.61410-5773)]cm2/sec = 9.010-24 cm2/sec, and 500 years = 1.581010 sec.

Hence, x j  1.42  10 13 cm  1.42  10 9 m .

Our assumption ( x j  x j ) in part (a) is correct.

3.9 (a) Junction depth is distance from surface where the dopant concentration equals the substrate concentration.

N sub  N d  N junction  1015 cm 3 (Required junction dopant concentration)

From Eq. (3.6.1), 2 N 0  x j 4 Dt N junction  e  1015 cm 3 Dt Using

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

D  D0 e  Ea / kT , D0  10.5 cm 2 / sec, N 0  1015 cm 2 , and E a  3.7eV , X j  1.97  10 4 cm .

(b) X j has the form 1/ 2

  k  X j  2 Dt ln  .   Dt  Taking the derivative,     1/ 2 dX j  Xj 1   k  1 1      . ln 1/ 2  dDt Dt   Dt    k   2 Dt X j   2ln    Dt    dX j 1.98  10 -4 1    9.39  10 4 cm 1 9 4 dDt 2  10 1.98  10 3.70 373

 

. 10 6 3600  3.996  10 40 cm 2 X j .Dt  3.75  10 35 cm  0.

Hence, X j  Dt Dt  10.5e

Visualization 3.10 (a)

(b)

(c)

1.0m resist

1.0m oxide

1.0m oxide

Silicon substrate

Silicon substrate

1.0m resist

(d)

1.0m oxide

1.0m oxide

Silicon substrate

Silicon substrate

(e)

(f)

1.0m poly

1.0m oxide

1.0m oxide

0.3 0.3

Silicon substrate

0.3 0.3

Silicon substrate

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(g)

(h)

1.0m resist 1.0m poly

1.0m resist 1.0m poly

1.0m oxide

1.0m oxide

0.3 0.3

0.3

Silicon substrate

(i)

0.3

(j)

Silicon substrate

1.0m oxide 1.0m poly

1.0m poly 1.0m oxide 0.3 0.3

(k)

1.0m oxide 0.3

Silicon substrate

0.3

(l)

1.0m resist 1.0m oxide

Silicon substrate

1.0m resist 1.0m oxide

1.0m poly

1.0m poly 1.0m oxide

0.3 0.3

1.0m oxide 0.3

Silicon substrate

0.3

Silicon substrate

(n)

(m) 1.0m resist 1.0m oxide

1.0m oxide

1.0m poly

1.0m poly 1.0m oxide

1.0m oxide 0.3 0.3

0.3

Silicon substrate

(o)

0.3

(p)

Silicon substrate

1.0m poly

1.0m oxide

1.0m oxide

1.0m poly

1.0m poly 1.0m oxide

1.0m oxide 0.3 0.3

Si substrate0.3

0.3

0.3 0.3

0.3

Si substrate0.3

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(q)

(r)

1.0m resist

1.0m resist 1.0m poly

1.0m poly 1.0m oxide

1.0m oxide

1.0m poly

1.0m poly 1.0m oxide

1.0m oxide 0.3 0.3

(s)

0.3

Si substrate0.3

0.3 0.3

0.3

Si substrate0.3

1.0m poly 1.0m oxide 1.0m poly 1.0m oxide 0.3 0.3

0.3

Si substrate0.3

3.11

um Poly 1 Mask Contact 1 Mask

(a)

Poly 2 Mask Contact 2 Mask

(c)

1.0um oxide Si-sub Si-s ub

1.0 m oxide

1.0 m resist 1.0 m oxide

(a)

Silicon substrate (b)

1.0um resist

1.0 m resist

Silicon substrate (d)

1.0um oxide

1.0 m oxideSi-sub

1.0 m oxide

Silicon(b)substrate

Silicon substrate

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Changing the polarity of the contact 1 mask results in the same cross section as problem 3.10 (d). The same cross-section is obtained by using a negative resist and the reversed mask of contact 1, which is opaque in the inside of contact 1 area.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 4. Problems

Part I: P Junction Electrostatics of P Junctions 4.1 The linearly graded junction is P-type for x<0 and n-type for x>0 with the junction at x = 0.

Nd-Na Slope = a

N-type

x 0 P-type

Since the doping levels are symmetric on either side of the junction, the depletion region widths into both sides must be the same (xn=xp=Wdep/2). (a) To find the electric field distribution, we utilize the following relation:

dε ρ q ( d − a ) q × a × x . = = = dx ε S εS εS Integrating the equation with the boundary conditions ε(x = xn= xp = Wdep/2) =0, we find 2 q × a  2 Wdep  ε( x) = x − .

2ε S  4 

(b) The potential distribution is 2 dV qa  x 3 Wdep x  +C. = −ε ⇒ V = − − dx 2ε S  3 4 

For convenience, if we define V(x=0) =0 as a reference, we find C = 0. Hence, 2 dV qa  x 3 Wdep x  = −ε ⇒ V = − − .

2ε S  3 4  dx

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(c) The built-in potential is the same as (Eip-Ein)/q at thermal equilibrium.

Energy-Band Diagram

Eip

Ec Ef Ein Ev x Wdep -xp

xn

The carrier concentrations at the edges of the depletion region are (E f − Ein )

n ( x n ) = ni e

kT

 aWdep   n( x )  ⇒ Ein = Ei ( x = x n ) = E f − kT ln  n  = E f − kT ln    ni   2n i 

and (Eip − E f )

p ( x p ) = pi e

kT

 p (x p )  aWdep  ⇒ Eip = Ei ( x = x p ) = E f − kT ln   = E f + kT ln  .  ni   2ni 

Therefore, φbi is

φ bi =

E ip − Ein q

=2

kT aWdep . ln q 2 ni

(d) The total voltage drop takes place in the depletion region. From part (b), we know that |V(xn)|=|V(xp)|=qaWdep3/(24εS). Remember that xn=xp=Wdep/2. Hence, the potential drop is φbi-Va = qaWdep3/(12εS). Solving for Wdep gives 12ε S Wdep =  (φ bi − Va )   qa

1/ 3

.

Note that if we substitute this in part (c), we can iteratively solve for φbi.

4.2 (a)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(

) (

)

 1016 cm − 3 × 5 × 1015 cm − 3   kT   a d   = 0 .7 V  ln  2  = (0.026 V ) × ln  −3 2 10   10 cm  q   ni   

φbi = 

(

)

(b) Wdep = =

2ε sφbi  1 1    + q  a d 

(

)

2 × 12 × 8.85 × 10 −14 × 0.7  1 1  ×  16 +  −19 1.6 × 10 5 × 1015   10

= 5.28 × 10 − 5 cm = 0.528 µm = 528nm

In order to calculate xn and xp we need to use Wdep = xn+xp and xn Nd=xp Na . xn = Wdep − x p = Wdep −

d xn a

 a    5 × 1015 cm − 3 Wdep =   × (0.528 µm ) = 0.176 µm ∴ xn =  15 −3 16 −3   5 × 10 cm + 10 cm   a + d 

(

) (

)

Likewise,  d    1016 cm − 3     × (0.528 µm ) = 0.352 µm xp =  Wdep =  15 −3 16 −3   × +

+

5 10 cm 10 cm   d   a

(

(c)

ε

max

=

q d xn

εs

) (

)

(1.6 × 10 C )× (10 cm )× (1.76 × 10 cm) = 2.652 × 10 V / cm = 12 × (8.85 × 10 F / cm ) −19

16

−3

−5

4

−14

Since the built-in potential is the integration of the electric field, the maximum electric field can also be calculated from the area of the triangle of the electric field profile.

1 φbi = Wdep max 2 2φ 2 × (0.7V ) ∴ max = bi = = 2.652 × 10 4 V / cm −5 Wdep 5.28 × 10 cm

ε

ε

(d)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Фbi

Ec Ef Ev x Wdep

xn

-xp Ф

Фbi

x

ε

x

xn x

-εmax ρ q d +

x

-

x

xn -q a

(e)

(

) (

)

 1016 cm − 3 × 1018 cm − 3   kT   a d   = 0.838V  ln  2  = (0.026 V ) × ln  10 −3 2   10 cm  q   ni   

φbi = 

Wdep = =

(

)

2ε sφbi  1 1    + q  a d 

(

)

2 × 12 × 8.85 × 10 −14 × 0.838  1 1  ×  16 + 18  −19 1.6 × 10 10   10

= 3.35 × 10 −5 cm = 0.335 µm

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 a    1018 cm −3 Wdep =  18 −3  × (0.335 µm ) = 0.332 µm xn =  16 −3   (10 cm ) + (10 cm )   a + d   d    1016 cm −3 Wdep =  18 −3  × (0.366 µm ) = 0.003µm x p =  16 −3   (10 cm ) + (10 cm )   a + d 

ε

max

=

2φbi 2 × (0.838 V ) = = 5.003 × 10 4 V / cm −5 Wdep 3.35 × 10 cm

The depletion width has decreased due to a higher doping and since the acceptor doping is now 100 times greater than the donor concentration, most of the depletion region is on the n-side.

4.3 (a) If we apply reverse bias to the sample, the depletion width will increase. From Equation (4.2.5), we know that Naxp = Ndxn. This means that the numbers of ionized dopants on the N and the P sides are equal. The side that has its dopants totally ionized is fully depleted. Hence, we need to find the total number of dopants per unit area on each side. The side with the smaller value will become depleted before the other. a × W P = 5 × 1016 cm −3 × 1.2 × 10 −4 cm = 6 × 1012 cm −2 and d × W = 1 × 1017 cm −3 × 0.4 × 10 −4 cm = 4 × 1012 cm −2 . where WP and WN are the widths of the P-type region and N-type region, respectively. Since the N-type region contains less dopants per unit area, the Ntype region will be fully depleted before the P-type region. (b) Repeating the calculation for Nd = 1×1016cm-3, a × W P = 1 × 1016 cm −3 × 1.2 × 10 −4 cm = 1.2 × 1012 cm −2 .

Hence, the P-type region will be fully depleted first.

(Note: You can also solve part (a) and (b) by finding the voltage at which each side depletes. However, this will make the problem unnecessarily complicated.) (c) From Equation (4.4.1), we know that C dep / unit area =

εS Wdep

.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

First, we need to find the total depletion region width.

For part (a), we know that the N-type region is fully depleted. And, from Equation (4.2.5), Naxp=Ndxn. Hence,

d W = a x p ⇒ x p =

C dep / unit area =

d W = 0.8 µm ⇒ Wdep = 0.4 µm + 0.8 µm = 1.2 µm . a

εS = 8.63 × 10 −9 F / cm 2 .

1.2 µm

For part (b), we know that the P-type region is fully depleted. And, from Equation (4.2.5), Naxp=Ndxn. Hence,

d x n = aW P ⇒ x n =

C dep / unit area =

aW P = 0.12 µm ⇒ Wdep = 0.12 µm + 1.2 µm = 1.32 µm . d

εS = 7.85 × 10 −9 F / cm 2 .

1.32 µm

4.4 (a) At thermal equilibrium, the Fermi level is constant throughout the system. Since this is the case for the given figure, equilibrium conditions prevail. (b) Carrier concentrations are given by:

n = ce

− ( Ec − E f ) / kT

and n = v e

− ( E f − EV ) / kT

.

Therefore, Log n Log p

pp1 nn1 pp2 pn1

np2

np1

x 0

L/4

L/2

3L/4

L

(c) The potential is the reverse of the band diagram. That is,

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1 q

φ = − ( Ei − E f ) .

For this problem, sketching the qualitative shape is sufficient.

φ ( x) x

0 L/4

L/2

3L/4

L

(d) We assume that the total energy is constant: ETOTAL = EKE + EPE = Constant.

E ETOTAL KE

PE x

0

L/4

L/2

3L/4

L

4.5 (a) The intrinsic region has no dopants. Consequently, there is no charge exposed in the I-region. On the N-type side, the ionized donor atoms are exposed due to the diffusion of the electrons. The same situation prevails on the P-type region due to the diffusion of holes. The field lines that start from the ionized donor atoms on the N-type side do not have any negative charge to terminate on until they reach the P-type region where ionized acceptors are present. Thus, the field is constant in the I-region. Alternatively, you may note that dε/dx=0 in the intrinsic region since ρ = 0. The free-carriers that are present in the I-region will be swept away by the built-in electric field. As a result, we may assume the I-region is fully depleted. To find the depletion-region width, we first need to calculate the builtin potential φbi using Equation (4.1.2):

φ bi =

kT  d a   = 0.78V . ln q  ni2 

The electric field distribution in the structure is

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ε xp

L=0.5 µm

xn x

0 Slope = -(qNa/εS), from Eq (4.2.2)

Slope = (qNd/εS), from Eq (4.2.4) |Emax| = (qNd/εS)(xn) = (qNa/εS)xp P

I

N

φbi is the voltage drop across the P-I-N structure, and is the sum of the area under the electric field in each region. That is,

φ bi =

1 ε max x p + ε max L + 1 ε max xn where ε max = q d xn .

2 2 εS

We also know from Equation (4.2.5) that

x p a = xn d ⇒ x p =

d xn . a

Hence, the equation becomes

 q d  d q d L  + 1 x n2 + x n − φ bi = 0 .

2ε S  a εS  Solving for xn yields 0.032 µm. Consequently, xp is 0.019 µm. Then, the depletion-region width Wdep is xn+xp+L = 0.551 µm. (b) We found the maximum electric field in part (a):

ε max = q d x n = q a x p = 148 V / cm . εS

εS

(c) We also found the built-in potential in part (a). It was 0.78 V. (d) The breakdown voltage is the sum of the areas under the electric field distribution with εmax = εcrit = 2 105 Vcm-1. The depletion-region width in each region can be written in term of εcrit as

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ε crit = q d x n = q a x p ⇒ xn = εS

εS

ε crit ε S q d

and x p =

ε crit ε S q a

.

For P-N junction, the breakdown voltage is

VB =

1 1  2  1   = 6.91 V . + ε crit x p + 1 ε crit xn = ε S ε crit 2 2 2q

a   d

For P-I-N junction, the breakdown voltage is 1 ε crit x p + ε crit L + 1 ε crit xn = 2 2 ε 2  1 1    + ε crit 0.5µm = 16.14 V . = S ε crit + 2q  d a 

VB =

Therefore, the breakdown voltage of the P-I-N structure is ~ 10V higher than the P-N structure with the same doping levels.

Diffusion Equation 4.6 (a) At x = -∞, all the holes generated by illumination in the region x > 0 are recombined well before reaching x = -∞ due to the finite lifetime. Hence, there is no excess holes exist at x = -∞, and the hole concentration is p = p0 =

ni2 = 210.25 cm −3 . a

(b) At x = +∞, excess hole concentration won’t have any position dependence.

According to the continuity equations, photo generation will be balanced by thermal recombination (since the steady-state conditions prevail), therefore GL = U = p’/τ.

p = p0 + p ' = 210.25 cm −3 + GLτ ≈ 109 cm −3 . (c) “Low-level injection” implies that the minority carrier concentration is much smaller than the majority carrier concentration. For this problem, since the maximum minority carrier concentration (p = 109 cm-3) is much smaller than the majority carrier concentration (n = Nd = 1018 cm-3), low-level injection conditions do prevail. (d) The continuity equation is

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

∂ 2 p' ( x) dp ' p' = Dp + GL − 2 ∂x dt τ

dp ' = 0 at steady state . dt

and

For x ≥ 0,

∂ 2 p' ( x) p' dp ' = Dp + GL − 2 dt ∂x τ ⇒ p ' ( x) = A1e

(x / D τ ) p

+ A2 e

where GL = 1015 cm −3 sec −1

(− x / D τ ) p

+ G Lτ .

Since p’≠∞ as x→∞, A2 = 0. Therefore,

p1' ( x) = A1e

(x / D τ ) p

+ G Lτ .

For x < 0,

p ' ( x) = C1e

(x / D τ ) p

+ C2e

(− x / D τ )

(G L = 0) .

p

Similarly C1 = 0. Therefore,

p 2 ' ( x) = C 2 e

(x / D τ ) p

.

Now, if we apply the boundary condition or continuity to the equations above, we find p1 ' ( x = 0 ) = p 2 ' ( x = 0 ) & A1 =

G Lτ = 5 × 10 8 cm −3 2

∂p1 ' (0) ∂p 2 ' (0) = ⇒ A1 = C 2 + G Lτ & A1 = −C 2 ∂x ∂x Gτ and C 2 = − L = −5 × 10 8 cm −3 .

2

Therefore, p1 ' ( x) = 10 9 cm −3 − 5 × 10 8 e p2 ' ( x ) = 5 × 108 e

(x / D τ ) p

(− x / D τ )

for

p

for x ≥ 0 and

x < 0 where D pτ = 4.07 × 10−6 cm 2 .

4.7 (a) Thermal equilibrium will be disturbed if any non-thermally generated excess carriers are present. If L (the distance from the origin to the excess carrier generation point) is much larger than the diffusion length of the holes (i.e., L >> Dp τ p ), then almost all the excess carriers generated due to the light would have recombined (i.e., pN’ ≈ 0).

In this case, we could assume thermal

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

equilibrium. Otherwise, if L is not much larger than

Dp τ p , then significant

number of excess carriers will be present at x=0 and the origin will not be at thermal equilibrium. (b) pN’(x) = nN’(x) everywhere in the semiconductor because of charge neutrality inside the semiconductor and all the excess carriers due to light are generated in pairs of electrons and holes. Excess concentrations at x = -L is γΝd. Total electron and hole concentrations at x=-L are n(x = -L) = d + γΝd and p(x = -L) = ni2/ d + γΝd. (c) The excess carriers are generated at x = -L and x = L. In the rest of the semiconductor the excess carriers decay due to recombination. That is,

pN’(-L) > pN’(x) for -L < x < L. So the maximum excess carrier concentration is at x = -L and x = L. Now, pN’(-L) = pN’(L) = γ d. At γ = 10-3, pN’(-L) << d in which Nd is the majority for carrier concentration at thermal equilibrium. Therefore, pN’(x) << Nd -L ≤ x ≤ L which means the excess carrier concentration is always much smaller than the equilibrium majority carrier concentration. This is the condition for lowlevel injection. (d) Inside the bar, carrier generation due to light is zero. In steady state, the continuity equation simplifies to terms involving diffusion and recombination: 0 = Dp

d 2 p ' p ' − . τn dx 2

Rearranging terms gives us

d 2 p ' p ' = .

D pτ dx 2 (e) The general solution of the differential equation in part (d) is

∆p ( x ) = A ⋅ e where Lp =

−

x Lp

+ B⋅e

x Lp

Dp τ , and A and B are integration constants.

The boundary conditions for this problem are

pN’(-L) = pN’(L) = γ d

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Although you do not have to solve the differential equation, the solution to this particular problem is

 x  p ' (x ) = ⋅ cosh   L   L   p   2 cosh L   p

γ d

(Please note that cosh( z) =

-W/2

e z + e −z .)

2

W/2

γ d

γNd

γNd

2 cosh (L / L p )

-L

L

This curve confirms what you would physically expect from carrier injection on both ends of a silicon bar.

4.8 (a) From Figure 2-5, µp is approximately 450 cm2V-1sec-1. τp is given to be 1 µsec.

Using Equation (2.5.6b) and (4.7.6), we calculate Dp and Lp, respectively: Dp =

kT µ p = 11.7cm 2 / sec and q

L p = D pτ p = 34.2 µm .

(b) Using Equation (4.6.3), we calculate the excess hole concentration:

[

] n [e

p ' (0) = p 0 e qV / kT − 1 =

2 i

qV / kT

]

−1.

d

Hence, pN’ = nN’ = 0 for (i), and pN’ = nN’ = 4.8×1010 cm-3 for (ii).

4.9 (a) From Figure 2-8,

ρ = 0.2 Ωcm, n − type ⇒ d = 3 × 1016 cm −3 , and ρ = 1 Ωcm, p − type ⇒ a = 1.5 × 1016 cm −3 .

Therefore, using Equation (4.1.2), we find that the built-in voltage is

φ bi = kT ln

d a = 0.76 V . ni2

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) Using Equation (4.8.2), n P (0 ) =

ni2 (qV / kT ) e − 1 = 0.65 × 1014 cm −3 , and a

p (0 ) =

ni2 (qV / kT ) e − 1 = 3.25 × 1013 cm −3 . d

(

)

(

)

(c) P-type side: Na = 1.5×1016cm-3 ⇒ µn = 1150 cm2V-1sec-1, τn=10-6sec, Dn=29.9 cm2/sec, and Ln=5.47×10-3cm.

Using Equation (4.9.2), we need to derive an expression for Jn(x):

J n ( x) = −q = −q

(

)

Dn n P 0 e qV / kT − 1 e x / Ln = Ln Dn ni2 qV / kT e − 1 e x / Ln = 54e x / Ln mA / cm 2 .

Ln a

(

)

N-type side: Nd = 3×1016cm-3 ⇒ µp = 400 cm2V-1sec-1, τp=10-8sec, Dp=10.4 cm2/sec, and Lp=3.22×10-4cm.

Using Equation (4.9.1), we need to derive an expression for Jp(x): J p ( x) = −q = −q

Dp Lp

(

)

− x / Lp

=

)

−x / Lp

= 168e

p 0 e qV / kT − 1 e

D p ni2

(e L p

qV / kT

−1 e

− x / Lp

mA / cm 2 .

d

And the total current density J is

J = J n (0) + J p (0) = 0.222 A / cm 2 .

J Jn p

n Jp -xp

xn

x1

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(d) At the point (x1) where Jn and Jp cross each other, Jn = Jp = ½ J. Therefore, Jp =

0.222 −( x ) / L A / cm 2 = 0.111A / cm 2 = 0.168e 1 p ⇒ x1 = 1.38 µm .

2

4.10 For this problem, please note that: (1) ∆pN = 0 for xb ≤ x ≤ xc because τp = 0. This yields the boundary condition ∆pN = 0 and x = xb.

(2) Because we have a P+N diode, we need only deal with the N-side of the junction in establishing an expression for I. Moreover, the depletion width is all but totally on the N-side of the junction given a P+N diode (i.e., xn ≅ Wdep).

(3) Because τp = ∞ for 0 ≤ x ≤ xb, there will be no IR-G current. (τp = ∞ implies that there are no R-G centers.) Thus, we only need to develop an expression for the diffusion current flowing in the diode.

Since we are interested in the static state, ∂∆pN/∂t =0. Also, GL = 0 (no light) and ∆pN/ τp 0 because τp = ∞. Thus the minority carrier diffusion equation reduces to the form ∂2∆pN/∂x2 = 0 for W ≤ x ≤ xb

which is subject to the boundary conditions ∆pN(xb) = 0 and ∆pN(W) =

2 qV  ni  kTA  e −1 .  d  

The general (narrow-base type) solution is ∆pN(x) = A1 + A2 x.

Applying the boundary conditions 0 = A1 + A2 xb and ∆pN(W) = A1 + A2 W giving ∆pN(W) = -A2(xb - W) or A2 = - ∆pN(W)/( xb - W) and A1 = -A2 xb = ∆pN(W)

xb . xb − W

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

  x − x  ni 2  xb − x   qVkTA  =   ⋅  e − 1 for W ≤ x ≤ xb, Thus ∆p ( x ) = ∆p (W ) b    xb − W  d  xb − W    2 qV  d∆p n ni  Dp   kTA   and   ⋅ − J p ≅ − qDp e =q 1  dx' d  x b − W   

I ≅ AJ p = qA

2 qV  ni  Dp   kTA .  ⋅  e  − 1  d  x b − W   

Please note that since ∆pN(x) is a linear function of x, Jp is constant throughout the narrow-base (Wdep ≤ xb ≤ xb) region.

Proof of Minority Drift Current Being egligible 4.11 (a) 0.5 Ω-cm n-type Si => Nd = 1016cm-3, Dn = 30 cm2sec-1, Ln=5.5 µm, Dp = 10 cm2sec-1, Lp=3.2 µm n i2 qVa / kT − x / Lp − x / Lp e −1 e ≈ 1014 e cm − 3 . d (b) Minority current:

(

p( x ) = p 0 +

J p ( x) = −qD p

)

(

)

dp( x) n2 −x / L −x / L = q i D p e qVa / kT − 1 e p = 0.5e p A / cm 2 . dx d Lp

Majority current:

(

J n ( x ) = J t ( x ) − J p ( x ) = J p ( 0) − J p ( x ) = 0.5 1 − e

− x / Lp

) A / cm

2

J (Acm-2) 0.6 Jt 0.5

0.4

Jn

0.3

0.2

Jp 0.1

0 0

2

4

6

8

10

12

14

16

18

20

x (µm)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(c) J ndiff ( x) = qDn

d (n0 + n' ( x) ) dn' ( x) D −x / L = qDn = − J p ( x) n = 3J p ( x) = −1.5e p dx dx Dp

(d) J ndrift ( x ) = J n ( x ) − J ndiff ( x ) = 0.5 + 1 × e

E ( x) =

− x / Lp

A / cm 2 .

J ndrift −x / L = 0.25 + 0.5 × e p Vcm −1 . nqµ n

J pdrift ( x) = pqµ p E ( x) ≈ p' qµ p E ( x) = 1.6 × 10 −3 e

−x / Lp

(1 + 2e

−x / Lp

) A / cm 2 .

J pdrift << J pdiff ⇒ J p ≈ J pdiff .

J (Acm-2)

J (Acm-2)

0.5

0.03

0.45

0.025

0.4 0.35

Jpdiff

0.02

Jpdiff

0.3

0.015

0.25

0.01

Zoom In

0.2

0.005

Jpdrift

0.15

0 0.1

-0.005

0.05

Jpdrift -0.01

0 0

2

4

6

8

10

12

14

16

18

20

0

2

4

6

8

10

x (µm)

12

x (µm)

(e) From E (x) in (d) and the Possion equation, Eq. (4.1.5), one the net charge density, ρ to be 1.7 ×10-9 e-x/Lp C/ cm-3 .

This ρ is negligibly small compared with the excess hole charge density, qp' in (a), which is 1.6 ×10-5 e-x/Lp C/ cm-3 .

This shows that p' = n' is a good assumption because ρ = qp' _ qn' .

Temperature Effect on IV 4.12 (a) From Equations (4.9.4) and (4.9.5),  Dp Dn    ≡ ηe − E g / kT . + L   p d Ln a  For simplicity, we will ignore the weak temperature dependence of η.

I ≈ I 0 (e qV / kT ) where

I 0 = Aq c v e

− E g / kT

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

V≈

kT  I  ln   ⇒ q  I 0 

E  dV k  I  kT d   ln I − ln η + g  ≈ ln   + dT q  I 0  q dT  kT 

≈

k qV kT d  E g    + q kT q dT  kT 

=

V Eg − T kT

.

(b) Assuming V=0.5V and T=300K, dV 0.5 − 1.1 = = 0.002 V/K. dT 300 (c) In Fig. 4-21, start at T=300K, V=0.5V, or I = 240 µA: At that fixed I, ∆V = -0.2 V between -25C and 75C (∆T = 50 K). Hence, ∆V / ∆T ~ -0.2 V/ 100 ºK = -0.002 V/ºK, in excellent agreement with the value found in part (b).

4.13  ni2 L p ni2 Ln     = Aq − p ' L p + − n' Ln  . + I = Aq  τ  τ  τn  d p a τ n   p 

4.14

(a) If we model the quasi-neutral regions as a resistor R, the voltage drop across the neutral regions is Vneutral = IR and that across the junction is V − Vneutral = V − IR .

Hence, Equation 4.9.4 becomes I = I 0 (e q (V − IR ) / kT − 1) . (b) Solving the above equation for V yields  kT  I V = IR + ln + 1 . q  I0  (c) Let’s use I0 = 1.8×10-12A from Figure 4-22.

With R = 0,

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

V (V)

R=0

I (A) With R = 200Ω,

V (V)

R = 200Ω

R=0 I (A) Clearly for the case R = 200Ω, the first term IR becomes dominant over ln(I/I0+1), and the IV curve becomes very linear.

Charge Storage 4.15 Please refer to Section 4.9, 4.10, 8.2, 8.3, and 8.7. (a) First, we need to find an expression for Qt: WB −xp Qt = Aq  ∫ p ' ( x)dx + ∫ nP ' ( x)dx  =  x n  −W E

 W '  x − xn   = Aq p 0 e qVa / kT − 1  − B 1 − WB '   2 

(

(

+ Aq nP 0 e

=

)

qVa / kT

x + xp   W '   − 1  E 1 + WE '   2 

)

2 WB

xn

2 −xp

= −W E

W ' W ' A ⋅ q qV A / kT 1 e − 1 ( p 0WB '+ nP 0WE ') = Aqni2 e qVa / kT − 1  B + E .

2 2  d a 

(

)

(

)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

WB’ and WE’ are the actual lengths of the base and emitter regions respectively excluding the depletion regions. It for a short-base diode is given by

D 1 D 1   . + n I t = Aqni2 e qVa / kT − 1  p  WB ' d WE ' a 

(

)

Hence,  WB ' WE '    + d a  Qt 1 (W 'W ')( aWB '+ dWE ') .  = = B E τS = It 2  Dp 1 D 1  2(D pWE ' a + DnWB ' d )   + n  WB ' d WE ' a  (b) s = 1.8 × 10 −10 sec (<< τS = τS = 1×10-6 sec) (c) A short-based diode has a shorter storage time, thus it can operate at a higher frequency.

Part II: Application to Optoelectronic Devices IV of Photodiode/Solar Cell 4.16 (a) Let us examine the minority carrier diffusion equation for hole. In general, ∆∂p ∂ 2 ∆p ∆p = Dp − + GL . ∂t ∂x 2 τp For the steady state problem at hand, ∂∆pN/∂t = 0. Also ∂2∆pN/∂x2 = 0 if one goes far from the junction on the N-side where carrier perturbation introduced by the junction has decayed to zero. Thus,

0=−

∆p ( x → ∞ )

τp

+ GL or ∆p ( x → ∞ ) = GLτ p . ⇐ boundary condition

(b) One simply parallels the ideal diode derivation to obtain the desired I-VA relationship. Given a P+N junction, however, we need consider only the lightly doped side of the junction. To simplify the mathematics, we set the origin of coordinates to the n-edge of the depletion region. Under steady state conditions and with x’ as defined as above, we must solve

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0 = Dp

∂ 2 ∆p ∆p − + GL ∂x 2 τp

subject to the boundary conditions ∆pN(x’=0) =

2 qV  ni  kTA e  and ∆p ( x → ∞ ) = GLτ p . −1  d  

The general solution is ∆p ( x' ) = GLτ p + A1e

− x' Lp

+ A2e

x' Lp

.

Because exp(x’/Lp) ∞ as x’ ∞, the only way the second boundary condition can be satisfied is for A2 to be identically zero. With A2 = 0, the application of the first boundary condition yields ∆p ( x' = 0) = GLτ p + A1 = A1 =

qV 2  ni  kTA  or − 1 e  d  

2 qV  ni  kTA e  − GLτ p and − 1  d   − x'

 L  ni 2  qV A  ∆p ( x' ) = GLτ p +   e kT − 1 − GLτ p  ⋅ e p .    d  The associated hole current density is then − x'

 L  D p  ni 2  qVkTA d∆p J p ( x' ) = −qD p =q   e − 1 − GLτ p  ⋅ e p dx' L p  d    and for a P+N diode, I = AJ = A[Jn(x = -xp) + Jp(x = xn)] ≅AJp(x’ = 0) I = qA

or

Dp ni 2  qVkTA  Dτ e  − qA p p GL . − 1  Lp d  Lp 

Finally noting Dp τp = Lp2, we conclude

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

  qVA I = I 0  e kT − 1 − I L   where

I 0 = qA

D p ni 2 Lp d

and I L = qAL p G L .

(c) Voltage that appears when the illuminated diode is open circuit is

V OC =

kT  I L  ln 1 +  . q  I0 

Current that flows when the illuminated diode is short-circuited is I SC = − I L .

I

VOC

GL = 0

VA

ISC

GL = GL0

Part III: Metal-Semiconductor Junction Ohmic Contacts and Schottky Diodes 4.17 (a) A very heavy P+ doping is important to forming an ohmic contact.

TiSi2

P+-Si

Ec

0.55 eV Ef

Ev, Ef 0.55 eV

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) A very heavy N+ doping is important to forming an ohmic contact.

N+-Si

TiSi2 0.55 eV

Ef

Ec 0.55 eV Ev, Ef

(c)

A very heavy doping is unacceptable to forming a rectifying contact.

P- Si

Ec

TiSi2 Ef Ev 0.55 eV

2 eV

Ef 0.55 eV

4.18

(a) χsi

Eo q(φbi-|Va|) qφs

qϕM qφBn q|Va|

Ec Ef

Ev © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ϕ M = 4.8V, χ Si = 4.05 eV 3.22 × 1019 = 4.26V 1016 φ Bn = 4.8 − 4.05 = 0.75V

φ s = χ si + 0.026 ln

φ bi = φ M − φ s = 0.54V φ bi − Va = 0.54 − 0.4 = 0.14V

(b) ρ(x)

φ(x) Va=0

Va=0,0.4

Va=0.4

ε(x) Va=0.4

Va=0

4.19 (a)

1/C2

Voltage at which Wdep=1µm

(b)

ε

1µm

2µm

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2 µm

(c) V = ∫ εdx = 1.

0

4 ε max + 1 .1. 1 ε max + 1 .1. 4 ε max = (1.3µm)ε max 5 2 5 2 5

To find εmax, we use the slope when xd<1um, − q (−1.6 × 10 −19 )(1016 ) = = 1.54 × 109 V/cm 2 εs 11.7ε o Using this slope, we need xd=5um for ε=0,

ε max = (1.54 × 109 )(5 × 10−4 ) = 7.723 × 105V/cm V = (1.3µm)(7.723 × 105V/cm) = 100.4V . (d) Express V (areas under electric field profile) in terms of Wdep, V = area = 30.9Wdep − 23.175

Wdep = C=

εs Wdep

V + 23.175 30.9 =

εs V + 23.175 30.9

.

Depletion-Layer Analysis for Schottky Diodes 4.20 (a) ΦB(Platinum to Si) = 5.3 eV, d = 1016 cm-3 and A = 10-5 cm2 Now, Ec – Ef = kT ln( c/ d) = 0.205 eV ∴ qΦs = 4.05 + 0.205 = 4.255 eV ∴ φbi = - (Φs - ΦM) = 1.045 V (Si to Pt) The capacitance is give by C = charge/voltage drop = where K = 2.88×10

-13

K q d 2ε Si (φbi − VA ) = 1/ 2 (φbi − VA ) φ bi − VA

(

)

1/2

FV

If VA = 0, C = 2.82×10-13 F = 0.282 pF

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 φ bi   (b) For a 25% reduction C/C(0) = 0.75 ⇒   φ bi + VA  ∴ VA = φbi(1- 0.75-2) = -0.813 V

1/ 2

= 0.75

4.21 The development of relationships for the electrostatic variables in a linearly graded MS diode closely parallels the uniformly doped analysis. (a) With d(x) = ax for x ≥ 0, invoking the depletion approximation yields ρ(x) = qax

0 ≤ x ≤ Wdep.

Substituting into Poisson’s equation gives dε qa ρ = = x 0 ≤ x ≤ Wdep and dx ε Si ε Si 0

qa

W

∫ d 'ε' = ε ∫ x' dx'

ε(x )

or

Si x

ε(x ) = − qa (Wdep2 − x 2 )

0 ≤ x ≤ Wdep.

2ε Si

Turning to the electrostatic potential, we can write dV qa 2 = −ε( x ) = Wdep − x2 and 2ε Si dx

(

)

Wdep

0

(

)

qa 2 dV ' = Wdep − x'2 dx' ∫ ∫ ε 2 Si x V (x ) V (x ) = −

(

qa 3 2 2Wdep − 3Wdep x + x3 6ε Si

or

)

0 ≤ x ≤ Wdep.

Finally, V = -(φbi - VA) at x = 0, and therefore qa 3 − (φbi − VA ) = − Wdep 3ε Si  3ε  Wdep =  Si (φbi − VA )  qa 

1/ 3

.

(b) Paralleling the developments for the linearly graded P-N junction, the expression for φbi will be 1 φbi = Φ B − (Ec − E f ) x =W 0 q

[

]

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

where W0 is the depletion width when VA = 0. Since approximate charge neutrality applies fro x > W0, it follows that n x =W = ni e

[( E − E ) f

i

x =W 0

]/ kT

0

(E − E ) c

f

≅ x =W 0

≅ d ( x = W0 ) = aW0

or

Eg  aW  − kT ln 0  2  ni 

Thus, to determine φbi, one must simultaneously solve the following two equations employing numerical techniques.

1/ 3

 3ε  W0 =  Si φbi   qa 

and φbi =

ε Si A

ε Si A

(c) C J =

Wdep

=

Eg  aW0  1  + kTln Φ B − q 2  ni 

 3ε Si   qa (φbi − VA )  

1/ 3

4.22 (a) From the graph, the built-in voltage is 0.8V. (b) From Equation 4.16.7, we know that the slope of the line is inversely proportional to the doping concentration. Hence, for Region I, 1 2 2 7 × 1014 cm 2 / F 2 = + ⇒ ≡ = ( ) . φ V Slope bi C 2 q d ε s q d ε s 5V Solving for Nd yields d = 8.62 × 1016 cm −3 .

For Region II, the slope is 0. This implies that Region II is a very heavily doped N++ region. Hence, Nd ≥1020cm-3.

In order to find the width of Region I, we calculate Wdep at Va = 5V since Region I becomes fully depleted. Using Equation 4.3.1, ≥ 1020cm-3

Nd 2ε s (φbi + 5V ) Wdep = = 0.30 µm . q d 8.62×1016cm-3 0

0.30µm

x

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Comparison between Schottky Diodes and P Junction Diodes 4.23 (a) Similarity: Both have the same slope of 1 decade / 60mV at room temperature.

Log(I) Schottky Diode

Difference: The schottky diode current is several orders of magnitude larger than the PN diode current.

PN Diode

V (b) For Schottky diode, use Equations 4.18.2 and 4.18.6: I 0 = AKT 2e − qφ Bn / kT = (0.01)(100 ) 300 2 e − q 0.55V / kT A = 58.5 µA

[

(

]

)

For PN diode, use Equation 4.9.5:

  Dp   = 0.01cm 2 q 1010 cm −3 2  I 0 = Aqni2  L    p d 

(

)(

)

  = 0.32 fA 4cm 2 / s (1µs ) 

(

4cm 2 / s

)

(c) For Schottky diode,  kT  I V= ln + 1 = 0.36V . q  I0  For PN diode,  kT  I V= ln + 1 = 1.03V . q  I0  (d) Use a metal or silicide that yields a smaller φBn. (e) I0 may be so large, especially at elevated temperatures, that the reverse leakage current a. interferes with the rectification property of the diode, and b. causes too large a power (heat) dissipation, I0×Vr, under reverse bias.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ohmic Contacts 4.24 (a) The semiconductor is uniformly doped with a doping concentration of d = 1017cm-3. Using depletion approximation, the energy band varies as Ec (x) = ½ (q d/εSi) (x - x0)2 where x0 is a constant and x is positive in the semiconductor and zero at the MS junction.

At x = 0, Ec(0) = 0.65 eV (using the equilibrium Fermi-level as the reference) and at x= 10 nm, Ec(0) ≈ 0 ⇒ x0 = 10 nm and d = 8.85×1018 cm-3 (b)

qφB Ec

Ef

x

4.25 (a) Rc =

50mV × 0.08µm 2 = 4 × 10 −8 Ω ⋅ cm 2 1mA

(b) Using Eq. 4.21.7, Rc ≡

d  Rc  2 V Hφ / Hφ / ln  = e Bn d = Be Bn d ⇒ φ Bn = J qvthx H d H B

( ) (m / m )(ε / ε ) = (5.4 × 10 )× 0.26 × 11.7

H = 5.4 × 1010 ×

n

0

s

0

10

= 9.418 × 1010 cm −3 / 2V −1 vthx = 2kT / πmn = 1.057 × 10 7 cm / s B=

2 = 1.256 × 10 −9 Ωcm 2 qvthx H d

∴ φ Bn =

d  Rc  ln  = 0.3675 V H B

This is the maximum ϕBn allowed since a larger ϕBn will result in a voltage

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

drop larger than 50 mV at the ohmic contact. (c) According to the graph, φ Bn ≈ 0.36 V .

4.26

(a) φ Bn = 0.67V

∴ RC ≈ 6 × 10 −7 Ω ⋅ cm 2

φBp = 0.43V

∴ RC ≈ 5 × 10 −8 Ω ⋅ cm 2

(b) φBp = 0.23V

∴ RC ≈ 7 × 10 −9 Ω ⋅ cm 2

(c) φ Bn = 0.28V

∴ RC ≈ 1 × 10 −8 Ω ⋅ cm 2

(d) According to the graph, this contact resistance can be achieved with a barrier height of 0.3V or lower and a doping concentration of 3 × 10 20 cm −3 or higher is required. Looking at Table 4-2, we can see that this cannot be achieved using only one type of metal silicide. So different silicides should be used for contact on N+ and P+ silicon. + :

ErSi1.7

≥ 3.5 × 10 20 cm −3

P+ :

PtSi

≥ 2.5 × 10 20 cm −3

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 5 Energy Band Diagram 5.1

Ec

E c ,E f

Ef

Ev

Ev

E c ,E f

Ef Ev

Accumulation

E c ,E f

Ev Flat-band

Ec Ef

Ev

Ec

Ev Depletion

E c ,E f Ec Ev

Ef Ev Threshold

E c ,E f Ev

Ec Ef Ev Inversion

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

5.2

(a) 1 -cm N-type silicon substrate : N d =51015 cm-3, E f = E c - 0.223eV

3.1eV

3.75eV

Ec

0.22eV

Ef=Ec W

Ox

N-Si n-Si

Ef

EE Vv

3.1eV

3.75eV

Ec

0.22eV

Ef=Ec W

Ox

Ef

EVv E

n-Si N-Si

VFB V fb = 3.75 - (3.1 + 0.22) = 0.43V

(b) 1 -cm P-type silicon substrate :N a =1.51016 cm-3, E f = E v + 0.168eV Thermal Equil.

Flat Band

3.1eV 3.75eV 0.93eV

Ef = Ec W

Ox

P-Si p-Si

3.1eV

3.75eV Ec Ef EV E

Ec

Ef=Ec

v

0.93eV

Ox

W

P-Si p-Si

Ef EEV

v

VFB fb = 3.75 - (3.1 + 0.93) = - 0.28V (c) Heavily doped P+-polycrystalline silicon gate with 1 -cm N-type silicon substrate.

Thermal Equil.

Flat Band

3.1eV

Ec

E 0.22eV c

Ef

P+ -poly

Ox

n-Si N-Si

3.1eV

3.1eV

3.1eV

EV v

0.22eV

EVv Ef=E

P+ -poly

Ox

N-Si n-Si

Ec Ef EEVv

VVFBfb = 3.1 + 1.12 - (3.1 + 0.22) = 0.88V

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(d) Heavily doped N+-polycrystalline silicon gate with 1 -cm P-type silicon substrate.

Thermal Equil.

Flat Band

3.1eV Ec

3.1eV

3.1eV

3.1eV Ec

0.93eV

Ef E EV

EVv E

E f= E c

v

N+ -poly Ox

E EvV

P-Si p-Si

0.93eV

N+-poly Ox

p-Si P-Si

Ef E E vV

VVFBfb= 3.1 - (3.1 + 0.93) = -0.93V

MOS System: Inversion, threshold, depletion, and accumulation 5.3

(a) 3 (b) 2 (c) 1 (d) 4 (e) 5

5.4

(a)  1018 cm 3   kT   N a    0.026 V   ln  10 3   0.479 V  ln   q   ni   10 cm 

 B  

  1  E        1  E   V fb   si    si    g    B      g    B   q   q  2  q    2  q  1.12V   0.479V  1.039V 2

(b) Wd max 

2 si  2 B  qN a





2  11.7   8.84  10 14  2  0.479   3.521  10 6 cm 1.6  10 19  1018



  

(c)

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Vt  V fb  2 B 

2 si qN a  2 B Cox

  1.039V   2  0.479V  



 

 



2  11.7   8.85  10 14 F / cm  1.6  10 19 C  1018 cm 3  2  0.479V  3.9   8.85  10 14 F / cm / 2  10 7 cm







 0.2455 V

(d) Only the flat-band voltage changes.     E     1  E   1E  V fb   si    g    si    g    B    g    B  q   q   q  2  q   2 q  1.12V   0.479V  0.081V 2  Vt  0.081V   2  0.479V 



 

 



2  11.7   8.85  10 14 F / cm  1.6  10 19 C  1018 cm 3  2  0.479V   3.9   8.85  10 14 F / cm / 2  10 7 cm  1.3655 V

5.5







(a) At V g -V fb = -1V, the MOS capacitor is in accumulation.

Cox 

QS  ox F   4  10 7 2 Vox Tox cm

Tox  8.62nm (b) At threshold, V g  V fb  1V   s  Vox

B 

1/ 2  2 s qN a 2 B   2  B

C ox

where

kT  N a  . ln q  ni 

Solving iteratively, we get

N a  2.9  1016 cm 3 . (c) Since  s  0, Vox  1V . (d) V g  V fb  1V   s  Vox   s 

2 s qN a s 1 / 2 C ox

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 s  0.245( s )1 / 2  0.5 Solving the equations above, we get

 s  0.354V .

5.6

s

Vt

INV

DEP

V fb

W dep ACC Vg

W dmax

-2 b INV

Vt

DEP

V fb

ACC

Vg

Q inv

Q acc slope=C ox V fb

Vt INV

DEP

ACC

Vg

slope=C ox

Vt

Q dep

V fb

Vg

V fb

Vg

Qs

qN a W dmax qN a W d V fb

Vt

5.7

Vg

Vt

(a) From Equation 5.3.2,

V g  V fb   s 

1 C ox

2qN a  s s where  s 

  .

2

s

Rearranging the terms, we obtain

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

       2qN   V  V   0 .  C 2

a

s



s

s



ox

fb

g

Solving for  s yields 

2qN a  s C ox

s 



2qN a  s  4V fb  V g  C ox2 2

.

Since  s cannot be less than 0,  2qN a  s

s 

2C ox

2qN a  s  2   V fb  V g     V fb  V g  2 4 4C ox2

where 2qN a  s

 

C ox

.

Hence,

s 

2



 2   2    V fb  V g    V fb  V g  4 4 2 4 

2 2

 V fb  V g  



2

2

4

 V fb  V g  .

Or, 2

  2  V fb  V g  . s    4  2  (b) From Equation 5.3.1, Vox 

2qN a  s

  2 s      V fb  V g  .  2

C ox

4



(c) We can qualitatively predict that  s V g since | |<<1. Also, V ox  V g .

 s (V) R

R

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

V g (V) V ox (V)

V g (V)

(d) From Equation 4.2.10, 2q s

Wdep 

5.8 (a) Cox 

C ox

 ox Tox

 3.45  10 7

 B  kT ln V fb  

Eg 2

s 

 2q s   2  V fb  V g  .   C ox  2 4 

F cm 2

Na  0.4V ni   B  0.96V

Vt  V fb  2 B 

1 2q s N a 2 B   0.96  0.8  0.34 V  0.18V Cox

(b) Qacc  C ox V g  V fb   3.45  10 7

C cm 2

(c) Qdep   2q s N a 2 B  1.1  10  7

C cm 2

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Qinv  C ox V g  Vt   6.3  10 7

C cm 2

(d)

Q S , 10-7 C/cm2 3.45 -1.96

-0.96

0.18

2 V g , volts

-1.1 -7.4

5.9

Parameters a b c d e

Increase

Accumulation Region Capacitance Flat-band Voltage, Vfb Depletion Region Capacitance Threshold Voltage, Vt Inversion Region Capacitance

Decrease

Unchanged X

X X X X

(a) At accumulation: Accumulation capacitance C  Cox   ox / Tox (b) At flat-band: Flat-band Voltage

  N  V fb   g   4.05  0.56  0.026 ln a    ni    N a , |  B |

(c) At depletion: Depletion Capacitance 1 / C  1 / Cox  Wdep /  s N a ,Wdep , C 

(d) At threshold: Threshold Voltage

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Vt  V fb  | 2 B | Qd / C ox N a , |  B |, Qd ,Vt  (e) At inversion: Inversion Capacitance 1 / C  1 / C ox  Wd max /  s N a ,Wd max , C  5.10 To find the value of the oxide capacitance,

Cox 

 ox Tox

 1.15 10 7

F . cm 2

The capacitance at V g =V fb is given by C fb 

1 1

C ox

 LD

 7.9  10 8

s

F cm 2

where   kT  LD   2s  q Na 

1/ 2

 4.09  10 6 cm .

To find the minimum value of the capacitance, 4 s B kT N a  3  10 5 cm ;  B  ln  0.348V qN a q ni

Wd max 

C min 

1

C ox



1 Wd max

 2.65  10 8

s

F . cm 2

V t is given by Vt  2 B 

1 C ox

2 s qN a 2 B  V fb  1.11V

where V fb =  g -  s = -2.00 V.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C ox C fb

C min

V fb

Vt

To find the effective oxide charge, we need to find

 Al  4.1V  s  4.05  0.562  0.3482  4.96V  Al  s  0.86V Vt  V fb   Al  s  2  (0.86)  1.14V 

 Qox C ox

Hence, Qox  1.31  10 7

C q  8.2  1011 .

2 cm cm 2

Field Threshold Voltage 5.11 (a) V fb   Al   Si  4.1V  (  Si  E g / 2q 

Ei  E f

) q  4.1V  (4.05V  1.12 / 2V  kT / q   ln( N a / ni ))  0.80V

(b)  B 

Ei  E f q

Wd max 

 kT / q  ln( N a / ni )  0.290V

2 s 2 B  0.866um qN a

Vt  V fb  2 B  qN aWd max / Cox  5V C ox  2.65  10 9 F / cm 2 (c) C ox 

 Tox

  0 / 1m  2.65  10 9 F / cm 2 where K  2.99

(d) In accumulation, V g - V fb = -1V,

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C ox = Q s /V ox Vt  5V C ox  2.65  10 9 F / cm 2 K  2.99 It is the maximum allowable K. (e) V g  Vt  

Qinv C ox

Qinv  C ox (V g  Vt )  (2.65  10 9 F / cm 2 )(2V )  5.3  10 9 C / cm 2

(f) 1 / C  1 / C ox  1 / C dep  1 / C ox  Wd max /  s  1 / 2.65  10 9 F / cm 2  0.866 m /  s C  2.17  10 9 F / cm 2 (g) V g  V fb  Vs  Vox (V g  Vt )  V fb  2 B  Vox 7V  0.8V  2(0.29V )  Vox Vox  7.22V

Oxide Charge 5.12 V  Q / C ox

C0 45  10 12 Q  C ox V   V    0.05  3.52  10 8 C / cm 2 5 A 6.4  10 V fb   g   s  Q f / C ox It should be negative since negative charges increase V fb .

5.13 Oxide charge will change the device characteristics. Mobile charges are particularly bad as they give the device instability. That is, as the charge moves from one side of the oxide to the other, it will change the threshold voltage. This is undesirable.

Mobile charges are generally introduced into the oxide during wafer cleaning or oxidation. Strict measures of cleanliness can be achieved by using ultra-clean chemicals. And, the impurities such as sodium can be immobilized by introducing a chlorine compound during oxidation.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C-V Characteristics 5.14 V g  V fb   s  Vox 2

qN aWd W qN a Using Vox  and  s  d , we solve for Wd C ox 2 s W 2 W  V g  V fb  qN a  d  d   2 s C ox   V g  V fb  2   2 s  4 s  4C ox 2 s C ox  qN a   Wd  2C ox

2C ox V g  V fb  2

Wd

s

1 1

q s N a



Since

C ox

1 1 1 1 Wd     , C C ox C dep C ox  s

2V g  V fb  1 1   .

2 C q s N a C ox

5.15 (a) The substrate doping is P-type since the threshold voltage is larger than V fb .

(b) Tox  A

 ox C ox



10 4  0.345  1012  6.9 nm 50  10 12

(c) Vt  V fb  2 B 

2 s qN a 2 B kT  N   2 ln a   Cox q  ni 

2  s qN a ln

Na ni

Cox

Using V fb =-1V and V t = 0.5V and solving iteratively, we obtain N a  3  1017 cm 2 . (e) At position C, MOS has the minimum capacitance.

 1 Wd max   C min     s   C ox

1

 12.7 pF where Wd max 

2 s 2kT N a ln qN a q ni

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(e) Ec

c) Vg =VT

a) Flat band

Ei Ef

Ec 0.45eV

Ev Substrate

Ei Ef

Substrate Ev

Gate

Ec

b) Vg=0

Gate Ec

d) & e) the same

Ei Ef

Ef

Ev

Ev Substrate

Substrate

Gate

Gate

(f) V g  0  V fb   s 

1 C ox

2 s qN a s

Substituting the numerical values and solving the quadratic equation, we obtain

 s  0.644  s  1  0  s  0.53V .

5.16 (a) P-type, since V t > V fb .

A (b) T ox : V fb : A W dmax : N sub : A Vt : A

B B A B B

(Cmax= ox /T ox (directly from the graph) B (1/C min = 1/C ox +W dmax / s and C ox A > C ox B) (W dmax ~(N sub )-1/2) (directly from the graph)

5.17 A = 100100 m2 = 10-4 cm-2

For the MOS capacitor, the field in the oxide  max = 8106 =

10V Tox

 T ox = 1.2510-6 cm or 12.5 nm

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 Cox 

 ox

A = 28.32 pF = C MOS Tox For P+N junction,

B 

kT  N d   = 0.356 V for N d = 1016 cm-3 ln q  ni 

Wdep  C

s Wdep

2 s  B  VB  = 8.3510-5 cm for V R = 5 V. qN d A = 1.25 pF = C P-N jn

C MOS / C P-N jn = 22.66 5.18 (a) The flatband voltage is 0V because the 2 silicon sides are equally doped

(i)Vg =0

(ii) Vg<0 and large

(iii) Vg>0 and large

(b)

As both sides are equally doped, the values of C 1 and C 2 will be equal.



1

C1 = C 2  C ox  C dep

C dep 

s Wdep



1 1

where

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Wdep 

 2 s 2kT  N d    C dep = 34.5 nF cm-2 and C ox  ox . ln qN d q Tox  ni 

(c) When the left side is P-type, both the silicon layers go into accumulation and depletion for the same type of bias.

The flat band voltage will be ( Si + E g /2 +  BL ) - ( Si + E g /2 -  BR ) =  BL +  BR = 0.713 V C

VGg V

5.19

Bias Condition accumulation flatband threshold inversion

Surface Potential ~0 =0 = 2 B  

MOS cap (LF) C ox C ox C ox C ox

MOS Cap (HF)

MOSFET

C ox C ox (C ox -1 +W dep,max / s )-1 (C ox -1 +W dep,max / s )-1

C ox C ox C ox C ox

5.20 (a) Accumulation -1 V < V g Depletion -3 V < V g < -1 V Inversion V g < -3 V The substrate is N-type.

(b) C 0 

 ox

A Tox Thus, T ox = 205 nm.



1

(c) From the high-frequency C-V curve, C min  C 0  C dep



1 1

Plugging in C min = 0.4C 0  C dep = 0.67C 0 = 54.9 pF Now, Cdep 

s

Wdep

A

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For uniform doping Wdep 

kT  N d  2 s 2 B   at threshold, where  B  ln qN d q  n i 

4 s kT  N d   ln 2 Wdep q 2  ni  which can be solved by iteration and we get the value of N d ~ 1015 cm-3 .

 Nd 

(d) V fb ~ .55V + (kT/q)ln(1015/1010) = .85V Since V fb from the plot is ~ -1V, then Q ox consists of positive charges. The shift in V fb is 1.85V. Thus, Q ox = 1.85V  C 0 = 1.85V  82pF / 4.75x10-3 cm2 = 32 nC/cm2.

Poly-gate Depletion 5.21 (a) Using Gauss Law, Q poly  qN poly Wdpoly   poly ε poly   ox ε ox ,

where  ox is the electric field inside the oxide and Wdpoly 

(b)  poly 

 ox ε ox qN poly

.

2 qN polyWdpoly

2 poly



2 qN poly  ox2 ε ox

2  ox2 ε ox .

2q poly N poly



2 2 poly q 2 N poly

(c) V g   poly  Vox   s  V fb .

Vox C ox  Qs  Q poly  Vox 

1 C ox

2q poly N poly  poly

V g   poly  V fb   s  Vox   poly  V fb  2  B 

 poly 

Tox

 ox 

 poly  

Tox

 ox

2q poly N poly poly

2q poly N poly  poly  V g  V fb  2  B   0

Tox

 ox

2q poly N poly 

Tox2

 ox2

2q poly N poly  4V g  V fb  2  B  2

Tox 2 ox

2q poly N poly 

2 ox 2 ox

T 4

2q poly N poly  V g  V fb  2  B 

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

We know that  poly >0. Also, if we set



Tox

 ox

2q poly N poly and   V g  V fb  2  B , then



 poly     2   and  poly    2 2 1  1  

Hence,

 poly

2 ox 2 ox

T q  V  V  2    g

(d) W dpoly  

fb

 ox ε ox qN poly



B



2 poly  poly qN poly

  .  2 

 2 ox2 V g  V fb  2  B    . poly N poly 1  1    Tox2 q poly N poly  



2 poly qN poly

       2

2 poly  Tox2 Tox   2 q  N V V 2       poly poly g fb B 2 ox qN poly  4 ox2

 2q poly N poly  . 

(e) Using the equation derived in part (c), we find  poly  0.28 V .

And, using the equation derived in part (d), we find Wdpoly  2.46 nm .   N    B  kT ln a   0.41 V , V fb  0.97 V , and  poly   s  .     q  ni    (f) Calculate V t using Equation 5.4.3: Vt  V fb  2  B 

1 C ox

4q s N a  B   0.97  0.82  0.095V  0.055 V

Using Equation 5.8.3 with  poly  0.28 V , Qinv  C ox V g  Vt   poly  

 ox Tox

1.5  0.055  0.28V  2.20  10 6 coul / cm 2 .

(g) First, we calculate C oxe .

C oxe 

 ox Tox  Wdpoly / 3

 1.22  10 6 F / cm 2 .

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Qinv  C oxe V g  Vt   1.90  10 6 coul / cm 2 .

This value is smaller than what we have found in part (f).

5.22 (a)

W dpoly

Ec E v ,E f  poly

Ec Ef Ev

V g is negative, V fb is positive,  st is negative, V ox is negative and  poly is negative

5.23 (a) Q poly  qN polyWdpoly  qN poly

(b) C poly 

s



Wdpoly

 s qN poly Q poly

1



1

(c) Ctotal  (Cox  C poly ) 1  (

2 sV poly qN poly

 2 s qN polyV poly

 s qN poly 2V poly 2V poly 1 1  ) Cox  s qN poly

Threshold Voltage Expression 5.24 Vt  V fb  s  Vox

 V fb  s 

qN aWdep Cox

 V fb  s 

qN a 2 ss Cox

 V fb  2B 

qN a 2 s 2B Cox

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 6 MOSFET V t 6.1

15 -cm =1015 cm-3, E f = E v + 0.26eV oxide trap density = 8x1010 cm-2, Z = 50 m, L = 2 m, T ox = 5nm (a) V’ fb = V fb + V. V fb = 3.1 - (3.1 + 0.86) = -0.86eV, V = Q f /C ox = 18.5 mV. Therefore V’ fb  V fb = -0.86 eV When oxide thickness is thin, the trap charge effect can be ignored. (b) V t = V’ fb + V ox + V s  V fb + 2 B + (2 s qN a 2 B )1/2/C ox = -0.86 +0.6 + 0.02V = -0.24V (c) To make V t = 0.5V, one should implant boron into silicon substrate such that  V t = Q imp /C ox . Therefore ion implant dose should be (0.5V+0.24V)  C ox q = 3.21012 cm-2.

6.2

(a) Using Equation 4.16.4 and referring to Table 1-4, we find  N (GaAs)   4.7  1017    1V  kT ln   0.96V . bi   Bn  kT ln c 17  Nd  110    Then, 2 sbi 1 213 0 bi 1 Wdep    0.12 m . q Nd q Nd 2 qN d Wdep 213 0  bi  V  1 (b) Wdep  0.2 m  V    bi  1.82V . q Nd 213 0  A negative V g is need to increase W dep and turn-off the channel. (A metal/Ntype semiconductor Schottky diode exhibits the same forward/reverse bias properties as an P+/N diode.)

(c) Yes. If the positive V g is kept small (say 0.5V), the forward current of the Schottky gate maybe comparable to the subthreshold drain leakage current. A positive V g would reduce W dep and therefore raise I ds . (d) The channel thickness or doping concentration must be reduced so that W dep  channel thickness at V g = 0.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6.3

kT N sub F , B  ln  0.47eV 2 q ni cm 1 2q s N sub 2 B (a)` Vt  V fb  2 B  C ox Eg 1   B  2 B  2q s N sub 2 B  0.09  0.61  0.52V Vt   2 C ox 1 2q s N sub 2 B (b) Vt  V fb  2 B  C ox Eg 1 Vt     B  2 B  2q s N sub 2 B  0.56  0.47  0.61  1.64V 2 C ox 1 (c) Vt  V fb  2 B  2q s N sub 2 B Cox E 1 Vt  g   B  2 B  2q s N sub 2 B  0.52V 2 Cox Cox  6.9  10 7

(d) Vb  0V Vs  0V Vd  2.5V V g  2.5V

(e) Vb  2.5V Vs  2.5V Vd  0V V g  0V (f) I dsat 

 nWC ox

(V gs  Vt ) 2 2L I dsatc (2.5  (0.52)) 2   5.3 I dsatb (2.5  (1.64)) 2

The transistor with the lower absolute value of threshold voltage has a higher saturation current. That is why P+ poly-gate PMOSFETs are typically used in IC. (g) The ratio of the current is the ratio of the mobilities.

To find  n , Vgs  Vt  0.2/ 6Toxe  2.5  0.52  0.2V / 6  5  10 7 cm  1.07 MV / cm and  n = 250cm2V-1s-1.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

To find  p ,  V gs  1.5Vt  0.25 / 6Toxe   2.5  1.5  (0.52)  0.25V / 6  5  10 7 cm  1.01MV / cm and  p = 63cm2V-1s-1.

I dsat ( c ) I dsat ( a )



p 1  n 4

Basic MOSFET IV Characteristics 6.4 (a) Due to the highly doped regions nearby, transistor C-V always approaches C ox in inversion. Hence, it is impossible to determine the frequency. Either high or low frequency could have been used.

(b) Since V t >V fb , this is a NMOS. (c) From the I d -V g curve, V t is 0.55V. More precisely, V  V W  I d  C ox Vds V g  Vt  ds   0.55  Vt  ds  0 2  2 L  V t  0 .5 V W (d) Slope of curve I d - V g line  C ox Vds  5  10 3  1 .

L Vds  0.1V From the CV curve, W F C ox WL  1pF  C ox  10 -4 L cm 2 cm 2 5  10 3 Thus,    500 Vs 0.1  10  4 (e) Vdsat  V g  Vt I dsat 

Vg V dsat I dsat

C oxW 2L

(Vdsat ) 2  0.025

1V 0.5V 6.25mA

A 2 .Vdsat 2 V

2.5V 2V 100mA

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

I d (mA) V g =2.5V

100

V g =1V

6.25 0.5

6.5

2

V d (V)

(a) For V gs =4V, V dsat ~ 3V=(V gs -V t ) and V t =1V (b) I dsat   n C oxW / 2 L  (V g  Vt ) 2 C ox  3.45  10 7 F / cm 2 2 I dsat n   361cm 2 / Vs W  Cox  (Vg  Vt ) 2 L (c) At V gs =3V, V dsat =(3-1)V=2V I dsat  361  3.45  10 7  (10 / 2)  (3  1) 2 / 2  1.25  10 3 A

6.6

(a) At saturation, V d = V dsat = V g -V t . V dd =2V. Therefore the transistor is in saturation mode when V g <2.5V. I dsat = 125(V g -0.5)2A. When V g >2.5V, the transistor is in linear region with I d = 500(V g -1.5) A.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

40

SQRT(Idsat, A)

35 30 25 20 15 10 5 0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Gate Voltage, Vi (V)

(b) & (c) Transconductance: solid line, Output Conductance: dotted line gd , gm gm 500 A/v

250 A/v

gd

0.5

6.7

1.5

2.5

3.0

Vi

(a) V gs  Vt  2V V gs  2.5V (b) Qn  C ox (V gs  Vt  Vc )  0 (Pinch-Off) (c) I ds @ Vds  4V Vdsat  V gs  Vt  3V (in saturation) I ds  (V gs  Vt ) 2 I ds  10 3 

32  2.25  10 3 A 2 2

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(d)

C Cox (Same for high and low frequencies)

Vfb

6.8

Vg

Vt

kT N a ln  0.297V q ni

(a)  B 

 ox

C ox 

 7.08  10 8 F

t ox

V fb   Si  (  Si 

Eg

cm 2

  B )  0.857 V

2

V g  V fb  Vs  Vox  Vt  V fb  2 B 

(b) I dsat 

 nCoxW

C ox

 0.064V

V  V   1.21mA 2

g

2L

2 s qN a 2 B

t

(c) Since V d is less than (V g -V t ), it is in the linear region.

Id  gd 

C oxW  L

V  V  V V   

2 d

g

t

d

2 

C oxW I d Vg  Vt   Vd  1.17mS  V D L





(d) Since V d is less than (V g -V t ), it is in the linear region.

2 C oxW  Vd  Id  V g  Vt Vd   L  2  C oxW I gd  d  Vd  1.13mS V g L

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Potential and Carrier Velocity in MOSFET Channel 6.9

I d  Q n  n

dVc dV W  (V g  Vt  Vc )C ox  n c W dx dx

x

Vc Id dx   (V g  Vt  Vc )dVc  n C oxW 0 0



 I d  x /(  n C oxW )  (V g  Vt )Vc  1 / 2Vc2

Solving this quadratic equation of V c , we get Vc ( x)  (V g  Vt )  (V g  Vt ) 2 

2I d x  n C oxW

Choosing “-” so that V c (0)=0,   2I d x Vc ( x)  (V g  Vt ) 1  1    n C oxW (V g  Vt ) 2     W 2 x n C ox (V g  Vt ) 2   2L   (V g  Vt ) 1  1   n C oxW (V g  Vt ) 2      (V g  Vt ) 1  1  x / L .





6.10 (a) I ds  WCoxe (Vgs  mVcs  Vt )  es d Vcs dx x

Vcs

 I dx  WC   (V  mV  V )dV ds

oxe

0

s

0

gs

I ds x  WC oxe  s (V gs  Vt 

cs

t

cs

mVcs )Vcs 2

Equating the expression above with I ds 

W m C oxe  s (V gs  Vt  Vds )Vds , L 2

we get

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

mVds  mVcs x )Vcs V gs  Vt  Vds  (V gs  Vt  L 2  2 x 2 mVcs  2(V g  Vt )Vcs  (2(V g  Vt )  mVds )Vds  0 L Solving the quadratic equation, we get

Vcs  Vcs 

V gs  Vt m V gs  Vt m



(V g  Vt ) 2  m

(1  1 

x (2(V g  Vt )  mVds )Vds L m

x ) L

 x (b) Qinv ( x)  C oxe (V gs  mVcs  Vt )  C oxe V gs  Vt  (V gs  Vt )(1  1   L     x  x  C oxe V gs  Vt 1  1  1    C oxe (V gs  Vt ) 1   L  L   

Vg  Vt   1  1  1  Vg  Vt   1  dVcs  ε( x)  (c)        dx m  2  1  x  L  2mL  1  x  L L      n V g  Vt   1  dVcs   n ε( x)   ( x)   n   dx 2mL  1 x  L      x   n V g  Vt   1 (d) WQinv  n ε  W .C oxe (V gs  Vt ) 1  .   L 2mL   1 x  L  WC oxe  n Vgs  Vt 2  I dsat  2mL

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(e)

V cs

x

IV Characteristics of Novel MOSFET dVc dV W  (V g  Vt  Vc )C ox ( x)  n c W dx dx dV   (V g  Vt  Vc ) 2ox  n c W dx Ax  B

6.11 (a) I d  Qn  n

L/2

Vds

L / 2

0

  I d  ( Ax 2  B)dx   (V g  Vt  Vc ) ox  nWdVc

A  I d  [ x 3  Bx] LL/ 2/ 2   ox  nW [(V g  Vt )Vc  1 / 2Vc2 ]V0ds 3 W  ox  n  Id    [(V g  Vt )Vds  1 / 2Vds2 ] 2 L AL B 12 I (b) Vdsat  Vds @ d |Vgs  0 Vdsat  V g  Vt Vds (c) It suggests a large W dmax . Vox  Qn / C ox

6.12 (a) I d  Qn  n

dVc dV W ( x)  (V g  Vt  Vc )C ox  n c W ( x) dx dx

L

Vds

0

0

  I d / W ( x)dx   (V g  Vt  Vc )  n C ox dVc  I d  ln[(W0  L) / W0 ]   n C ox [(V g  Vt )Vds  1 / 2Vds2 ]  Id 

 n C ox ln(1  L / W0 )

[(V g  Vt )Vds  1 / 2Vds2 ]

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) Vdsat  Vds @  I dast 

I d |V  0 Vds gs

 n C ox ln(1  L / W0 )



Vdsat  V g  Vt

(V g  Vt ) 2 2

CMOS 6.13 (a) V fb,NMOS = -(E g /2) – (kT/q * ln(5e15/1e10)) = - 0.55 -0 .4 = -0.95V V fb,PMOS = -0.55 + 0.4 = -0.15V Not symmetrical

(b) V fb,NMOS = 0.55 - 0.4 = 0.15V V fb,PMOS = 0.55 + 0.4 = 0.95V Not symmetrical (c) Since V ox and V s will be symmetrical, I would use a mid-gap gate material such as tungsten.

So the workfuction will be 4.05 eV + E g,Si /2 = 4.6eV. However, processing issues makes tungsten (or any metal gates for that matter) a challenge to implement. (d) In the same process, the NMOS and PMOS will have same oxide thickness. If the substrate doping levels for n and p flavors are the same, then I would use P+ gates for PMOS devices and N+ gates for NMOS devices. In this way, the flatband voltages will be symmetrical and the resulting |V t | small.

6.14 (a) PMOS, N-type substrate: N  n  kT ln d  0.38V ni Eg Vfb   m   si    n  4.1  4.05  0.55  0.38  0.12V .

2

NMOS, N-type substrate: N  p  kT ln a  0.42V ni Eg Vfb   m   si    p  4.1  4.05  0.55  0.42  0.92V 2 (b) C ox 

 ox t ox

 6.9  10 7

F cm 2

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

PMOS: Vt  V fb  2 n 

1 C ox

2 s qN d 2 n  0.12  0.76  0.10  0.98V

NMOS: V fb  2 p 

1 C ox

2 s qN d 2 p  0.92  0.84  0.24  0.16V

(c) The threshold voltage must be changed by Qimpl Vt    0.82V .

C ox Hence, C Qimpl  5.7  10 7 . cm 2 6.15 I ds  WC oxe (V gs  mVcs  Vt )

 s dVcs dx 1

dVcs dx

ε sat

 I dx   WC  V  mV  V   I ε dV L

0

Vds

ds

I ds L  I ds

0

Vds

 sat

oxe

s

gs

cs

 WC oxe  s (V gs  Vt 

t

ds

sat

cs

m Vds )Vds 2

 V   L  ds  ε sat    V  I (Long channel) W m .

I ds  C oxe  s (V gs  Vt  V ds )V ds 1  ds   ds L 2 Lε sat  1  V ds ε sat L 

I ds  WC oxe  s (V gs  Vt 

m Vds )Vds 2

6.16

A B C D

NFET Operation Mode Cut-off Saturation Linear Linear

PFET Operation Mode Linear Linear Saturation Cut-off

A: Vgs<Vth for NFET, therefore it is cut off. For PFET |Vgs| > |Vth| and |Vds|<|Vdsat| (|Vds|~0V, |Vdsat| ~ 1.05V), so it operates in linear mode.

B: For NFET Vgs > Vth and Vds>Vdsat (Vds~1.75V, Vdsat ~ 0.3V), so it operates in saturation mode. For PFET |Vgs| > |Vth| and |Vds|<|Vdsat| (|Vds|~0.25V, |Vdsat| ~ 0.6V), so it operates in linear mode.

The answers to C and D can be worked out through the same procedure.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6.17 (a) cut-off { NMOS PMOS linear

Vo

saturated { NMOS PMOS linear

A B

saturated { NMOS PMOS saturated

C

linear { NMOS PMOS saturated linear { NMOS PMOS cut-off

D

0

VTn

Vx

(VDD+VTp) V DD

Vi

(b) At the point B where V i =V x , the NMOS is just becoming saturated from the linear region. Since NMOS is in the linear region  V x  Vtn 2  I dn  K N V x  Vtn V x  Vtn    2   Since PMOS is saturated K 2 I dp  P Vdd  V x  Vtp  2 But I DN = I DP 2  Vx  1  35  2 2  5  Vx  1  40V x  1   2 2     40(V x - 1)2 = 40(4 - V x )2  V x = 2.45 V Thus, Point A B C D

V i (V) 1 2.45 2.45 4

V o (V) 5 3.45 1.45 0

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Body Effect 6.18 For a P-channel MOSFET, we have 2 s qN d (2 B  Vbs ) Vt  V fb  2 B  C ox 2 s qN d

( 2 B  Vbs  2 B ) C ox (a) For 100 nm oxide, C ox = 3.4510-8 F/cm2.

If V bs = 5V, V t = -0.8V. By iteration, using initial guess of  B = 0.3V, we obtain N d = 8.91014 /cm3 and  B = 0.284V. V t 

(b) If V sb is -2.5 V, V t = -0.497V. V t = -1.5 - 0.497 = 2.0 V

Velocity-Saturation Effect 6.19 In all 3 cases, use the general equation I=WQ inv v drift .

Case A: The NMOS is in the triode region.

On source side, Q inv =C ox (V g -V t ) = 138e-9(5-.7) = 593 nC/cm2.

So  drift = I/(WQ inv ) = 1.5e-3/(15e-4  593e-9) = 1.7 x 106 cm/sec.

On drain side, Q inv = C ox (V g -V t -V d ) = 138e-9(5-.7-.5) = 524 nC/cm2.

Thus,  dr = 1.5e-3/(15e-4  524e-9) = 1.9 x 106 cm/sec.

Case B: The NMOS enters saturation region.

On source side,  dr = 3.75e-3/(15e-4593e-9) = 4.2 x 106 cm/sec.

On drain side, the electron velocity is saturated.

Thus,  dr =  sat = 8 x 106 cm/sec.

Case C: Similar to case B. On source side,  dr = 4e-3/(15e-4593e-9) = 4.5 x 106 cm/sec.

On drain side,  dr =  sat = 8 x 106 cm/sec.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6.20

T ox  

V dsat I dsat

W L No change   

Vt  

Vg  

Reducing T ox means smaller V t => larger V dsat (1/(V g -V t ) + 1/(E sat L))-1 & larger I dsat (Q inv  C ox ).

Reducing W has no effect on V dsat and decreases I dsat since I dsat = WQ inv v sat .

Reducing L reduces V dsat (as discussed in lecture) and increases I dsat . If you want to consider very short-channel length devices (L => 0), then essentially I dsat is independent of L. Reducing V t => larger V dsat & larger I dsat .

Reducing V g => smaller V dsat & smaller I dsat .

V   6.21 I d   s C oxW V gs  Vt  m ds Vds 2    WC ox V gs  Vt  mVds Vsat  WC ox V gs  Vt  mVds  s

 V   L  ds  ε sat  

ε sat 2

 V  V   V gs  Vt  m ds Vds  V gs  Vt  mVds ε sat / 2 L  ds  2  ε sat    ε L V  V gs  Vt  mVds  sat  V gs  Vt  mVds  ds 2 2

V gsVds  VtVds  V gs  Vt  mVds ε sat L

Vds V gs  Vt  mε sat L   (V gs  Vt )ε sat L

(V g  Vt )ε sat L

 1  m Vds    Vgs  Vt  mε sat L   (Vgs  Vt ) ε sat L 

1

  1  m W W m      C oxe  ns  V gs  Vt  C oxe  ns V gs  Vt  Vds Vds  L 2 L  Vds  2     6.22 I ds   Vds 1  1  1 1   ε sat L Vds2  ε sat L  Vds 1 Vdsat



1 m  V gs  Vt ε sat L

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6.23 (a) We know that 1   2     V V  g t    1 Vdsat  ε c L 1  2   E c L    1   2 4     c L = 0.1 V  Vdsat  0.1  1  2    1 = 0.54 V   0.1   



1   2 4    (b)  c L = 10 V  Vdsat  10  1  2    1 = 1.83 V   10    (c) We know that 10fF  C Z 2 .

I dsat  n0 ox Vdsat and Cox Z  L 2L  n0 10fF 7mA  0.54 2 2 L2   n0 = 480 cm2 V-1 s-1

6.24 V dsat =  sat L(V g -V t ) / (V g - V t +  sat L)

What is  sat ? 2v sat / s =  sat .  s is given by the universal mobility curve.

At T ox =60A, (V g +V t +0.2)/6T ox =  eff = .9MV/cm.

From the curve,  s ~ 250 cm2V-1s-1.

This yields  sat ~ 2(8x106)/250 V/cm = 6.4x104V/cm. Plug this back into the expression for V dsat to get L ~ 0.19um.

I dsat /W = Q inv v sat = C ox (V g -V t -V dsat )v sat = (3.9 o /60e-8)  (2.5-.5-.75)  8106 = 575uA/um width.

Note: You will often find in literature that the saturation current is stated in units of uA/um instead of amperes. Also, notice that the Q inv at V c =V dsat is not zero.

That is, I dsat is limited by velocity saturation instead of pinch-off.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6.25 (a)

1

V gs  Vt mE sat L

2

V gs  Vt  mE sat L 

2mv sat L

 ns



2  1.2  8  10 6 cm / s  1  10 5 cm  0.64V 300cm 2 / V  s

(b) V gs  Vt  mE sat L L

Vgs  Vt mE sat



 ns Vgs  Vt  2mv sat



300cm 2 / Vs  0.2V  3.13  10 6 cm  31.3nm 2  1.2  8  10 6 cm / s

Effective Channel Length 6.26 (a) For very small V ds , V L Rchannel  ds  I ds  s C oxW (V g  Vt ) In a short-channel device, S/D resistance can seriously degrade saturation current. Note that series resistance is worse for higher currents because R channel is the lowest under these bias conditions.

(b)

Rtotal  Rs  Rd  Rchannel  Rsd  Rchannel

  Leff Lgate  L )  Rsd     Rsd  (  s C oxW (V g  Vt )   s C oxW (V g  Vt )  Think of R total as the y-value, L gate as the x-value, and ( s C ox W(V g -V t ))-1 as the slope. This fits nicely into the standard equation of the line: y = mx + b. You can choose devices with several gate lengths and measure the current from these devices at discrete gate voltages. Remember, that you are assuming V ds is small (<100mV) in these measurements.

From the current, you can plot R total versus L gate . One sample data line is taken with the same V g at different gate lengths. For example, if you measure your current at 5 different V g ’s, you will get 5 separate curves. Ideally, all the lines will intersect at the same point on your plot. This intersection point occurs at L gate = L and R total = R sd .

In practice, it is not always straightforward to make such a plot. For instance, V t can be difficult to determine accurately. Also, there is a strong dependence of mobility on gate voltage for thin-oxide MOSFETs. It’s a good idea to check your data by taking measurements at several different V g instead of at 2 or 3 gate voltages.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(c) I dsat  k(Vgs -I dsat Rs -Vt ) , where k is a constant of proportionality I dsat ( 1  kRs )  k(Vgs -Vt )  I dsat 0 , notice here that k  I dsat0 /(V gs - Vt ) I dsat  I dsat 0 / ( 1  kRs )  I dsat 0 / ( 1  I dsat 0 Rs /(V gs -Vt )) (d)  sat = (V gs +V t +0.2)/6T ox = 1.1 MV/cm.  s ~225cm2V-1s-1 is picked out from the universal mobility plot.

 sat = 2v sat /( s ) = 7x104V/cm.

I dsat0 = (long channel I dsat ) / (1+ (V gs -V t )/( sat L)) = 1.6mA / (1 + (1.1/.7)) = 0.622mA. Plug in 0.622mA into the expression derived in part c and get the following: @ R s = 0ohms, I dsat = .62mA @ R s = 100ohms, I dsat = .59mA @ R s = 1000ohms, I dsat = .40mA 6.27 (a) Choose three transistors with same channel width, Z, and different channel length, L 1 , L 2 ,and L 3 . Measure I dsat at saturation condition for the 3 transistors to get I d1 , I d2 , and I d3 . Solve the 3 equations to get C ox , and L eff .

(b)L = L- L eff when gate oxide thickness is 4.5nm. Z =10 m,  = 300 cm2/Vs. Using approach of (a), L  0.1 m.

3.0 -1

Idsat = const(Ldrawn - L)

Idsat

-1

2.5 2.0 1.5 1.0

0.1 m : L 0.5 0.0 0

1

2

3

4

5

Channel Length(m)

(c) 2.59mA(L 1 - L eff ) = ZC ox (V gs -V t ), V t = 0.5V.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 7 Subthreshold Leakage Current 7.1 a) Cox=3.45E-13/11E-8 = 3.138uF/cm2 Vt = -1.04 + 0.95 + Qdep/Cox = -1.04 + 0.95 + 0.180 = 0.109V b) S = 60(1+Cdep/Cox) = 60(1+2.953E-7/3.138E-6) = 65.64mV/dec c) 1/S = [log(W/L*100nA) – log(I leakage )]/Vt I leakage = 121nA

Field Oxide Leakage 7.2 a) Cox=3.45E-13/0.3E-4 = 1.151E-8F/cm2 Vt = -1.01 + 0.92 + Qdep/Cox = -1.01 + 0.92 + 33.96 = 33.87V b) S = 60(1+Cdep/Cox) = 60(1+2.119E-7/1.151E-8) = 1164.6mV/dec c) 1/S = [log(W/L*100nA) – log(I leakage )]/Vt I leakage = 2.81E-26nA

Vt Roll-off 7.3 log(I ds )

L=0.2um, N a =1E17

L=0.2um, N a =1E15

L=1um, N a =1E15 L=1um, N a =1E17

Vg

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Trade-off between I off and I on .

7.4 i) Larger V t : Decrease I off and decrease I on ii) Larger L : Decrease I off and decrease I on iii) Shallower junction : Decrease I off and somewhat decrease I on iv) Smaller V dd : Decrease I off and decrease I on v) Smaller T ox : Decrease I off and increase I on A smaller T ox contributes to leakage reduction and increases the precious I on .

7.5 There is a lot of concern that we will soon be unable to extend Moore’s Law. In your own words explain this concern and the concerns for high I on and low I off . (a) Answer this question using 1 paragraph of less then 50 words. : Methods of manufacturing very small objects such as photolithography will run into limitations. Transistors may not be able to provide circuit functions with high speed and low power. Specifically, means of increasing speed (I on ) tend to worsen leakage (I off ). Other relevant comments include the following. V t must not be too low because of subthreshold current but small L reduces V t . Since I on (speed) is roughly proportional to V gs -V t (V dd -V t ), I on and speed can not be raised by lowering V t especially because V dd needs to be reduced to reduce the power consumption. (b) Support your description in (a) with 3 hand drawn sketches of your choice. : Possible choices are Log(I ds ) vs. V gs , Vt vs. L and Vds, Ion vs. Ioff, gate leakage vs. Tox, etc. (c) Why is it not possible to achieve high I on and small I off by picking optimal T ox , X j , W dep etc?

Please explain in your own words.: Decreasing T ox is good for Ion but bad for I off . So is increasing X j . W dep is basically fixed by the choice of T ox and V t . (d) Provide three equations that help to quantify the issues discussed in part (c). : Some relevant equations include Eq. (6.9.14), Eq.(6..10.1), (7.2.8), Eq.(7.3.3), Eq.(7.5.2).

7.6 (a) Final equation:

ld  3

 ox Vt  V fb  2B  qN a

Xj 3

X j Tox Toxe 2 s 2B

 ox Vt  V fb  2B 

(b) The minimum acceptable L is several times of ld. In order to support the reduction of L at each new technology node, ld must be reduced in proportion to L. According to the derived equation, in order to reduce the minimum acceptable channel length, Vt  V fb  2B should be increased by gate workfunction engineering, Xj can be





decreased, and the oxide permittivity can be increased .

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

7.7 (a) A smaller W dep Reduces DIBL (V t is not as low)  decreases I off (b) I dsat 

W 2 Coxe  ns Vgs  Vt  where m  1    1  3Toxe / Wdmax 2mL

Reducing W dep increases m which decreases I dsat .

MOSFET with Ideal Retrograde Doping Profile 7.8 (a) 3.1eV

V ox

P+ T rg

Ec

Eg

3.1eV

Vt

Ef, Ev

Ec, Ef

(b) The figure above assumes that the gate is N+ polysislicon and the substrate (below the undoped layer) is P+ silison. In that case , it can be seen that V t = V ox . In general, V t is determined by Eq. (5.2.2) and

V t = V fb + φ st + V ox Note – You need to apply a positive gate bias to reach inversion. Thus the Fermi level of the poly gate is below the Fermi level of the substrate and the difference is V t . (c) ε ox

ε =ε ε

 ox

ox

si

si

Eg / q  Vox   si st   si Tox Trg Trg

The last equality is the result of a further approximation for st . Using the result of part (b),

 Vt  V fb  st  st

 si Tox  ox Trg

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4

(d) First derive Eq. (7.5.2) using Eq. (7.5.1) and eq. (5.5.1). By comparing Eq. (7.5.2) and Eq. (7.5.3) one concludes that, for the same Vt, the depletion region width of the “ideal retrograde doping” MOSFET (Trg) is half of that of a MOSFET with uniformly doped substrate. (e) A small W dep reduces the V t roll-off caused by DIBL which in effect decreases I off . (f) I dsat 

W 2 Coxe  ns Vgs  Vt  where m  1    1  3Toxe / Wdmax 2mL

Reducing W dep increases m which decreases I dsat .

A smaller I dsat causes a longer inverter delay. However, it is important to reduce W dmax so that L can be reduced to increase the device integration density and to reduce chip cost.

I dsat and speed still get improved due to the reduction in L.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 8 Energy Band Diagram of BJT 8.1

(a) & (b) For the given doping concentrations, one computes E f -E i = -0.521 eV, 0.419 eV, and –0.299 eV in the emitter, base and collector, respectively. Also with N aE >>N dB , the E-B depletion width will lie almost exclusively in the base.

Likewise, the majority of the C-B depletion width will lie in the collector.

Energy Band Diagram Ec x

Ei Ef E

Charge Density

x x

(c) The built-in potential between the collector and emitter is VCE 

 5  1018  kT  N aE    0.026  ln   0.221 V . ln 15  q  N aC   1  10 

(d) W  Wn  xnEB  xnCB 1

 2  2 x nEB   Si VbiEB   0.112m  qN dB  1

 2  2 N aC x nCB   Si VbiCB   0.01m  qN dB N aC  N dB  Therefore,

W  2.878m .

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 qN  (e) Electric Field max ( E  B)   dB  x nEB  1.72  10 5 Vcm 1 .   Si   qN  Electric Field max (C  B )   dB  x nCB  1.54  10 4 Vcm 1 .   Si 

(f)

Ec Energy Band Diagram Ei Ef E x

Charge Density x

x

IV Characteristics and Current Gain 8.2

From Eq. 8.4.1 and Eq. 8.4.2, I I F  C  IB  C F IB I C   F I E   F I C  I B    F I C   F I B .

Substituting I C / F into I B , we obtain I  IC   F IC   F C  1   F  F .

F

F

Solving for  F yields

IC 

F .

1F

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

8.3

(a) N+: Emitter E f,n+ E f,p N: Collector P: Base V BE is forward biased and V BC is reverse biased.

(b)

P

EMITTER

E f,n

N

P

BASE

COLLECTOR

The P,N,P on the diagram refer to the minority carrier type in each region. The horizontal dotted lines refer to the equilibrium minority concentration (i.e. p N0 , n P0 ). The remaining dotted curves correspond to the excess minority carrier concentrations.

Assumptions made here: W E & W B are shorter than the diffusion lengths of the holes & electrons respectively, resulting in a linear decay of excess minority carriers in the emitter and base. You should also notice that the scale for the yaxis differs for each region. (c) Base current consists of injection of holes into the emitter and recombination with a very small part of the collector current (remember that I E ~ I C ). The collector current consists almost entirely of electrons emitted from the forwardbiased BE junction which travel across the CB junction. Incidentally, an easy way to remember how a BJT works is to associate the names emitter and collector with the physical emission and collection of the minority electrons in the base.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

8.4

 x  nB ( x)  nB (0)1   and  xB 

 x'  pE ( x' )  pE (0' )1   .  xE 

In the base region,  dn  qDB nB (0) .

J n ( x)  qDB    xB  dx  In Emitter region,  dp  qDE pE (0) J p ( x)   qDE    . xE  dx  n B (0) = 1013 cm-3 , p E (0’) = 1011 cm-3, and p C (x B +) = p CO = 105 cm-3 D B = 20.8 cm2s-1, D E =1.8 cm2s-1, and D C =11.9 cm2s-1. (a) J n , B ( x  0)  qDB nB (0) / xB  0.666 A / cm 2 , J p , E ( x'  0' )  qDE pE (0) / xE  3.6  10 4 A / cm 2 ,

and J p ,C ( x  0.5m  )  qDC pCO / xC  8.7  1010 A / cm 2 .

Therefore, J TOTAL , BE  0.666 A / cm 2  3.6  10 4 A / cm 2  0.666 A / cm 2

and J TOTAL , BC  J n ( xB )  J p ( xB ) .

For short diode approximation, we assume J n (x B -) = J n (x=0). More accurate relationship is J n (X B -) =  T J n (x=0). However, since  T 1, we can still say J n (X B -) = J n (x=0).

Hence, J TOTAL , BC  0.666 A / cm 2 J n , E  J TOTAL , BE  J p (x'  0' )  0.666 A / cm 2 J n , C  J TOTAL , BC  J p ( x'  xB )  0.666 A / cm 2 J p , E  J p , B  3.6  104 A / cm 2

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Jn

1 1 0 -2 1 0 -4 1 0 -6

Jp E

C

B

1 0 -8 m

1 0 -1 0 0 .8

1 .3

3 .5

1  0.99875 LB  10 m  1  WB2 1 2 L2B 0.666 J n ( 0) E    0.999325 J n (0)  J p (0) 0.666  3.6  10 4

(b)  T 

F 

8.5

F  519 1F

   T  E 

(a) To improve the emitter injection efficiency, and to reduce the back injected carriers from B-E. (b) “Small” means WB  LB , typically  1m (c) W B would become a larger percentage of W B . Therefore, it would increase the slope of output characteristics.

(d) At very small values of I C , the recombination current (excessive base junction current) in the E-B depletion region becomes a less significant part of I B as I C increases. (e) At larger values of I C , due to the high level injection in base, I C does not increase exponentially as it does in the moderate level injection region.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

high level injection in base

IC (A )

,IB(A)

10 -2

IC

10 -4

IB

10 -6

F

10 -8

10

Excessive base junction current

-10

10 -12

0.2

0.4

0.6

0.8

1.0

1.2

V BE

(f) Region

VEB

VCB

Active Saturation Cutoff

+ + -

+ -

Schottky Emitter and Collector 8.6 (a) As you will find out in the latter part of this problem the injection of majority carrier of the semiconductor into the metal is much higher than the injection of minority carrier into the semiconductor region from the metal. This would make the emitter efficiency in the BJT very small! Hence, it would not be desirable to use a metal as an emitter in a BJT.

(b) We know that the hole diffusion current is given by 2 D n I diff  I diff 0 e qVA / kT  1  qA P i e qVA / kT  1 .

LP N d From Section 4.17 and 4.18, I te  I te0 e qVA / kT  1  AKe q B / kT e qVA / kT  1 .







Noting that



DP  LP







DP  p

 kT / q  p



p

,

we have

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

I diff I diff 0   I te I te0

q

kT / q  p ni 2

 0.0259437   1.6  10   p Nd 10 6      B / kT  0.72  / 0.0259  140e Ke



19



1/ 2

 10 20   16   10 

 5.05  10 7 . (c) A Schottky base-collector junction (the collector being the metal) would be functional in a BJT. The energy diagram of the base-collector junction would be similar to Fig. 4-34(b). It would be effective collecting the electrons arriving at this junction from the P type base to the metal. The field at the Schottky junction sweeps the electrons into the metal collector just as in the PN base-collector junction shown in Fig. 8-1(b).

Gummel Number and Gummel Plot 8.7 (a)

=J C /J B =100.

(b) Intercept of J C is 10-10 A/cm2= qni2 Dn / N BWB

qni2 Dn N B  10  8  1016 cm 3 .

10  WB (c) Peak concentration qV ni2 kTBE  e  1017 NB qVBE

e kT  8  1013 V BE  ln(8  1013 )  26mV  832mV .

(d)  n 

WB2 (0.2  10 4 ) 2   2  10 11 sec.

2 Dn 2  10

Ebers-Moll Model 8.8 (a)

Consider that n’ = n P0 (eqVa/kT-1) and p’ = p N0 (eqVa/kT-1). Next, take a look at the BC junction ,  n’/p’ = N B /N C (1) Similarly, at the BE junction,  p’/n’ = N E /N B (2) Multiply (1) and (2) to get N E /N C = (8/4)(10/2) = 10.

Thus, N C = 0.1N E = 1017cm-3.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b) The BJT is operating in saturation because both BE and BC junctions are forward-biased and the resulting minority carrier concentrations are larger than the equilibrium values. (c) The stored minority charge is equal to the area under the curve in the base.

Aq  p' x  1m   p' x  2m WB  2  10 5  1.6  10 19  0.5  14  1014  10 4  1.12  10 13 coul .

Q

(d) I E  I nE  I pE   Aq[ Dn

n' ( x  1m)  n' ( x  0 m) p' ( x  1m)  p' ( x  2m)  Dp ] WE WB

  2  1014   6  1014     mA  0.192 mA .  10 5  1.6  10 19  30  10 4   4   1  10    1  10  DW N 10  10 4  N E 1 5 (e)   B E E    1  5   1.7 (Not much gain.) 4 DEWB N B 30  10  N B 3 3 8.9 If the NPN BJT is biased at the boundary between active mode and saturation mode, then forward-biased emitter-base junction (V BE >0) and unbiased collector-base junction (V BC =0). So I R =0.

(a) At the given operating point, we can simplify the Ebers-Moll model as follows:

IE

I F  I ES ( e qVBE / kT  1)

FIF

IC

E IB

B

(b) Since I C  I B  I F  0  I B  I F (1   F )

 I B  I ES ( e qVBE  1)(1   F )  VBE 

 IB kT  ln   q  I ES (1   F ) 

V EC  V BC  V BE  0 

 kT  I ES (1   F )  IB kT  ln  ln   . q  I ES (1   F )  q  IB 

Drift-Base Transistors 8.10 (a) I B is independent of changes in base parameters. I C is dependent on base parameters. You should convince yourself that this is the case by referring to

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

the current equations in the reader. If a SiGe base is used, I c increases as a result of the base bandgap narrowing. Coupled with a graded base that shortens base transport time, SiGe-base BJTs are a simple and attractive alternative to conventional Si-base BJTs. * This solution ignores the case of increased hole barrier between base and emitter. If you wish to include this effect, then I B (SiGe) is increased by eEg/kT over I B (Si) (b) In this case, n iB (SiGe) varies along the base region. This requires that we do an integration to find J c0 (SiGe).

qDB ni2 J c0 (SiGe) = NB

=

1 WB  E g ,SiGe x exp( )dx 0 kT WB E g ,SiGe / kT qDB ni2

N BWB 1  exp(  E g ,SiGe / kT )

.

Divide J c0 (SiGe) by J c0 (Si) and you get the following: (SiGe)/ (Si) =

8.11

(a)

E g ,SiGe / kT

1  exp(  E g ,SiGe / kT )

= 4.

Find where N dE (x) = N aB (x) to obtain the first junction.

20

10 e



x 0.106

 4  10 e 18



x 0.19

 x1  0.77 m .

To obtain the second junction, equate N aB (x) to the background concentration. 

x

4  1018 e 0.19  5  1019

 x2  1.27 m .

Therefore, the base width is x 2 -x 1 = 0.5 m.

(Please note that the depletion widths have been ignored in this case. In general, you must subtract the depletion region widths in the base in both the junctions from the metallurgical base width.) (b) Base Gummel Number:



x2

x1





 

N aB ( x)dx   4  1018 0.19  10  4 exp  x

 cm  1.23  10 cm .

0.19 x2

2

12

2

x1

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

From Eq. 8.2.12, the base Gummel number is the number above divided by the electron diffusivity in the base, which we shall assume to be around 30 cm2/s.

The result is 41010 s-cm-4 Emitter Gummel Number:



x1

0



 



N dE ( x)dx   10 20 0.106  10  4 exp  x

 cm  1.06  10 cm .

0.106 x1

2

15

2

0

From Eq. 8.3.2, the emitter Gummel number is the number above divided by the hole diffusivity in the base, which we shall assume to be around 2 cm2/s. The result is 21014 s-cm-4 (c) Since the doping level is not constant, we use the average doping densities to estimate the diffusivities.

Average base doping density: GN B 1.23  1012  cm 3  2.46  1016 cm 3 .

4 WB 0.5  10 Average emitter doping density: GN E 1.06  1019   1.38  1019 cm 3 . xE 0.77  104 Average electron diffusivity in the base:





D' nB   n 2.46  1016 kT / q  1150  0.026 cm 2 s 1  29.9 cm 2 s 1 .

Average hole diffusivity in the emitter:





D' pE   p 1.38  1019 kT / q  70  0.026  cm 2 s 1  1.82 cm 2 s 1 .

  1

1  GN B D' pE GN E D'nB

1  0.99993 .

1.23  1012  1.82 1 1.06  1015  29.9

(d) Diffusion current:

J diff   q  qD dp

dx

 

D dn d ln n  D . n dx dx

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

p  N B  4  1018 e  x /  ,   0.19 m    

D

( x direction) .



Drift current: J drift  q p pE  qp  J diff  q   q   D and    D    p E  E   D  kT   kT 

p  

Therefore, Ebi 

kt 0.06 V   3.16  105 V / m . q 0.19 m

Note: E bi should be “-x” direction so that diffusion current and drift current will balance.

Ec

Ef

Ev

Kirk Effect -

8.12

Base

N Collector

N+ Collector

Original Base Witdth x Effective Base Witdth

Depletion Layer, W dep

Clearly, W B_Effective = W B_Original + (W C -W dep ). W B_Original and W C are assumed to be known. So, in order to find W B_Effective , we need to calculate W dep .

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

IC where v sat is the saturation velocity. The length of the depletion AE v sat region becomes 2 s VBC   i  Wdep  .

IC  qN C AE v sat

  qN C 

Therefore,    WB _ Effective  WB _ Original  WC   

  2 s V BC   i   .

IC  qN C  AE v sat 

Charge Control Model 8.13

The equation describing the system is dQF (t ) Q (t )  I B (t )  F . dt  F F Since I B (t) = 0 for t<0 and I B (t) =I B0 for t≥0, the equation becomes dQF (t )  1  Q F (t )  I B 0    dt   F F  with the initial condition Q F (0) = 0. Solving this equation yields QF (t )   F  F I B 0 1  e t /  F  F .





Hence, I C (t )  QF (t ) /  F   F I B 0 1  e  t /  F  F .



8.14



dQB QB t /  F  iB   QB  A  Be dt  F Boundary Conditions: F QB ()  A    F iB 2 , QB (0 )    F iB1    F iB 2  B Hence, B= F (i B1 -i B2 )

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Therefore, QB    F iB 2    F iB1  iB 2 e t /   F Q iC  B   iB 2   iB1  iB 2 e  t /   F ,

F    F  100,   F  1ns (Chracteristic time T ) 1F

And, iC  1 mA  (9 mA)e t / 1ns for t  0, and iC  10 mA

for t  0.

iC(t) T=1s T = 1ns

10mA

1mA

t

Cutoff Frequency 8.15

Consider the following figure: ib

ic

+ Signal Source

C

r

v be

g m v be

Load

-

i b and i c are given by ib  vbe / input impedance  vbe  input admittance  vbe 1 / r  jC  ic  g m vbe .

The gain is i gm 1     c   ib 1 / r  jC 1 / g m r  j   jC dBE / g m 

1 .

1 /  F  j   jC dBE kT / qI C

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

If  F >>1 so that 1/ F becomes negligible, the equation above shows that  F ()  1/, and  becomes 1 at fT 

1

2  F  C dBE kT / qI C 

.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.