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

Solutions to Chapter 2 Exercises

2.1.

From Eq. (2.4), f (E − E) = f

eE / kT

1+ e−E / kT = eE / kT +1 ,

and f (Ef + E) =

1 . 1 + eE / kT

Adding the above two equations yields f (E f − E) + f (E f + E) =

2.2.

eE / kT +1 = 1. eE / kT +1

Neglecting the hole (last) term in Eq. (2.19), one obtains N c e−( Ec − E f )/ kT + 2N c e−( Ec + Ed −2E f )/ kT = N d .

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 / N )e( Ec − Ed )/ kT =

4e− Ed / 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 / N )e( Ec − Ed )/ kT N E / kT d c = de c , − Ed / kT 4e 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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

Ec

1/ 2 −( E − E f )/ kT

dE dE .

Applying integration by parts to the numerator yields (3 / 2)kT  ( E − E c)1/2 e−( E − E f



)/ kT

K.E. =



 (E − E ) e

1/ 2 −( E − E f )/ kT

3 = kT . 2

(b) For a degenerate semiconductor at 0 K, f(E) = 1 if E < Ef and f(E) = 0 if E > Ef. Ef > Ec. Therefore,

Here

( E − E )3/2 dE K.E. =  ( E − E )1/2 dE = 3 ( E − E ). 5  E Ec f

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r2E , S

where E is the magnitude of the electric field on S. E=

Q 4sir 2

3-D Gauss’s law then gives

which is Coulomb’s law. Since E =−dV/dr, the electric potential at a point on the sphere is V=

Q 4sir

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

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:

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

For x < 0, d 2 (i )

−

q2Nd

sikT

 = 0 i

General solution: i = Aex / LD where LD 

si kT q2 Nd

For x > 0, q d 2 (i ) q 2 N d =−  −  i Nd  kT dx2  

General solution: i = Be− x / LD +

kT Nd qNd

Matching i and di/dx at x = 0, A = −B =

kT Nd 2qNd

2.7 −( E − E f ) / kT Electron density per energy = N (E) f (E)  E − EC e

Differentiate the above and set to zero gives

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

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.

At point A:

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

2.9 (a) Eq. 2.20,

E − E = kT ln c

 Nc  .    Nd 

Table 2.1,

Nc = 2.9 × 1019 cm-3

Therefore,

Ec − E f = 0.266 eV

(b) Eq. 2.67

LD 

sikT = 0.13 m q2 Nd

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

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

(d)

 2kT  h2  

3/2

 T 3/2

Eq. (2.10)

Nc = 2g

Therefore,

Nc(100C) = (373/300)3/2 Nc(300 K) = 4.0 × 1019 cm-3 N  E − E = kT ln c = 0.274 eV at 100C. c f    Nd 

2.10 (a) E f − Ei = 0.2 eV Eq. (2.14),

n = ni e(E f −Ei )/kT = 2.26 1013 cm−3

Eq. (2.15),

p = ni e(Ei −E f )/kT = 4.43106 cm−3

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

Solutions to Chapter 3 Exercises 3.1

Equation (3.138) gives 1 W W x M = exp − 0( p −  n )dx − 0  n exp − 0( p −  n )dx dx .



(1)

0 = exp− W( p −  n )dx − W  n exp− x( p −  n )dxdx  0  0  0 

(2)





As Mp → , 1/Mp → 0, the above equation gives

) ( p−n n ) exp(−( p − n )W ) − 1 .

(

= exp −( p − n )W +

(

) W = ln( n /  p ) / (n −  p ) . n =  p exp −( p − n )W ,

Therefore, or

(4)

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

3.2

u( x) = − f ( x' )dx' .

Let

(1)

du( x)

Therefore

= − f ( x)

(2)

dx u( x ) deu( x ) = − f ( x)e . dx

and

(3)

Using (3), we have

 f ( x) exp− f ( x' )dx'dx =  f ( x)eu( x)dx W

 0



(4)

0 W

= −

deu( x )

dx = 1 − eu(W ) .

 f ( x) exp− f ( x' )dx'dx = 1 − exp− f ( x' )dx' .

Therefore,

 0



 0



A similar procedure can be used to show that W W W f ( x) exp− f ( x' )dx'dx = 1 − exp− f ( x' )dx'











 0



(5)

(6)

by letting u( x) = −x f ( x' )dx' .

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

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,

). 1 2(N a + N d ) = ( bi −V app 2 Cd  si qNa Nd

(1)

A plot of 1/ C 2d 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 = kT ln n0 p0 = kT ln d a .     bi  n 2i   ni2 

(2)

Therefore, from the plot of 1/ C d2 versus Vapp plot, we have Slope =

− 2(N a + N d )

(3)

 si qNa Nd 1 and

C (V d

= 0)

 Na Nd  2(Na + Nd )   qN N . ln q  n2i  si a d

= kT

app

(4)

The two equations (3) and (4) can be solved for the two unknowns Na and Nd. 2

1/C d

bi

Vapp

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

3.4

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 = qNa (W )dW ,

(1)

where the depletion-layer width W is a function of V. C=

The depletion-layer capacitance is

dW dQd , = qNa (W ) dV dV

(2)

C . q(dW / dV )

(3)

which gives Na (W ) = C=

From

 si

W we have

=−

 si dC C2 dV

 siC d(1 / C2 ) 2

(4)

Substituting (4) into (3), we have N (W ) = a

3.5

q si [d(1 / C ) / dV ] 2

Referring to Fig. 3.6 and Eq. (3.20), we have Em (Wd + d ) ,  m =  bi + V = 2

where V is the applied voltage.

(1)

From Gauss’s law we have

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

Qd =  siEm . Therefore, 1

(2)

dEm (Wd + d ) / 2 d E 

si m

1  dWd  . (Wd + d ) +Em  2 dE  si



(3)



Now, from Eq. (3.19), we have d( xn + x p ) dxp dWd dxn E =E =E +E = (x − d ) + x . m n p m m m 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

slope = qE

Ev Eg 0

x Ec Ev n-region

p-region

The WKB tunneling coefficient is, from Eq. (3.127), − 4 W  T (E) = exp  2m* − q (x) − Edx  h 0  with W = Eg / qE.

(1)

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* E − qxE dx (3)  h





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

− 4 2m* E3g/ 2  . = exp 3 qE   3.7 (a) p-type Ec Ef Ev 2V Ec Ef Ev n-type

 Na Nd   = 0.82 V ln  bi = q  ni 2  kT

(b)

Wd =

(c)

Emax =

2 si (Na + Nd )( bi +VR ) = 0.27 m qNa Nd

qNa (Wd / 2) = 2.1105 V/cm  si

3.8 (a) Wd Ec

bi

Ei Ef Ev

Ec Ef

Ev n+

p-type x=0

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

 Na  eV Ei − E f = kT ln n  = 0.38  i 

(b)

 bi =

(c)

Wd =

(d)

E −E = f

3.9 −

+ 0.38 eV = 0.94 eV

2 si bi

5 = 0.20 m = 210− cm

 Nv  increases as temperature increases.  kT ln N  a

Ei − Ef

And

 bi

decreases as temperature increases. decreases as temperature increases.

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

 [np (0) − np 0 ]

= eq[ i (0)− n(0)] kT − eq[ i(0)− n(−W d)] kT

ni =

np (0)

−

(1) −q i kT

Equation (3.43) gives e−q i kT =

n2 i

Nd Na

qVapp kT

np0

eqVapp kT .

(2)

Therefore, Eq. (1) can be reduced to np (0) = eqVapp kT +  . np0

1+ 

Using Eqs. (3.45), (2) and (3), we have kT  Nd   (−W ) −  (0) =  − ln   n d n i q n p (0)  kT kT 1+ e−qVapp  = ln  . q  1+  

(3)

(4)

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

3.10 0

xm x

qBn q Ef

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

image (x) =

q , 160x

and the electrostatic potential associated with the electric field in the silicon is given by d 2 field (x) qNd . − =  si dx2

(1)

(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 

Therefore, the total electrostatic potential energy is  q qN d  x2  Wd x −  . PE (x) = −q[ image (x) + field (x)] = −q + 2  16 si x  si  The peak of PE(x) is given by dPE(x)/dx = 0, that is − q + qNdWd  xm  1 − =0. 16 x2  si m si  Wd 

(3)

(4)

(5)

We will first make the assumption that xm/Wd << 1, and then come back to verify that our assumption is correct. With this assumption, (5) gives

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

where

xm 

q , 16 siEm

(6)

Em =

− d field (x) qN W = d d  si dx x =0

(7)

is the absolute value of the electric field in the silicon at the interface.

The energy barrier

lowering is  q qNd  xm2  Wd xm −  q =| PE (xm ) |= q + 16  x  2  si m si    q qN W x   q d d m + = 16  x  si  m si 

(8)

q3Em . 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 2 si bi , qNd

Wd =

(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.

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

3.11

The multiplication factor for hole-initiated impact ionization is 1 Mp

(

) )

( (

) )

(1)

(

) )

(2)

= exp − W  p − n dx − 0W  exp − x  p − n dx dx,  n   0

and that for electron-initiated impact ionization is 1 Mn

(

) )

(

= exp − W  n −  p dx − 0W  exp − W  n −  p dx dx.   p  0

From Ex. 3.2, we have

 f ( x ) exp ( −  f ( x) dx)dx = 1− exp(− f ( x ) dx ) W

(3)

and

 f ( x)exp(− W

)

(

)

f ( x) dx dx = 1− exp −  f ( x ) dx . W 0

(4)

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

Applying Eq. (3) to Eq. (1), we obtain W x 1 1− =   p exp −  (  p −  n )dx dx. 0 0 Mp

(5)

Similarly, applying Eq. (4) to Eq. (2), we have W W 1 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

(

)

0 p exp − 0( p − n )dx dx = 1. W

(7)

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

  p exp (−  (  p − n )dx)dx = 1− exp(− ( p − n )dx) W

(

)

+ 0 n exp −  0 (  p −  n )dx dx. W

(8)

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

( ) = exp(− (  −  ) dx ) .

0 n exp − 0 ( p −  n )dx dx W

(9)

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

  exp(− ( −  )dx)dx = 1, W

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