Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

SOLUTION SET to Exercises in

FUNDAMENTALS OF MODERN VLSI DEVICES, 3rd ed.

published by Cambridge University Press

Yuan Taur and Tak H. Ning

SOLUTION

1

Solutions to Chapter 2 Exercises

2.1.

From Eq. (2.4),

f ( E f  E ) 

1 eE / kT ,  1  e E / kT eE / kT  1

f ( E f  E ) 

1 .

1  eE / kT

and

Adding the above two equations yields

eE / kT  1 f ( E f  E )  f ( E f  E )  E / kT  1. e 1

2.2.

Neglecting the hole (last) term in Eq. (2.19), one obtains N ce

 ( E c  E f )/ kT

 2N c e

 ( E c  E d  2 E f )/ kT

 Nd .

Treating exp(Ef/kT) as an unknown, the above equation is a quadratic equation with the solution E / kT

e f



1  1  8( N d / N c )e( E c  E d )/ kT .

4e E d / kT

Here only the positive root has been kept. For shallow donors with low to moderate concentration at room temperature, (Nd/Nc)exp(Ec  Ed)/kT  1, and the last equation can be approximated by E / kT

e f



4( N d / N c )e( E c  E d )/ kT N d E c / kT  e , 4e E d / kT Nc

which is the same as Eq. (2.20). If we compare the above relation with Eq. (2.19), it is clear that in this case, exp[(Ed  Ef)/kT] << 1, and Nd+  Nd or complete ionization.

If the condition for low to moderate concentration of shallow donors is not met, then exp[(Ed  Ef)/kT] is no longer negligible compared with unity. That means Nd+ < Nd (Eq. (2.19)) or incomplete ionization (freeze-out). [Note that incomplete ionization never occurs for shallow impurities: arsenic, boron, phosphorus, and antimony at room temperature, even for doping concentrations higher than Nc or Nv. This is because in heavily doped silicon, the

SOLUTION

2

impurity level broadens and the ionization energy decreases to zero, as discussed in Subsection 9.1.1.2.]

2.3. (a) Substituting Eqs. (2.5) and (2.3) into the expression for average kinetic energy, one obtains 

 (E  E ) e K.E.   (E  E ) e

3 / 2  ( E  E f )/ kT

c

Ec 

1/ 2  ( E  E f )/ kT

c

Ec

dE

. dE

Applying integration by parts to the numerator yields 

K.E. 

(3 / 2)kT  ( E  Ec )1/ 2 e 

 ( E  E f )/ kT

Ec

 (E  E ) e

1/ 2  ( E  E f )/ kT

c

Ec

dE

dE



3 kT .

2

(b) For a degenerate semiconductor at 0 K, f(E) = 1 if E < Ef and f(E) = 0 if E > Ef. Here Ef > Ec. Therefore, ( E  E ) dE 3  K.E.   ( E  E ).

5 ( E  E ) dE  Ef

3/ 2

c

Ec Ef

1/ 2

f

c

c

Ec

2.4. With the point charge Q at the center, construct a closed spherical surface S with radius r.

By symmetry, the electric field at every point on S has the same magnitude and points outward perpendicular to the surface. Therefore,

 E  dS  4r E , 2

S

where E is the magnitude of the electric field on S. 3-D Gauss’s law then gives E

Q 4 si r 2

,

which is Coulomb’s law.

Since E =dV/dr, the electric potential at a point on the sphere is V

SOLUTION

Q 4 si r

,

3

if one defines the potential to be zero at infinity.

2.5.

(a) Construct a cylindrical Gaussian surface perpendicular to the charge sheet as shown:

E

A Qs E The cross-sectional area is A. At the two ends of the cylinder, the electric field E is perpendicular to the surface and pointing outward. Along the side surface of the cylinder, the field is parallel to the surface, so EdS = 0. For an infinitely large sheet of charge, E is uniform across A from symmetry. Therefore,

 E  dS  2AE .

S

The charge enclosed within the surface is QsA. From Gauss’s law, one obtains E = Qs/2. (b) The field due to the positively charged sheet is Qs/2 pointing away from the sheet.

The field due to the negatively charged sheet is also Qs/2, but pointing toward the negatively charged sheet. In the region between the two sheets, the two fields are in the same direction and the total field adds up to Qs/, pointing from the positively charged sheet toward the negatively charged sheet. In the regions outside the two parallel sheets, the fields are equal and opposite to each other, resulting in zero net field.

2.6

SOLUTION

4

For x < 0,

d 2 (  i ) q 2 N d   i  0  si kT dx 2 General solution:

 i  Ae x / LD where

LD 

 si kT q2 Nd

For x > 0,

d 2 (  i ) q 2 N d q   i   N d 2  si kT  si dx General solution:

 i  Be  x / LD 

kT N d qN d

Matching i and di/dx at x = 0,

A  B 

kT N d 2qN d

2.7 Electron density per energy  N ( E ) f ( E )  E  EC e

 ( E  E f ) / kT

Differentiate the above and set to zero gives

SOLUTION

5

E  EC  kT / 2  13 mV

2.8 (a)

The bands are flat in the region to the right Electron density:

At point A:

charge neutral.

n = NC  exp[(EC – Ef)/kT] = 1.3  1016 cm-3.

Therefore, n-type with doping density (b)



Nd = Nd+ = 1.3  1016 cm-3.

n = NC  exp[0.3/kT] = 2.8  1014 cm-3. p = ni2/n = 3.5  105 cm-3.

At point B:

n = NC  exp[0.4/kT] = 6  1012 cm-3. p = ni2/n = 1.7  107 cm-3.

(c) At point B:

At point A:

Net charge = Nd+  n  1.3  1016 cm-3.

Net charge = Nd+  n  1.3  1016 cm-3.

2.9 (a) Eq. 2.20,

N  Ec  E f  kT ln  c  .  Nd 

Table 2.1,

Nc = 2.9 × 1019 cm-3

Therefore,

Ec  E f  0.266 eV

(b) Eq. 2.67

LD 

 si kT  0.13  m q2 N d

(c) ρnεsi is the dielectric relaxation time.

-3

From Fig. 2.10, ρn = 4.3 -cm for n-type, 1015 cm density. So ρnεsi = 4.5 ps.

SOLUTION

6

(d) Eq. (2.10)

Nc  2g

 2 kT  mt2ml  2   h



3/2

 T 3/ 2

Nc(100C) = (373/300)3/2 Nc(300 K) = 4.0 × 1019 cm-3 Therefore,

N  Ec  E f  kT ln  c   0.274 eV  Nd 

at 100C.

2.10 (a)

E f  Ei  0.2 eV ( E f  Ei ) / kT

 2.26  1013 cm3

( Ei  E f )/ kT

 4.43  106 cm3

Eq. (2.14),

n  ni e

Eq. (2.15),

p  ni e

Ionized donor density

Nd+ = 1016 cm-3.

Therefore,

not charge neutral, net positive charge.

(b) See above. (c) The one in the middle so electrons fall toward positive charge in the center.

SOLUTION

7

Solutions to Chapter 3 Exercises 3.1

Equation (3.138) gives









W W x 1  exp   ( p   n )dx    n exp   ( p   n )dx  dx .

0 0 0 Mp

(1)

As Mp  , 1/Mp  0, the above equation gives W W x 0  exp   ( p   n )dx     n exp   ( p   n )dx  dx  0  0  0 

(2)

 (   ) exp(   )W   1 .



 exp ( p   n )W 

n

p

p

Therefore,

 n   p exp ( p   n )W  ,

or

W  ln  n /  p /  n   p .



 

n

n

(4)



Similarly, Eq. (3.139) gives the same expression for W as Mn  .

x

u( x )    f ( x ' )dx ' .

(1)

Therefore

du( x )   f (x) dx

(2)

and

deu( x )   f ( x )e u ( x ) . dx

(3)

3.2

Let

0

Using (3), we have

 f ( x )exp  f ( x' )dx' dx   f ( x )e W

x

W

0

0

0

u( x )

dx

(4)

deu( x ) dx  1  eu(W ) .

0 dx

= 

W

 f ( x )exp  f ( x' )dx' dx  1  exp  f ( x' )dx' .

Therefore,

W

x

W

0

0

0

(5)

A similar procedure can be used to show that  W f ( x ' )dx ' dx  1  exp  Wf ( x ' )dx ' f ( x )exp 0  x   0  W

(6)

W

by letting u( x )    f ( x ' )dx ' . x

SOLUTION

8

3.3

From Eqs. (3.10), (3.11), and (3.13), we have 2 si ( Na  Nd ) m , qNa Nd

Wd 

 m   bi  Vapp , dQd (p - side)  si .  dVapp Wd

and

Cd 

Therefore,

2( N a  N d ) 1  bi  Vapp  .  2  si qN a N d Cd

(1)

A plot of 1/ Cd2 versus Vapp gives a straight line, which intercept the horizontal Vapp axis at Vapp = bi and has a slope of 2(Na + Nd)/siqNaNd. Now, from Eq. (3.3), we have n p  N N  q bi  kT ln  n 0 2 p 0   kT ln  d 2 a  .  ni   ni 

(2)

Therefore, from the plot of 1/ Cd2 versus Vapp plot, we have

Slope 

and

 2( N a  N d ) ,  si qN a N d

1 C (Vapp  0) 2 d



(3)

kT  N a N d  2( N a  N d )  . ln q  ni2   si qN a N d

(4)

The two equations (3) and (4) can be solved for the two unknowns Na and Nd. 1/Cd2

0

SOLUTION

 bi

Vapp

9

3.4 Na (x)

0 n+ region

W(V)

x

p region

For the one-sided n+p diode, the incremental increase in the magnitude of the depletion-layer charge dQd on the p-side as a result of an incremental increase in the applied voltage dV is

dQd  qN a (W )dW ,

(1)

where the depletion-layer width W is a function of V. The depletion-layer capacitance is

dQd dW ,  qN a (W ) dV dV

(2)

C . q dW / dV 

(3)

From

C

we have

dW  dC  si C d (1 / C 2 )   si2  . dV C dV 2 dV

(4)

C which gives

N a (W ) 

 si W

,

Substituting (4) into (3), we have

N a (W ) 

3.5

2 . q si [d (1 / C 2 ) / dV ]

Referring to Fig. 3.6 and Eq. (3.20), we have

E (Wd  d ) , 2

 m   bi  V  m

(1)

where V is the applied voltage. From Gauss’s law we have

SOLUTION

10

Qd   siEm .

(2)

Therefore, dEm (Wd  d ) / 2 1 dV 1  dWd  .    (Wd  d )  Em  Cd dQd d siEm  2 si  dEm 

(3)

Now, from Eq. (3.19), we have Em

d ( xn  x p ) dx dWd dx  Em = Em n  Em p  ( xn  d )  x p . dEm dEm dEm dEm

(4)

Substituting (4) into (3), we have

1 Wd  .

Cd  si

3.6

The band-to-band tunneling can be represented as tunneling through a triangular barrier

of height Eg, slope qE and tunneling distance Eg / qE. This is illustrated in the figure below.

Energy

Ec

slope = qE

Ev Eg 0

W

x Ec Ev

p-region

n-region

The WKB tunneling coefficient is, from Eq. (3.127),

  4 W  T ( E )  exp  2m*  q ( x)  E dx  0 h  

(1)

with W = Eg / qE. For the coordinates indicated in the figure, we have

 q ( x)  Eg  qxE .

(2)

Also, for the tunneling electron being considered, E = 0. Substituting into (1), we have

  4 W  T ( E  0)  exp  2m* Eg  qxE dx   h 0 

SOLUTION

(3)

11

  4 2m* E 3 / 2  g .  exp  3qE  

3.7 (a) p-type Ec Ef Ev 2V Ec Ef Ev n-type

 bi 

(b)

kT  N a N d    0.82 V ln q  ni 2  Wd 

(c)

Emax 

qN a (Wd / 2)

 si

2 si ( N a  N d )( bi  VR )  0.27 m qN a N d

 2.1105 V/cm

3.8 (a) Wd Ec

 bi

Ei Ef Ev

Ec E f

Ev n+

p-type x=0

SOLUTION

12

N  Ei  E f  kT ln a   0.38 eV  ni 

Eg

(b)

 bi 

(c)

Wd 

(d)

N  E f  Ev  kT ln v  increases as temperature increases.  Na 

3.9 

2q

 0.38 eV = 0.94 eV

2 si bi 5  0.20 m = 210 cm qN a

So

Ei  E f

And

 bi

decreases as temperature increases. decreases as temperature increases.

Using Eq. (3.45), Eq. (3.47) can be rearranged to give

 [n p (0)  n p 0 ] ni

 e q[ i (0) n (0)] kT  e q[ i (0) n ( Wd )] kT 

n p (0) ni

N  d e  q i kT . ni

(1)

Equation (3.43) gives e q i kT 

ni2 qVapp kT n p 0 qVapp kT e  e .

Nd Na Nd

(2)

Therefore, Eq. (1) can be reduced to

n p (0) np0

e app    .

1  qV

kT

(3)

Using Eqs. (3.45), (2) and (3), we have

n (Wd )  n (0)   i 

kT  N d  ln   q  n p (0) 

 qV kT kT 1   e app   ln  . q  1  

SOLUTION

(4)

13

3.10 xm

0

x

q (x) field

q Bn q

q image(x) Ec

Ef

Total

Ev Metal

Silicon

Referring to the figure, the electrostatic potential associated with the image force is

 image ( x) 

q 16 0 x

,

(1)

and the electrostatic potential associated with the electric field in the silicon is given by 

d 2 field ( x) dx

2



qN d

 si

.

(2)

Integrating (2) twice, subject to the boundary condition that the electric field is zero at x = Wd, where Wd is the depletion layer width, and the boundary condition that field(0) = 0, we have

 field ( x)  

 qN d  x 2   Wd x  .  si  2 

(3)

Therefore, the total electrostatic potential energy is  q qN d  x 2  Wd x   .

PE ( x)  q[ image ( x)   field ( x)]  q   2  16 si x  si 

(4)

The peak of PE(x) is given by dPE(x)/dx = 0, that is q 16 x

2 si m



qN dWd  x  1  m   0 .  si  Wd 

(5)

We will first make the assumption that xm/Wd << 1, and then come back to verify that our assumption is correct.

SOLUTION

With this assumption, (5) gives

14

q

xm 

16 siEm

(6)

 d field ( x)

Em 

where

,

dx

 x 0

qN dWd

(7)

 si

is the absolute value of the electric field in the silicon at the interface. The energy barrier lowering is  q qN d  x 2  Wd xm  m  q | PE ( xm ) | q    si  2  16 si xm

 q qN dWd xm   q    si 16 si xm 

(8)

q 3Em .

4 si

To show that xm/Wd << 1, we note that Wd is related to the total band bending. For simplicity we assume there is no external applied voltage, so that Wd 

2 si bi , qN d

(9)

where bi is the built-in potential. For bi = 0.4 V and Nd = 11016 cm3, we have Wd = 2.3105 cm and xm = 3107 cm.

3.11

That is, xm/Wd << 1 as assumed.

The multiplication factor for hole-initiated impact ionization is





 

 











W W x 1  exp    p   n dx    n exp    p   n dx dx, (1) 0 0 0 Mp

and that for electron-initiated impact ionization is











W W W 1  exp    n   p dx    p exp    n   p dx dx. (2) 0 0 x Mn

From Ex. 3.2, we have

 f  x  exp   f  x dx dx  1  exp   f  x  dx  W

x

W

0

0

0

(3)

and



W

0









f  x  exp   f  x  dx dx  1  exp   f  x  dx .

SOLUTION

W

x

W

0

(4)

15

Applying Eq. (3) to Eq. (1), we obtain

1





W x 1    p exp    p   n  dx dx.

0 0 Mp

(5)

Similarly, applying Eq. (4) to Eq. (2), we have

1





W W 1    n exp    n   p  dx dx.

0 x Mn

(6)

Avalanche breakdown occurs when carrier multiplication by impact ionization runs away, i.e., when the multiplication factor becomes infinite. Thus, Eq. (5) gives the hole-initiated breakdown condition as





  p exp   p   n  dx dx  1.

W

0

x

0

(7)

Using Eq. (3) the left-hand side of Eq. (7) can be rearranged to give

  exp       dx dx  1  exp       dx  W

x

p

0

W

p

0



n



0

   n exp    p   n  dx dx.

W

0

x

0

p

n

(8)

Substituting Eq. (7) into Eq. (8) gives

  exp       dx dx W

0

x

n



0

 exp  

W

p

    dx  . p

0

n

(9)

n

Dividing both sides of Eq. (9) by its RHS gives





  n exp   n   p  dx dx  1, W

0

W

x

(10)

which, according to Eq. (6), is simply the condition for electron-initiated breakdown. Thus, the condition for avalanche breakdown is the same whether the breakdown process is initiated by electrons or by holes.

SOLUTION

16

3.12

On p-side, Ei (xp) – Ef = kT ln(Na /ni) = 0.400 eV, on n-side, Ef – Ei (xn) = kT ln(Nd /ni) = 0.417 eV. Wd  xn  x p 

2 m

Em



So bi = m = 0.817 V.

2 si m ( N a  N d )  1.782  105 cm qN a N d

xn = [Na /(Na + Nd)]Wd = (1/3)Wd,

xp = Wd  xn = (2/3)Wd,

Ei (0) – Ei (xn) = (1/2) xn  Em = (1/2)(Wd /3)(2m /Wd) = m /3 = 0.272 V. At x = 0, Ei – Ef = [Ei (0) – Ei (xn)] – [Ef – Ei (xn)] = 0.272 – 0.417 = –0.145 eV. Ei crosses Ef on the p-side when Ei (xp) – Ei (x) = Ei (xp) – Ef = 0.400 eV. Since Ei (xp) – Ei (x)  (x + xp)2 and Ei (xp) – Ei (0) = (1/2) xp  Em = 2m /3 = 0.545 V, solve x from (x + xp)2/xp2 = 0.400/0.545

SOLUTION



x = –0.170105 cm.

17

3.13 depletion region

n-type (Nd

p-type (Na

=1e18 cm-3)

=1e17 cm-3)

x 0

At x = 0,

10m

J n (0)   qDn ni2 / ( N a Ln )  exp qV / kT = 0.44 A/cm2.

Hole current equals the diffusion current on the n-side, i.e., J p (0)   qD p ni2 / ( N d L p )  exp qV / kT = 0.088 A/cm2.

At x = 10 m, J n (10 m)  J n (0)e  x / Ln = 0.27 A/cm2.

The total current is continuous, so

J p (10 m)  J n (0)  J p (0)  J n (10 m) = 0.26 A/cm2.

3.14 (a)

Built-in potential  bi  Eg / 2q  (kT / q) ln( N a / ni ) = 0.98 V.

Depletion layer width Wd   2 si ( bi  0.6) / qN a 

1/2

= 70 nm.

Depletion capacitance = si /Wd = 0.148 F/cm2.

(b)

The total current is dominated by the electron diffusion current:

J n ( x  0)  [qDn ni2 / N a Ln ]exp( qVapp / kT ) = 9.2 mA/cm2.

(c)

SOLUTION

J n ( x)  J n ( x  0) exp( x / Ln ) = 5.6 mA/cm2.

18

3.15

 bi 

(a)

Wd 

kT  N a N d    0.804 V ln q  ni 2 

2 si ( bi  0.5)( N a  N d )  1.31  10 5 cm qN a N d

Qd ( p  side)  qN a x p  qN aWd

(b)

Cd 

 si Wd

Nd  4.8  108 C/cm 2 , negative Nd  Na

 7.9  108 F/cm2

3.16 (a)

 bi 

Eg 2q



kT  N a    0.56  0.42  0.98 V ln q  ni 

Wd 

2 si ( bi  VF )  7  10 6 cm qN a

Emax 

qN aWd

 si

 1.1  105 V/cm

2

(b)

SOLUTION

np 

ni qV F / kT e  1.15  1013 cm -3 Na

19

3.17 (a)

p-type

n-type

Ec

0.5 V

Ef Ev

(b)

 bi 

Ec Ef Ev

kT  N a N d    0.90 V ln q  ni 2  Wd 

Emax 

2 si ( N a  N d )( bi  0.5)  0.075 m qN a N d qN aWd N d /( N a  N d )

 si

 1.1105 V/cm

2

(c)

np 

ni qV / kT e  4.41  1011 /cm3 Na

3.18 (a)

 bi 

kT  N a N d    0.70 V ln q  ni 2 

Wd 

2 si ( N a  N d ) bi  0.522 m qN a N d

(b)

SOLUTION

xn 

Na Wd  0.17 m Na  Nd

xp 

Nd Wd  0.35 m Na  Nd 20

(c)

SOLUTION

Em 

2 bi  2.68  104 V/cm Wd

21

Solutions to Chapter 4 Exercises 4.1.

(a) Using the integral expressions of Qd and Qi, one has Cd  

dQd qN a (1  e q s / kT )  , d s Es

and dQ q( ni / N a )( eq s / kT  1) Ci   i  . d s Es 2

Here Es = E ( = s). (b) Take the square of Eq. (4.16) and differentiate with respect to s, we obtain 2Qs

2   dQs n  2 si qN a  1  e q s / kT  i 2 eq s / kT  1  . d s Na  









Using Qs =siE s (Gauss’s law) and the above two equations in (a), it is straightforward to show that Csi  

2  dQs qN a  ni  q s / kT q s / kT  1  e   1   Cd  Ci .  2 e d s Es  Na 









(c) When s = 2B, exp(qs/kT) = (Na/ni)2 >> 1 and exp(qs/kT) << 1. From (a), one has Ci  Cd 

qN a .

Es

(d) It is clear from Fig. 4.7 that Qs and therefore Es takes off rapidly beyond strong inversion. This means that Cd  E s1 decreases rapidly beyond strong inversion. We say that the depletion layer (charge) is “screened” by the inversion layer. Note that Ci, on the other hand, increases rapidly beyond strong inversion because of the exp(qs/kT) factor.

4.2 (a)

SOLUTION

Cox 

 ox tox

 Cmax  3.4  10 7 F/cm2



tox = 10 nm.

22

(b)

Cd 

Cmin 

CoxCd  7.3  108 F/cm2 Cox  Cd

 si Wdm



Wdm 



2 si (2 B )  1.1  10 5 cm qN a

Take 2B  0.9 V (c)

Cd = 9.410-8 F/cm2

1 1 L 1 kT   D   C fb Cox  si Cox  si q 2 N a

 

Na = 9.61016 cm-3, p-type.

Cfb = 2.410-7 F/cm2

From C-V, Vfb = 0.4 V. (d)

s    Eg / 2q   B  5.03 V

4.3.



m  s  V fb  4.63 V

Eq. (4.33),

 s  Vg  V fb 

 si qN a 2Cox

2



 si qN a / 2 Cox

.

Solve for Vg, 2

  si qN a / 2  Vg  V fb       . s Cox 2Cox 2  

 si qN a

Take a differential,   qN / 2  1  qN / 2 s Vg /  s  2   s  si a   1  si a .

Cox Cox   2  s

This is simply s = Vg/(1 + Cd/Cox), where Cd is the depletion charge capacitance given by Eq. (4.38).

4.4. By keeping only the dominant inversion charge term in Eq. (4.16) and substituting it in Eq. (4.7), one obtains

SOLUTION

23

Vg  V fb   s 

2 si kTN a  ni  q s / 2 kT .  e Cox  Na 

When Vg increases beyond the point where s = 2B, most of that increase appears as inversion charge in the third term. There is very little change in the band bending beyond s = 2B. A good approximate solution for s is then obtained by letting s = 2B for the second term and solving s from the third term:

 s  2 B 

2kT  Cox (Vg  V fb  2 B )  ln  .  q 2 si kTN a  

Here exp(qB/kT) = Na/ni has been applied.

It is clear from the above equation that s  2B is a weak function of Vg. As an example, for Na = 1017 cm3, tox = 100 Å, and Vg  Vfb = 4 V, one has 2B = 0.82 V, and s  2B  7.2kT/q  0.19 V.

4.5. In Fig. 4.13(b), the total capacitance seen by the gate equals Cox and (Ci + Cd) connected in series, i.e., C

d ( Qs ) Cox ( Ci  Cd )  . dVg Cox  Ci  Cd

Since Qs = Qi + Qd and Qi/Qd = Ci/Cd, it follows that d ( Qi ) Cox Ci  . dVg Cox  Ci  Cd

When Vg is below the threshold of strong inversion, Ci << Cox, therefore d(Qi)/dVg  Ci/(1 + Cd/Cox) << Cox. Above strong inversion, Ci >> Cox (> Cd) (see Exercise 4.1), and one has d(Qi)/dVg  Cox. Such a behavior is sketched schematically below.

d(-Qi )

Cox

dVg

Vg

SOLUTION

24

The sharp transition takes place at the threshold voltage. Qi(Vg) is simply the integrated area under the above curve. Beyond threshold, Qi increases linearly with Vg with a slope equal to Cox (see Fig. 5.4 in the text).

4.6.

  Eg  kT N d    (0.56  0.35)  0.21 V. V fb    B    0.56  ln q ni   2q  

C

Vfb

Vg

Band diagram at Vg = 0:

Ef

Ef

n+

n-type

At s = 2B = 0.7 V:

Vox 

2 si qN d s 4.8 108 Qd    0.14 V. Cox Cox Cox

4.7.

SOLUTION

25

B 

(a)

kT N a ln  0.42 V q ni Eg V fb    B  0.98 V 2q

Vt  V fb  2 B 

2 si qN a (2 B )  0.98  0.84  0.97  0.83 V. Cox

(b)

C Cmax

Cmin Vfb

(c)

Vt

Vg

Cmax  Cox  1.73 107 F/cm2

Wdm 

2 si (2 B )  0.104 m qN a

1 1 Wdm   Cmin Cox  si

Cmin  0.63 107 F/cm2

4.8.

Let the surface potential at zero gate voltage be s.

Then

 V fb   s 

2 si qN a s Cox

Solve the above quadratic equation in

SOLUTION

s :

26

s  

2 si qN a 2 si qN a   V fb 2 2Cox 4Cox

2 si qN a  0.278 V, 2 4Cox Therefore,

s = 0.353 V 2

Electron density at the surface: n 

hole density at the surface:

ni q s / kT e  8.3 108 cm-3 Na

p  N a e  q s / kT  1.2 1011 cm-3

4.9 (a)

s = 0.5 V,

Qd = (2si q Na s)1/2 = 4.110-8 C/cm2

Cox = ox/tox = 3.4510-7 F/cm2 Vg = s + Qd/Cox = 0.62 V. (b)

W d = (2si s/ q Na)1/2 = 0.25 m

si/W d = 4.110-8 F/cm2 C = Cox(si/W d)/[Cox + si/W d] = 3.6610-8 F/cm2. (c) Vg = 2 V  inversion  s (max) = 2 B = 2 (kT/q) ln (Na/ni) = 0.72 V Vox = Vg – s(max) = 1.28 V. (d)

Vg = -2 V



accumulation

In the bulk, Ef is (Eg/2q) – B = 0.2 V above Ev. This is the most upward band bending at the surface,

i.e., s (min) = 0.2 V

SOLUTION

so Vox = Vg  s(min) = 1.8 V.

27

4.10.

B 

kT  N a    0.4083(V ) ln q  ni 

Qs  Cox (Vg  V fb  2 B )  1.09  106 (C / cm 2 )

x poly 

Since p 

 Qs / q  0.680(nm) N poly

qN poly x 2poly

 35.5(mV ) , the band-bending in the gate can be neglected to the first 2 si order in the charge calculation.

4.11.

Electron energy level under triangular approximation, the electron energy levels are given by

 3hqEs  3  Ej    j   4   4 2mx 

B 

2/3

where j  0,1,2, 

kT  N a    0.4679(V ) ln q  ni 

2 si (2 B )  34.9(nm) qN a q Wd qN aWd Es  N ( x)dx   0.537( MV cm)  0  si  si

Wd 

Substituting Es and j gives, E0  0.1143(eV ) , E1  0.2010(eV ) , E2  0.2717(eV ) , …

4.12. We will do this by showing that







nT ( t )  N T 1  exp  N inj ( t ) ,

jG ( t  ) dt  , q is a solution to the differential equation. From (1), we have

where

N inj ( t )  

t

0





dN inj ( t ) dnT .  N T exp N inj ( t ) dt dt SOLUTION

(1)

(2)

(3)

28

From (2), we have dN inj ( t ) jG ( t )  . dt q

(4)

Substituting (4) into (3) and using (1), we have dnT j  G  ( N T  nT ) . dt q 1E+13

nT (cm-3)

NT = 5E12 cm-3  = 1E-13 cm2 1E+12

1E+11 2E+10 1E+11

1E+12

1E+13

5E+13

Ninj (cm-2)

SOLUTION

29

Solutions to Chapter 5 Exercises

5.1. kT N a ln  0.42 V q ni Eg V fb    B  0.98 V 2q

B 

(a)

Vt  V fb  2 B 

2 si qN a (2 B )  0.98  0.84  0.97  0.83 V. Cox

Ec Ei Ef Ev Ef

p-type silicon

Oxide

SOLUTION

30

(b)

Vt  V fb  2 B 

2 si qN a (2 B  Vbs )  0.98  0.84  1.44  1V.30 Cox

Ec Ei Ef Ev Efn

Ef p-type silicon

Oxide

5.2.

Eq. (5.14) is an implicit equation relating s and V. Taking a differential yields 2 si kTN a  q s ni 2 q( s V )/ kT  0  d s   2e   2Cox  kT N a 

1/ 2

2  q  ni q( s V )/ kT d   e ( d s  dV ) .  s 2 kT  Na 

It is straightforward to re-group the terms and show that

d s ( ni / N a )eq s / kT  dV 1  ( ni 2 / N a 2 )eq s / kT  ( Cox 2 /  si qN a )( Qs / Cox ) 2

2

in the limit of V  0, where 1/ 2

2  q  n Qs  2 s kTN a  s  i 2 eq s / kT  .  kT N a 

SOLUTION

31

5.3. Applying 1-D Gauss’s law to a box bounded by a plane at depth x from the surface and another plane deep in the substrate where E = 0, one can write xi 1   Qd  qx n( x)dx  ,   si 

E( x ) 

where Qd is the depletion charge density and xi is the inversion layer depth.

Qd

q

Therefore,

 n( x )E ( x )dx    n( x )dx    n( x ) n( x)dxdx . xi

0

si

xi

0

si

xi

xi

0

x

If we let n(x) = df(x)/dx, the second term on the RHS becomes x i df df q  1 xi d ( f 2 )  f ( x )  f ( x ) dx  f ( x ) dx  dx  ,  i   si 0 dx  si  i 0 dx 2 0 dx 

q

xi

which can be further simplified to 2 q  1 2 q q  xi 2  2   f ( xi )  f (0)  0 n( x )dx  .  f ( x ) f ( xi )  f ( 0)  f ( xi )  f ( 0)    si  i 2 2 si  2 si





Substituting it back, one has

 n( x )E ( x )dx  Q  Q ,  2  n( x )dx xi

Eeff  0

d

i

si

si

xi

0

where Qi  q  n( x )dx . xi

0

5.4.

This exercise is closely related to Exercise 4.5, in which Eq. (5.25) is derived: d ( Qi ) Cox Ci  . dVg Cox  Ci  Cd

When Vg is below the threshold of strong inversion, d(Qi)/dVg  Ci/(1 + Cd/Cox) << Cox. Above strong inversion, Ci >> Cox (> Cd), and one has d(Qi)/dVg  Cox. This behavior is sketched below.

SOLUTION

32

d(-Qi )

Cox

dVg

Vg Qi(Vg) is simply the integrated area under the curve. Beyond threshold, Qi increases linearly with Vg with a slope equal to Cox (see Fig. 5.4 in the text).

The sharp transition above can be used to define a kind of inversion charge threshold voltage, Vtinv, where Ci = Cox. Since Cd  0 due to screening by inversion charge (Exercise 4.1), dQi/dVg  CoxCi/(Cox+Ci) = Cox/2 at Vg = Vtinv. In the region where Boltzmann approximation is valid, Qs  Qi  exp(qs/2kT), hence Ci  (q/2kT)Qi and Qi  (2kT/q)Cox. Such a threshold voltage is slightly higher than the “2B” threshold where Ci = Cd.

5.5 2 2 2 kT  N a  Cox (Vgs  V fb   s ) q s   V  s  ln  2    q  ni  2 si kTN a kT    

The implicit eq. for s,s and s,d is with V = 0 for s,s and V = Vds for s,d.

For s,s,

 ss 

2 2 2 2 2 kT  q ss   kT  Cox (Vgs  V fb   ss ) q ss   N  C (V  V fb   ss )  ln a2  ox gs   2   ln     B q  ni  2 si kTN a kT   q  2 si kTN a kT   

The s,s in the square bracket of ln can be approximated by 2B = 2 (kT/q) ln(Na/ni) = 0.87 V. Cox = 4.910-7 F/cm2, therefore

s,s = 0.87 + 0.0259ln[178.6  33.6] = 0.87 + 0.13 = 1.0 V.

For s,d in saturation, pinch-off or Qi  0, Vds  ,

Cox (Vgs  V fb   sd ) 2 2

2 si kTN a

SOLUTION



q sd 0 kT

33

The solution is

1 2 

 sd  

 2 si qN a 2 si qN a   4 ( V  V )  gs fb 2 Cox Cox 

s,d = 1.38 V.

2siqNa/Cox2 = 0.28 V, so

5.6

B 

Wdm 

kT N a ln  0.435 V q ni

2 si (2 B )  0.075 m qN a

m = 1 + 3tox/Wdm = 1.4 Inverse SS = m60 mV/dec = 84 mV/decade

Qd  qN aWdm  2.4  107 C/cm2 Qi  Qs  Qd  7.6  107 C/cm2 The effective field is

 Qd  Qi / 2 /  si  6.0  105 V/cm

From the universal mobility curve in Fig. 5.14, eff  400 cm2/V-s

5.7 Integrating from 0 to y after multiplying both sides of Eq. (5.8) by dy yields V

I ds y  eff W  [Qi (V )]dV 0

m    eff CoxW (Vgs  Vt )V  V 2  , 2  

where the simplified Qi(V), Eq. (5.32), has been used.

Substituting Ids from Eq. (5.27) into the equation above, one can solve for V(y):

V ( y) 

Vgs  Vt

SOLUTION

m

 Vgs  Vt  y  Vgs  Vt  y 2    2   Vds  Vds .

L m  L  m  2

34

5.8 A more general form of the saturation voltage is obtained by letting Qi = 0 in Eq. (5.16) with ψs = 2ψB + V and solving for V = Vdsat [equivalent to solving dIds/dVds = 0 by differentiating Eq. (5.22)].

Qi  Qs  Qd  Cox (Vgs  V fb  s )  2 si qN a s

Charge Sheet model,

Pinchoff happens when Qi = 0. Solve the quadratic eq. for s1/2:

 s, max  Vgs  V fb 

 s, max  Vgs  V fb 

Vdsat  Vgs  V fb  2 B 

 si qN a Cox2



 si qN a Cox 2

1

 si qN a 2Cox

 si qN a Cox

2

2





 si qN a 2Cox

 si qN a Cox

 si qN a

1

2

2(Vgs  V fb )Cox 2

2

2(Vgs  V fb )Cox

2

 si qN a

.

It turns out that Eq. (5.27) serves as a good approximation to the drain current over a much wider range of voltages than expected. Even for a drain voltage several times greater than 2ψB, the current Idsat is only slightly (≈5%) underestimated.

5.9.

Let V = 0 in Eq. (5.12): Qi 

qni  s eq / kT d .

N a  B E ( ) 2

In the subthreshold region, both the band bending  and the electric field E are mainly set by the depletion charge. Because of the exponential factor in , the above integration of inversion charge is dominated by that over a thin surface layer within which E() is essentially constant and can be taken outside the integral. Thus one obtains Qi 

2

kTni q s / kT e , Es N a

where Es is the surface electric field.

SOLUTION

Note that exp(qB/kT) << exp(qs/kT) in weak inversion.

35

This equation is generally valid for nonuniform (vertically) dopings with Na being the ptype concentration at the edge of the depletion layer. (Note that the factor Na merely reflects the fact in Fig. 4.6 that the band bending s is defined with respect to the bands of the neutral bulk region of doping Na.)

5.10. From Eq. (5.8),

I ds ( y )  eff WQi ( y )

dV . dy

Current continuity requires that Ids be a constant, independent of y. For a short-channel or a nonuniformly-doped (laterally) MOSFET, dy

eff W

Vds

 Q ( y )  I  dV   W I .

L

0

i

ds

V ds

eff

0

ds

Generalizing the result of the last exercise to laterally varying s and Es, one obtains L dy L Vds 1 Na   Es ( y )e q s ( y )/ kT dy .

2   0 0 I ds eff W Qi ( y ) eff WkTni

Since Es(y)  Vg  Vfb  s(y)/3tox is not a strong function of s, the exponential factor dominates. This implies that the subthreshold current is controlled by the point of highest barrier (lowest s) in the channel.

5.11

Charge Sheet model,

Qi  Qs  Qd  Cox (Vgs  V fb  s )  2 si qN a s

Pinchoff happens when Qi = 0. Solve the quadratic eq. for s1/2:

 s, max  Vgs  V fb 

So,

SOLUTION

 s, max  Vgs  V fb 

 si qN a 2Cox

 si qN a Cox

2

2





 si qN a 2Cox

 si qN a Cox

2

2

1

2(Vgs  V fb )Cox

2

 si qN a

36

Solutions to Chapter 6 Exercises 6.1 For the MOSFET scale length given by Eq. (6.23), show that (a) λ ≥ Wdm and λ ≥ tox, i.e., λ is larger than the larger of Wdm, tox. (b) In the special case of εox = εsi, λ = Wdm + tox. (c) If tox ≪ Wdm (lower right corner of Fig. 6.7), λ ≈ Wdm + (εsi/εox)tox. (a) Let

f (k ) 

k = /

1

 ox

tan  ktox  

1

 si

tan  kWdm 

Assume Wdm > tox with no loss of generality.

As k goes up from 0 to [/(2Wdm)], f(k) stays positive. Once k = [/(2Wdm)]+, the 2nd tangent is –∞ so f(k) becomes negative.

As k increases to [/Wdm], the 2nd tangent goes to zero.

If 2tox < Wdm: The 1st tangent is positive /Wdm 



f(k) changed sign between k = [/(2Wdm)]+ and

There is a  solution  (Wdm, 2Wdm).

If 2tox > Wdm: The 1st tangent goes to +∞ at k = [/(2tox)] < [/Wdm]



f(k) changed

sign between k = [/(2Wdm)]+ and /Wdm. 

In either case, a solution of  exists > Wdm, the larger of (Wdm, tox).

(b)

 πt   πWdm  tan  ox  + tan   = 0.  λ   λ 

εox = εsi,

So tan[tox/ + Wdm/] = 0 which means [tox/ + Wdm/] = , 2, 3, etc.

The largest is λ = Wdm + tox. (c) the first angle  0, the second angle  

If tox ≪ Wdm, So

1

 ox

tan  tox /   

which gives

1

 si

tan  Wdm /   

1  tox

 ox 



1   Wdm  0    si   

λ ≈ Wdm + (εsi/εox)tox.

6.2.

SOLUTION

37

-9

1

x 10

0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 -8

x 10

f (n )   si tan(ti / n )   i tan(Wd / n ) From the figure above, 1 ~ 15.9(nm) , 2 ~ 7.3(nm) , 1 ~ 5.0(nm) .

By substituting the numbers,

exp L 21  ~ 40.14 , when L ~ 21 . exp L 22 

6.3 Show that Eq. (6.38) for short-channel subthreshold current is reduced to Eq. (5.36) of longchannel subthreshold current if 2(x, y) is taken to be the long channel potential v2(x) of Eq. (6.10). [Hint: Convert dx to d/E and approximate E by Es, the field at the surface.]

ni2 I ds  eff kTW Na

Eq. 6.38:



1  e  qVds / kT L dy

0



Wdm

0

.

e q 2  x , y / kT dx

The integral over x can be approximated as



Wdm

0

e 2 q

x , y  / kT

s

dx   e 2  q

x , y  / kT

0

d 2 1 kT q s / kT  e E Es q

For long channel, there is no y-dependence. So the y-integral is just multiplication of L.

 x   2 ( x)   s 1    Wd 

With Eq. 6.10 for long channel,

Es 

Therefore,

SOLUTION

W I ds  eff L

2

2 s 2qN a s  Wd  si 2

2

 si qN a  kT   ni  q / kT 1  e  qV / kT  .   e    2 s  q   N a  s

ds

38

6.4 Under n = 1 velocity saturation, Eq. 6.36 becomes

I ds  W ( Qi )

eff dV / dy .

1  ( eff / vsat )dV / dy

Solve for dV/dy: I ds / eff dV  , dy WQi  I ds / vsat

With -Qi given by Eq. 6.37, Qi ( y )  q

ni2  qV  Wdm  q 2 ( x, y )  exp    0 exp   dx, Na kT  kT   

We have I ds / eff dV  .

Wd q 2 ( x , y )/ kT  qV / kT dy qW ( n 2 / N )   e dx e  I ds / vsat i a   0 

6.5. (a)

I ds 



eff Cox (W / L ) (Vg  Vt )Vds  ( m / 2)Vds 2



(m = 1)

1  ( eff Vds / vsat L ) 8

Drain current (mA)

Vg -Vt =5.0 V

6

4.0 V

4

3.0 V 2.0 V

2

1.0 V

0

(b)

SOLUTION

0

1

2 3 Drain voltage (V)

I dsat  CoxWvsat (Vg  Vt )

4

5

1  2eff (Vg  Vt ) / ( mvsat L )  1 1  2eff (Vg  Vt ) / ( mvsat L )  1

(m = 1)

39

Saturation current (mA)

15 Vg -Vt =3.0 V no velocity saturation

10

5 velocity saturation

0

6.6.

Make use of z 

I dsat  CinvWvsat

0

2 eff (Vgs  Vt ) mvsat L

1

2 3 Channel length (um)

4

5

, Eq. (6.49) becomes

mvsat L 1  z  1 mv L z  CinvWvsat sat 2 eff 2 eff 1 z 1

 1  z  1 .

2

Differentiating the above with respect to z, one obtains

gmsat 

dI dsat dI dsat 2 eff   CoxWvsat 2 dVgs dz mvsat L





 1  z  1 2 11 z  C Wv 1  1  2 (V 1 V ) / mv L  ox

sat



eff

g

t

sat

When L  0, gmsat  CoxWvsat (Eq. (6.79)).

When vsat  , the square root factor can be expanded in a Taylor series to yield

   (V  Vt )  W Vg  Vt , gmsat  CoxWvsat 1  1  eff g   eff Cox mvsat L  L m   which can also be derived from the long-channel current expression, Eq. (6.50).

6.7.

Since Ids = WQiv and Qi = Cox(Vg  Vt) at the source, one has v  vsat

SOLUTION

1  2 eff (Vg  Vt ) / ( mvsat L )  1 1  2 eff (Vg  Vt ) / ( mvsat L )  1

40



for the carrier velocity at the source end of the channel.

When L  0, v  vsat. When vsat  , the denominator approaches 2 and the squareroot expression in the numerator can be expaned in a Taylor series to yield v

 eff (Vg  Vt ) 2mL

,

which also follows from Ids = WQiv = WCox(Vg  Vt)v and the long-channel current, Eq. (6.50).

6.8.

By definition, the total integrated inversion charge under the gate is Qi ( total )   Qi ( y )Wdy .

L

0

Change variable from y to V:

Qi ( total )  

Vds

0

Qi (V )WdV . dV / dy

Substituting dV/dy from Eq. (3.74), one obtains

Qi ( total )   

Vds

0

 eff W eff W 2 2  Qi (V )  Qi (V ) dV .  I ds  vsat 

From Eq. (3.76), Qi(V) = Cox(Vg  Vt  mV), the above integral can be evaluated to yield

CoxWeff Cox W 2 eff 2 2 Qi ( total )  (Vg  Vt )  (Vg  Vt  mVds )  (Vg  Vt )3  (Vg  Vt  mVds )3 .

2mvsat 3mI ds





2





In the saturation region, Vds and Ids are given by Eqs. (6.48) and (6.49). It can be shown that the above equation reduces to Qi ( total )  WLCox (Vg  Vt )

6.9.

(a)

SOLUTION

1  2 eff (Vg  Vt ) / ( mvsat L )  (1 / 3)

Vt  V fb  2 B 

1  2 eff (Vg  Vt ) / ( mvsat L )  1

.

qN a (Wdm  xs ) Cox

41

4 si B 2  xs qN a Eliminate Na from the above eqs, 4 si B xs   Wdm Cox (Vt  V fb  2 B ) Wdm 

Vfb = -Eg/2q – B = -1.06 V, so xs = 17 nm.

Substituting in the 2nd eq., Na = 1.31017 cm-3. (b)

m = 1 + 3tox/Wdm = 1.21 Inv. Subth. Slope = m60mV/decade = 73 mV/decade

(c)

 = Wdm + 3tox = 0.12 m Lmin  2 = 0.24 m

6.10 The depletion charge density in silicon is Qd = q Na (Wd – xs) = 6.410-8 C/cm2.

Surface field:

Es = Qd/si = 61800 V/cm.

Surface potential:

s 

xN dx  W  x   2 E (W  x )  0.37 V.   2 q

si

SOLUTION

qN a

Wd

xs

a

2 d

2 s

1

s

d

s

si

42

Solutions to Chapter 7 Exercises

7.1. (a) The expression for the drain current is given by s

2

W 4 si  2kT    2  si ti 2 2      tan   I ds     tan   .

L tsi  q   2  itsi 

(1)

d

The boundary conditions at the source and drain are:  2 2 si kT  q(Vgs  mi ) 2 si ti  ln   tan  s (2)   ln  s  ln  cos  s   2 2kT q ni   i tsi s  tsi  2 2 si kT  q(Vgs  mi  Vds ) 2 si ti  ln   tan  d . (3)   ln  d  ln  cos  d   2 2kT q ni   i tsi d  tsi Here miV. If the values of βs and βd are solved for given biases Vgs and Vds, the value of the current is obtained by Eq. (1). The problem is then to solve the implicit equations (2) and (3) with given boundary conditions.

1.0

Vg=0.0(V) Vg=0.5(V)

0.8

Vg=1.0(V)

Ids (mA)

Vg=1.5(V) Vg=2.0(V)

0.6

0.4

0.2

0.0 0.0

0.5

1.0

1.5

2.0

Vds (V)

Various numerical methods can be employed: (i) Scan βs over 0   s   2 and use Eq. (2) to generate a look-up table mapping βs to Vgs. Pick βs values that give Vgs = 0, 0.5, 1.0, 1.5, 2.0 V. Interpolation may be needed. For a given Vgs and the corresponding βs, scan βd over (0, /2) and use Eq. (3) to calculate corresponding Vds. An IdsVds curve can then be generated for each Vgs (βs). (ii) The implicit equations can be solved by numerical methods such as Newton Raphson method. Use of internal functions of mathematical tools is also acceptable. (e.g., fsolve or solve of Matlab.)

SOLUTION

43

tan(ti /  ) tan(tsi / 2 )   i /  si

(b)

-9

f=sitan(tox/))-oxcot(tsi/2)

1.0x10

-10

5.0x10

=19.80(nm)

0.0

-10

-5.0x10

5

10

15

20

25

 (nm)

By numerically plotting the function in the graph above, 1  19.80(nm) .

Therefore, the device can be scaled down to Lmin  21  39.6(nm) without severe SCEs. 7.2

 ( x)  V 

(a)

2kT  tsi ln  q  2

 2 x   q 2 ni cos   , 2 si kT  tsi  

Center potential at the source:

 (0)  

2kT  tsi ln  q  2 

q 2ni  2kT  2  ln   2 si kT  q  tsi

2 si kT   q 2ni 

 cannot exceed /2, hence the maximum center potential is

 (0)max 

2kT   ln  q  tsi

2 si kT   q 2ni 

(b) From the boundary condition, q(Vgs  mi  V ) 2kT

 2 2 si kT  1  2 si   tox  1    ln  .   ln(1   )  ln     ln  1   2 R    ox    R q ni  2

At the source, V = 0. For Vgs >> threshold,  << 1.

SOLUTION

44

The potential solution is (V = 0)

 ( ) 

 kT  8 kT (1   ) ln  2 2 si , 2 2  q  q ni R [1  (1   )(  / R ) ] 

At center, with  << 1,

 (   0) 

kT  8 si kT  ln  2 2  , q  q ni R 

7.3 (a)

Let r  sitox/(oxtsi) and a  [2sikT/(q2ni)]1/2 to compact the expressions.

 ( x)  V 

2kT  2  ln  q  t si

2 si kT  2kT lncos2 x / t si   q q 2 ni 

 (tsi / 2) 

At the source, V = 0. Surface potential:

2kT  2  2kT ln  a   lncos  q q  tsi 

Condition for :

q(Vgs  mi  V ) 2kT

2  ln   tsi

2 si kT  2 si tox  tan    ln   ln cos    2 q ni   ox tsi

q(Vgs  mi )

At the source,

2kT Vg >> threshold:

  /2 

tan term dominates.

q(Vgs  mi )

 2a  r  ln    2r  tan   cos   tsi 

2kT Then

 (tsi / 2) 

 2a   ln    ln   ln cos    2r  tan   tsi 

 2a   2kT 2kT    2kT  q(Vgs  mi ) ln  a   ln   ln     ln( r ) q q 2 kT t q si  tsi     

Simplification (optional):

 (tsi / 2) 

 2a    2kT  a  q(Vgs  mi ) ln   ln     q 2kT  tsi     rtsi 

(b) From the boundary condition,

SOLUTION

45

 R q 2n  1    si  tox  i  ln  ln 1     ln 1    ln   2   2kT   ox  R  2 2 si kT  At the source, V = 0. For Vgs >> threshold,  << 1, so the last term dominates and q(Vgs  mi  V )

q(Vgs  mi ) 2kT

 R  2  si  tox   ln    ln  1   R  2a    ox 

The potential solution is (V = 0)

 ( )  

 (1   )  2   2kT  R ln  1   q R2 2 a 1     

At surface,

2kT  R  kT  4a 2  2kT  1   (   R)   ln     ln  2   ln   q q  2a  q  R    Substituting the  solution from the above boundary condition:

 (   R) 

 2 2kT   q(Vgs  mi )  R   2a   t     ln     ln    ln  si ln  1  ox    ln  q   2kT R     2a    R   ox 

7.4. gm  I ds  

q(Vgs  mi  V ) 2kT

I ds  s I ds  d   s Vgs  d Vgs

W Vds W d dV [Qi (V )]dV    [ Qi (  )] d, L 0 L s d

2  ln   tsi

2 si kT  2 si tox  tan   f (  ).   ln   ln  cos    2 q ni   oxtsi

 2kT  I ds W dV W   Qi (  s )   Qi (  s )   f (  s )  ,  s L d  s L q    2kT  I ds W dV W    Qi (  d )   Qi (  d )  f (  d )  ,  d L d  d L q  

q(Vgs  mi ) 2kT gm  

2  ln   tsi

2 si kT    f (  s ), q 2ni 

q(Vgs  mi  Vds ) 2kT

2  ln   tsi

2 si kT    f (  d ). q 2ni 

 2kT   2kT  W 1 W 1 W Qi (  s )   f (  s )    Qi (  d )  f (  d )     Qi (  s )  Qi (  d )  L q L L   (2kT / q) f (  s )  q  (2kT / q) f (  d )

Since

SOLUTION

Qi (  )  2 si

d 2kT 2  tan   2 si , dx x tsi /2 q tsi

46

g m  8

W kT  si   s tan  s   d tan  d  L q tsi

Likewise, gd 

 2kT  I ds  s I ds  d I ds  d W 1 W     Qi (  d )  f (  d )     Qi (  d )   s Vds  d Vds  d Vds L q (  2 kT / q ) f (  ) L   d

g d  8

W kT  si  d tan  d L q tsi

7.5 At the source, Qi  2 si

2kT 2  s tan  s , q tsi

where

q(Vgs  mi ) 2kT

2  ln   tsi

 s << 1,

Subthreshold,

2 si kT  2 si tox  tan  s .   ln  s  ln  cos  s   2 q ni   oxtsi s ln  s dominates.

s  e

q (Vgs mi )/2 kT

tsi 2

q 2ni 2 si kT

Therefore,

Far above threshold, Qi  2 si

Qi  8 si

kT  s 2 q (V  )/ kT  qni tsi e gs mi q tsi

 s  /2,

 s tan  s dominates.

2kT 2  s tan  s   2kT  2  2 ox Vgs  mi  ln  q tsi tox  q  tsi

7.6

Let

SOLUTION

2 si kT    q 2ni  

 t  t   tan  ox  tan  si   ox     2   si  =  tox/,

then

 tsi/(2) =  (4tox)/(2) = 2.

47

So

tan  tan 2  tan 

tan  

Solve

1 1  2 si /  ox

Then



Check:

If ox/si = 1,



2 tan    ox 2 1  tan   si

 tox   1 tan 1   1  2 /   si ox  

 tox  tox   6tox  15 nm. tan 1 1 / 3  / 6





Ifox/si = ,  

 tox t  ox  4tox  10 nm. 1 tan 1  / 4

7.7 (a)

q(Vgs  mi ) 2kT

2  ln   tsi

2 si kT  2 si tox  tan    ln   ln cos    2 q ni   ox tsi

mi = m – ( + Eg/2q) = 0.16 V.

SOLUTION

The LHS = 2.41  in subthreshold.

48

The RHS is dominated by ln , so

 = e-2.41 = 0.09.

Electron charge density

Qi = (8kT/q) (si/tsi)  tan  = 1.7410-9 C/cm2.

Acceptor charge density

q Na tsi = 4.810-10 C/cm2.

Both charges are negative,

so total net charge density = 2.2210-9 C/cm2.

(b)

The threshold shift is

q Na tsi / (2 Cox) = 0.139 V, more positive.

(c)

Number of dopants is

N = Na tsi W L = 27

The number fluctuation is

sqrt (N) = 5.2

It is equivalent to an avg. doping of sqrt (N)/(tsi W L) = Na = 5.81017 cm-3.

It causes a 1 threshold voltage uncertainty of q Na tsi / (2 Cox) = 0.027 V.

7.8 q(Vgs  mi  V ) 2kT

2  ln   tsi

2 si kT  2 si tox  tan  .   ln   ln  cos    q 2ni   oxtsi

Vt  mi 

2kT  2 ln  q  tsi

2 si kT   q 2ni 

So, Vgs  Vt  V 

 2kT  2 si tox  tan   ln   lncos    q   oxt si  Qi 

8kT  si  tan  q t si

Therefore, 2(Vgs  Vt  V ) 1   Cinv Qi

ln   lncos   

2 si tox  tan   oxt si

2 si  tan  t si



tox

 ox



ln   lncos   2 si  tan  t si

and

SOLUTION

49

Csi 

2 si  tan  t si ln   lncos  

7.9 Integration of current continuity eq. from 0 to L yields s

2

W 4 si  2kT    2  si tox 2 2  I ds    tan     tan   .

L tsi  q   2  ox tsi  d

Similar integration of current continuity eq. from 0 to y yields s

2

W 4 si  2kT    2  si tox 2 2  I ds    tan     tan   . y tsi  q   2  ox tsi 

For  << 1, the function in the square bracket  2/2.

Ratio of the two eqs. gives 2 2 L s  d  y s2   2 Also, for  << 1,  2 2 si kT  q(Vgs  mi  V ) 2 si tox  ln   tan   ln  .   ln   ln  cos    2 2kT  ox tsi  tsi q ni  where V=0 for s and V=Vds for d.

Substituting , s, and d into the ratio-ed eq., 2 2 L s  d 1  e qVds / kT .   y  s 2   2 1  e qV / kT Solving V, we obtain  kT  1 .

V ( y)  ln   qVds / kT  q 1  ( y / L)(1  e )

7.10 Use the subthreshold model for SOI MOSFETs. Apply Eq. 7.10,

 s1 

(tsi  3tox 2 )(Vgs1  mi1 )  3tox1 (Vgs 2  mi 2 ) tsi  3tox1  3tox 2

.

If Vgs2 = constant, SOLUTION

50

Vg1

t  3tox1  3tox 2  si  s1 t si  3tox 2

Therefore,

t  3tox1  3tox 2 SS  si  60 mV/dec  75 mV/dec t si  3tox 2

For Vt, solve Vgs1 with s1 = Eg/2q. Or use Eq. 7.13: Vt  mi1  m

with

Eg 2q

  m  1 (Vgs 2  mi 2 )

t  3tox1  3tox 2 m  si  1.25. tsi  3tox 2

mi1 = 4.35 – (4.05 + 0.56) = –0.26 V, mi2 = 4.95 – (4.05 + 0.56) = 0.34 V, Vgs2 = 0.5 V  Vt = 0.40 V. 7.11 tsi = 5 nm, tox1 = tox2 = 2 nm, mi1 = Eg/2q, mi2 = 0,

 s1 

so

t  3tox1  3tox 2 17 m  si  . tsi  3tox 2 11

1 1 11 6  (Vgs1  mi1 )   1   (Vgs 2  mi 2 )  (Vgs1  0.56)  Vgs 2 m m 17 17 

When both gates switch, Vgs1 = Vgs2. s1 = 0.56 when Vgs1 = Vgs2 = Vt = 0.56(6/17) = 0.20 V. Vgs2 = 0.  s1 = 0.56 when Vgs1 = Vt1 = 0.56(17/11 – 1) = 0.31 V. From the above, the Vgs1/s1 ratio is 17/11.

Thus the inverse subthreshold slope is (17/11)60 mV/dec = 93 mV/dec.

From Eq. 7.8,

Qi   ox

Vgs  mi1  (0) tox

  ox

Vgs  mi 2  (tsi ) tox

.

Above the threshold in (a), (0) is pinned to Eg/2q. Thus Qi  Cox Vgs  Cox  Vgs   (tsi ) .

Assume negligible charge in silicon except near the front surface, we have

SOLUTION

51

 (tsi ) tsi 5   . Vgs tsi  3tox 11

Therefore, Qi / Vgs  Cox  Cox 1   (tsi ) / Vgs   (17 / 11)Cox  2.67  F/cm2 .

SOLUTION

52

Solutions to Chapter 8 Exercises 8.1. Under the scaling transformation, W  W/, L  L/, tox  tox/, Vds  Vds/, Vg  Vg/, and Vt  Vt/, Eq. (5.23) becomes I ds   eff (Cox )

W /   Vg Vt  Vds I ds   ; L /       

and Eq. (5.31) becomes 2

W /  1  Vg Vt  I I ds  eff (Cox )    ds .  L /  2m     

Note that both m = 1 + 3tox/Wdm and eff which is a function of Eeff given by Eq. (5.53) are nearly invariant under constant field scaling.

8.2.

Under the same scaling rules as above, Eq. (5.39) becomes 2

 kT  q(V V )/ mkT W / I ds  eff (Cox ) ( m  1)  e g t .

L/  q 

The exp(qVds/kT) term has been neglected since typically Vds >> kT/q. In subthreshold, Vg < Vt and exp[q(Vg  Vt)/mkT] > exp[q(Vg  Vt)/mkT] (note that  > 1), therefore, the subthreshold current increases with scaling faster than Ids. If the temperature also scales down by the same factor, i.e., T  T/, then Ids  Ids/, same as the drift current in Exercise 8.1.

8.3. Since the factor z = 2eff(Vg  Vt)/(mvsatL) of Eq. (6.47) is invariant under the scaling transformation, W  W/, L  L/, tox  tox/, Vds  Vds/, Vg  Vg/, and Vt  Vt/, the square-root of (z + 1) and therefore the fraction in Eq. (6.49) is unchanged after scaling. The saturation current Idsat of Eq. (6.49) then scales the same way as the fully saturation-velocity limited current, Eq. (6.51), i.e., I dsat  (Cox )

8.4.

V V  I vsat  g  t   dsat .    

W

From Eqs. (8.8) and (8.9), the average CMOS inverter delay is

SOLUTION

53



 n  p 2



CVdd  1 1    .

4  Wn I nsat Wp I psat 

If the inverter is driving another stage with the same n- to p-width ratio and if both the n- and pdevices have the same capacitance per unit width, the load capacitance C is proportional to (Wn + Wp), i.e.,



W / Wn Wn / Wp 1 1  1 1 .    p    W I W I I I I I n nsat p psat nsat psat nsat psat  

  (Wn  Wp )

The delay is a minimum when the n- to p-width ratio is such that the last two terms are equal, i.e., when Wp/Wn = (Insat/Ipsat)1/2. Note that n < p for minimum delay.

8.5.

For the following RC circuit,

Vdd R I(t)

V(t) C

if the voltage source is switched to Vdd at t = 0, then

Vdd  V (t )  RI ( t )  V ( t )  RC

dV . dt

With the initial condition V(t = 0) = 0, the solution for V(t) is





V ( t )  Vdd 1  e t / RC .

The energy dissipated in R is 

E   RI 2 dt  0

SOLUTION

2

Vdd  2t / RC 1 2 e dt  CVdd ,  0 R 2

54

which is independent of R. The same amount of energy is stored in C. If the voltage source is now switched to 0, one has

V (t )  RC

dV  0. dt

With the initial condition V(t = 0) = Vdd, the solution for V(t) is

V (t )  Vdd e t / RC .

The energy stored in C is now all dissipated in R: 

2

 V 1 2 E   RI dt  dd  e2t / RC dt  CVdd .

0 0 R 2

8.6.

2

From Eq. (5.27) and the inversion charge expression above Eq. (5.70), the transit time is

WLCox (Vg  Vt  mVds / 2) L2  tr    eff Cox (W / L)(Vg  Vt  mVds / 2)Vds  eff Vds for a MOSFET device biased in the linear region.

From Eq. (5.31) and the inversion charge expression above Eq. (5.72), the transit time is

 tr 

( 2 / 3)WLCox (Vg  Vt ) 4mL2   eff Cox (W / L)(Vg  Vt )2 / 2m 3 eff (Vg  Vt )

for a long-channel MOSFET biased in saturation.

8.7.

From Eq. (6.49) and the inversion charge expression in Exercise 6.8, the transit time is

 tr 

Qi L 1  2eff (Vg  Vt ) / ( mvsat L )  1 / 3  I dsat vsat 1  2eff (Vg  Vt ) / ( mvsat L )  1

for a short-channel MOSFET biased in saturation. The limiting value of tr is L/vsat when the device becomes fully velocity saturated as L  0.

8.8.

When Ids is considered as a function of the external voltages: Ids(Vg,Vds), one has

SOLUTION

55

 I  I  dI ds   ds  dVg   ds  dVds .   Vg    Vds V g  Vds

From Eqs. (8.13) and (8.14), dVds  dVds  ( Rs  Rd )dI ds ,

and dVg  dVg  Rs dI ds .

Substituting them into the top equation and solving for dIds, one obtains dI ds 

( I ds /  Vg )Vds

1  Rs ( I ds /  Vg )Vds  ( Rs  Rd )( I ds /  Vds )V g

dVg  (a similar term in dVds ) .

Now Ids is expressed as a function of the internal voltages: Ids(Vg,Vds). One can identify the first coefficient as the intrinsic transconductance, i.e.,  I  gm gm   ds   ,   Vg   Vds 1  gm Rs  gds ( Rs  Rd )

where  I  gm   ds    Vg   Vds

is the extrinsic transconductance, and I  gds   ds    Vds V g

is the extrinsic output conductance.

8.9. The transmission line model of contact resistance in a planar geometry is represented by the distributed network below. The current flows from a thin resistive film (diffusion with a sheet resistivity sd) into a ground plane (metal) with an interfacial contact resistivity c between them (Fig. 8.18).

SOLUTION

56

I Metal contact

Rdx

I

x=0

V(x)

Gdx

I(x)

x=lc

Following a similar approach as in Eqs. (8.29)-(8.31), one can write

V ( x  dx )  V ( x ) 

dV dx   I ( x ) Rdx , dx

and

I ( x  dx )  I ( x ) 

dI dx  V ( x )Gdx . dx

Here R = sd/W and G = W/c. From the above two equations, one obtains d2 f   RGf  sd f , 2 dx c

where f(x) = V(x) or I(x).

8.10. The solution to the second-order differential equation in the above exercise is of the sinh[(sd/c)1/2x] and cosh[(sd/c)1/2x] form. With the boundary condition I(x=lc) = 0 where x = 0 is the leading edge and x = lc is the far end of the contact window (see figure in Exercise 5.5), the solution is

   I ( x )  I0 sinh  sd (lc  x ) ,   c  where I0 is a constant multiplying factor. From the second equation in Exercise 8.9, the voltage is

V (x)  

c dI W dx



c sd W

   I0 cosh  sd (lc  x ) .  c 

The contact resistance is then

SOLUTION

57

Rco 

 c  sd

V ( x  0)  I ( x  0)

W

  sd  coth lc .  c  

8.11. From Eqs. (8.49) and (8.51), bmin <  if

Cout  CL  2Cout  2 CinCL .

This inequality is quadratic in CL1/2, which can be solved to yield CL 

 C  C C  .

2

in

in

out

Under this condition, the insertion of one or a few properly designed buffer stage(s) will help reduce the overall delay.

8.12. For an n-stage buffer with successive width ratios of k1, k2, k3,     , kn, the switching resistances are Rsw, Rsw/k1, Rsw/k1k2, Rsw/k1k2k3,     , etc. The output capacitances are Cout, k1Cout, k1k2Cout, k1k2k3Cout,     , etc.; and the input capacitances are k1Cin, k1k2Cin, k1k2k3Cin, k1k2k3k4Cin,     , etc. The delay of the first stage is

 b,1  Rsw (Cout  k1Cin ) ; the delay of the second stage is

 b, 2 

Rsw ( k1Cout  k1k2Cin )  Rsw ( Cout  k2Cin ) ; k1

    , and so on, until the n-th stage:

 b, n  Rsw ( Cout  kn Cin ) .

The delay of the last stage is

 b, n 1 

  Rsw CL ( k1k2    kn Cout  CL )  Rsw  Cout  . k1k2   kn k1k2    kn  

The total buffer delay is the sum of all the stage delays, i.e., 

 b ( n)  Rsw  ( n  1)Cout  ( k1  k2       kn )Cin  

SOLUTION

 CL . k1k2    kn 

58

[Note that delay equations of the form Eq. (8.45) used above are defined in terms of propagation delays through cascaded stages of identical width and loading (Fig. 8.32), and as such are not strictly applicable to a chain of drivers with varying widths and rise times. When the input rise time is shorter than that of the output as is the case with the above tapered drivers, the switching resistance Rsw  d/dCL is somewhat less than that defined in Section 8.4.1.3.]

8.13. Take any pair of k’s (positive numbers), e.g., k1 and k2, for a given product k1k2, the sum k1 + k2 is a minimum when k1 = k2. This can be generalized to the result that for a given product k1k2k3    kn, the sum k1 + k2 + k3 +    + kn is a minimum when k1 = k2 =      = kn. Let k1 = k2 =      = kn = k, the delay in Exercise 8.12 becomes  

 b ( n )  Rsw  ( n  1)Cout  nkCin 

CL  . kn 

Differentiating it with respect to k, one finds that the above expression has a minimum 1/ ( n 1)    CL  ,  b min ( n)  Rsw ( n  1)Cout  ( n  1)Cin      Cin 

when k = (CL/Cin)1/(n+1). [Note that the same result can be obtained alternatively by partially differentiating the delay in Exercise 8.12 with respect to k1, k2, k3,     , kn respectively and solving the n coupled equations.] bmin(n), as expected, is reduced to Eq. (8.51) if n = 1.

8.14. To find the optimum n which gives the shortest overall delay, we let k = (CL/Cin)1/(n+1) and treat it as a continuous variable. Then n + 1 = ln(CL/Cin)/ln(k) and the delay in Exercise 8.13 becomes

 b min ( k )  Rsw ( Cout  kCin )

ln( CL / Cin ) . ln( k )

This has a minimum when dbmin/dk = 0, which yields an equation for k: kln( k )  1 

Cout .

Cin

n is then the closest integer to

n

SOLUTION

ln( CL / Cin )  1. ln( k )

59

For typical Cout/Cin ratios not too different from unity, k is in the range of 3-5. Note that k also gives the optimum width ratio between the successive buffer stages, i.e., k1 = k2 =      = kn = k. Since Cout + kCin = kln(k)Cin, the minimum buffered delay can be written as

 b min  kRswCin ln( CL / Cin ), which only increases logarithmically with load capacitance.

8.15.

Power = CVdd2f So

where f = 1/(2)

and C = Cin+Cout+CL

Power = Vdd2/2Rsw

If

(Wn, Wp)  (kWn, kWp), Rsw  Rsw/k Cin+Cout  k(Cin+Cout)

Therefore, Delay   Rsw(Cin+Cout+CL/k) Power  k times original value.

SOLUTION

60

Solutions to Chapter 9 Exercises 9.1

Consider hole current flow in an n-region.

J p   qpn  p

Eq. (2.78) gives

d p . dx

(1)

Now, as shown in Fig. 3.10, dn(n-region)/dx  0 when there is negligible IR drop in the n-region.

Therefore, (1) can be rewritten as

J p ( n  region)   qpn  p





d  p  n . dx

(2)

Eq. (9.15) gives 



     kTq ln pnn  p

for the n-region.

n

n n 2 ie

(3)

Substituting (3) into (2), we obtain

 kT  1 dnn 1 dnie2   dp J p ( n  region )  qpn  p   2     qDp n , dx  q  nn dx nie dx  

(4)

where we have used the Einstein relation Dp = kTp/q. If we compare (4) with Eq. (2.69), i.e.,

J p ( n  region)  qpn  pE ( n  region)  qDp

(5)

 kT  1 dnn 1 dnie2   E( n  region )    2   .  q  nn dx nie dx  

we have

9.2

dpn , dx

Eq. (2.78) gives, for an n-region, J p   qpn  p

d p . dx

(1)

As shown in Fig. 3.10, when there is negligible IR drop in the n-region, we have

dn (n  region)  0. dx

(2)

Therefore, (1) and (2) give J p ( n  region )   qpn  p

d ( p   n ) . dx

(3)

From Eq. (9.15), for the n-region,

SOLUTION

61

( p   n ) 

kT  pn nn  ln  . q  nie2 

(4)

Substituting (4) into (3), and using Dp = kTp/q, we have J p ( n  region )   qDp

nie2 d  nn pn   . nn dx  nie2 

9.3 pn (x) WE1

silicon

polysilicon

WE + WE1)

(a)

W E

x

0

E-B junction

Let pn(WE) be the excess hole density at the polysilicon-silicon interface.

Also let

x'  ( x  WE ) . That is, x’ is the distance between a point inside the polysilicon layer and the polysilicon-silicon interface. The steady-state equation governing the excess hole density is given by the hole-equivalent of Eq. (3.36), i.e.

d 2 pn pn  2  0. dx 2 Lp

(1)

Applying (1) to hole distribution inside the polysilicon layer, we have

d 2 pn pn  2  0. dx '2 LpE1

(2)

The general equation for (2) has the form

pn ( x' )  Ae

x ' /L pE 1

 Be x ' /LpE1 .

(3)

At the ohmic contact at x’ = WE1, we have

0  Ae E1 W

SOLUTION

/ L pE 1

 Be

WE 1 / L pE 1

.

(4)

62

At the polysilicon-silicon interface (x’ = 0), we have

pn ( WE )  A  B .

(5)

pn ( WE ) 2W / L 1  e E1 pE1

(6)

(4) and (5) give

A

and

 pn ( WE )e E 1 B 2W / L 1  e E 1 pE 1 2W

/ L pE 1

.

(7)

Substituting (6) and (7) into (3), we have pn ( x ' )  pn ( WE )

(b)



sinh (WE1  x' ) / LpE1 sinh(WE1 / LpE1 )

.

(8)

The hole current density entering the polysilicon layer is given by Eq. (9.59), i.e.,

J p ( x  WE )  qpn ( WE )S p .

(9)

Using (8), this current density is also J p ( x  WE )   qDpE1

dpn ( x ) dx x  WE

 qDpE1

dpn ( x ' ) dx ' x '  0

  qDpE1

Sp 

Equations (9) and (10) give

9.4

(10)

pn ( WE ) .

LpE1 tanh(WE1 / LpE1 )

DpE1 .

LpE1 tanh(WE1 / LpE1 )

Neglecting parasitic resistances, Eqs. (9.182) and (9.183) give

and

I E   I EBO exp( qVBE  / kT )  1   R IC

(1)

IC   ICBO exp( qVBC  / kT )  1   F I E ,

(2)

which can be rearranged to give VBE  

SOLUTION

kT  I E   R IC  ln 1  q  I EBO 

(3)

63

kT  IC   F I E  . ln 1  q  ICBO 

and

VBC  

Now,

VCE   VC  VE  (VC  VB )  (VB  VE )  VBE   VBC  .

(4) (5)

Substituting (3) and (4) into (5) and rearranging, we obtain VCE  

Now, Eq. (9.144) gives

kT  ICBO ( I EBO  I E   R IC )  . ln q  I EBO ( ICBO  IC   F I E ) 

(6)

 R I R0   F IF 0 ,

(7)

and Eqs. (9.180) and (9.181) give

I EBO  I F 0 (1   R F ) ,

(8)

and

ICBO  I R 0 (1   R F ) .

(9)

Therefore,

ICBO I R 0  F   .

I EBO I F 0  R

(10)

Substituting (10) into (6) gives

9.5

VCE  

kT   F  I EBO  I E   R IC   ln  . q   R  ICBO  IC   F I E  

From Exercise (9.4), we have VCE  

kT   F  I EBO  I E   R IC   ln  . q   R  ICBO  IC   F I E  

(1)

Now, when emitter and collector resistances are included, we have

VCE  VC  VE  VC  rc IC  (VE  re I E )  VCE   rc IC  re ( I B  IC ) ,

(2)

where we have used the fact that

I B  IC  I E  0 .

(3)

Substituting (1) into (2), we obtain VCE 

kT   F  I EBO  I E   R IC   ln    rc IC  re ( I B  IC ) . q   R  ICBO  IC   F I E  

(4)

If we neglect the saturation currents IEBO and ICBO, then using (3) we can rearrange (4) to give VCE 

SOLUTION

kT   F  I B  IC   R IC   ln    rC IC  re ( I B  IC ) q   R  F ( I B  I C )  IC  

(5)

64



 kT  I B  IC (1   R ) ln    re ( I B  IC )  rc IC . q   R  I B  IC (1   F ) /  F  

For open-circuit collector, IC = 0, and (5) reduces to VCE ( IC  0) 

9.6

kT  1  ln    re I B . q R 

From Eq. (3.65), we have sinh(WB  x ) / LnB  . sinh(WB / LnB )

(1)

dnp cosh(WB  x ) / LnB   1   np 0  exp( qVBE / kT )  1  . dx sinh(WB / LnB )  LnB 

(2)

n p  n p 0  n p 0 exp(qVBE / kT )  1

If we assume diffusion current only, then J n ( x )  qDnB

Therefore, the base transport factor is 1

J ( x  WB )  W  T  n   cosh B  .

J n ( x  0)  LnB 

(3)

For a base region of WB = 100 nm and NB = 1018 cm3, we have, from Fig. 3.14, LnB  15 m, and  

 T   cosh

9.7

01 .  15 

1

 0.9998 .

Eq. (9.188) give the condition for collector-emitter breakdown as

 T M ( BVCEO )  1 .

(1)

Now, from Eqs. (9.78) and (9.178), we have

0   T , 1 0

(2)

where 0 is the common-emitter current gain at negligible base-collector junction avalanche. (1) and (2) give M ( BVCEO )  1 

SOLUTION

1

0

.

65

9.8

The equation

d 2 n p ( x) dx



2

n p ( x) 2 LnB

 0,

(1)

where

  LnB

 nB DnB LnB  1  j nB 1  j nB

(2)

has a general solution

 )  B exp( x LnB  ), n p ( x)  A exp( x LnB

(3)

where A and B can be determined from the boundary conditions for Δnp(x).

Expanding the Shockley diode equation, Eq. (3.61), and keeping only the first order term in vbe, we have q   n p (0, t )  n p 0 exp  qvBE (t ) kT   n p 0 exp(qVBE kT ) 1  vbee jt  .  kT 

(4)

Therefore, the boundary condition for Δnp(x) at x = 0 is

n p (0)  n p 0 exp(qVBE / kT )

qvbe . kT

(5)

At the ohmic base contact, there are no excess electrons. Therefore, the boundary condition for Δnp(x) at x = WB is

n p (WB )  0.

(6)

Applying these boundary conditions to Eqs. (1) and (3), we obtain n p ( x) 

9.9

n p 0 qvbe kT

   qV  sinh  (WB  x) LnB exp  BE  .  )  kT  sinh(WB LnB

From Eq. (2.97), the net recombination rate in the p-type base region of an npn

transistor is U ( x) 

n p ( x)  n p 0

 nB

.

(1)

From Eq. (3.65), the electron distribution in the p-type base is n p ( x)  n p 0  n p 0 [e qVBE / kT  1]

SOLUTION

sinh  (WB  x) / LnB  sinh(WB / LnB )

.

(2)

66

For a typical transistor in operation, WB/LnB

<< 1, and qVBE/kT

>> 1, and (2) reduces to

 x  n p ( x)  n p 0  n p 0eqVBE / kT 1  .  WB 

(3) The base current component due to recombination in the base region is J B,rec  

WB

0

(4)

qU ( x)dx.

2 Substituting (3) into (4), and using the fact that p p 0 n p 0  nieB for the base region [see Eq.

(9.14)], and that pp0 = NB, we have

J B ,rec 

2 qWB nieB exp(qVBE / kT ).

2 nB N B

(5)

Eq. (9.71) gives the base current due to hole injection from base into the emitter as

J B ,inj  J B,rec

Therefore,

J B,inj

2 qD pE nieE exp  qVBE kT 



N E L pE tanh WE L pE 



.

2 WB L pE tanh(WE / L pE ) nieB NE

2 nB D pE

2 NB nieE

(6)

.

(7)

For WE = 100 nm, NE = 1 × 1020 cm-3, NB = 1 × 1018 cm-3, and WB = 50 nm, we have, from Fig. 3.14, nB = 3×107 s, LpE = 0.38×104 cm, tanh(WE/LpE) = 0.26, p = 120 cm2/V-s. That is, from Eq. (2.51), DpE = (kT/q)p = 0.026×120 cm2/s =3.12 cm2/s. Also, from Eq. (9.14), we have 2 nieB 2 nieE

 exp[(EgB  EgE ) / kT ].

(8)

From the Unified model in Fig. 9.4, EgB = 31 meV and EgE = 92 meV. Therefore 2 2 nieB / nieE  exp(61/ 26) = 0.096. We have

J B,rec J B,inj



2 WB L pE tanh(WE / L pE ) nieB NE

2 nB D pE

2 NB nieE

= [50×107 × 0.38×104 × 0.26/ (2 × 3×107 × 3.12)] × 0.096 × 100 = 2.5×104.

9.10

Consider a wide-emitter but narrow-base n+-p diode, with emitter doping concentration

NE = 1020 cm3, base doping concentration NB = 1017 cm3, and base width WB = 100 nm.

SOLUTION

67

(a) Ignoring heavy-doping effect, the diffusion capacitance ratio is C Dn 2 N E WB .  C Dp 3 N B L pE

(1)

From Fig. 3.14, we have LpE = 0.38 m for holes in the n+ emitter. Substituting all these values into (1) gives CDn/CDp = 175.

(b) When heavy-doping effect cannot be ignored, Eq. (9.133) gives

C Dn 

2 q 2 qA I nt B  J n tB , 3 kT 3 kT

(2) where A is the cross-sectional area of the diode. From Eq. (9.119), for a p-region with WB << LnB, we have Jn   

qDnB n p 0 WB

exp(qVBE / kT )

qDnB p p 0 n p 0 WB p p 0

exp(qVBE / kT )

(3)

2 qDnB nieB exp( qVBE / kT ).

WB N B

From Eq. (9.106), we have

WB2 tB  .

2 DnB

(4)

Substituting (3) and (4) into (2), we obtain

CDn 

2 WB 2 qA qnieB exp(qVBE / kT ) .

3 kT 2 N B

(5)

Similarly, for a n-type emitter region with WE << LpE, we have

C Dp 

SOLUTION

1 q 1 qA I p pE  J p  pE 2 kT 2 kT

(6)

68

1 qA qD p nieE pE exp(qVBE / kT ) 2 kT N E L pE 2



2 1 qA qnieE L pE  exp(qVBE / kT ) 2 kT NE

where we have used the fact that L2 = D. (5) and (6) give 2  N  W C Dn 2  nieB   2  E  B .

C Dp 3  nieE  N B  L pE

Now,

2 nieB  ni2 exp( EgB / kT )

and

2 nieE  ni2 exp( EgE / kT ) .

Therefore,

2  E  EgE  nieB  exp gB .

2   nieE kT

(7)

(8)

From Fig. 9.4, we have EgB  2 meV for p-type base with doping concentration of 1017 cm3, and EgE  92 meV for n-type emitter with doping concentration of 1020 cm3. Also, at room temperature (300 K), kT = 26 meV. Therefore 2  E  EgE  nieB  exp gB  = 0.031.

2   nieE kT

(9)

That is, the effect of heavy-doping reduces the ratio CDn/CDp from 175 to 5.4.

9.11

I C (VBE )lowinjection 

2 AE qDnB nieB exp(qVBE / kT ) .

N BWB

1

A qD n 2 exp(qVBE / kT )  1  4n 2 exp(qVBE / kT )   1   1  ieB I C (VBE )all injection  E nB ieB  1  .

2  N BWB  4 N B   

Therefore,

Ratio 

Eq. (9.14) gives

SOLUTION

 1 4n 2 exp(qVBE / kT )    1   1  ieB  1  .

2  IC (VBE )all injection  4  N B   

I C (VBE )lowinjection

(1)

2 nieB  ni2 exp  EgB ( N B ) / kT  . Therefore, (1) becomes

69

  4ni2 exp{[ qVBE  EgB ( N B )] / kT }   1  Ratio  1  1  1  .  4  N B2   

(2)

For Eg, we use the unified model given in Eq. (9.18), namely 2   N   N      Eg  N   6.92 ln   ln  0.5  meV.   17  17   1.3  10 1.3  10          

Ratio = IC(low inj) / IC(all inj)

15

10 NB=1E17 NB=1E18 NB=1E19 5

0 0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

VBE (V)

SOLUTION

70

Solutions to Chapter 10 Exercises 10.1

For the polysilicon-emitter model described in Exercise 9.3,





LpE1 tanh WE1 / LpE1   n2   W .

GE (WE ,WE1 )  N E  2i   E   DpE1  nieE   DpE 

(1)

pn (x) WE1

silicon

polysilicon

WE + WE1)

W E

x

0

E-B junction

Assume the lifetime in polysilicon layer is the same as in silicon, and assume the mobility in polysilicon is smaller than that in silicon.

(a) For non-polysilicon-emitter:

NE = 1020 cm3, WE = 300 nm, WE1 = 0. From Eq.

(9.14), we have ni2  exp(  E gE / kT ) .

2 nieE

(2)

From Fig. 9.4, we have, for NE = 1020 cm3, EgE = 92 meV. Using kT = 26 meV, we have ni2  exp(  E gE / kT )  0.029 .

2 nieE

(3)

From, Fig. 3.14, we have pE = 130 cm2/V-s. Therefore, DpE 

Therefore,

SOLUTION

kT pE q

 338 .

cm2/s.

(4)

 n2  W GE (300 nm, 0)  N E  2i  E  2.6  1013 cm4-s.  nieE  DpE

71

With pE1/pE = 1/3, we

(b) For polysilicon-emitter: NE = 1020 cm3 and WE = 30 nm. have DpE1 

kT pE1 q

 113 . cm2/s.

(5)

Now, from Fig. 3.14, we have LpE = 0.38 m. Therefore, we have

LpE1 

L kT  pE1 pE  pE  0.22 m. q 3

(6)

Substituting all these values into (1) we have WE1 (nm)

GE(30 nm, WE1) (cm4-s)

50

1.521013

100

2.661013

200

4.361013

300

5.221013

GE(poly)/GE(non-poly)

(c) 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4

0

50

100

150

200

250

300

350

Polysilicon-layer thickness (nm)

10.2

For a box-like profile, Eq. (10.5) gives WB RSbi   q  p p ( x ) p ( x )dx 0

SOLUTION

1



1 qN B  pWB



B WB

,

(1)

72

where B is the resistivity of the p-type base layer. Setting RSbi = 104 /, and using Fig. 2.10 for B, we obtain the values in the following table.

SOLUTION

73

NB (cm3)

B (-cm)

WB (nm)

61016

0.3

300

11017

0.2

200

2.91017

0.1

100

71017

0.05

50

Base doping (cm-3)

1E+18 5E+17 3E+17 2E+17 1E+17 5E+16 30

50

100

200

300

500

Base width (nm)

10.3

From Exercise 10.2, and Fig. 3.14, we have the following table.

WB (nm)

NB (cm3)

nB (cm2/V-s)

DnB (cm2/s)

tB (ps)

50

71017

37.5

9.75

1.3

100

2.91017

505

13.1

3.8

200

11017

760

19.8

10.1

where we have used DnB = kTnB/q and tB = WB2/2DnB. Also, since Fig. 3.14(a) does not have values for NB < 11017 cm3, we have dropped the corresponding base width and base doping concentration values in the table and the following plot.

SOLUTION

74

Base transit time (ps)

12 10 8 6 4 2 0

0

50

100

150

200

250

Base width (nm)

10.4

From Eq. (9.94), we have VA 

QpB CdBC



qN BWB .

CdBC

(1)

To avoid significant base widening, collector current density is kept at

J C  0.3qvsat N C ,

or

NC 

JC , 0.3qvsat

(2)

with vsat = 107 cm/s.

(a) For a one-sided BC junction, with VCB = 2 V, Eqs. (3.13) and (3.15) give CdBC 

 si WdBC



qN C si  2( bi  VCB )

J C si .

0.6vsat ( bi  VCB )

(3)

We can obtain bi from Fig. 3.4 (or set it to about 0.95 V as an approximation) and use si = 1.041012 F/cm to generate the following table.

JC (mA/m2)

NC (cm3)

bi (V)

CdBC (fF/m2)

0.1

2.081016

0.925

0.243

0.2

4.16

0.945

0.343

0.3

6.24

0.95

0.420

0.4

8.32

0.96

0.484

0.5

1.041017

0.97

0.540

SOLUTION

75

0.8

1.66

0.98

0.659

1.0

2.08

0.99

0.761

2.0

4.16

1.0

1.07

3.0

6.24

1.02

1.31

4.0

8.32

1.02

1.52

5.0

1.041018

1.02

1.69

3

CdBC (fF/um2)

2 1 0.5 0.3 0.2 0.1 0.1

0.2

0.3

0.5

1

2

3

5

JC (mA/um2)

(b) For a base design with qNBWB = 1.6106 C/cm2, we have the following table

SOLUTION

JC (mA/m2)

CdBC (fF/m2)

VA (V)

0.1

0.243

65.8

0.2

0.343

46.6

0.3

0.420

38.1

0.4

0.484

33.1

0.5

0.540

29.6

0.8

0.659

24.3

1.0

0.761

21.0

2.0

1.07

15.0

3.0

1.31

12.2

76

4.0

1.52

10.5

5.0

1.69

9.5

100

VA (V)

50 30 20 10 5 0.1

0.2 0.3

0.5

1

2

3

5

JC (mA/um2)

10.5

An npn transistor with WB = 500 nm and negligible heavy doping effect in base layer,

and a graded base profile of

N B ( x )  A  x ,

(1)

A  N B (0)  2  1017 cm3,

(2)

N B (WB )  2  1016 cm3

(3)



and

A  N B (WB )  3.6  1021 cm4.

WB

(4)

From Eq. (10.11), the electric field due to the dopant distribution is E( dopant ) 

kT 1 dN B kT   . q N B ( x ) dx q NB(x)

(5)

We have the following table and plot of results.

x (nm)

NB (cm3)

E (V/cm)

0

2.0E17

-468

50

1.82E17

-514

100

1.64E17

-571

SOLUTION

77

150

1.46E17

-641

200

1.28E17

-731

250

1.10E17

-851

300

9.2E16

-1017

350

7.4E16

-1265

400

5.6E16

-1671

450

3.8E16

-2463

500

2.0E16

-4680

Electric field (V/cm)

0 -1,000 -2,000 -3,000 -4,000 -5,000

0

100

200

300

400

500

600

x (nm)

10.6

An npn transistor with WB = 100 nm and non-negligible heavy doping effect in base

layer, and a graded base profile of

and

N B ( x )  A  x ,

(1)

A  N B (0)  5  1018 cm3,

(2)

N B (WB )  5  1017 cm3

(3)



A  N B (WB )  4.5  1023 cm4.

WB

(4)

From Eq. (10.11), the electric field due to the dopant distribution is E( dopant ) 

kT 1 dN B kT   , q N B ( x ) dx q NB(x)

(5)

and the field due to heavy-doping effect is

SOLUTION

78

E( HD )  

1 dE gB . q dx

(6)

From Eq. (9.17), the bandgap narrowing in the base layer is



Eg ( N a )  9  F  F 2  0.5

where

F  ln( N B / 1017 ) .

Therefore, we have

 E( HD )  9  103 1  

 meV

(7) (8)

  V/cm.  F 2  0.5  N B F

(9)

We have the following table and plot of the results.

x (nm)

NB (cm3)

E (dopant)(V/cm)

E (HD)(V/cm)

E (Tot)(V/cm)

0

5.0E18

-2340

1607

-733

10

4.55

-2571

1765

-806

20

4.1

-2854

1958

-895

30

3.65

-3205

2198

-1007

40

3.2

-3656

2506

-1151

50

2.75

-4255

2913

-1342

60

2.3

-5087

3479

-1608

70

1.85

-6324

4317

-2008

80

1.4

-8357

5687

-2670

90

9.5E17

-12316

8330

-3985

100

5.0

-23400

15516

-7884

SOLUTION

79

Electric field (V/cm)

20,000 10,000 0 -10,000 E(dopant) E(HD) E(Total) -20,000 0

20

40

60

80

100

120

x (nm)

10.7

Polysilicon emitter with NE = 1020 cm3, AE = 1 m2, single-crystal region thickness WE =

30 nm, polysilicon layer thickness WE1 = 200 nm, metal-poly contact resistivity contact = 2107 -cm2. From Fig. 2.10, we have for the emitter doping concentration single-crystal = 8104 cm.

Therefore, we have polysilicon = 24104 -cm.

From the information, we can calculate

the emitter resistance components. rsin gle  crystal   sin gle  crystal rpolysilicon   polysilicon

WE  0.24 AE

WE1  4.8 AE

rcontact   contact / AE  20

JC 0 

10.8



WB

0

q pp 2 DnB nieB

.

dx

(1) For uniformly doped base with NB = 1018 cm3, WB = 100 nm, and assuming low current levels so that pp = NB, we have JC0 

2 qDnB nieB .

N BWB

(2)

(a) Neglecting heavy doping effect, we have nieB = ni = 1.01010 cm3. Using the mobility values in Fig. 3.14, we have

SOLUTION

80

DnB 

kT nB  8.58 cm2/s. q

Putting these values into (1), we have JC0 (neglecting HD effect) = 1.371011 A/cm2.

(b) When heavy doping effect is included, we have from Eq. (9.14) and Fig. 9.4, 2 nieB  ni2 exp( E gB / kT )  ni2 exp( 43 / 26)  5.23ni2 .

(3)

That is, JC0 is 5.23 times larger, or JC0 = 7.191011 A/cm2.

SOLUTION

81

10.9

n+

n+ emitter poly

Metal or metal silicide

xjE

Ge

p

Eg0 = Egmax x

WB0

0 WB Wcap

Fig. 10.11.

Schematic illustrating the emitter and base regions of a polysilicon-emitter SiGe-base bipolar

transistor having a constant Ge distribution in the base.

The emitter depth xjE is assumed to be smaller

than the thickness Wcap of the starting Ge-free layer, resulting in a small region of thickness Wcap xjE near the emitter end without Ge.

 W  x jE   WB 0  Wcap     exp Eg 0 / kT     cap J C 0 (Si, x jE )  WB 0  x jE   WB 0  x jE   x jE Wcap

1

J C 0 (SiGe, x jE )

 Wcap  x jE   (WB 0  x jE )  (Wcap  x jE )     exp E g 0 / kT      WB 0  x jE  WB 0  x jE    

(1) 1

1

 Wcap  x jE   Wcap  x jE     1   exp E g 0 / kT  .     WB 0  x jE   WB 0  x jE   200 dE/kT=0.5 dE/kT=1 dE/kT=2.5 dE/kT=5

Current Ratio

100 50 20 10 5 2 1

SOLUTION

0

0.2

0.4

0.6

(WcapxjE )/(WB0 xjE)

0.8

1

82

 Wcap  x jE   E   W  Wcap   exp g 0    B 0    kT   W  x  W x  V A (Si, x jE ) jE  jE     B0  B0 x jE Wcap

V A (SiGe, x jE )

(2)

 Wcap  x jE   E   (W  x jE )  (Wcap  x jE )   exp g 0    B 0    kT   W x   W  x jE  B0 jE     B0   Wcap  x jE   E   W  x jE   exp g 0   1  cap    kT   W  x  W x  jE  B0 jE      B0 200 dE/kT=0.5 dE/kT=1 dE/kT=2.5 dE/kT=5

Early Voltage Ratio

100 50 20 10 5 2 1

0.2

0.4

0.6

(WcapxjE )/(WB0 xjE)

0.8

1

The base bandgap narrowing parameter corresponding to Fig. 10.37 is given by

n+ emitter poly

10.10

0

n+

Ge

Egmax = Eg0 p WB

0

x

WB1

E gB ,SiGe ( x) 

x E g 0 WB1

 E g 0

x < WB1

(1)

WB > x > WB1.

From Eq. (10.27), we have

SOLUTION

83

J C 0 (SiGe)  J C 0 (Si)

 WB WB

 exp[E 0

gB , SiGe

.

(2)

( x) / kT ]dx

Using (1), we have WB1 WB   E gB ,SiGe ( x)    xE g 0    E g 0  dx   exp dx   exp dx 0 WB1 kT   WB1kT   kT 

WB

 exp 0





(3)



kTWB1 1  exp E g 0 / kT   WB  WB1 exp E g 0 / kT  . E g 0

Substituting (3) into (2), we have 1

 kT WB1   W  J C 0 (SiGe)    1  exp E g 0 / kT   1  B1  exp E g 0 / kT  . (4) J C 0 (Si)  E g 0 WB   WB 





Similarly, the VA ratio can be obtained from Eq. (10.32), namely V A (SiGe) exp(E g max kT ) WB  0 exp[E gB,SiGe ( x) / kT ]dx .

V A (Si) WB

(5)

Substituting (3) into (5), we have  V A (SiGe) exp(E g 0 / kT )  kTWB1  1  exp E g 0 / kT   WB  WB1 exp E g 0 / kT   V A (Si) WB  E g 0 







(6)

 W  kT WB1 expE g 0 / kT   1  1  B1  . E g 0 WB  WB 





The tB ratio can be obtained from Eq. (10.36), namely WB t B (SiGe) 2 WB  exp[  E ( x ) / kT ] gB , SiGe x exp[EgB, SiGe ( x' ) / kT ]dx' dx . t B (Si)  WB2 0

(7)

The integration wrt x can be written as the sum of two parts, with one part for x < WB1 and another part for x > WB1. That is, we have WB

 exp[E 0

WB

gB , SiGe

WB1



0

( x) / kT ] exp[E gB ,SiGe ( x' ) / kT ]dx' dx x

(8)

 E gB ,SiGe ( x)  WB   E gB ,SiGe ( x' )    exp dx' dx exp x kT kT     WB  E gB ,SiGe ( x)  WB   E gB ,SiGe ( x' )    exp dx' dx .   exp WB1 kT kT  x  

For the first integral on RHS of (8), we have SOLUTION

84



WB1

0

( x)  WB  E   EgB , SiGe ( x' )    exp dx' dx exp gB , SiGe x kT kT    



W B1



W B1

0

0



(9)

( x )  W B1  E   EgB , SiGe ( x' )  WB   EgB , SiGe ( x' )     exp dx' dx exp gB , SiGe  exp kT kT kT   x   WB1  

( x) W kT   Eg 0 x    E   Eg 0  W kT    (WB  WB1 )  B1  exp dx exp gB , SiGe  B1 exp kT Eg 0    Eg 0  kT   kTWB1  

  Eg 0  WB21kT  W kT  W kT    .  (WB  WB1 )  B1   B1  1  exp Eg 0  Eg 0   Eg 0    kT 

For the second integral on RHS of (8), we have  EgB , SiGe ( x)  WB   EgB , SiGe ( x' )    dx' dx exp exp WB1  kT x  kT  WB

(10)

( x)   E   Eg 0  WB  x dx exp gB , SiGe  exp W B1 kT    kT 





WB

1 WB  WB1 2 .

2

Therefore, (7) gives    Eg 0  1 t B (SiGe) 2 WB21kT  WB1kT  WB1kT   2       ( W  W )  1  exp  W  W         B B 1 B b 1  kT  2 t B (Si)  WB2  Eg 0  Eg 0   Eg 0      2 2 1 2kT  WB1    WB1         1    WB  E g 0  WB    



SOLUTION

  E g 0   1  2kT WB1  WB1 kT WB1     .

1  1  exp     E g 0 WB  WB E g 0 WB    kT  

85

The base bandgap narrowing parameter corresponding to Fig. 10.13 is given by Egmax xje n+ emitter poly

Metal or metal silicide

10.11

Ge

n+

Eg0 p WB0

0

x

WB Wcap

 x  Wcap  Eg max  Eg 0  EgB , SiGe ( x)  Eg 0   W W  cap   B0

 Eg 0

x > Wcap

(1)

x < Wcap.

From Eq. (10.27) we have

J C 0 (SiGe, x jE ) J C 0 (Si, x jE )



  (WB 0  x jE ) WB 0



x jE

exp[E gB ,SiGe ( x) / kT ]dx

.

(2)

The integral in (2) can be written as Wcap WB 0   E gB ,SiGe ( x)    E gB ,SiGe ( x)    E gB ,SiGe ( x)       dx . (3) exp dx  exp dx  exp x jE       x jE Wcap kT kT kT      WB 0

The first integral on RHS of (3) is   E gB ,SiGe ( x)    E g 0  dx  (Wcap  x jE ) exp  . exp x jE kT    kT  Wcap



(4)

The second integral in (2) is   E g 0 (E g max  E g 0 )( x  Wcap )  WB 0   E gB ,SiGe ( x)    exp dx  exp dx Wcap  Wcap  kT   kT kT (WB 0  Wcap )    WB 0

(5)

     (E g max  E g 0 )     E g 0   kT  exp    .  WB 0  Wcap   1 exp     E  kT    kT   g max  E g 0   

Substituting (4) and (5) into (3), we have from (2) SOLUTION

86

J C 0 (SiGe, x jE ) J C 0 (Si, x jE )



 exp(E g 0 / kT )  Wcap  x jE   WB 0  Wcap    E  E g max  kT     1  exp g 0    W  x   W  x  E  kT jE   B0 jE  g max  E g 0      B0

.

10.12 For evaluating the transit time ratio, it is convenient to write the base bandgap narrowing

Egmax xje n+ emitter poly

Metal or metal silicide

parameter in the form

Ge

n+

Eg0 p WB0

0

x

WB Wcap

 ,SiGe ( x) , E gB ,SiGe ( x)  E g 0  E gB  x  Wcap  Eg max  Eg 0   , SiGe ( x)   EgB W W  cap   B0

where

0

(1) x > Wcap

(2)

x < Wcap.

The transit time ratio is, from Eq. (10.54) t B ( SiGe, x jE ) t B ( Si, x jE )   

(3) x jE Wcap

WB 0 WB 0 2 exp[E gB , SiGe ( x) / kT ] exp[E gB , SiGe ( x' ) / kT ]dx' dx 2 x x  (WB 0  x jE ) jE WB 0 WB 0 2 exp[E ' gB , SiGe ( x) / kT ] exp[E ' gB , SiGe ( x' ) / kT ]dx' dx 2 x x  (WB 0  x jE ) jE Wcap WB 0 2 exp[  E ' ( x ) / kT ] gB , SiGe x exp[E ' gB,SiGe ( x' ) / kT ]dx' dx  (WB 0  x jE ) 2 x jE

SOLUTION

87



WB 0 WB 0 2 exp[E ' gB , SiGe ( x) / kT ] exp[E ' gB , SiGe ( x' ) / kT ]dx' dx .

2 W x  (WB 0  x jE ) cap

The first double integral in (3) is for x < Wcap, and its value is Wcap WB 0 2 exp[E ' gB , SiGe ( x) / kT ] exp[E ' gB , SiGe ( x' ) / kT ]dx' dx 2 x x  (WB 0  x jE ) jE

 

Wcap WB 0 2 exp[E ' gB , SiGe ( x' ) / kT ]dx' dx 2 x  (WB 0  x jE ) jE x Wcap Wcap 2 exp[E ' gB , SiGe ( x' ) / kT ]dx' dx 2 x  (WB 0  x jE ) jE x

 

Wcap W B 0 2 exp[E 'gB , SiGe ( x' ) / kT ]dx' dx  (WB 0  x jE ) 2 x jE Wcap

Wcap 2 (Wcap  x)dx 2 x  (WB 0  x jE ) jE





(4)

 ( x'Wcap )(E g max  Eg 0 )  Wcap W B 0 2 exp   dx' dx  (WB 0  x jE ) 2 x jE Wcap (WB 0  Wcap )kT  

2(Wcap  x jE ) 2

 (WB 0  x jE ) 2 

 2(WB 0  Wcap )(Wcap  x jE )   Eg 0  Eg max  kT     .

1  exp   E   (WB 0  x jE ) 2 kT    g max  Eg 0  

The second double integral in (3) is for x > Wcap, and its value is WB 0 WB 0 2 exp[E 'gB , SiGe ( x) / kT ] exp[E 'gB , SiGe ( x' ) / kT ]dx' dx 2 W x  (WB 0  x jE ) cap

(5)



 ( x  Wcap )(E g max  E g 0 )  WB 0  ( x'Wcap )(E g 0  E g max )  WB 0 2 exp   x exp   dx' dx 2 W (WB 0  Wcap )kT (WB 0  Wcap )kT  (WB 0  x jE ) cap    



   2(WB 0  Wcap ) 2   E  E g max   kT kT  1    1  exp g 0   .

2   kT  (WB 0  x jE )  E g max  E g 0   E g max  E g 0     

Substituting (4) and (5) into (3), we have t B ( SiGe, x jE ) t B ( Si, x jE )

SOLUTION

 x jE Wcap

2(Wcap  x jE ) 2

 (WB 0  x jE ) 2

88



 2(WB 0  Wcap )(Wcap  x jE )  E  Eg max  kT   1  exp g 0  2   E   (WB 0  x jE )   E kT   g max g 0  

   2(WB 0  Wcap ) 2   E  E g max   kT kT  1    1  exp g 0   .   kT  (WB 0  x jE ) 2  E g max  E g 0   E g max  E g 0     

10.13

n+ emitter poly

n+

Ge

Egmax = Eg0 p x

WB

0 WB1

E gB ,SiGe ( x) 

x E g 0 WB1

x < WB1

 E g 0

(1)

WB > x > WB1.

 kT WB1   W  J C 0 (SiGe)    1  exp E g 0 / kT   1  B1  exp E g 0 / kT  J C 0 (Si)  E g 0 WB   WB 





1

(2)

100000 dE/kT=2.5 dE/kT=5 dE/kT=10

Current Ratio

10000 1000 100 10 1

0

0.2

0.4

0.6

0.8

1

WB1/WB

 W  V A (SiGe) kT WB1  expE g 0 / kT   1  1  B1  .

V A (Si) E g 0 WB  WB 



SOLUTION



(3)

89

Early Voltage Ratio

10000 3000 1000

dE/kT=2.5

300 dE/kT=5

100

dE/kT=10

30 10 3 1

0

0.2

0.4

0.6

0.8

1

WB1/WB

2 2 t B (SiGe) 1  2kT  WB1    WB1         1  t B (Si)   WB  E g 0  WB    

  E g 0   1  2kT WB1  WB1 kT WB1     .

1  1  exp     E g 0 WB  WB E g 0 WB    kT  

Transit Time Ratio



(4)

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

dE/kT=2.5 dE/kT=5 dE/kT=10

0

0.2

0.4

0.6

0.8

1

WB1/WB

SOLUTION

90

10.14

The Ge profile of interest, from Fig. 10.19, is

Electron flow (normal)

Electron flow (reverse) Egmax

Ge

B p

n+

n+

x

WB

0

The current ratio, from Eq. (10.27) is J C 0 (SiGe)  J C 0 (Si)

 WB WB

 exp[E 0

gB , SiGe

.

(1)

( x) / kT ]dx

For the forward current ratio, we use in (1) xE g max

E gB , SiBe , FORWARD ( x) 

WB

.

(2)

For the reverse current ratio, we should use in (1) E gB , SiBe , REVERSE ( x) 

(WB  x)E g max WB

(3)

Substituting (3) into (1), we have J C 0 (SiGe, reverse)  J C 0 (Si, reverse)

  WB

WB



0



SOLUTION

 ( x  WB )E g max  dx exp WB kT  

(4)

  WB   E g max  WB  xE g max    exp dx exp 0 kT W kT    B 

91





  E g max  WB kT    E g max    exp   1 exp  E    kT   kT   g max     

  E g max / kT 1  exp( E g max / kT )

.

From Fig. 10.21, the two transistors of interest are:

B n+ p

0

WB

(a)

x

Egmax Ge

n+ emitter poly

Metal or metal silicide

Egmax

Ge

n+ emitter poly

Metal or metal silicide

10.15

  WB

B n+ p

WB

0

x

(b)

From Eq. (10.32), we have for transistor b,

 E gB ,SiGe (WB )   exp kT V A (SiGe, b) DnB (SiGe, WB )  (WB )   WB exp[E  gB , SiGe ( x ) / kT ]dx 0 V A (Si, b) DnB (SiGe)  WB

SOLUTION

(1)



 ( x  WB )E g max  DnB (SiGe,WB )  (WB ) 1 WB dx exp  DnB (SiGe)  WB 0 WB kT  



 xE g max  DnB (SiGe,WB )  (WB ) exp(E g max / kT ) WB dx exp  0 DnB (SiGe)  WB  WB kT 



 xE g max  DnB (SiGe,WB )  (WB ) exp(E g max / kT ) WB exp 0  WB kT dx DnB (SiGe)  WB



DnB (SiGe, WB )  (WB ) exp(E g max / kT ) WB kT exp(E g max / kT )  1 DnB (SiGe)  WB E g max





92







kT exp(E g max / kT ) exp(E g max / kT )  1 . E g max

The VA ratio for transistor a is given by Eq. (10.33), namely





V A (SiGe, a) kT  exp(E g max / kT )  1 .

V A (Si, a) E g max

(2)

Since VA(Si, a) = VA(Si, b), we have   E g max  V A (SiGe, b)  .  exp V A (SiGe, a) kT  

(3)

From Eq. (10.36), we have for transistor b WB t B (SiGe, b) 2 WB  exp[  E ( x ) / kT ] gB , SiGe x exp[E gB,SiGe ( x' ) / kT ]dx' dx t B (Si, b)  WB2 0



WB  ( x'WB )E g max  2 WB exp[  E ( x ) / kT ] exp   dx' dx gB , SiGe x WB kT  WB2 0  



  E g max   x E g max  2 WB kTWB   exp  exp[E gB , SiGe ( x) / kT ]dx 1  exp 2 0 E g max   WB  kT   WB kT 



 (WB  x)E g max  2 WB kTWB 2 WB kTWB  dx exp dx   WB kT  WB2 0 E g max  WB2 0 E g max  



  E g max     E g max  2 kTWB  kTWB  1  exp   WB  exp  2  WB E g max  E g max   kT    kT 



 E g max     E g max   kT  kT  1  exp   1 . exp   E g max  E g max kT kT      

(4)

2

From Eq. (10.37), we have for transistor a  t B (SiGe, a ) 2kT  kT  1  exp E g max / kT   . 1  t B (Si, a )  E g max  E g max 





(5)

Therefore, we have

SOLUTION

93

t B (SiGe, b)  t B (SiGe, a )

 E g max     E g max  kT  1  exp   1 exp E g max  kT    kT    E g max  kT   1 1  exp E g max   kT 

.

(6)

Solutions to Chapter 11 Exercises 11.1

Ignoring the collector-substrate capacitance term, Fig. 9.26 gives the short-circuit-load

equivalent circuit

C

rb

rc C

B CdCS,tot C

r' (1+g'm re)

r'0

g'm vbe /(1+g'm re) re E

E

From Eq. (9.161), we have for the small-signal baseemitter voltage

vbe   vbe  ic re  ib (rb  re )  vbe  ic re ,

(1)

where we have assumed the base current to be small compared to the collector current.

Similarly, we have

vcb   vcb  ic rc  ibrb  vcb  ic rc   vbc  ic rc .

(2)

Now, for the short-circuit load, vbc = vbe, so (2) gives

vcb    vbe  ic rc .

(3)

From the equivalent circuit, we have ic 

 g m  g m   vbe  jC  vcb  jC   vbe  jC  rc ic , 1  g m re 1  g m re 

(4)

which can be rearranged to give

SOLUTION

94

1  jC r i  1 gg  r  jC v .  c c

m





m e

(5)

be



For a transistor having collector doping concentration of 2.51016 cm3, Fig. 3.5 shows that CdBC is about 0.45 fF/m2. If the basecollector junction area is 2 m2 and the frequency is 1 GHz, then CCdBC,tot103 mhos. For a bipolar transistor biased to have IC = 2 mA, gm = qIC/kT = 7.7102 mhos. That is, we can assume C to be small compared to gm. We can also show that Crc is small compared to unity. Therefore, (5) gives ic 

g m vbe .

1  g m re

(6)

Similarly, we have from the equivalent circuit vbe   jC  vbc  .  jC vbe rb  re  r (1  g m re )

ib 

(7)

Since 1/gm is on the order of 13 , r = 0/gm is on the order of 1300 , which is large compared to rb and re. Therefore, using (1) and (3), we can rewrite (7) as ib 

vbe  jC (vbe  ic re )  jC  (vbe  ic rc ) r (1  g m re )

j C  rc  C re g m   1   j C  C     vbe 1  g m re  r (1  g m re ) 

and

i ib

 ( )  c 

g m

1  (1  g m re ) j C  C    j C  rc  C re g m r

(8)

.

(9)

(9) can be rearranged to give

 ( ) 

g m r .

1  jr (C  C )  g m (re  rc )C





(10)

In the high-frequency limit, (10) becomes

 ( ) 

g m , j (C  C )  g m (re  rc )C





(11)

which drops to unity at a frequency fT given by 1

SOLUTION

g m 2f T (C  C )  g m (re  rc )C





(12)

95

1

or

2 f T

F 

kT (C dBE ,tot  C dBC ,tot )  C dBC ,tot (re  rc ) , qI C

(13)

where we have used Eqs. (9.153), (9.156), (9.157), and (9.170), i.e.  IC qI  C,   VBE kT

g m 

C  CdBC ,tot , C  CdBE ,tot  CDE C DE   F

and

11.2

qI C   F g m . kT

NPN transistor with AE = 2 m2, ABC = 10 m2, deep emitter with NE = 1020 cm3, NB =

1018 cm3, WB = 100 nm, 0 = 100, and NC = 51016 cm3. Assume one-sided junction approximation, with VBE = 0.8 V and VCB = 2 V.

(a) From Eq. (9.134), the diffusion capacitance due to holes in the deep emitter is

CDp  qI p pE / 2kT . For an npn transistor with deep emitter, Ip is the base current IB. From Eq. (9.170), E is given by the relation CDp   E qIC / kT . That is E given by

 E (deep emitter) 

 pE I B 2 IC



 pE .

20

(1)

From Fig. 3.14(b), we have pE = 51010 s. Therefore

 E (deep  emitter) 

 pE  2.5 10 12 s.

2 0

(2)

From Eq. (9.133), the diffusion capacitance due to electrons in the narrow base is

CDn  2qI ntB / 3kT . For an npn transistor, In is the collector current IC. Therefore, from Eqs. (9.166) to (9.171), we have

2 3

 B  tB .

From Fig. 3.14(a), we have nB = 330 cm2/V-s.

(3) Therefore DnB = kTnB/q = 8.58 cm2/s.

Therefore, Eq. (10.21) gives

SOLUTION

96

tB 

WB2  5.83  1012 s, 2 DnB

(4)

2 si ( bi  VCB ) . qN C

(5)

or  B  3.89 1012 s.

From Eq. (3.15), we have WdBC 

From Fig.3.4, we have bi  0.95 V. Therefore WdBC 

2 si ( bi  VCB )  2.77  105 cm. qN C

(6)

As discussed towards the end of Section 9.6.2.3, the space-charge-region delay time is Wd/2vsat, where Wd is the space-charge-region width and vsat is the saturated electron velocity. Therefore, we have t BC 

WdBC  138 .  1012 s, 2vsat

(7)

where we have used vsat = 107 cm/s. Similarly, we have WdBE 

t BE 

and

2 si ( bi  VBE )  171 .  106 cm, qN B

WdBE  0.86  1012 s, 2 v sat

(8)

(9)

where we have used bi = 1.025 V from Fig. 3.4. Therefore, we have

 F   E   B   BE   BC   E   B  t BE  t BC  8.63 10 12 s.

(10)

(b) From Eq. (3.13), we have

and

C dBE ,tot 

AE  si  1.22 10 14 F, WdBE

(11)

C dBC ,tot 

ABC  si  3.75  10 15 F. WdBC

(12)

(c) For JC = 0.3qvsatNC = 2.4104 A/cm2, we have

SOLUTION

97

and

1 2 f T

 F 

IC  AE J C  4.8  104 A

(13)

kT (C dBE ,tot  C dBC ,tot )  (8.63 10 12  0.86  10 12 ) s qI C

(14)

or fT = 16.9 GHz. (d) For (re + rc) = 50 , we have (re  rc )C dBC ,tot  0.19 10 12 s,

which is very small compared to F in this example.

1 2 f T

F 

(15) From (14) and (15), we have

kT (C dBE  C dBC )  C dBC (re  rc )  (9.59 10 12  0.19  10 12 ) s, qI C

(16) or fT = 16.3 GHz.

11.3

Transistor in Exercise 11.2 is replaced by a pedestal-collector design, with NC(int) =

51016 cm3, and NC(ext) = 51015 cm3, i.e., NC(ext) is reduced by 10 times. The corresponding areas are ABC(int) = AE = 2 m2 and ABC(ext) = (ABC  AE) = 8 m2. From (5) of Ex. 11.2, we have WdBC (int) 

2 si ( bi  VCB )  qN C (int)

2 si ( 0.95  2)  2.77  105 cm. qN C (int)

(1)

2 si ( bi  VCB )  qN C (ext )

2 si (0.89  2)  8.67  10 5 cm. qN C (ext )

(2)

Similarly, we have WdBC (ext ) 

Therefore,

C dBC ,tot (int) 

AE  si  0.75110 15 F, WdBC (int)

(3)

ABC (ext ) si  0.960  10 15 F WdBC (ext )

(4)

CdBC ,tot (ext ) 

and

CdBC ,tot  CdBC ,tot (int)  C dBC ,tot (ext )  1.7110 15 F.

(5)

To repeat Ex. 11.2, we simply use this value of CdBC,tot in Ex. 11.2.

SOLUTION

98

(a) Compared to Ex. 11.2, E, B, tBE and tBC all remain unchanged by the pedestal collector design, since NC(int) is unchanged. Therefore, F is also unchanged.

(b) CdBE,tot is also unchanged, and CdBC,tot is given by (5) above.

(c)

JC is determined by NC(int) and hence is unchanged.

Therefore, IC is also

unchanged, and 1 2 f T

 F 

kT (C dBE ,tot  C dBC ,tot )  (8.63 10 12  0.75  10 12 ) s qI C

(6)

or fT = 17.0 GHz. (d) For (re +rc) = 50 , (re  rc )C dBC ,tot  0.09 10 12 s,

and

1 2 f T

F 

(7)

kT (C dBE ,tot  C dBC ,tot )  C dBC ,tot (re  rc )  (9.38  10 12  0.09  10 12 ) s (8) qI C

or fT = 16.8 GHz.

11.4

Same as Ex. 11.3, but with NC(int) = 11017 cm3. In this case, WdBC(int) is reduced and

JC is increased.

WdBC (int) 

2 si ( bi  VCB )  qN C (int)

C dBC ,tot (int) 

and

2 si ( 0.97  2)  196 .  105 cm qN C (int)

AE  si  1.06 10 15 F WdBC (int)

CdBC ,tot  CdBC ,tot (int)  CdBC ,tot (ext )  (1.06  0.960)  10 15 F  2.02  10 15 F.

(1)

(2) (3)

(a) Compared to Ex. 11.3, E, B, and tBE all remain unchanged by the pedestal collector design. But tBC is reduced because of the larger NC(int). We have from (1) t BC 

WdBC (int)  0.98  10 12 s.

2 v sat

(4)

Therefore,

SOLUTION

99

 F   E   B  t BE  t BC  (2.5  3.89  0.86  0.98)  10 12 s  8.23  1012 s.

(5)

(b) CdBE,tot is unchanged, and CdBC,tot is given by (3) above.

(c) JC is increased to JC = 0.3qvsatNC(int) = 4.8104 A/cm2, and IC is increased to

IC  AE J C  9.6  104 A. Therefore,

1 2 f T

 F 

(6)

kT (C dBE ,tot  C dBC ,tot )  (8.23 10 12  0.38  10 12 ) s qI C

(7)

or fT = 18.5 GHz. (d) For (re + rc) = 50 , (re  rc )C dBC ,tot  0.10 10 12 s,

1

and

2 f T

F 

(8)

kT (C dBE ,tot  C dBC ,tot )  C dBC ,tot (re  rc )  (8.6110 12  0.10 10 12 ) s qI C

(9) or fT = 18.3 GHz.

11.5

Fig. 10.6 is shown here for use in this exercise.

SiGe/Si Improvement Factor

1E+4 Current gain Early voltage Base transit time

1E+3 1E+2 1E+1 1E+0 1E-1

0

2

4

Egmax/kT

6

8

10

Fig. 10.6. Relative improvement factors for current gain, Early voltage, and base transit time of a SiGebase bipolar transistor over a Si-base bipolar transistor, as a function of the maximum base bandgap narrowing. A linearly graded Ge profile is assumed. Also,  and  are set to unity.

SOLUTION

100

For Egmax = 100 meV, we have Egmax/kT  4. The incorporate of Ge does not change the capacitances or the depletion layer thickness.

It increases the current gain and reduces the base

and emitter transit times.

(a) Compared to Ex. 11.4, tBE and tBC are unchanged, but

or

 0 (Si)  E (SiGe) 1    E (Si)  0 (SiGe) 4

(1)

2.5 10 12  E (SiGe)  s  0.625 10 12 s, 4

(2)

where we have used E(Si) = 2.51012 s from (2) of Ex. 11.2. Also, from the plot above, we have

 B (SiGe) t B (SiGe)   0 .4 ,  B (Si) t B (Si)

(3)

 B (SiGe)  0.4  3.89 1012 s  1.56 1012 s

or (4)

where we have used B(Si) = 3.891012 s from (4) of Ex. 11.2. Therefore,

 F   E   B  t BE  t BC  (0.625  1.56  0.98  0.86)  1012 s  4.03  1012 s. (5)

(b) Compared to Ex. 11.4, CdBE,tot and CdBC,tot are unchanged.

(c) Compared to Ex. 11.4, JC and IC are unchanged, and 1 2 f T

 F 

kT (C dBE ,tot  C dBC ,tot )  (4.03  10 12  0.38  10 12 ) s, qI C

(6) or fT = 36.1 GHz.

(d) Compared to Ex. 11.4, (rc + re)CdBC,tot is unchanged, and

SOLUTION

101

1 2 f T

F 

kT (C dBE ,tot  C dBC ,tot )  C dBC ,tot (re  rc )  (4.41 10 12  0.10 10 12 ) s qI C

(7)

or fT = 35.3 GHz.

11.6

Two-port network for an intrinsic transistor (parasitic resistances ignored) biased in

forward-active mode, adapted from Fig. 9.25 (small-signal hybrid-π model of a bipolar transistor when parasitic resistances are neglected).

C C

B

v'1 i1

C

r'

r'0

g'm v'be

v'2 i2 E

E

(a)

The matrix representation for the two-port network is

    i1  ˆ  v1   Y11 Y12  v1  .   Y i    v  Y  Y   v   2  2   21 22   2 

(1)

Referring to the network, we can write

and

i1 

1 v1  j (C  C )v1  jC v2 , r

i2 

1 v2  g m v1  jC v2  jC v1. r0

Compare with the matrix representation in (1), we can write

Y 11 

g 1  j (C  C )  m  j (C  C ), r  0

(2)

Y 12   jC ,

(3)

Y 21  g m  jC ,

(4)

and

Y 22 

1  jC . r 0

SOLUTION

(5)

102

In Eq. (2), we have used the fact that 1/ r  gm / 0 where β0 is the static current gain [see Eq. (9.154)]. Note that the real part of Y 12 is zero because the base current is not a function of collector voltage for a transistor biased in the forward-active mode.

(b)

To derive the admittance matrix for an extrinsic transistor, we need to include the

IR drops in the resistors rb (= rbi + rbx), re and rc. The small-signal forms of Eqs. (9.161) and (9.162) are

vbe  vbe  ib (rb  re )  ic re

(6)

and

vce  vce  ib re  ic (rc  re ),

(7)

which, in matrix notation, imply  v1   v1   rb  re  v    v    r  2  2  e

re   i1  . rc  re  i2 

(8)

Inverting the matrix in (1), and substituting into (8), we have   v1    ˆ 1  rb  re re   i1   .  v    Y    r r  r   2  e c e     i2 

(9)

The admittance matrix for the extrinsic transistor is

i1  ˆ v1  Y11 Y12  v1  i   Y  v   Y Y  v  .  2  2   21 22   2 

(10)

Comparing (9) and (10) with Eqs. (8.71) and (8.72) for the case of a MOSFET, we see the equivalencies between rb and Rg, between re and Rs, and between rc and Rd. Therefore, from Eq. (8.73) for an extrinsic MOSFET, we can write, to first order in parasitic resistance, the corresponding matrix elements for an extrinsic bipolar transistor as

Y11  Y 11  Y 112 rb  Y 12 Y 21 rc  (Y 11  Y 12 )(Y 11  Y 21 )re ,

(11)

Y12  Y 12  Y 11Y 12 rb  Y 12 Y 22 rc  (Y 11  Y 12 )(Y 12  Y 22 )re ,

(12)

Y21  Y 21  Y 11Y 21 rb  Y 21Y 22 rc  (Y 11  Y 21 )(Y 21  Y 22 )re ,

(13)

Y22  Y 22  Y 12 Y 21 rb  Y222 rc  (Y 12  Y 22 )(Y 21  Y 22 )re .

(14)

SOLUTION

103

(c)

To obtained the cutoff frequency, we use the short-circuit-load configuration. For

the intrinsic transistor, that means v2  0. The intrinsic device current gain is i Y  21 . i1 v 0 Y 11

  2

(15)

2

Therefore, the cutoff frequency of the intrinsic device is given by

Y 1  21  Y 11

g 2m  T2 C2 g 2m

 02

(16)

,

  (C  C ) 2 T

2

which can be rearranged to give

T  2 f T 

g 2m 

g 2m

 02

(C  C )  C 2

2



g m (C  C ) 2  C2

,

(17)

where we have use the approximation 1/β0 << 1. For a typical modern bipolar transistor which has reduced Cμ (= CdBC,tot), we have Cπ (= CdBE,tot + CDE) >> Cμ. Therefore, (17) can be simplified to give the commonly used approximation, i. e., Eq. (11.8), of

T  2 f T 

g m .

C  C

(18)

For the extrinsic device, the short-circuit-load extrinsic current gain, to the first order of rb, rc, and re, is  1  rbY 11  rcY 22  re (Y 11  Y 21 )(Y 21  Y 22 ) Y 21 i Y Y   21  21   i1 v 0 Y11 Y 11  1  rbY 11  rcY 12 Y 21 Y 11  re (Y 11  Y 12 )(Y 11  Y 21 ) Y 11 

 2

2

Y  21 1  (rc  re )(Y 22  Y 12 Y 21 Y 11 )  re (Y 12  Y 11Y 22 Y 21 ) .

Y 11

(19)

Notice that β is independent of rb. Noting that Re(Y 12 )  0 and that (rc  re )Re(Y 22 )

1 , we

have, to the first order of rb, rc, and re,

 

Y 21 1  (rc  re )Re(Y 12 Y 21 Y 11 )  re Re(Y 11Y 22 Y 21 ) Y 11

Y  21 1  (rc  re )Re(Y 12 Y 21 Y 11 ) .

Y 11

SOLUTION

(20)

104

where we have used the fact that re Re(Y 11Y 22 / Y 21 )

1 for a typical transistor in writing the

last equation. The cutoff frequency is obtained by setting |β| = 1, i.e.,  (rc  re )T2 g m (C2  0  C C  C2 )  1  . ( g m  0 ) 2  T2 (C  C ) 2  ( g m  0 ) 2  T2 (C  C ) 2  g 2m  T2C2

1

(21)

Noting that 1/β0 ≪ 1 and keeping only the terms to the first order of rc and re, we can rearrange (21) to give

g 2m  T2C2  T2 (C  C )2  2(rc  re ) g m T2 (C  C )C ,

(22)

which in turn can be rearranged to give

(C  C ) 2  C2  g m (C  C )C   1  (re  rc ) . T g m (C  C ) 2  C2   1

(23)

Using (17) and (18) and the approximations involved in them, we can simplify (23) to the commonly used approximation, i. e., Eq. (11.9), of

1

T



11.7

1 1  (re  rc )C   (re  rc )CdBC ,tot .  T  T

(24)

From Eq. (8.69) and the discussion in Section 8.5.4, fmax is given by the equation 2

4 Re(Y11 ) Re(Y22 )  Y12  Y21* .

(1)

From Ex. 11.6, we have

and

Y12  Y 12  Y 11Y 12 rb  Y 12 Y 22 rc  (Y 11  Y 12 )(Y 12  Y 22 )re ,

(2)

Y21  Y 21  Y 11Y 21 rb  Y 21Y 22 rc  (Y 11  Y 21 )(Y 21  Y 22 )re .

(3)

Keeping only the terms to the first order of rb, rc, and re, we have Re(Y11 )Re(Y22 )  g m  0 r 0  rb  g m2  02 r 0  g m  2C2  0   2 (C  C ) 2 r 0   rc  g m  0 r02  g m  2C2  0   2C2 r0 

(4)

 re  2 g m2  0 r 0   2C2 r 0  g m  0 r 02  .

2

A similar expansion and approximation can be made for Y12  Y21* , and (1) can then be used to determine fmax. However, if we make the large Early voltage approximation early on, the procedure and the final expression for fmax can be greatly simplified. For a typical modern bipolar

SOLUTION

105

transistor operating at IC = 1 mA, Cμ (= CdBC,tot) > 1 fF, β0 > 100, r0  105  , g m  4 102 1 and fmax > 30 GHz. Thus, for a typical modern bipolar transistor, the terms of the second order in

1 r0 in (4) can be ignored compared to those of the first order in 1 r0 . That is, (4) can be approximated by

Re(Y11 )Re(Y22 )  g m  0 r 0  rb  g m2  02 r 0  g m  2C2  0   2 (C  C ) 2 r0   rc  g m  2C2  0   2C2 r0   re  2 g 2m  0 r0   2C2 r0  .

(5)

Again, keeping only the terms to the first order of rb, rc, and re, we have 2

Y12  Y21*  g 2m  rb  4 g m  2 (C  C )C  2 g 3m  0   rc  4 g m  2C2  2 g m2 r0   re  2 g 3m  2 g m2  0 r 0  2 g m2 r0 

(6)

 g 2m  rb  4 g m  2 (C  C )C  2 g 3m  0   4rc g m  2C2  2re g 3m ,

1 r0 , re r0

where in writing the last equation, we have used the fact that g m

rc r0 re

1 r0 , rc

1 . Substituting (5) and (6) into (1), and again noting that g m

1 and

r0 , rb

r0 ,

r0 and β0 ≫ 1, we have the equation for fmax in the form

g m2 (1  2 g m rb  0  2 g m re )





2  max 4 g m rb (C  C )C  (C  C ) 2 g m r0   rcC2  reC2 g m r0 .

(7)

To the first order of rb, rc, and re, (7) can be rearranged to give

max 



g m (1  g m rb  0  g m re )



4 g m rb (C  C )C  (C  C ) 2 g m r0   rcC2  reC2 g m r0

.

(8)

For a typical modern bipolar transistor, g m r0  qVA kT  4 103 (see Section 9.3.2.1). While it is true that Cπ = (CdBE,tot + CDE) ≫ Cμ, Cπ is not larger than Cμ by a factor of g m r0 without severe base widening occurring. In other words, for a transistor operated at typical current densities of interest, the terms proportional to 1 gm r0 in the denominator can be ignored. That is, we have

max 

g m (1  g m rb  0  g m r e ) 4 g m  rb (C  C )C  rcC2 

SOLUTION

.

(9)

106

Equation (9) is equivalent to that obtained by making the large Early voltage approximation ( 1/ r0  0 ) from the very beginning. The familiar approximation for ωmax is obtained if we approximate the numerator by g m and keep only the term containing rb in the denominator, i.e.,

max 

g m 4 g m (C  C )C rb



f T , 8 C rb

(10)

where we have used Eq. (11.7) for fT .

11.8

A symmetric lateral npn transistor on SOI having the following physical parameters:

WE = 45 nm, WEC = 40 nm, NE = 4×1020 cm3, NB = 5×1018 cm3, LE = 2 m and tsi = 100 nm (i.e., AE = 0.1 × 2 m2).

 F   E   B   BE   BC ,  E (shallow emitter) 

B 

W W (V  ,V  ) N n 2 2 I B WE2  E B BE BC 2 B ieE , 3I C 2 D pE 3DnB N E nieB

 , VBC  ) 2t B WB2 (VBE  , 3 3DnB

 BE  t BE 

 ) WdBE (VBE 2vsat

 BC  t BC 

 ) WdBC (VBC .

2vsat

(a) To calculate WB, WdBE, WdBC, E, B, BE, BC, and F for a given WEC, we need to use the following equations (the equation numbers are those in the book) :

 ,VBC  )  WdBE (VBE  )  WdBC (VBC  ).

WE C  WB (VBE  )  2 si ( bi  VBE  ) / q( N B  n), WdBE (VBE

(10.72) (10.78)

 )  2 si ( bi  VBC  ) / q ( N B  n), WdBC (VBC

(10.79)

J C  qvsat n.

(10.80)

where, n is given by

[However, in the low-injection approximation, we ignore mobile space charge and assume n = 0 in this exercise.]

SOLUTION

107

Cd 

 si Wd

.

(3.26)

 )  LE CBC , fringe , C dBC ,tot  AE CdBC (VBC

(11.19)

 )  LE CBE , fringe , CdBE ,tot  AE CdBE (VBE



 







p0 Eg n0 Eg  nie2  ni2 exp Eg / kT .

(11.20)

(9.14)

The Klaassen unified model for the apparent bandgap-narrowing parameter 2   N   N      Eg  N   6.92 ln   ln  0.5  meV.   17  17   1.3  10 1.3  10          

(9.18)

DnB for the p-type base can be obtained from Fig. 3.14.

The calculated results are shown in the table in (c).

(b)

The collector current can be calculated from Eqs. (10.74) and (10.75), namely

 ,VBC  )  IC 0 (VBE  ,VBC  ) exp(qVBE  / kT ), IC (VBE with 1

2  1  / kT )   qAE DnB nieB 4n 2 exp(qVBE 1   1  ieB  ,VBC  ) I C 0 (VBE  1  .

2   ,VBC  )  4 N BWB (VBE N B   

The calculated results are shown in table in (c).

(c)

The calculated values of fT and fT , assuming (re  rc )  200  , are shown in

table.  (V) WB VBE

WdBE

WdBC

E

B

BE

BC

F

IC

fT

(nm)

(nm)

(nm)

(ps)

(ps)

(ps)

(ps)

(ps)

(mA)

(GHz) (GHz)

0.88

9.56

7.19

23.2

0.0274

0.0453

0.0359

0.116

0.225

0.18

197

156

0.90

9.94

6.82

23.2

0.0285

0.0490

0.0341

0.116

0.228

0.38

305

217

0.92

10.4

6.43

23.2

0.0297

0.0531

0.0321

0.116

0.231

0.77

418

269

SOLUTION

fT

108

0.94

10.8

6.01

23.2

0.0310

0.0578

0.0300

0.116

0.235

1.5

508

303

0.96

11.3

5.56

23.1

0.0325

0.0635

0.0278

0.116

0.239

3.0

564

322

0.98

12.0

5.07

23.0

0.0343

0.0707

0.0253

0.115

0.245

5.6

589

329

Fig. 11.5 (vertical SiGe-base transistor) and comparison with calculated fT of symmetric lateral transistor. The solid black line is fmax of SiGe device. The dash black line is fT of SiGe device.

The red dots are calculated fT of lateral device.

(d)

A symmetric lateral transistor on SOI has significantly smaller capacitance than a

vertical transistor. At low currents, fT is determined by capacitance, and not by transit times. At high currents, fT is determined by the transit times, which are not too different for a vertical transistor and a lateral transistor. This explains the significantly higher fT for a lateral transistor at lower currents.

SOLUTION

109

Solutions to Chapter 12 Exercises 12.1 Consider a bipolar transistor biased to operate in the saturation region.

Ignoring parasitic

resistances, the collector current is given by Eq. (12.5) and the base current is given by Eq. (12.6), namely,   I I C  I B 0 F e qVBE / kT   0 (1  e  qVCE / kT )  B 0 R e qVCE / kT  , I B0F  

and   I I B  I B 0 F e qVBE / kT 1  B 0 R e  qVCE / kT  .  I B0F 

The current gain is



IC  IB

 0 (1  e  qV

CE / kT

)

I B 0 R  qVCE / kT e I BoF

I 1  B 0 R e  qVCE / kT I B0F

.

Plot the current gain  as a function of qVCE/kT for qVCE/kT = 0 to qVCE/kT = 4, using 0 = 100 and IB0R/IB0F = 1. Repeat using 0 = 100 and IB0R/IB0F = 10. This exercise shows that for

IC / IB

practical transistors the current gain is less than 1 only for very small VCE values.

SOLUTION

100 90 80 70 60 50 40 30 20 10 0 -10

IB0R/IB0F = 1 IB0R/IB0F = 10

0

0.5

1

1.5

2 qVCE / kT

2.5

3

3.5

4

110

IC / IB

10 9 8 7 6 5 4 3 2 1 0 -1

IB0R/IB0F = 1 IB0R/IB0F = 10

0

SOLUTION

0.05

0.1 qVCE / kT

0.15

0.2

111