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

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

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

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

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

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 =

d

c

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

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

c

Ec

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

dE

E

K.E. =



c

 (E − E ) e

1/ 2 −( E − E f )/ kT

c

Ec

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. Ef > Ec. Therefore,

Here

Ef

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

5  E Ec f

c

Ec

c

f

c

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

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:

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

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

For x < 0, d 2 (i )

−

q2Nd

sikT

2

dx

 = 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  

si

si

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

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.

(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,

E − E = kT ln c

f

 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

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

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

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)





p



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

x

u( x) = − f ( x' )dx' .

Let

(1)

0

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

x

 0

W



(4)

0 W

= −

deu( x )

dx = 1 − eu(W ) .

dx

0

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

Therefore,

W

x

0

 0

W



 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





x





 0



(5)

(6)

W

by letting u( x) = −x f ( x' )dx' .

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

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

2

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

0

bi

Vapp

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

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

3.4

x

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

dW

=−

dV

 si dC C2 dV

=

 siC d(1 / C2 ) 2

.

(4)

dV

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

3.5

2

.

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

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

Qd =  siEm .

Therefore, 1

=

dV

=

C

dQ

d

d

(2)

dEm (Wd + d ) / 2 d E 

=

si m

1  dWd  .

(Wd + d ) +Em  2 dE  si



m

(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

Ec

slope = qE

Ev Eg 0

W

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



0

g



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

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

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 

Eg

(b)

 bi =

(c)

Wd =

(d)

E −E = f

3.9 −

2q

+ 0.38 eV = 0.94 eV

2 si bi

v

5 = 0.20 m = 210− cm

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

So

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)

−

ni

Nd

(1) −q i kT

e

.

ni

Equation (3.43) gives e−q i kT =

n2 i

Nd Na

e

qVapp kT

=

np0

eqVapp kT .

Nd

(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

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

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

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

x

From Ex. 3.2, we have

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

0

x

W

0

0

(3)

and

 f ( x)exp(− W

W

0

x

)

(

)

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

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

x

(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

x

W

0

0

0

(

)

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

W

x

(8)

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

( ) = exp(− (  −  ) dx ) .

0 n exp − 0 ( p −  n )dx dx W

x

(9)

W

0

p

n

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

  exp(− ( −  )dx)dx = 1, W

0

W

n

x

n

p

(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

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

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. W =x +x = d

n

p

2 m

=

Em

So bi = m = 0.817 V.

= 1.782 10−5 cm qNa Nd

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

→

x = –0.17010−5 cm.

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

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

3.13 depletion region

n-type (Nd

p-type (Na

=1e18 cm-3)

=1e17 cm-3)

x 0

At x = 0,

10m

Jn (0) = qD n2 / (N L ) expqV /kT = 0.44 A/cm2. a n   n i

Hole current equals the diffusion current on the n-side, i.e., J p (0) = qD n2 / (N L ) expqV /kT = 0.088 A/cm2. d p   p i At x = 10 m, Jn (10m) = Jn (0)e−x/Ln = 0.27 A/cm2.

The total current is continuous, so J p (10m) = Jn (0) + J p (0) − Jn (10m) = 0.26 A/cm2.

3.14 (a)

Built-in potential bi = Eg / 2q + (kT / q) ln(Na / ni ) = 0.98 V.

Depletion layer width W = 2 ( − 0.6) / qN 

1/2

d

si

bi

= 70 nm.

a

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

(b)

The total current is dominated by the electron diffusion current:

Jn (x = 0) = [qD nn2i / N a Ln ]exp(qV app / kT ) = 9.2 mA/cm2.

(c)

Jn (x) = Jn (x = 0) exp(−x / Ln ) = 5.6 mA/cm2.

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

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

3.15  Na Nd   ln  = 0.804 V  bi = q  ni 2  kT

(a)

Wd =

−5 2si ( bi − 0.5)(Na + Nd ) = 1.3110 cm qNa Nd

Nd

Q ( p − side) = qN x = qN W d

a

(b)

Cd =

p

 si Wd

a

d

= 4.8 10−8 C/cm2 , negative

N d + Na

= 7.9 10−8 F/cm2

3.16 (a)

 bi =

kT  Na  + q ln n  = 0.56 + 0.42 = 0.98 V 2q  i 

Eg

Wd = Emax =

= 7 10−6 cm qNaWd = 1.1105 V/cm  si

2

(b)

n p = n i eqVF / kT = 1.15 1013 cm-3 Na

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

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

3.17 (a)

p-type

n-type

Ec

0.5 V

Ef Ev

(b)

Ec Ef Ev

 Na Nd  V  = 0.90 ln  bi = 2 q  ni  kT

Wd =

2si (Na + Nd )( bi − 0.5) = 0.075 m qNa Nd

Emax =

qNaWd Nd /(Na + Nd ) = 1.1 105 V/cm  si

2

(c)

n p = ni eqV / kT = 4.411011 /cm3 Na

3.18 (a)

 bi =

N N  ln a d  = 0.70 V q  ni2 

kT

(b) Wd =

2si (Na + Nd ) bi = 0.522 m qNa Nd

xn =

Na Wd = 0.17 m N a + Nd

x =

Nd Wd = 0.35 m + N Na d

p

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

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

(c)

Em =

2 bi = 2.68 104 V/cm Wd

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

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

Solutions to Chapter 4 Exercises 4.1.

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

dQd d s

qN a(1 − e−q s / kT )

=

,

Es

and Ci  −

dQ i

d s

=

q(n 2 / N )(eq s / kT − 1) i

a

.

Es

Here Es = E ( = s). (b) Take the square of Eq. (4.16) and differentiate with respect to s, we obtain   dQ n2 − s q s / kT i q s / kT 2Q = 2 qN  1 − e + e −1 . si a 2 s d s Na  

(

)

(

)

Using Qs =−siE s (Gauss’s law) and the above two equations in (a), it is straightforward to show that  dQ qN  n2 − s a q s / kT i q s / kT C − = + e −1  = C + C .  1− e d i si 2 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 C C  i

d

qNa

.

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)

Cox =

ox = Cmax = 3.4 10−7 F/cm2 tox

→

tox = 10 nm.

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

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

(b)

Cd =

Cmin =

CoxCd Cox + Cd

 si Wdm

= 7.3 10−8 F/cm2

→

=

W dm

→

2 si (2 B ) = 1.110−5 cm qNa

Take 2B  0.9 V (c)

Cd = 9.410-8 F/cm2

1 + LD 1 1 kT + = =  si Cox Cfb 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 + Vfb = 4.63 V

Eq. (4.33),

siqNa − siqNa /2 . s = V − V + Cox 2C Solve for Vg, 2

  qN / 2  si a Vg −V fb + =   s + si a  .

2 Cox 2Cox  

 qN

Take a differential,

siqNa / 2  1 V /  = 2  + = 1 + siqNa / 2 s .  s g s  Cox  2 s Cox   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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

V = V + + g

fb

s

 ni  q s / 2kT

2sikTNa  e Cox N  a

.

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 − Vfb − 2 B )  ln  . q 2sikTNa  

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 ) dVg

=

Cox (Ci + Cd ) .

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

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

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 Nd   = −(0.56 − 0.35) = −0.21 V. Vfb = − − B  = − 0.56 − ln 2q q n    i 

C

Vfb

Vg

Band diagram at Vg = 0:

Ef

Ef

n+

n-type

At s = 2B = 0.7 V: V = ox

Qd Cox

=

2 siqNd s 4.810−8 = 0.14 V. = Cox Cox

4.7.

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

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

(a)

 =

kT

B

q

ln

Na

= 0.42 V

ni

Vfb = −

Eg 2q

− = −0.98 V B

Vt = Vfb + 2 B +

2 siqNa (2 B ) = −0.98 + 0.84 + 0.97 = 0.83 V. Cox

(b)

C Cmax

Vfb

(c)

Vt

Vg

Cmax = Cox = 1.7310−7 F/cm2 Wdm =

2 si (2 B ) = 0.104 m qNa

1 1 Wdm = + C  si Cmin ox Cmin = 0.6310−7 F/cm2

4.8.

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

Then

−Vfb =  s +

2 siqNa s Cox

Solve the above quadratic equation in

s :

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

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

2 siqNa

s = −

2Cox 2 si qNa 4Cox

+ 2 siqN2 a −V fb 4Cox

= 0.278 V,

2

Therefore,

s = 0.353 V 2

ni eq s / kT = 8.3108 cm-3 Electron density at the surface: n = Na hole density at the surface:

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)

Wd = (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

so Vox = Vg − s(min) = −1.8 V.

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

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

4.10.

kT  Na   B = q ln n  = 0.4083(V )  i  Qs = −Cox (Vg − Vfb − 2 B ) = −1.09 10−6 (C / cm2 ) − Qs / q = 0.680(nm) xpoly = N poly

qN poly x 2poly Since p = = 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 2/3  3hqE   3 s  Ej =  where j = 0,1,2,  j +  4 4 2m    x  kT  Na   B = q ln n  = 0.4679(V )  i  2 si (2 B ) = 34.9(nm) Wd = q qNWad qN W E = N (x)dx = a d = 0.537(MV cm) s

 si 0

si

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 ) = NT 1 − exp − Ninj (t ) where

N (t )  inj

t

jG (t) dt ,

(1) (2)



q is a solution to the differential equation.

0

From (1), we have

dNinj (t ) dnT . = N T exp −Ninj (t ) dt dt

(

)

(3)

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

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

From (2), we have dNinj (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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

Solutions to Chapter 5 Exercises

5.1.

 =

kT

Na

= 0.42 V q ni Eg Vfb = − − B = −0.98 V 2q

(a)

ln

B

Vt = Vfb + 2 B +

2 siqNa (2 B ) = −0.98 + 0.84 + 0.97 = 0.83 V. Cox

Ec Ei Ef Ev Ef

p-type silicon

Oxide

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

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

(b)

Vt = Vfb + 2 B +

2siqNa (2 B + Vbs ) = −0.98 + 0.84 + 1.44 = 1V. 3 0 Cox

Ec Ei Ef Ev Efn

Ef p-type silicon

Oxide

5.2.

Eq. (5.14) is an implicit equation relating s and V. 0 = d + s

Taking a differential yields

−1/ 2 n2 2 kTN q  q  d ni 2 q( −V )/ kT si a q( −V )/ kT − dV ). s i + e + e (d    s 2 s 2Cox Na Na2  kT    kT s

s

It is straightforward to re-group the terms and show that d (n 2 / N 2 )eq s / kT s i a dV = 1 + (n 2 / N 2 )eq s / kT + (C 2 /  qN )( Q / C ) i

a

ox

si

a

s

ox

in the limit of V → 0, where 1/ 2 q n2  q / kT s i Qs = 2skTN + 2e  . a   kT N a  s

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

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

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 E(x) =  Q + q n(x)dx,

 si 

d





x

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

q

xi

xi

0

si 0

x

 n(x)E (x)dx = Q  n(x)dx + xi

d

si

0

Therefore,

 n(x) n(x)dxdx .

If we let n(x) = df(x)/dx, the second term on the RHS becomes q

xi

df

 si 0 dx

 f (xi ) − f (x)dx =

q 

xi

df

1

xi

dx − 0 f (x ) 2  si  i 0 dx

d( f 2 )



dx

dx , 

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





Substituting it back, one has n(x)E (x)dx Q Q E  = + ,  2si n(x)dx  xi

0

xi

eff

d

i

si

0

where xi

Qi = q  n(x)dx .

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

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

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 N 2 C 2 (V − V − )2 q  fb s V = − s ln a2  ox gs −  s 2 sikTNa kT  q n  i    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,

N 2 C 2 (V − V − )2 q  kT C 2 (V − V − )2 q  fb ss fb ss ss ss  ss = ln a2  ox gs − ln ox gs −  = 2 B +  q kT q kT 2  kTN 2  kTN n  si si    i  a 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 2 (Vgs − Vfb − sd ) 2 2 sikTNa

−

q sd

=0

kT

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

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

The solution is

=

 sd

1 − 2 

 2 siqNa 2 siqNa + 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

Na

ln

q

= 0.435 V

ni

2 si (2 B ) = 0.075 m qNa

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

Q = qN W d

Qi = Qs − Qd = 7.6 10

a

−7

= 2.4 10−7 C/cm2

dm 2

C/cm

The effective field is

 Q + Q / 2 /  = 6.0 105 V/cm i  d  si

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

Ids y = eff W 0 [−Qi (V )]dV m   =  C W (V −V )V − V 2 , eff ox  gs t 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 −V t m

2

 Vgs −Vt  y  V −Vt  y − − 2  gs V + V2.    ds L ds L m   m 

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

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

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 ) +

Charge Sheet model,

2 siqNa s

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

 s,max = Vgs −V fb +

 s, max = Vgs − V fb +

V dsat

= V −V − 2 + gs

fb

B

si qN a

 si qN a

−

2 ox

C

Cox

1+

siqNa siqNa − 2 2 2Cox 2Cox

 qN si

Cox

−

a 2

 qN si

Cox

2

2(Vgs −V fb )Cox2

2

 qN si

1+

a

2(Vgs − V fb )C ox

2

siqNa

.

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 =

qn 2  eq / kT i

Na



 s B

d .

E ( )

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 =

kTn 2 i

EsNa

eq / kT ,

where Es is the surface electric field.

s

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

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

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

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 ( y) =  WQ ( y)

dV

ds

dy

eff

i

.

Current continuity requires that Ids be a constant, independent of y. nonuniformly-doped (laterally) MOSFET,  W Vds V L dy = eff dV =  W ds .

 Q ( y) 0

i



Ids

0

eff

For a short-channel or a

Ids

Generalizing the result of the last exercise to laterally varying s and Es, one obtains L L dy Vds 1 Na E( y)e−q ( y )/ kT dy . = =   2 I  W 0 Q ( y)  WkTn 0 s s

ds

eff

i

eff

i

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 siqNa s

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

 s,max = Vgs −V fb +

So,



= V −V + s, max gs fb

siqNa siqNa − 2 2 2Cox 2Cox

 qN si

Cox

a 2

−

 qN si

Cox

a 2

1+

2(Vgs − V fb )C ox

2

siqNa

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

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

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 1  tan (ktox ) +  tan (kWdm ) ox

si

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) εox = εsi,

tan

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

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

The largest is λ = Wdm + tox. (c) If tox ≪ Wdm, So

1



the first angle  0, the second angle   1 1 tox 1  Wdm  tan (t /  ) + tan (Wdm /  )  −  − = 0 ox

ox

which gives





si



ox



si







λ ≈ Wdm + (εsi/εox)tox.

6.2.

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

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

x 10

-9

1 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) .

exp(− L 21) ~ 40.14 , when L ~ 21 . exp(− L 22 )

By substituting the numbers,

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.]

2 Ids = eff kTW nNi a

Eq. 6.38:

L

1 − e− qVds /kT dy

0

Wdm





0

.

eq 2 ( x, y )/kT dx

The integral over x can be approximated as



Wdm q x, y /kT ) 2(

e

0

s

d 2

0

E

dx =  eq 2 ( x, y )/kT



1 kT

eq s /kT

Es q

For long channel, there is no y-dependence. So the y-integral is just multiplication of L. 2

 x   2 ( x) =  s 1 −   Wd 

With Eq. 6.10 for long channel,

2 s

= 2qNa s si Wd 2 2  kT   n  siqNa    i  eqs / kT 1− e−qVds / kT . ( ) 2 s  q   Na  E= s

Therefore,

I = W ds eff L

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

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

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

I = W (−Q ) ds

i

eff dV / dy .

1 + ( eff / vsat )dV / dy

Solve for dV/dy: Ids / eff dV = , dy −WQi − Ids / vsat

With -Qi given by Eq. 6.37,  q ( x, y)   qV  Wdm 2 exp exp −    dx, 0 kT kT Na     n2

−Q ( y) = q i

i

We have dV dy = qW (n2 / N )  i

6.5.

a

Wd

0



Ids / eff eq ( x, y )/kT dx e− qV /kT − I / v . ds sat  2

 C (W / L) (V − V )V − (m / 2)V 2 (a)

I ds =

eff

ox

g

t

ds

1 + ( V / v L) eff

ds

ds



(m = 1)

sat

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)

0

1

2 3 Drain voltage (V)

Idsat = CoxWvsat (Vg − Vt )

4

5

1 + 2eff (Vg − Vt ) / (mvsat L) − 1 1 + 2eff (Vg − Vt ) / (mvsat L) + 1

(m = 1)

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

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

Saturation current (mA)

15 Vg -Vt=3.0 V no velocity saturation

10

5 velocity saturation

0

6.6.

Make use of z 

0

2eff (Vgs −Vt )

1

2 3 Channel length (um)

= C Wv dsat

inv

5

, Eq. (6.49) becomes

mvsat L I

4

mvsat L 1 + z −1 mv L z = C invWv sat sat sat 2eff 2eff 1+ z +1

( 1+ z ) . −1

2

Differentiating the above with respect to z, one obtains 2eff = C Wv 2 dI dI g  dsat = dsat msat ox sat dV dz mv L gs

(

1+ z

sat

−1

)

2

1

= C Wv

1+ z

ox

sat

 1 

−

1 1 + 2 (V −V ) / mv L eff

g

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

When vsat → , the square root factor can be expanded in a Taylor series to yield   eff (Vg − Vt )  W Vg − Vt g = C Wv − − = C , msat ox sat 1 1  eff ox L m mvsat L    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

1 + 2eff (Vg − Vt ) / (mvsat L) − 1 1 + 2eff (Vg − Vt ) / (mvsat L) + 1

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

t

sat

  

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

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)  0 Qi ( y)Wdy .

L

Change variable from y to V: Q (total) = Vds Qi (V )WdV .



i

dV / dy

0

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

Q (total) = −

 v W V 0 ds

eff

 

i

Q (V ) +

 W2 effI ds

sat

 Q (V )dV .  i

2

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

C W ox

eff

2mv sat

(V − V ) − (V − V − mV ) − 2

g

2

t

g

t

ds

C 2W 2 ox

3mIds

eff

(V − V ) − (V − V − mV ) .

3

g

t

3

g

t

ds

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 1 + 2eff (Vg − Vt ) / (mvsat L) + (1 / 3)

Qi (total) = −WLCox (Vg − Vt )

6.9.

(a)

1 + 2eff (Vg − Vt ) / (mvsat L) + 1

V = V + 2 + t

fb

B

qNa Cox

.

(W − x ) dm

s

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

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

4 si B 2 + xs qN a Eliminate Na from the above eqs, 4 si B x = −W s dm Cox (Vt −Vfb − 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: Surface potential:

Es = Qd/si = 61800 V/cm. q Wd qN 1  = xN dx = a W 2 − x2 = E(W + x ) = 0.37 V.

(



s

)

x

si

a s

d

2si

s

s

d

s

2

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

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

Solutions to Chapter 7 Exercises

7.1. (a) The expression for the2 drain current is given by 2 s I =  W 4 si  2kT    tan  −  +  siti  2 tan2   .

(1)  ds  L t si  q   2 itsi  d The boundary conditions at the  source and drain are:  2 2sikT = ln  − ln (cos  q(Vgs − mi ) (2) ) + 2siti  tan  − ln   s s 2 s q ni  2kT  i tsi s  tsi   2 2sikT = ln  − ln (cos  q(Vgs − mi −Vds ) ) + 2siti  tan  . (3) − ln   d d 2 d q ni  2kT  it si 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

V =0.0(V) V =0.5(V) 0.8

V =1.0(V) V =2.0(V)

ds

I (mA)

V =1.5(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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

(b)

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

-9

si

f= tan(t /))- cot(t /2)

1.0x10

-10

ox

ox

5.0x10

=19.80(nm)

si

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 (a)

 (x) = V − 2kT ln t si q 2

 q2n i cos 2 x  , 2sikT  tsi 

Center potential at the source:

 (0) = −

t si ln  2 q  

2kT

 q2n i = 2kT  2 ln   2sikT  q  t si

2sikT   q2n i 

 cannot exceed /2, hence the maximum center potential is  2kT   ln  (0) = 2sikT   max q2n  q  tsi (b) From the boundary condition,

 1  2si   tox  1 −   2 2 kT = ln(1 −  ) − ln  + q(Vgs − mi −V ) ln 1 + . si − ln      R  2kT R q2n 2   ox   i  

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

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

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

The potential solution is (V = 0)  ( ) = At center, with  << 1,

kT   ln  2 2 8sikT (1 −  ) q q n R [1 − (1 −  )( / R)2 ]2 ,  i 

 ( = 0) =

kT  8sikT  ln  2 2 , q qnR  i 

7.3 (a)

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

 (x) = V +

2kT q

 2 ln   t si

At the source, V = 0. Surface potential:

Condition for :

q(Vgs − mi − V ) 2kT At the source,

 2 t  2 2sikT = ln  − lncos   + si ox  tan  − ln   q2ni  oxtsi  tsi q(Vgs − mi )  2a  − ln = ln  − lncos   + 2r tan    2kT  tsi    /2 →

Vg >> threshold:

Then

 2kT − lncos(2x / t )  si q q 2n i  2kT  2  2kT ln a − lncos( )  (t / 2) =   si q q  tsi 

q(Vgs − mi )

 tan  term dominates.

− ln

 2a 

 2r tan  

r

  2kT cos   tsi   2a  2kT 2kT    2kT  q(Vgs − mi )  ln(r ) ln  ln  a + (tsi / 2)  − ln   − t 2kT t q q q  si   si  

Simplification (optional):

 2a  2kT ln  a  q(V −  ) gs mi  (tsi / 2)  q rt  2kT − ln  t    si   si 

(b) From the boundary condition,

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

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

q(Vgs − mi −V )

+ ln

R

 1−   t  q2n i = ln − ln  + 2 si ox ln 1−  1 +  2sikT   ox  R

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

The potential solution is (V = 0)



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

At surface,

 4a2  2kT  1  ( = R) = − ln    = q ln  R2  + q ln    q  2a      Substituting the  solution from the above boundary condition:  2si  tox  2kT   q(Vgs − mi )  R   2a   ( = R) = ln + ln + ln − ln ln 1 +   2a   R    2kT  R  q         ox  R

2kT





kT

7.4. I  I  g m = ds V s + ds Vd I =

W

ds

s



Vds

gs

[−Q (V )]dV = 

W

i

0

L

d

d



L



gs

[−Q ( )]

dV

d ,

i s

d 2sitox

 tan   f ( ).  2 2 kT = ln  − ln (cos  ) + si − ln   2 qn  2kT oxtsi  tsi Ids W dV W  2kT  =  Q ( ) =  Q ( ) − f ( s ) ,  L i s d L i s  q s    Ids W  W = − Q ( dV =  Q (  2kT f ( ) ) i d )  d ,  L i d d L q d      q(Vgs − mi )  2 2 kT = f ( ), q(Vgs − mi − Vds )  2 2 kT = f ( ). si si − ln  − ln    s d 2 2 q n 2kT t 2kT t q n   si   si  1 W W  2kT  1 W =  −Q ( ) + Q ( ) g =  Q ( ) − f ( s )  +  Q (  2kT f ( d )  m i s  ( i d ) ( (2kT / q) f  ) (2kT / q) f  ) i s i d L q L q L     s d q(Vgs − mi − V )

s

d

Since

d

−Q ( ) = 2 i

si

= 2 2kT 2 tan  , si dx x=t /2 q tsi si

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

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

g = 8

W kT si

m

( tan  −  tan  )

L q t si

s

s

d

d

Likewise,  1 W I  I  I  W  2kT f ( d )  = − Q ( ) g d = ds V s + ds Vd = ds Vd =  L Q i( d )  ( (−2kT / q) f  ) q L i d   s ds d ds d ds d

g = 8 d

W kT si L q tsi

 d tan d

7.5 At the source, −Q = 2 2kT 2s tan s , i si q tsi where

q(Vgs − mi ) 2kT



2 t 2sikT = ln  − ln (cos  ) + si ox  tan  .  s s s q2ni   oxtsi s

2 − ln   tsi

 s << 1,

Subthreshold,

ln  s dominates. q(V gs −mi )/2kT

s  e

tsi q2n i 2 2sikT

Therefore, kT  s2

−Q  8 i

si

q tsi

= qn t e

q(V − )/kT gs

mi

i si

 s tan  s dominates.  2kT  2 −Q = 2 2kT 2s tan s  2 ox ln Vgs − mi −   i si q q tsi tox   tsi 

Far above threshold,

 s  /2,

7.6

Let

tan

 =  tox/,

then

 2sikT  q2n i 

 tox   tsi  ox tan =     2       si

 tsi/(2) =  (4tox)/(2) = 2.

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

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

So

tan tan 2 = tan

2 tan  = ox 2 1 − tan   si

1

tan =

Solve

1 + 2si / ox

Then

= tan

Check:

tox

=

If ox/si = 1,

If ox/si = ,

=

−1

tox



 1    1 + 2si / ox  =

tox = 6tox = 15 nm.  /6

tox t = ox = 4tox = 10 nm. −1 tan (1)  / 4

7.7 (a)

q(V −  ) gs

2kT

mi

 − ln  2 t

 2 t 2sikT  = ln  − lncos   + si ox  tan  q2n  oxtsi

mi = m – ( + Eg/2q) = −0.16 V.

The LHS = −2.41 → in subthreshold.

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

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

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sitox  tan  .

2sikT = ln  − ln (cos  ) +  q2n i  oxtsi

V   + 2kT ln  2  t mi q  tsi

So,

V −V −V = gs

t

2kT 

 − lncos  +

ln q  Q = i

Cinv



2(Vgs −Vt −V )

8kT  si

Qi

 tan 

  

 tan 

q t si

ln  − lncos  + =

2sitox

oxtsi

Therefore, 1

 2sikT  q2n 

2 si tsi

2 sitox  tan  oxtsi

 tan 

t = ox + ln  − lncos   2 si ox  tan  tsi

and

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

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

Csi =

 tan  2 si t si ln  − lncos  

7.9 Integration of current2 continuity eq. from 0 to L yields W 4 si  2kT    tan  −  2 +  si tox  2 tan2    .  I ds =  2  t L t  q     si  ox si  d s

Similar integration of2 current continuity eq. from 0 to y yields W 4 si  2kT    tan  −  2 +  sitox  2 tan2    .  I ds =  2  t y t  q     si  ox si  s

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

Ratio of the two eqs. gives L  2− 2 d = s y  s2 −  2 Also, for  << 1, 2 q(Vgs − mi −V ) − ln  2kT  tsi



2sitox = ln  − ln cos  +  tan   ln  . ( ) 2sikT  q2ni  oxtsi

where V=0 for s and V=Vds for d.

Substituting , s, and d into the ratio-ed eq., L  2 −  2 1− e−qVds / kT d . = s = y  s2 −  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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

Vg1

t + 3tox1 + 3tox 2 = si tsi + 3tox 2  s1

Therefore,

t + 3tox1 + 3tox 2 SS = si  60 mV/dec = 75 mV/dec tsi + 3tox 2

For Vt, solve Vgs1 with s1 = Eg/2q. Or use Eq. 7.13: V = t

+m

mi1

Eg

− (m − 1)(V

− gs 2

2q

)

mi 2

with 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 t = 5 nm, t si

= 2 nm, 

=t ox1

ox2

 = s1

1 m

= −E /2q,  mi1

(V gs1

g

= 0,

so

mi2

t + 3tox1 + 3tox 2 17 m = si = . tsi + 3tox 2 11

1 11 6  − ) + 1− (V −  ) = (V + 0.56) + V mi1   gs 2 mi 2 gs1 m 17 17 gs 2  

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,

−Q =  Vgs − mi1 − (0) +  Vgs − mi 2 − (tsi ) . i ox ox tox 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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

 (tsi ) Vgs

=

5 tsi = . tsi + 3tox 11

Therefore, −Qi / Vgs = Cox + Cox 1−  (t si ) / Vgs  = (17 /11)C ox = 2.67 F/cm2.

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

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

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 W /   Vg Vt  Vds Ids − = ; ox   L/      

I →  (C ) ds

eff

and Eq. (5.31) becomes 2 I →  (C ) W /  1  Vg − Vt  = Ids . ds eff ox   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

I →  (C ) W /  (m − 1) kT  eq(Vg −Vt )/ mkT . ds eff ox  q  L/   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.,  Vg Vt  Idsat W I → (C ) v −  . dsat ox  sat       

8.4.

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

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

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

=

 n +  p CVdd  1 1  2 = 4  W I + W I . p psat   n nsat

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.,

  (Wn + Wp ) 1 + 1  = 1 + 1 + Wp / Wn + Wn / Wp . W I WI  I I I I nsat psat nsat psat p psat   n nsat 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

V(t)

I(t)

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 dt = 0

2

V 2 dd

R



 e

−2t / RC

0

1

2

dt = CVdd , 2

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

C

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

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 dt

= 0.

With the initial condition V(t = 0) = Vdd, the solution for V(t) is V (t) = V dde−t/ RC .

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

E =  RI 2dt = 0

8.6.

V 2



1 2 −2t / RCdt = CV e . dd  0 R 2

dd

From Eq. (5.27) and the inversion charge expression above Eq. (5.70), the transit time is WLC (Vg − Vt − mVds / 2) L2  tr =  C (W ox/ L)(V = − V − mV / 2)V  V eff

ox

g

t

ds

ds

eff

ds

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 (2 / 3)WLCox (Vg − Vt ) 4mL2  tr =  C (W / L)(V = 2 − V ) / 2m 3 (V − V ) eff

ox

g

t

eff

g

t

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 Q L  tr  I i = v dsat

sat

1 + 2eff (Vg − Vt ) / (mvsat L) + 1 / 3 1 + 2 (V − V ) / (mv L) − 1 eff

for a short-channel MOSFET biased in saturation. device becomes fully velocity saturated as L → 0.

8.8.

g

t

sat

The limiting value of tr is L/vsat when the

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

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

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

dI = ds

  Ids   V  g V  ds

dV + g

  Ids   V   ds Vg

dV . ds

From Eqs. (8.13) and (8.14), dVds = dVds + ( Rs + Rd )dIds,

and dVg = dVg + RsdIds .

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

( Ids /  Vg )V ds 1 − R ( I /  V ) − ( R + R )( I /  V ) s

ds

g Vds

s

d

ds

dV  + (a similar term in dV  ). g

Now Ids is expressed as a function of the internal voltages: Ids(Vg,Vds). first coefficient as the intrinsic transconductance, i.e., gm 

ds

ds Vg

One can identify the

  Ids  = gm V 1− g R − g (R + R ) , g V  m s ds s d  ds

where  I  g m   ds    Vg V ds is the extrinsic transconductance, and I  gds   ds  V  ds Vg 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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

I Metal contact

I x=0

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 sd dx2 = RGf =  f , 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   dI V (x) = − c = csd I cosh  sd (l − . x) c  0 W dx W   c The contact resistance is then

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

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

R =

 csd cothlc W 

=

I(x = 0)

co

8.11.

V (x = 0)

sd .  c 

From Eqs. (8.49) and (8.51), bmin <  if Cout + CL  2Cout + 2

.

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



=

Rsw

b,2

k1

(k C 1

out

+ k k C ) = R (C 1 2

in

sw

+ k C ); out

2

in

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

 b,n = Rsw (Cout + knCin ).

The delay of the last stage is



= b, n +1

Rsw k1k2    kn

(k k    k C 1 2

n

out

+C )= R L

sw



C

 out 

+

 .  k1k2    kn  CL

The total buffer delay is the sum of all the stage delays, i.e.,   CL  (n) = +1)C + (k + k +      + k )C + .

Rsw (n b  out 1 2 n in k1k2      kn  

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

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

[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

 (n) = R b

C   (n + 1)C + nkC + L . sw  out in k n  

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



(k ) = R (C b min

sw

+ kC ) out

in

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=

ln(CL / Cin ) −1. ln(k)

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

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

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.

So

Power = CVdd 2f

where f = 1/(2)

and C = Cin+Cout+CL

Power = Vdd2/2R sw

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

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

Solutions to Chapter 9 Exercises 9.1

Consider hole current flow in an n-region. d p .

J p = −qpn p dx

Eq. (2.78) gives (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 Eq. (9.15) gives

(

)

d  p − n . dx

( −  ) = kT ln p n 

(3)

n n

p

n

q

(2)

 2   nie 

for the n-region.

Substituting (3) into (2), we obtain  kT  1 dn 1 dn2   dp J p (n − region) = qpn p − n ie n − − qD ,   p 2 dx n dx dx  q n  n ie  

(4)

where we have used the Einstein relation Dp = kTp/q.

we have

9.2

If we compare (4) with Eq. (2.69), i.e., dpn J (n − region) = qp  E(n − region) − qD , (5) p n p p dx  kT  1 dn 1 dn2   E(n − region) = − n − 2 ie   .   q  nn dx nie dx  

Eq. (2.78) gives, for an n-region, d p .

J p = −qpn  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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

( −  ) = p

n

kT q

 pnnn 

ln

 2   nie 

.

(4)

Substituting (4) into (3), and using Dp = kTp/q, we have nie2 d  nn pn  J p (n − region)  −qDp  . nn dx  nie2 

9.3

(a)

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p n pn − = 0. dx2 L2p

(1)

Applying (1) to hole distribution inside the polysilicon layer, we have d 2p p n n − = 0.

2 2 dx' LpE1

(2)

The general equation for (2) has the form pn (x') = Aex'/LpE1 + Be−x'/LpE1 .

(3)

At the ohmic contact at x’ = WE1, we have 0 = AeWE1 pE1 + Be−WE1 pE1 . /L

/L

(4)

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

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

At the polysilicon-silicon interface (x’ = 0), we have pn (−WE ) = A + B .

(4) and (5) give

A=

and

B=

(5)

pn (−WE ) 1 − e2WE1 / LpE1 /L −p (−W )e2WE 1 pE1 n

E

1− e

2WE 1 / LpE 1

(6) .

(7)

Substituting (6) and (7) into (3), we have p (x' ) = p (−W ) n

(b)

n



sinh (WE1 − x' ) / LpE1

E

.

(8)

sinh(WE1 / LpE1 )

The hole current density entering the polysilicon layer is given by Eq. (9.59), i.e., J p (x = −WE ) = −qpn (−WE )Sp .

(9)

Using (8), this current density is also J (x = −W ) = −qD p

E

pE1

dpn (x) dx x =−WE

= qDpE1

dpn (x' ) dx' x '= 0

= −qDpE1 Sp =

Equations (9) and (10) give

9.4

and

(10)

pn (−WE ) LpE1 tanh(WE1 / LpE1 )

DpE1 LpE1 tanh(WE1 / LpE1 )

.

.

Neglecting parasitic resistances, Eqs. (9.182) and (9.183) give IE = − IEBOexp(qVBE / kT ) − 1 −  R IC

(1)

IC = − ICBOexp(qVBC / kT ) − 1 −  F IE ,

(2)

which can be rearranged to give kT V = BE

 IE +  R IC  − ln1  q  IEBO 

(3)

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

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

and

V = BC

 IC +  F IE  − . ln1  q  ICBO 

kT

VCE = VC − VE = (VC − VB) + (VB − VE ) = VBE − VBC .

Now,

(4) (5)

Substituting (3) and (4) into (5) and rearranging, we obtain kT  I ( I − I −  I )  VCE  = q ln  ICBO ( IEBO − IE −  R IC ) .  EBO CBO C F E 

 R IR0 =  F IF 0 ,

Now, Eq. (9.144) gives

(6) (7)

and Eqs. (9.180) and (9.181) give IEBO  IF 0 (1 −  R  F ) ,

(8)

and

ICBO  IR0 (1 −  RF ) .

(9)

Therefore,

ICBO IR0 =  F . = IEBO IF 0  R

(10)

Substituting (10) into (6) gives

9.5

kT  F (IEBO − IE −  R IC )  VCE  = q ln  (I − I −  I )  .

C F E   R CBO

From Exercise (9.4), we have kT  F (IEBO − IE −  R IC )  VCE  = q ln  (I − I −  I )  .

C F E   R CBO

(1)

Now, when emitter and collector resistances are included, we have VCE = VC − VE = VC + rc IC − (VE + re IE ) = VCE + rc IC + re ( IB + IC ) ,

(2)

where we have used the fact that IB + IC + IE = 0 .

(3)

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

kT  F (IEBO − IE −  R IC )  + r (I + I ) . ln  (I − I −  I ) + rcI C e B C q C F E   R CBO

(4)

If we neglect the saturation currents IEBO and ICBO, then using (3) we can rearrange (4) to give  kT  V  ln   F ( I B + IC −  R IC )  + r I + r ( I + I ) (5)  ( ( I + I ) − I ) CE C C e B C q C   R F B C

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

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

 kT  + r (I IB + IC (1 −  R ) = q ln   I − I (1 −  ) /   e B + I C ) + rc I C .

F F   R B C For open-circuit collector, IC = 0, and (5) reduces to kT  1  ln  + re IB .

VCE ( IC = 0) = q R 

9.6

From Eq. (3.65), we have n −n p

exp(qV

=n p0

p0

/ kT ) − 1

BE

sinh(WB − x) / LnB  . sinh(W / L ) B

(1)

nB

If we assume diffusion current only, then cosh(WB − x) / LnB   −1  dnp Jn (x) = qDnB = np0exp(qVBE / kT ) − 1  . dx sinh(WB / LnB )  LnB 

(2)

Therefore, the base transport factor is

T 

−1  = cosh WB  .

Jn (x = 0)  LnB 

Jn (x = WB )

(3)

For a base region of WB = 100 nm and NB = 1018 cm−3, we have, from Fig. 3.14, LnB  15 m, and −1 0.1  T =  cosh  = 0.9998 .  15 

9.7

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. and (2) give 1 M (BVCEO ) − 1 =  .

0

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

(1)

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

9.8

The equation

d 2np (x) 2

dx

−

np (x) 2

LnB

= 0,

(1)

where LnB 

nBDnB 1+ j nB

=

LnB

(2)

1+ j nB

has a general solution np (x) = Aexp(x LnB ) + B exp(−x LnB ),

(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

 kT ) 1+ q v e jt . p p0 BE p0 BE   kT be Therefore, the boundary condition for Δnp(x) at x = 0 is n (0, t) = n

exp(qv

(t) kT )  n

np (0) = np0 exp(qVBE / kT )

qvbe

exp(qV

.

(4)

(5)

kT

At the ohmic base contact, there are no excess electrons. Therefore, the boundary condition for Δnp(x) at x = WB is np (WB ) = 0.

(6)

Applying these boundary conditions to Eqs. (1) and (3), we obtain np0qvbe  qV  sinh(WB − x) LnB  . n p (x) = kT exp  BE  sinh(W L ) kT   B nB

9.9

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

transistor is U (x) =

np (x) − np0 .  nB

(1)

From Eq. (3.65), the electron distribution in the p-type base is np (x) − np0 = np0[e qVBE /kT −1]

sinh(WB − x) / LnB  . sinh(WB / LnB )

(2)

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

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

For a typical transistor in operation, WB/LnB

<<

1, and qVBE/kT  x  n (x) − n = n eqVBE /kT 1− .  W  p p0 p0  B 

>> 1, and (2) reduces to

(3) The base current component due to recombination in the base region is JB,rec = 

WB

qU (x)dx.

(4)

0

Substituting (3) into (4), and using the fact that p p0 n p0 = n2ieB for the base region [see Eq. (9.14)], and that pp0 = NB, we have JB,rec =

qW n2

B ieB

2 nB NB

exp(qVBE / kT ).

(5)

Eq. (9.71) gives the base current due to hole injection from base into the emitter as JB,inj = JB,rec

Therefore,

JB,inj

qD n

exp qV

(

kT )

.

(6)

NE LpE tanh WE =

WBLpE tanh(WE / LpE ) n2

N

2nBDpE

NB

ieB 2 nieE

E .

(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 n2 / n2 ieB

= exp(−61/ 26) = 0.096. We have

ieE

JB,rec JB,inj

=

WBLpE tanh(WE / LpE ) n2

N

2nBDpE

NB

ieB 2 nieE

E

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

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

(a) Ignoring heavy-doping effect, the diffusion capacitance ratio is CDn 2 NE WB = .

CDp 3 NB LpE

(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 2 q qA CDn = IntB = J n tB , 3 kT 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 qDnBnp0 J = exp(qV / kT ) n BE WB qDnB p p0 n p0 = exp(qV / kT ) BE WB pp0 qD n2 nB ieB = exp(qVBE / kT ).

WB NB

(3)

From Eq. (9.106), we have tB =

WB2 .

2DnB

(4)

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

2 2 qA qnieB WB

3 kT

exp(qVBE / kT ) .

(5)

2NB

Similarly, for a n-type emitter region with WE << LpE, we have 1 q C = I  = qA J  p pE p pE Dp 2 kT kT

(6)

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

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

= =

2 1 qA qD p nieE pE

2 kT NE LpE 2 1 qA qnieE L pE 2 kT

exp(qV

BE

exp(qV

NE

BE

/ kT )

/ kT )

where we have used the fact that L2 = D. (5) and (6) give 2  C Dn 2  nieB N W  E  B . =  2  C 3 n N L Dp  ieE  B  pE n2 = n2 exp(E

Now,

ieB

i

/ kT ) gB

n2 = n2 exp(E

and

ieE

n2 Therefore,

n

ieB 2 ieE

(7)

i

/ kT ) . gE

 E − EgE  = exp gB .  kT 

(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 n2

ieB 2 ieE

n

 E − EgE  = exp gB  = 0.031.  kT 

(9)

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

9.11 IC (VBE )low−injection =

IC (VBE )all −injection =

AEqD n2 exp(qV nB ieB

NBWB

Therefore,

Ratio 

IC (VBE )low−injection IC (VBE )all −injection

Eq. (9.14) gives

2 AEqDnB nieB exp(qV BE / kT ) .

NBWB

 −1 / kT )  1  2 4n exp(qV BE / kT ) −1 . 1+  BE 1+ ieB   N B2  4     1 = 1+   4  

4n

2

exp(qV N2

 / kT ) −1 .  

(1)

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

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

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

   2 Ratio = 1+ 1  1+ 4ni exp{[qV BE + E gB (N B )] / kT} −1 .  4  N B2   

(2)

For Eg, we use the unified model given in Eq. (9.18), namely 2   N       N  + ln Eg ( N ) = 6.92 ln    17  + 0.5 17 meV.     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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

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  i  E + 2   DpE1  n ieE   DpE 

(

)

(1)

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 n2i = exp(−EgE / kT ) .

2 nieE

(2)

From Fig. 9.4, we have, for NE = 1020 cm−3, EgE = 92 meV.

Using kT = 26 meV, we have

n2i = exp(−EgE / kT ) = 0.029 .

2 nieE From, Fig. 3.14, we have pE = 130 cm2/V-s.

DpE = Therefore,

kT pE = 3.38 q

(3)

Therefore, cm2/s.

(4)

 n2  W 13 4 GE (300 nm, 0) = N E  i  E = 2.6  10 cm -s.

2  nieE  D pE

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

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

With pE1/pE = 1/3, we

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

kT pE1 q

= 1.13 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 R = q WB p ( x) ( x)dx−1 = = B , 1  Sbi p qN  W W  0 p B

p

B

(1)

B

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

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

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

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

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

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

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 = qNBWB .

CdBC CdBC

(1)

To avoid significant base widening, collector current density is kept at JC = 0.3qvsat NC ,

or

NC =

JC , 0.3qv sat

(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

qNC si = 2( bi + VCB )

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

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

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

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

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

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

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 NB (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.

(4)

WB

From Eq. (10.11), the electric field due to the dopant distribution is kT  kT 1 dN .

E(dopant ) = B = − 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 Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

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

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

NB (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.

(4)

WB

From Eq. (10.11), the electric field due to the dopant distribution is kT  kT 1 dN , E(dopant ) = B = − q N B (x) dx q N B (x)

(5)

and the field due to heavy-doping effect is

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

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

E( HD) = −

1 dEgB q

.

(6)

dx

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

(

Eg (Na ) = 9  F +

) meV

F2 + 0.5

F = ln(N B / 1017 ) .

where Therefore, we have

E( HD) = 9 

 + 10 1  −3

F

 

(7) (8) V/cm.

(9)

 F 2 + 0.5  N B

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

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

Electric field (V/cm)

20,000 10,000 0 -10,000 -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

WE

= 0.24 AE WE1 r = = 4.8 polysilicon polysilicon 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 JC 0 = (a)

qD n2 nB ieB

NBWB

.

Neglecting heavy doping effect, we have nieB = ni = 1.01010 cm−3.

(2) Using the

mobility values in Fig. 3.14, we have

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

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

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, n2 = n2 exp(E / kT ) = n2 exp(43 / 26) = 5.23n2 .

(3) ieB

i

gB

i

i

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

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

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

10.9

n+

n+ emitter poly

Metal or metal silicide

xjE

p

Ge Eg0 = Egmax x

0

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.

−1 W − x   W − W  ( ) =   Wcap − x jE  +  WB0 − x cap exp − Eg 0 / kT   jE jE  jE   B0 x jE Wcap  B0 −1 ) W − x   (W − x ) − (W − x jE )  ( =   cap jE   B0 jE exp − E / kT  cap  +  g0  WB0 − x jE  WB0 − x jE       −1  W  − x jE   Wcap − x jE   + 1 −  exp(− E / kT ) . =   cap g0  WB0 − x jE   WB0 − x jE  

J (SiGe, x ) C0 jE J (Si, x ) C0

(1)

200 100

dE/kT=0.5 dE/kT=1 dE/kT=2.5 dE/kT=5

Current Ratio

50 20 10 5 2 1 0

0.2

0.4

0.6

0.8

1

(Wcap − xjE )/(WB0 − xjE)

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

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

 Wcap − x jE   Eg0   WB0 −Wcap  exp VA (SiGe, x jE ) =    W −x  kT  +  W − x  VA (Si, x jE ) jE     B0 jE   B0 x jE Wcap  Wcap − x jE   Eg0   (WB0 − x jE ) − (Wcap − x jE )  exp =  +     WB 0 − x jE   WB0 − x jE   kT   Wcap − x jE  Eg0   Wcap − x jE  exp   + 1− =    WB0 − x jE   WB 0 − x jE   kT  

(2)

200

Early Voltage Ratio

100 50 20 10 5 2 1 0

0.2

0.4

0.6

0.8

1

(Wcap − xjE )/(WB0 − xjE)

10.10

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

n+

n+ emitter poly

Ge Egmax = Eg0 p WB

0

x

WB1 EgB,SiGe (x) =

x Eg 0 WB1 = Eg 0

x < WB1

(1)

WB > x > WB1.

From Eq. (10.27), we have

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

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

JC 0 (SiGe)

WB

 W

 exp[−E

JC 0 (Si)

B

gB,SiGe

0

.

(2)

(x) / kT ]dx

Using (1), we have

 − EgB,SiGe (x)  WB1  − xEg0  WB  − Eg 0  dx dx dx exp = + exp exp 0 0 WB1 W kT kT kT  B1      kTWB1 = 1− exp(− E / kT ) + (W −WB1 )exp(− E / kT ). g0 B g0 Eg 0 WB



(3)



Substituting (3) into (2), we have J (SiGe)  kT W  W   −1 C0 B1 B1  exp(− Eg 0 / kT ) . (4) =   1 − exp(− Eg 0 / kT ) + 1 − J C 0 (Si) W   WB  Eg0 B





Similarly, the VA ratio can be obtained from Eq. (10.32), namely WB VA (SiGe) exp(Eg max kT ) exp[−E (x) / kT ]dx . = gB,SiGe 0 VA (Si) WB Substituting (3) into (5), we have

(5)



 / kT ) + (W −W )exp(− E / kT ) (6) VA (SiGe) exp(Eg 0 / kT ) kTWB1 = 1− exp ( − E  g0  g0 B B1  E g 0 VA (Si) WB  kT WB1  WB1  = . exp(Eg 0 / kT )−1 + 1− E W W  B  g0 B







The tB ratio can be obtained from Eq. (10.36), namely WB WB tB (SiGe) 2 exp[E (x) / kT ] exp[−E (x') / kT ]dx'dx . = gB, SiGe gB, SiGe x  WB2 0 tB (Si)

(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



0

exp[E gB,SiGe (x) / kT ] WB1

= 0

WB

 exp[−E x

gB,SiGe

(x') / kT ]dx'dx

(8)

 EgB,SiGe (x)  WB  − EgB,SiGe (x')  exp dx'dx  exp x kT kT     WB  EgB,SiGe (x)  WB  − EgB,SiGe (x')  +  exp dx'dx .  exp WB1 kT  x kT  

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

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

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

 EgB, SiGe (x)  WB  − EgB,SiGe (x')  exp dx'dx (9) 0 exp kT x kT   WB1  EgB,SiGe (x)  WB1  − EgB,SiGe (x')   − EgB, SiGe (x')  W exp exp exp dx'dx  + W = 0  x B1 kT kT kT         EgB, SiGe (x)  WB1kT WB1 WB1kT   − Eg0x    − Eg 0    = 0 exp E exp + (W − W ) − exp dx  B  kTW B1 kT E kT g 0   B1    g 0    2 W kT  W kT W kT   − Eg0   . = B1 + (WB − WB1 ) − B1  B1 1 − exp Eg 0  Eg 0  Eg0  kT   WB1

B

For the second integral on RHS of (8), we have WB  EgB, SiGe (x)  WB  − EgB,SiGe (x')  exp exp dx'dx WB1  kT x kT   WB  EgB, SiGe (x)   − Eg0  (WB − x)dx = W exp exp kT B1    kT  1 2 = (W − W ) .

B1 2 B

(10)

Therefore, (7) gives  2 WB12kT  2  WB1kT W B1 kT  t (SiGe)  − Eg 0  1 ( W − W ) B + (W − W ) − 1 − exp  +   B  b1   kT  2 B B1 E  t (Si) =  W 2  E   E   g 0  g 0    g0 B B    2 2 1  WB1  2kT  W    + = 1 −  B1     WB  Eg 0  WB   1  2kT WB1  WB1  − Eg0  kT WB1  1 − 1 − exp  . +  −   kT   Eg0 WB  WB Eg 0 WB 

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

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

The base bandgap narrowing parameter corresponding to Fig. 10.13 is given by Egmax Ge

n+ emitter poly

Metal or metal silicide

10.11

n+ Eg0 p x

0 WB

E gB, SiGe

 x − Wcap (E − E )  + W −W g max g0 cap   B0

(x) = E g0

= 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 )

=

 (WB0 − x jE ) WB 0



x jE

.

(2)

exp[−EgB,SiGe (x) / kT ]dx

The integral in (2) can be written as WB 0 Wcap WB 0  − EgB,SiGe (x)   − EgB,SiGe (x)   − EgB,SiGe (x)  dx dx dx . (3) = x exp + W exp xjE exp jE cap kT kT kT       The first integral on RHS of (3) is Wcap  − EgB,SiGe (x)   − Eg0  dx . exp = (W − x ) exp cap jE x jE kT    kT 

(4)

The second integral in (2) is WB 0  − EgB,SiGe (x)  WB 0 − Eg0 (Eg max − Eg 0 )(x −Wcap )  dx exp = exp − (5)  dx W W cap cap kT (W −W ) kT kT   B0 cap     −    Eg0   − (Eg max − Eg 0 )   − kT exp = (WB0 −Wcap )  −1exp .   kT    kT    E g max − E g 0  Substituting (4) and (5) into (3), we have from (2)

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

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

J C 0 (SiGe, x jE )  exp(Eg 0 / kT ) =   W − x   cap jE W −Wcap   Eg0 − Eg max  .

J C 0 (Si, x jE ) kT   +  B0  1− exp       kT    WB0 − x jE   WB0 − x jE  E g max − E g 0 

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

EgB,SiGe (x) = Eg 0 + EgB,SiGe (x) , − E ) (x) =  x − Wcap (E W −W gB, SiGe g max g0 cap   B0

E

where

=0

(1) x > Wcap

(2)

x < Wcap.

The transit time ratio is, from Eq. (10.54) tB (SiGe, x jE ) tB (Si, x jE ) =

=

x jE Wcap

2

WB 0

  (W − x )

x jE

exp[E gB,SiGe (x) / kT ] 

2 2



x jE

exp[E' gB,SiGe (x) / kT ] 

2



exp[E' gB,SiGe (x) / kT ] 

WB 0

2

B0

=

(3)

jE

 (WB0 − x jE )

WB 0

2

 (WB0 − x jE )

Wcap

x jE

x

exp[−E gB,SiGe (x') / kT ]dx' dx

WB 0 x

WB 0 x

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

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

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

+

2

  (W − x )

WB 0

2

B0

jE

Wcap

WB 0

exp[E' gB,SiGe (x) / kT ] 

x

exp[−E' gB,SiGe (x') / kT ]dx' dx .

The first double integral in (3) is for x < Wcap, and its value is WB 0 Wcap 2 (x) / kT ] exp[E' gB,SiGe x exp[−E' gB,SiGe (x') / kT ]dx' dx  (WB0 − x jE ) 2 x jE

=

2

jE

2

jE

+

x

Wcap

Wcap

x jE

x

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

2 2

 (WB0 − xjE )

Wcap

WB 0

x jE

Wcap

 

exp[−E' gB, SiGe (x') / kT ]dx'dx

Wcap 2 (Wcap − x)dx 2 x  (WB0 − xjE ) jE

+

=

x jE

   (W − x ) 2

B0

=

WB 0

   (W − x ) 2

B0

=

Wcap

(4)

 (x'−Wcap )(Eg max − Eg 0 )  exp dx'dx −    (WB0 − x jE ) 2 x jE Wcap (W − W )kT  B0 cap  2

Wcap

WB 0

2(Wcap − x jE ) 2 2  (WB0 − x jE ) +

 2(WB0 − Wcap )(Wcap − x jE )   E − Eg max  kT  1 − exp g 0  .

2  (WB0 − x jE ) kT    E g max − E g 0    

The second double integral in (3) is for x > Wcap, and its value is WB 0 WB 0 2 (x) / kT ] exp[E' gB, SiGe x exp[−E' gB, SiGe (x') / kT ]dx'dx  (WB0 − xjE ) 2 Wcap

(5)

(x −Wcap )(Eg max − Eg 0 )  WB 0 (x'−Wcap )(Eg 0 − Eg max )  exp  exp  dx'dx   2 W x cap  (WB0 − x jE ) (WB0 −Wcap )kT (WB0 −Wcap )kT      E   2(W − W )2  kT kT 1 − exp g0 − E g max  1 −  cap B0   . = 2       kT − E E − E  (WB0 − x jE )  E g max   g max g 0   g 0   =

2



WB 0

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

= x jE Wcap

2(Wcap − xjE ) 2 2  (WB0 − xjE )

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

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

+

 2(WB0 − Wcap )(Wcap − x jE )   Eg 0 − Eg max  kT   1 − exp    (WB0 − x jE ) 2 kT    E g max − E g 0     

 E   − W )2  kT kT 1 − exp g0 − E g max    cap 1 −    . +    kT  (WB0 − x jE ) 2  E g max − E g 0   E g max − E g 0    2(W

B0

n+ emitter poly

10.13

n+

Ge Egmax = Eg0 p

0

x

WB WB1

EgB,SiGe (x) =

x Eg 0 WB1

x < WB1

= Eg 0

(1) WB > x > WB1.

 kT W  W   −1 B1 B1  exp(− Eg 0 / kT ) (2) =   1 − exp(− Eg 0 / kT ) + 1 − J C 0 (Si)   WB  Eg0 WB

J (SiGe)



C0



100000

Current Ratio

10000 1000 100 10 1 0

0.2

0.4

0.6

0.8

1

WB1/WB

kT WB1 exp(E / kT ) − 1 + 1 − WB1  VA (SiGe) . = g0 VA (Si) Eg 0 WB  WB 





(3)

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

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

10000

Early Voltage Ratio

3000 1000

dE/kT=2.5

300 100 30 10 3 1 0

0.2

0.4

0.6

0.8

1

WB1/WB

1  WB1  2kT  W 2  1−  B +   B1   = tB (Si)   WB  Eg 0  WB   1  2kT WB1  WB1  − Eg0  kT WB1  1 − 1 − exp  . +  −   kT   Eg0 WB  WB Eg 0 WB  2

Transit Time Ratio

t (SiGe)

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

(4)

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

0.2

0.4

0.6

0.8

1

WB1/WB

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

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

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 JC 0 (SiGe) = JC 0 (Si)

WB

 exp[−E WB

gB,SiGe

0

.

(1)

(x) / kT ]dx

For the forward current ratio, we use in (1) xEg max .

W

EgB,SiBe ,FORWARD (x) =

(2)

B

For the reverse current ratio, we should use in (1) EgB,SiBe ,REVERSE (x) =

(WB − x)Eg max W

(3)

B

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

=

 WB  (x − WB )Eg max  dx 0 exp WB kT  WB

(4)

WB  − Eg max  W  xEg max   exp  exp W kT dx  kT  B  0 B

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

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

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

=

=

.

1 − exp(−Eg max / kT )

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

B n+ p

0

WB

x

(a)

Egmax Ge

n+ emitter poly

Metal or metal silicide

Egmax

Ge

n+ emitter poly

Metal or metal silicide

10.15

 Eg max / kT

B n+ p

WB

0

x

(b)

From Eq. (10.32), we have for transistor b,  EgB,SiGe (WB )  exp  WB exp[−E VA (SiGe, b) DnB (SiGe,WB )  (WB ) kT (x) / kT ]dx  = gB,SiGe 0 DnB (SiGe)  VA (Si, b) WB

(1)

 (x − WB )Eg max  dx = exp DnB (SiGe)  WB 0 W kT B    xEg max  DnB (SiGe,WB )  (WB ) exp(−Eg max / kT ) WB  = 0 exp WB kT dx DnB (SiGe)  WB  xEg max  DnB (SiGe,WB )  (WB ) exp(−Eg max / kT ) WB = 0 exp WB kT dx DnB (SiGe)  WB  DnB (SiGe,WB )  (WB ) 1

=

WB





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

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

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





kT exp(−E g max / kT ) exp(E g max / kT ) −1 . Eg max



The VA ratio for transistor a is given by Eq. (10.33), namely VA (SiGe, a) kT = exp(E / kT ) −1 . g max Eg max VA (Si, a)





(2)

Since VA(Si, a) = VA(Si, b), we have  − Eg max  VA (SiGe,b) = exp .

VA (SiGe, a)  kT 

(3)

From Eq. (10.36), we have for transistor b WB WB tB (SiGe, b) 2 exp[E (x) / kT ] exp[−E (4) (x') / kT ]dx' dx = gB,SiGe gB,SiGe x  W B2 0 tB (Si, b) WB (x'−WB )Eg max  2 WB (x) / kT ] exp dx' dx = exp[E gB,SiGe x  WB kT   W B2 0  x Eg max   − Eg max  = 2 2  WB kTWB  0 exp W  exp[EgB,SiGe (x) / kT ]dx W E 1 − exp kT kT g max    B B   (WB − x)Eg max  2 W kTWB 2 W kTWB = exp dx − dx    W 2 0 E W kT  W 2 0 E   B B g max B g max    − Eg max   Eg max  2 kTWB  kTWB  − WB  =  W 2 E E exp kT 1 − exp kT  g max  g max B       Eg max   − Eg max   = 2 kT  kT    −1 .  1 − exp  E E exp  kT kT g max      g max  B

B

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





(5)

Therefore, we have

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

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

t (SiGe,b)

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

kT E

tBB(SiGe, a) =

(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 C

B C

r'(1+g'mre)

r'0

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

E

E

From Eq. (9.161), we have for the small-signal base−emitter voltage vbe = vbe − icre − ib (rb + re )  vbe − icre ,

(1)

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

Similarly, we have vcb = vcb − icrc + ibrb  vcb − icrc = −vbc − icrc .

(2)

Now, for the short-circuit load, vbc = vbe, so (2) gives vcb = −vbe − icrc .

From the equivalent circuit, we have  gm  gm i = v + jC v =  − j − jC r i , C vbe c 1+ g r  cb  c c 1+ g r be  m e  m e

(3)

(4)

which can be rearranged to give

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

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

1 + jCrcic =  g

 − j .

C vbe 1 + g r  m e  m

(5)

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 ib =

vbe + r + r (1+ g r ) + jC vbe + jC vbc . r b



e

(7)

m e

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 vbe ib  r (1+ g r ) + jC (vbe − icre ) + jC (vbe + icrc )  m e  1 = + j(C + C )+ j(C rc − C re )gm  v   r (1+ g r )  be 1+ g r  m e m e  and

i gm  () = c = 1 . + (1+ g r ) j(C + C )+ j(C r − C r )g ib m e    c  e m r

(8)

(9)

(9) can be rearranged to give

 () =

gmr

1 + jr (C + C ) + gm (re + rc )C 

.

(10)

In the high-frequency limit, (10) becomes

 () 

,   j (C + C ) + g (r + r )C gm





m

e

c

(11)



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

g 2f (C + C ) +m g (r + r )C  T





m

e

c

(12)



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

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

or

1 2 f T

=F +

kT (CdBE ,tot + CdBC ,tot ) + C dBC ,tot (re + rc ) , qIC

(13)

where we have used Eqs. (9.153), (9.156), (9.157), and (9.170), i.e. gm =

 IC qI = C ,  VBE kT

C = CdBC ,tot , C = CdBE ,tot + CDE and

11.2

CDE =  F

qIC =  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 = EqIC / kT. That is E given by

 (deep− emitter) = E

 pE IB

=

2IC

 pE .

20

From Fig. 3.14(b), we have pE = 510−10 s. Therefore   E (deep− emitter) = pE = 2.510−12 s.

2

(1)

(2)

0

From Eq. (9.133), the diffusion capacitance due to electrons in the narrow base is CDn = 2qIntB / 3kT. For an n−p−n transistor, In is the collector current IC. Therefore, from Eqs. (9.166) to (9.171), we have 2  = t .

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

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

WB2 tB  = 5.83  10 −12 s, 2DnB

(4)

or  B = 3.8910−12 s.

From Eq. (3.15), we have WdBC =

2 si ( bi + VCB ) . qNC

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

(5)

Therefore

2 si ( bi + VCB ) = 2.77  10−5 cm. qNC

(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 W tBC = 2vdBC = 1.38  10−12 s,

(7)

sat

where we have used vsat = 107 cm/s.

WdBE = and

Similarly, we have

2 si ( bi − VBE ) = 1.71  10−6 cm, qNB

W tBE = 2vdBE = 0.86  10−12 s,

(8)

(9)

sat

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

Therefore, we have

 F =  E +  B + BE + BC =  E +  B + tBE + tBC = 8.6310−12 s.

(10)

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

and

CdBE ,tot =

AE si = 1.2210−14 F, WdBE

(11)

CdBC ,tot =

ABC  si = 3.7510−15 F. WdBC

(12)

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

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

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

IC = AE JC = 4.8 10 and

−4

A

(13)

kT 1 =  F + qIC (CdBE ,tot + CdBC ,tot ) = (8.6310−12 + 0.8610−12 ) s 2 fT

(14)

or fT = 16.9 GHz. (d) For (re + rc) = 50 , we have (re + rc )CdBC ,tot = 0.1910

−12

s,

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

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

1 kT =F + (CdBE + CdBC ) + C dBC (re + rc ) = (9.5910−12 + 0.1910−12 ) s, qI 2 fT 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 ) = qNC (int)

2 si (0.95 + 2) = 2.77  10−5 cm. qNC (int)

(1)

2 si ( bi + VCB ) = qNC (ext)

2 si (0.89 + 2) = 8.67 10−5 cm. qNC (ext)

(2)

Similarly, we have WdBC (ext) = Therefore,

A  si CdBC ,tot (int) = W E (int) = 0.75110−15 F,

(3)

dBC

(ext) = ABC (ext) si = 0.960 10−15 F WdBC (ext)

(4)

C dBC ,tot = C dBC ,tot (int) + C dBC ,tot (ext) = 1.7110−15 F.

(5)

C dBC ,tot

and

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

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

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

(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 kT 1 =  F + qIC (CdBE ,tot + CdBC ,tot ) = (8.6310−12 + 0.7510−12 ) s 2 fT

(6)

or fT = 17.0 GHz. (d) For (re +rc) = 50 , (re + rc )CdBC ,tot = 0.0910

and

−12

s,

(7)

1 kT (CdBE ,tot + CdBC ,tot ) + C dBC ,tot (re + rc ) = (9.3810−12 + 0.0910−12 ) s (8) =F + qIC 2 fT

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 ) = qNC (int)

2 si (0.97 + 2) = 1.96  10−5 cm qNC (int)

A  si CdBC ,tot (int) = W E (int) = 1.0610−15 F

(1)

(2)

dBC

and

C dBC ,tot = C dBC ,tot (int) + C dBC ,tot (ext) = (1.06 + 0.960) 10−15 F = 2.02 10−15 F.

(a) design.

(3)

Compared to Ex. 11.3, E, B, and tBE all remain unchanged by the pedestal collector

But tBC is reduced because of the larger NC(int). We have from (1) WdBC (int) = 0.98  10−12 s. tBC = 2v

(4)

sat

Therefore,

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

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

 F =  E +  B + tBE + tBC = (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 JC = 9.6 10−4 A. Therefore,

(6)

kT 1 =  F + qIC (CdBE ,tot + CdBC ,tot ) = (8.2310−12 + 0.3810−12 ) s 2 fT

(7)

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

(8)

kT 1 =  F + qIC (CdBE ,tot + CdBC ,tot ) + C dBC ,tot (re + rc ) = (8.6110−12 + 0.1010−12 ) s 2 fT

and (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

6

8

10

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

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

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

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

or

(1)

2.510−12 s = 0.62510 −12 s, 4

where we have used E(Si) = 2.510−12 s from (2) of Ex. 11.2.

(2) Also, from the plot above, we

have

 B (SiGe) tB (SiGe) =  0.4 ,  B (Si) tB (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 + tBE + tBC = (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 kT 1 =  F + qIC (CdBE ,tot + CdBC ,tot ) = (4.0310−12 + 0.3810−12 ) s, 2 fT (6) or fT = 36.1 GHz.

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

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

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

kT 1 (CdBE ,tot + CdBC ,tot ) + C dBC ,tot (re + rc ) = (4.4110−12 + 0.1010−12 ) s =F + qIC 2 fT

(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  v1  Y11 Y12  v1  i = Yˆ v  = Y  Y   v  .  2  2   21 22   2 

(1)

Referring to the network, we can write 1 i = v + j(C + C )v − jC v ,   1  2 1 r 1 1 i = v + g v + jC v − jC v. and 2

r0

2

m 1

 2

 1

Compare with the matrix representation in (1), we can write g 1 Y = + j(C + C ) = m + j(C + C ), (2)     11 0 r

Y12 = − jC ,

(3)

Y21 = gm − jC ,

(4)

and Y 22 =

1 + jC . r0

(5)

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

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

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 ) + icre

(6)

and

vce = vce + ibre + ic (rc + re ),

(7)

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

re  i1  . r + r  i  c e 2

(8)

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

(9)

The admittance matrix for the extrinsic transistor is i1  v1  Y11 Y12  v1  i  = Yˆ 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

Y = Y − Y2 r − Y Y r − (Y + Y )(Y + Y )r , 11

11

11 b

12

21 c

11

12

11

21

(11)

e

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)

Y = Y − Y Y r − Y 2r − (Y + Y )(Y + Y )r .

(14)

22

22

12

21 b

22 c

12

22

21

22

e

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

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

(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

  = i2 i1

= v2 =0

Y 21

.

(15)

Y 11

Therefore, the cutoff frequency of the intrinsic device is given by Y 1=

21

Y 11

=

g 2m + T2 C 2 g 2m + T2 (C + C  ) 2 2 0

,

(16)

which can be rearranged to give g2m 02 gm  = 2 f  =  , T T 2 2 2 2 (C + C  ) − C  (C + C ) − C g m2 −

(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  Y Y  1− r Y  − r Y  − r (Y  + Y  )(Y  + Y  ) Y  = Y21 = Y 21  1− r Y  b− r11Y  cY  22 Y  e − r11(Y  21+ Y  21)(Y  22+ Y  21) Y   1 v2 =0 11 11  b 11 c 12 21 11 e 11 12 11 21 11  (19) Y 21  1− (r + r )(Y  − Y  Y  Y  ) + r (Y  − Y  Y  Y  ). c e 22 12 21 11 e 12 11 22 21 Y 11

i  = i2

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+ (r + r )Re(Y  Y  Y  ) − r Re(Y  Y  Y  ) c e 12 21 11 e 11 22 21 Y 11  Y 21 1+ (r + r )Re(Y  Y  Y  ). c e 12 21 11 Y 11

(20)

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

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

where we have used the fact that reRe(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.,  (r + r )2g (C 2  + C C + C 2 )  c e T m  0    1= 1− . (g  )2 + 2 (C + C )2 2 2 2 (g m 0 ) + T (C + C  )  m 0 T    g 2m + T2C 2

(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2 + 2 C 2 = 2 (C + C )2 + 2(r + r )g 2 (C + C )C , m

T



T





c

e

m

T





(22)



which in turn can be rearranged to give g (C + C )C  1 (C + C  ) 2 − C 2 1+ (r m    =  + r )  g e c + C )2 − C 2 .

(C T m     

(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 1 1 + (r + r )C = + (r + r )C .  e c  e c dBC ,tot     T T T

11.7

(24)

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

4Re(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2m 02r0 − g m 2C 2 0 −  2 (C  + C  )2 r 0  − rc gm 0 r20 − g m 2C 2 0 − 2C 2 r 0 

(4)

− r e 2g2m 0 r0 − 2C2 r 0 + g m  0r20 .

2 A similar expansion and approximation can be made for Y + Y * , and (1) can then be used to 12

21

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

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

−2

−1

transistor operating at IC = 1 mA, Cμ (= CdBC,tot) > 1 fF, β0 > 100, r 0  105  , g m  410 

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(Y )Re(Y )  g 11

22

m

 r − r g2  2r − g 2C 2  − 2 (C + C )2 r 0

0

b

m

0

0

m





0



0

+ rc  gm 2C 2  0+ 2 C 2 r0 − r e2g2m  0r0− 2C 2 r0 .

(5)

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

Y12 + Y 21* = g2m − r b 4gm 2 (C + C )C + 2g3m  0 3 2 2 − rc 4g m 2C 2 + 2g2m r 0  − re 2gm + 2g m  0 r 0 + 2g m r 0

(6)

 g2m − r b4gm 2 (C + C  )C  + 2g3m  0 − 4rc g m2C 2 − 2r eg3m, 1 r0 , re r0 1 and

where in writing the last equation, we have used the fact that gm rc r0

1 r0 , rc

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

r0 , rb

r0 ,

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

re

g 2 (1− 2g r  − 2g r ) m

m b

0

m e





(7)

= 2max4g m rb (C + C )C + (C + C  )2 g mr0  + rcC 2 + reC 2 g mr0 .

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 )

4g m rb (C + C )C + (C + C )2 gm r0  + rcC  + reC g m 0 2

.

(8)

2

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 gm 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 ) 4g m rb (C + C )C  + rcC 2 

.

(9)

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

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

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 gm and keep only the term containing rb in the denominator, i.e.,

max 

gm f T , = 4gm (C + C )C rb 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 =

2t

B =

W W (V  ,V  )N n2 2IB WE2 = E B B E B C B ieE , 3IC 2DpE 3DnB NE n2ieB

W 2 (V  ,V  ) B

BE

BC

3

3DnB

BE = tBE =

WdBE (VBE ) 2vsat

BC = tBC =

WdBC (VBC ) .

2v

,

sat

(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) : WE−C = WB (VBE ,VBC ) +WdBE (VBE ) +WdBC (VBC ).

WdBE (VBE ) = 2si ( bi −VBE ) / q(NB + n), WdBC (VBC ) =

2si ( bi −VBC ) / q(N B + n),

(10.72) (10.78) (10.79)

where, n is given by JC = qvsat n.

(10.80)

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

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

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

C = d

C

 si

.

(3.26)

= AECdBC (VBC ) + LECBC , fringe ,

(11.19)

Wd

dBC ,tot

CdBE,tot = AECdBE (VBE ) + LECBE, fringe ,

(

) (

)

(

)

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      + ln Eg ( N ) = 6.92 ln    17  + 0.5 (9.18) 17 meV.     1.310    1.310   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 IC (VBE ,VBC ) = IC0 (VBE ,VBC ) exp(qVBE / kT ),

with  1 1+  IC0 (VBE ,VBC ) = nB ieB NBWB (VBE ,VBC )  4    qAE D n2

2

4n

 −1 exp(qV  / kT ) −1 .  N2 

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.

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

VBE (V) WB

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

fT

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

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

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

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

B0 F

  I eqVBE / kT  (1 − e−qVCE / kT ) − B0 R e− qVCE / kT  ,  0 IB0 F  

and   I IB = I B0 Fe qVBE / kT 1 + B0 R e− qVCE / kT  .  I B0 F  The current gain is



IC IB

=

I R − qVCE / kT 0 (1 − e− qVCE / kT ) − IB0 e BoF 1+

.

I B0 R − qVCE / kT e IB0 F

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 practical transistors the current gain is less than 1 only for very small VCE values.

100 90 80 70

IC / IB

60

IB0R/IB0F = 1

50 IB0R/IB0F = 10

40 30 20 10 0 -10

0

0.5

1

1.5

2 qVCE / kT

2.5

3

3.5

4

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

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

IC / IB

10 9 8 7 6 5 4

IB0R/IB0F = 1 IB0R/IB0F = 10

3 2 1 0 -1 0

0.05

0.1 qVCE / kT

0.15

0.2

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