Microelectronics Circuit Analysis and Design, Solutions Manual

665 Pages·2009·9.1 MB·English

by  D. A. Neamen

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

Chapter 1

1.1
n =BT3/2e−Eg/2kT
i
(a) Silicon
⎡ ⎤
−1.1
(i) n =(5.23×1015)(250)3/2exp⎢ ⎥
i ⎢2(86×10−6)(250)⎥
⎣ ⎦
=2.067×1019exp[−25.58]
n =1.61×108 cm−3
i
⎡ ⎤
−1.1
(ii) n =(5.23×1015)(350)3/2exp⎢ ⎥
i ⎢2(86×10−6)(350)⎥
⎣ ⎦
=3.425×1019exp[−18.27]
n =3.97×1011 cm−3
i

(b) GaAs
⎡ ⎤
−1.4
(i) n =(2.10×1014)(250)3/2exp⎢ ⎥
i ⎢2(86×10−6)(250)⎥
⎣ ⎦
=(8.301×1017)exp[−32.56]
n =6.02×103 cm−3
i
⎡ ⎤
−1.4
(ii) n =(2.10×1014)(350)3/2exp⎢ ⎥
i ⎢2(86×10−6)(350)⎥
⎣ ⎦
=(1.375×1018)exp[−23.26]
n =1.09×108 cm−3
i
______________________________________________________________________________________

1.2
⎛−Eg⎞
a. n =BT3/2exp⎜ ⎟
i ⎝2kT ⎠
⎛ −1.1 ⎞
1012 =5.23×1015T3/2exp⎜ ⎟
⎝2(86×10−6)(T)⎠
⎛ 6.40×103⎞
1.91×10−4 =T3/2exp⎜− ⎟
⎝ T ⎠
By trial and error, T ≈368 K
b. n =109 cm−3
i
⎛ ⎞
−1.1
109 =5.23×1015T3/2exp⎜ ⎟
⎜2(86×10−6)(T)⎟
⎝ ⎠
⎛ 6.40×103⎞
1.91×10−7 =T3/2exp⎜− ⎟
⎝ T ⎠
By trial and error, T ≈268° K
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1.3
Silicon
⎡ ⎤
−1.1
(a) n =(5.23×1015)(100)3/2exp⎢ ⎥
i ⎢2(86×10−6)(100)⎥
⎣ ⎦
=(5.23×1018)exp[−63.95]
n =8.79×10−10 cm−3
i

⎡ ⎤
−1.1
(b) n =(5.23×1015)(300)3/2exp⎢ ⎥
i ⎢2(86×10−6)(300)⎥
⎣ ⎦
=(2.718×1019)exp[−21.32]
n =1.5×1010 cm−3
i

⎡ ⎤
−1.1
(c) n =(5.23×1015)(500)3/2exp⎢ ⎥
i ⎢2(86×10−6)(500)⎥
⎣ ⎦
=(5.847×1019)exp[−12.79]
n =1.63×1014 cm−3
i

Germanium.
⎡ ⎤
−0.66
(a) n =(1.66×1015)(100)3/2exp⎢ ⎥=(1.66×1018)exp[−38.37]
i ⎢2(86×10−6)(100)⎥
⎣ ⎦
n =35.9 cm−3
i
⎡ ⎤
−0.66
(b) n =(1.66×1015)(300)3/2exp⎢ ⎥=(8.626×1018)exp[−12.79]
i ⎢2(86×10−6)(300)⎥
⎣ ⎦
n =2.40×1013 cm−3
i
⎡ ⎤
−0.66
(c) n =(1.66×1015)(500)3/2exp⎢ ⎥=(1.856×1019)exp[−7.674]
i ⎢2(86×10−6)(500)⎥
⎣ ⎦
n =8.62×1015 cm−3
i
______________________________________________________________________________________

1.4
n2 (2.4×1013)2
(a) n-type; n =1015 cm−3; p = i = =5.76×1011 cm−3
o o n 1015
o
n2 (1.5×1010)2
(b) n-type; n =1015 cm−3; p = i = =2.25×105 cm−3
o o n 1015
o
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1.5
n2 (1.8×106)2
(a) p-type; p =1016 cm−3; n = i = =3.24×10−4 cm−3
o o p 1016
o

n2 (2.4×1013)2
(b) p-type; p =1016 cm−3; n = i = =5.76×1010 cm−3
o o p 1016
o
______________________________________________________________________________________

1.6
(a) n-type
(b) n =N =5×1016 cm−3
o d
n2 (1.5×1010)2
p = i = =4.5×103 cm−3
o n 5×1016
o
(c) n =N =5×1016 cm−3
o d
From Problem 1.1(a)(ii) n =3.97×1011 cm−3
i
(3.97×1011)2
p = =3.15×106 cm−3
o 5×1016
______________________________________________________________________________________

1.7
n2 (1.5×1010)2
(a) p-type; p =5×1016 cm−3; n = i = =4.5×103 cm−3
o o p 5×1016
o
n2 (1.8×106)2
(b) p-type; p =5×1016 cm−3; n = i = =6.48×10−5 cm−3
o o p 5×1016
o
______________________________________________________________________________________

1.8
(a) Add boron atoms
(b) N = p =2×1017 cm−3
a o
n2 (1.5×1010)2
(c) n = i = =1.125×103 cm−3
o p 2×1017
o
______________________________________________________________________________________

1.9
(a) n =5×1015 cm−3
o
n2 (1.5×1010)2
p = i = ⇒ p =4.5×104 cm−3
o n 5×1015 o
o
(b) n > p ⇒n-type
o o
(c) n ≅ N =5×1015 cm−3
o d
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1.10
a. Add Donors
N =7×1015 cm−3
d
b. Want p =106 cm−3 =n2/N
o i d
So n2 =(106)(7×1015)=7×1021
i
⎛−Eg⎞
=B2T3exp⎜ ⎟
⎝ kT ⎠
⎛ ⎞
7×1021 =(5.23×1015)2T3exp⎜ −1.1 ⎟
⎜(86×10−6)(T)⎟
⎝ ⎠
By trial and error, T ≈324° K
______________________________________________________________________________________

1.11
(a) I = AσΕ=(10−5)(1.5)(10)⇒I =0.15 mA
AΕ Iρ (1.2×10−3)(0.4)
(b) I = ⇒Ε= = ( ) =2.4 V/cm
ρ A 2×10−4
______________________________________________________________________________________

1.12
J 120
J =σΕ⇒σ= = =6.67(Ω−cm)−1
Ε 18
σ (6.67)
σ≅eμnNd ⇒Nd = eμ = (1.6×10−19)(1250)=3.33×1016 cm−3
n
______________________________________________________________________________________

1.13
1 1 1
(a) ρ≅ eμN ⇒Nd = eμρ= (1.6×10−19)(1250)(0.65)=7.69×1015 cm−3
n d n
Ε
(b) J = ⇒Ε=ρJ =(0.65)(160)=104 V/cm
ρ
______________________________________________________________________________________

1.14
σ 1.5
(a) σ≅eμnNd ⇒Nd = eμ = (1.6×10−19)(1000)=9.375×1015 cm−3
n
σ 0.8
(b) Na = eμ = (1.6×10−19)(400)=1.25×1016 cm−3
p
______________________________________________________________________________________

1.15
(a) For n-type, σ≅eμN =(1.6×10−19)(8500)N
n d d
For 1015 ≤ N ≤1019 cm−3 ⇒1.36≤σ≤1.36×104(Ω−cm)−1
d
(b) J =σE=σ(0.1)⇒0.136≤J ≤1.36×103 A/cm2
______________________________________________________________________________________
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1.16
D =(0.026)(1250)=32.5 cm2/s; D =(0.026)(450)=11.7 cm2/s
n p
J =eD dn =(1.6×10−19)(32.5)⎜⎛1016 −1012 ⎟⎞=−52 A/cm2
n n dx ⎜⎝ 0−0.001 ⎟⎠
J =−eD dp =−(1.6×10−19)(11.7)⎜⎛1012 −1016 ⎟⎞=−18.72 A/cm2
p p dx ⎜⎝ 0−0.001 ⎟⎠
Total diffusion current density
J =−52−18.72=−70.7 A/cm2
______________________________________________________________________________________

1.17
dp
J =−eD
p p dx
⎛−1⎞ ⎛−x⎞
=−eD (1015)⎜ ⎟exp⎜ ⎟
p ⎜L ⎟ ⎜L ⎟
⎝ p ⎠ ⎝ p ⎠
(1.6×10−19)(15)(1015) ⎛−x⎞
J = exp⎜ ⎟
p 10×10−4 ⎜L ⎟
⎝ p ⎠
J =2.4 e−x/Lp
p
(a) x = 0 J =2.4 A/cm2
p
(b) x=10 μm J =2.4 e−1 =0.883 A/cm2
p
(c) x=30 μm J =2.4 e−3 =0.119 A/cm2
p
______________________________________________________________________________________

1.18
a. N =1017 cm−3 ⇒ p =1017 cm−3
a o
n2 (1.8×106)2
n = i = ⇒n =3.24×10−5 cm−3
o p 1017 o
o
b. n=n +δn=3.24×10−5+1015 ⇒n=1015 cm−3
o
p= p +δp=1017 +1015 ⇒ p=1.01×1017 cm−3
o
______________________________________________________________________________________

⎛N N ⎞
1.19 V =V ln⎜ a d ⎟
bi T ⎜⎝ ni2 ⎟⎠
⎡(5×1015)(5×1015)⎤
(a) (i) V =(0.026)ln⎢ ⎥=0.661 V
bi ⎢⎣ (1.5×1010)2 ⎥⎦
⎡(5×1017)(1015)⎤
(ii) V =(0.026)ln⎢ ⎥ =0.739 V
bi ⎢⎣ (1.5×1010)2 ⎥⎦
⎡(1018)(1018)⎤
(iii) V =(0.026)ln⎢ ⎥=0.937 V
bi ⎢⎣(1.5×1010)2⎥⎦
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

⎡(5×1015)(5×1015)⎤
(b) (i) V =(0.026)ln⎢ ⎥=1.13 V
bi ⎢⎣ (1.8×106)2 ⎥⎦
⎡(5×1017)(1015)⎤
(ii) V =(0.026)ln⎢ ⎥=1.21 V
bi ⎢⎣ (1.8×106)2 ⎥⎦
⎡(1018)(1018)⎤
(iii) V =(0.026)ln⎢ ⎥=1.41 V
bi ⎢⎣(1.8×106)2⎥⎦
______________________________________________________________________________________

1.20
⎛N N ⎞
V =V ln⎜ a d ⎟
bi T ⎜⎝ ni2 ⎟⎠
or
(n2) ⎛V ⎞ (1.5×10)2 ⎛0.712⎞
N = i exp⎜ bi ⎟= exp⎜ ⎟=1.76×1016 cm−3
a Nd ⎜⎝VT ⎟⎠ 1016 ⎝0.026⎠
______________________________________________________________________________________

1.21
⎛N N ⎞ ⎡ N (1016) ⎤
V =V ln⎜ a d ⎟=(0.026)ln⎢ a ⎥
bi T ⎝ ni2 ⎠ ⎢⎣(1.5×1010)2⎥⎦
For N =1015 cm−3, V =0.637 V
a bi
For N =1018 cm−3, V =0.817 V
a bi

______________________________________________________________________________________

1.22
⎛ T ⎞
kT =(0.026)⎜ ⎟
⎝300⎠
kT (T)3/2
200 0.01733 2828.4
250 0.02167 3952.8
300 0.026 5196.2
350 0.03033 6547.9
400 0.03467 8000.0
450 0.0390 9545.9
500 0.04333 11,180.3
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

⎛ ⎞
−1.4
n =(2.1×1014)(T3/2)exp⎜ ⎟
i ⎜2(86×10−6)(T)⎟
⎝ ⎠
⎛N N ⎞
V =V ln⎜ a d ⎟
bi T ⎝ n2 ⎠
i
T n V
i bi
200 1.256 1.405
250 6.02 × 103 1.389
300 1.80 × 106 1.370
350 1.09 × 108 1.349
400 2.44 × 109 1.327
450 2.80 × 1010 1.302
500 2.00 × 1011 1.277

______________________________________________________________________________________

1.23
−1/2
⎛ V ⎞
C =C ⎜1+ R ⎟
j jo⎝ V ⎠
bi
⎡(1.5×1016)(4×1015)⎤
V =(0.026)ln⎢ ⎥=0.684 V
bi ⎢ (1.5×1010)2 ⎥
⎣ ⎦
⎛ 1 ⎞−1/2
(a) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.255 pF
⎛ 3 ⎞−1/2
(b) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.172 pF
⎛ 5 ⎞−1/2
(c) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.139 pF
______________________________________________________________________________________

1.24
−1/2
⎛ V ⎞
(a) C =C ⎜1+ R ⎟
j jo⎝ V ⎠
bi
⎛ 5 ⎞−1/2
For VR = 5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.00743 pF
⎛ 1.5⎞−1/2
For VR = 1.5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.0118 pF
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

0.00743+0.0118
C (avg)= =0.00962 pF
j 2
v (t)=v (final)+(v (initial)−v (final))e−t/τ
C C C C
where
τ=RC=RC (avg)=(47×103)(0.00962×10−12)
j
or
τ=4.52×10−10 s
Then v (t)=1.5=0+(5−0)e−ti/τ
C
5 ⎛ 5 ⎞
=e+r1/τ⇒t =τln⎜ ⎟
1.5 1 ⎝1.5⎠
t =5.44×10−10 s
1
(b) For V = 0 V, C = C = 0.02 pF
R j jo
⎛ 3.5⎞−1/2
For VR = 3.5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.00863 pF
0.02+0.00863
C (avg)= =0.0143 pF
j 2
τ=RC (avg)=6.72×10−10 s
j
v (t)=v (final)+(v (initial)−v (final))e−t/τ
C C C C
3.5=5+(0−5)e−t2/τ=5(1−e−t2/τ)
so that t =8.09×10−10 s
2
______________________________________________________________________________________

1.25
⎛ V ⎞−1/2 ⎡(5×1015)(1017)⎤
C =C ⎜1+ R ⎟ ; V =(0.026)ln⎢ ⎥ =0.739 V
j jo⎜⎝ Vbi ⎟⎠ bi ⎢⎣ (1.5×1010)2 ⎥⎦
For V =1V,
R
0.60
C = =0.391 pF
j
1
1+
0.739
For V =3 V,
R
0.60
C = =0.267 pF
j
3
1+
0.739
For V =5 V,
R
0.60
C = =0.215 pF
j
5
1+
0.739
1 1
(a) f = = ⇒ f =6.57 MHz
o ( )( ) o
2π LC 2π 1.5×10−3 0.391×10−12
1
(b) f = ⇒ f =7.95MHz
o ( )( ) o
2π 1.5×10−3 0.267×10−12
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1
(c) f = ⇒ f =8.86 MHz
o ( )( ) o
2π 1.5×10−3 0.215×10−12
______________________________________________________________________________________

1.26
⎡ ⎛V ⎞ ⎤ ⎛V ⎞
a. I =I ⎢exp⎜ D ⎟−1⎥−0.90=exp⎜ D ⎟−1
S ⎣ ⎝VT ⎠ ⎦ ⎝VT ⎠
⎛V ⎞
exp⎜ D ⎟=1−0.90=0.10
⎝V ⎠
T
V =V ln(0.10)⇒V =−0.0599 V
D T D
b.
IF = IS ⋅⎢⎡⎣exp⎛⎜⎝VVTF ⎞⎟⎠−1⎥⎤⎦ = exp⎝⎛⎜00.0.226⎠⎞⎟−1
I I ⎡ ⎛V ⎞ ⎤ ⎛ −0.2 ⎞
R S ⎢exp⎜ R ⎟−1⎥ exp⎜ ⎟−1
⎣ ⎝VT ⎠ ⎦ ⎝0.026⎠
2190
=
−1
I
F =2190
I
R
______________________________________________________________________________________

⎡ ⎛V ⎞ ⎤
1.27 ID =IS⎢⎢⎣exp⎜⎜⎝VTD ⎟⎟⎠−1⎥⎥⎦
( ) ⎛ 0.3 ⎞
(a) (i) I = 10−11 exp⎜ ⎟⇒1.03μA
D ⎝0.026⎠
( ) ⎛ 0.5 ⎞
(ii) I = 10−11 exp⎜ ⎟⇒2.25 mA
D ⎝0.026⎠
( ) ⎛ 0.7 ⎞
(iii) I = 10−11 exp⎜ ⎟⇒4.93 A
D ⎝0.026⎠
( )⎡ ⎛−0.02⎞ ⎤
(iv) ID = 10−11 ⎢⎣exp⎜⎝ 0.026 ⎟⎠−1⎥⎦ =−5.37×10−12 A
( )⎡ ⎛−0.20⎞ ⎤
(v) ID = 10−11 ⎢⎣exp⎜⎝ 0.026 ⎟⎠−1⎥⎦≅−10−11 A
( )
(vi) I =−10−11 A
D

( ) ⎛ 0.3 ⎞
(b) (i) I = 10−13 exp⎜ ⎟⇒0.0103μA
D ⎝0.026⎠
( ) ⎛ 0.5 ⎞
(ii) I = 10−13 exp⎜ ⎟⇒22.5μA
D ⎝0.026⎠
( ) ⎛ 0.7 ⎞
(iii) I = 10−13 exp⎜ ⎟⇒49.3 mA
D ⎝0.026⎠