1. Chapter 1
Problem Solutions
1.1
− E / 2 kT
ni = BT 3 / 2 e g
(a) Silicon
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 250 )
3/ 2
(i) exp ⎢ ⎥
⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥
⎣ ⎦
= 2.067 × 1019 exp [ −25.58]
ni = 1.61× 108 cm −3
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 350 )
3/ 2
(ii) exp ⎢ ⎥
⎢ 2 ( 86 × 10−6 ) ( 350 ) ⎥
⎣ ⎦
= 3.425 × 1019 exp [ −18.27 ]
ni = 3.97 ×1011 cm −3
(b) GaAs
⎡ −1.4 ⎤
ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
(i) exp ⎢ ⎥
⎢ 2 ( 86 × 10 ) ( 250 ) ⎥
⎣
−6
⎦
= ( 8.301× 1017 ) exp [ −32.56]
ni = 6.02 × 103 cm −3
⎡ −1.4 ⎤
ni = ( 2.10 × 1014 ) ( 350 )
3/ 2
(ii) exp ⎢ ⎥
⎢ 2 ( 86 × 10−6 ) ( 350 ) ⎥
⎣ ⎦
= (1.375 × 1018 ) exp [ −23.26]
ni = 1.09 × 108 cm −3
1.2
⎛ − Eg ⎞
a. ni = BT 3 / 2 exp ⎜ ⎟
⎝ 2kT ⎠
⎛ −1.1 ⎞
1012 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟
⎝ 2(86 × 10−6 )(T ) ⎠
⎛ 6.40 × 103 ⎞
1.91× 10−4 = T 3 / 2 exp ⎜ − ⎟
⎝ T ⎠
By trial and error, T ≈ 368 K
b. ni = 109 cm −3
⎛ −1.1 ⎞
109 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟
⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝ ⎠
⎛ 6.40 × 103 ⎞
1.91× 10−7 = T 3 / 2 exp ⎜ − ⎟
⎝ T ⎠
By trial and error, T ≈ 268° K
1.3
Silicon

2. ⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) (100 )
3/ 2
(a) exp ⎢ ⎥
⎢ 2 ( 86 × 10 ) (100 ) ⎥
⎣
−6
⎦
= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 300 )
3/ 2
(b) exp ⎢ ⎥
⎢ 2 ( 86 × 10−6 ) ( 300 ) ⎥
⎣ ⎦
= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 500 )
3/ 2
(c) exp ⎢ ⎥
⎢ 2 ( 86 × 10 ) ( 500 ) ⎥
⎣
−6
⎦
= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3
Germanium.
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) (100 ) ⎥ = (1.66 × 1018 ) exp [ −38.37 ]
3/ 2
(a) exp ⎢
⎢ 2 ( 86 × 10−6 ) (100 ) ⎥
⎣ ⎦
ni = 35.9 cm −3
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) ( 300 ) ⎥ = ( 8.626 × 1018 ) exp [ −12.79]
3/ 2
(b) exp ⎢
⎢ 2 ( 86 × 10−6 ) ( 300 ) ⎥
⎣ ⎦
ni = 2.40 × 1013 cm −3
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) ( 500 ) ⎥ = (1.856 × 1019 ) exp [ −7.674]
3/ 2
(c) exp ⎢
⎢ 2 ( 86 × 10−6 ) ( 500 ) ⎥
⎣ ⎦
ni = 8.62 ×1015 cm −3
1.4
a. N d = 5 × 1015 cm −3 ⇒ n − type
n0 = N d = 5 × 1015 cm −3
n 2 (1.5 × 10 )
10 2
p0 = i = ⇒ p0 = 4.5 × 10 4 cm −3
n0 5 × 1015
b. N d = 5 × 1015 cm −3 ⇒ n − type
no = N d = 5 × 1015 cm −3
⎛ −1.4 ⎞
ni = ( 2.10 × 1014 ) ( 300 )
3/ 2
exp ⎜ −6 ⎟
⎝ 2(86 × 10 )(300) ⎠
= ( 2.10 × 1014 ) ( 300 ) (1.65 ×10 )
3/ 2 −12
= 1.80 × 106 cm −3
ni2 (1.8 × 10 )
6 2
p0 = = ⇒ p0 = 6.48 × 10 −4 cm −3
n0 5 × 1015
1.5

3. (a) n-type
(b) no = N d = 5 × 1016 cm −3
n 2 (1.5 × 10 )
10 2
po = i = = 4.5 × 103 cm −3
no 5 × 1016
(c) no = N d = 5 × 1016 cm −3
From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3
po =
( 3.97 × 10 ) 11 2
= 3.15 × 106 cm −3
5 × 1016
1.6
a. N a = 1016 cm −3 ⇒ p − type
p0 = N a = 1016 cm −3
n 2 (1.5 × 10 )
10 2
n0 = i = ⇒ n0 = 2.25 × 10 4 cm −3
p0 1016
b. Germanium
N a = 1016 cm −3 ⇒ p − type
p0 = N a = 1016 cm −3
⎛ −0.66 ⎞
ni = (1.66 × 1015 ) ( 300 )
3/ 2
exp ⎜ ⎟
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝ ⎠
= (1.66 × 1015 ) ( 300 ) ( 2.79 × 10 )
3/ 2 −6
= 2.4 × 1013 cm −3
n 2 ( 2.4 × 10 )
13 2
n0 = i = ⇒ n0 = 5.76 × 1010 cm −3
p0 1016
1.7
(a) p-type
(b) po = N a = 2 × 1017 cm −3
ni2 (1.5 × 10 )
10 2
no = = = 1.125 × 103 cm −3
po 2 × 1017
(c) po = 2 × 1017 cm −3
From Problem 1.1(a)(i) ni = 1.61 × 108 cm −3
no =
(1.61×10 ) 8 2
= 0.130 cm −3
2 × 1017
1.8
(a) no = 5 × 1015 cm −3
ni2 (1.5 × 10 )
10 2
po = = ⇒ po = 4.5 × 104 cm −3
no 5 × 1015
(b) no po ⇒ n-type
(c) no ≅ N d = 5 × 1015 cm −3
1.9
a. Add Donors
N d = 7 × 1015 cm −3

4. b. Want po = 106 cm −3 = ni2 / N d
So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21
⎛ − Eg ⎞
= B 2T 3 exp ⎜ ⎟
⎝ kT ⎠
⎛ −1.1 ⎞
7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜
2
⎟
⎜ ( 86 × 10 ) (T ) ⎟
−6
⎝ ⎠
By trial and error, T ≈ 324° K
1.10
I = J ⋅ A = σ EA
I = ( 2.2 )(15 ) (10−4 ) ⇒ I = 3.3 mA
1.11
J 85
J =σE ⇒σ = =
E 12
σ = 7.08 (ohm − cm) −1
1.12
1 1 1
g≈ ⇒ Na = =
eμ p N a eμ p g (1.6 × 10 −19 ) ( 480 )( 0.80 )
N a = 1.63 × 10 16 cm −3
1.13
σ = eμ n N d
σ ( 0.5)
Nd = =
eμ n (1.6 ×10 ) (1350 )
−19
N d = 2.31× 1015 cm −3
1.14
(a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d
For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm )
−1
(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 × 103 A / cm2
1.15
dn Δn
J n = eDn = eDn
dx Δx
⎡10 15 −10 2 ⎤
= (1.6 × 10 −19 ) (180 ) ⎢ −4 ⎥
⎣ 0.5 × 10 ⎦
J n = 576 A/cm 2
1.16

5. dp
J p = −eD p
dx
⎛ −1 ⎞ ⎛ −x ⎞
= −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟
⎜ Lp ⎟ ⎜ Lp ⎟
⎝ ⎠ ⎝ ⎠
Jp =
(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞
−19 15
⎜ ⎟
⎜L ⎟
10 × 10 −4 ⎝ p⎠
− x / Lp
J p = 2.4 e
(a) x=0 J p = 2.4 A/cm2
(b) x = 10 μ m J p = 2.4 e−1 = 0.883 A/cm 2
(c) x = 30 μ m J p = 2.4 e−3 = 0.119 A/cm 2
1.17
a. N a = 1017 cm −3 ⇒ po = 1017 cm −3
ni2 (1.8 × 10 )
6 2
no = = ⇒ no = 3.24 × 10−5 cm −3
po 1017
b. n = no + δ n = 3.24 × 10 −5 + 1015 ⇒ n = 1015 cm −3
p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3
1.18
⎛N N ⎞
(a) Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
⎡ (10 16 )(10 16 ) ⎤
= ( 0.026 ) ln ⎢ ⎥ = 0.697 V
⎢ (1.5 × 10 10 )2 ⎥
⎣ ⎦
⎡ (10 18 )(10 16 ) ⎤
(b) Vbi = ( 0.026 ) ln ⎢ ⎥ = 0.817 V
⎢ (1.5 × 10 10 )2 ⎥
⎣ ⎦
⎡ (10 18 )(10 18 ) ⎤
(c) Vbi = ( 0.026 ) ln ⎢ ⎥ = 0.937 V
⎢ (1.5 × 10 10 )2 ⎥
⎣ ⎦
1.19
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
⎡ (1016 )(1016 ) ⎤
a. Vbi = ( 0.026 ) ln ⎢ ⎥ ⇒ Vbi = 1.17 V
⎢ (1.8 × 10 ) ⎥
6 2
⎣ ⎦
⎡ (1018 )(1016 ) ⎤
b. Vbi = ( 0.026 ) ln ⎢ ⎥ ⇒ Vbi = 1.29 V
⎢ (1.8 × 10 ) ⎥
6 2
⎣ ⎦
⎡ (1018 )(1018 ) ⎤
c. Vbi = ( 0.026 ) ln ⎢ ⎥ ⇒ Vbi = 1.41 V
⎢ (1.8 × 10 ) ⎥
6 2
⎣ ⎦
1.20
⎛ Na Nd ⎞ ⎡ N a (1016 ) ⎤
Vbi = VT ln ⎜ 2 ⎟ = ( 0.026 ) ln ⎢ ⎥
⎢ (1.5 × 10 ) ⎥
10 2
⎝ ni ⎠ ⎣ ⎦

6. For N a = 1015 cm −3 , Vbi = 0.637 V
For N a = 1018 cm −3 , Vbi = 0.817 V
Vbi (V)
0.817
0.637
1015 1016 1017 1018 Na(cmϪ3)
1.21
⎛ T ⎞
kT = (0.026) ⎜ ⎟
⎝ 300 ⎠
T 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
⎛ −1.4 ⎞
ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜ ⎟
⎜ 2 ( 86 × 10 ) (T ) ⎟
−6
⎝ ⎠
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
T ni Vbi
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
Vbi (V)
1.45
1.35
1.25
200 250 300 350 400 450 500 T(ЊC)
1.22
−1/ 2
⎛ V ⎞
C j = C jo ⎜ 1 + R ⎟
⎝ Vbi ⎠

7. ⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤
Vbi = ( 0.026 ) ln ⎢ ⎥ = 0.684 V
⎢
⎣ (1.5 ×10 10 ) 2 ⎥ ⎦
−1/ 2
⎛ 1 ⎞
(a) C j = ( 0.4 ) ⎜ 1 + ⎟ = 0.255 pF
⎝ 0.684 ⎠
−1/ 2
⎛ 3 ⎞
(b) C j = ( 0.4 ) ⎜ 1 + ⎟ = 0.172 pF
⎝ 0.684 ⎠
−1/ 2
⎛ 5 ⎞
(c) C j = ( 0.4 ) ⎜ 1 + ⎟ = 0.139 pF
⎝ 0.684 ⎠
1.23
−1 / 2
⎛ V ⎞
(a) C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
−1 / 2
⎛ 5 ⎞
For VR = 5 V, C j = (0.02) ⎜ 1 + ⎟ = 0.00743 pF
⎝ 0. 8 ⎠
−1 / 2
⎛ 1. 5 ⎞
For VR = 1.5 V, C j = (0.02) ⎜1 + ⎟ = 0.0118 pF
⎝ 0. 8 ⎠
0.00743 + 0.0118
C j (avg ) = = 0.00962 pF
2
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ
where
τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 )
or
τ = 4.52 ×10−10 s
Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e − ti / τ
5 + r /τ ⎛ 5 ⎞
= e 1 ⇒ t1 = τ ln ⎜ ⎟
1.5 ⎝ 1.5 ⎠
−10
t1 = 5.44 × 10 s
(b) For VR = 0 V, Cj = Cjo = 0.02 pF
−1/ 2
⎛ 3.5 ⎞
For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 + ⎟ = 0.00863 pF
⎝ 0.8 ⎠
0.02 + 0.00863
C j (avg ) = = 0.0143 pF
2
τ = RC j ( avg ) = 6.72 ×10−10 s
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ
(
3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ )
−10
so that t2 = 8.09 × 10 s
1.24
⎡ (1018 )(1015 ) ⎤
Vbi = ( 0.026 ) ln ⎢ ⎥ = 0.757 V
⎢ (1.5 × 1010 )2 ⎥
⎣ ⎦
a. VR = 1 V
−1/ 2
⎛ 1 ⎞
C j = (0.25) ⎜ 1 + ⎟ = 0.164 pF
⎝ 0.757 ⎠

8. 1 1
f0 = =
2π LC 2π ( 2.2 ×10 )( 0.164 ×10 )
−3 −12
f 0 = 8.38 MHz
b. VR = 10 V
−1/ 2
⎛ 10 ⎞
C j = (0.25) ⎜ 1 + ⎟ = 0.0663 pF
⎝ 0.757 ⎠
1
f0 =
2π ( 2.2 × 10 )( 0.0663 × 10−12 )
−3
f 0 = 13.2 MHz
1.25
⎡ ⎛V ⎞ ⎤ ⎛ VD ⎞
a. I = I S ⎢ exp ⎜ D ⎟ − 1⎥ − 0.90 = exp ⎜ ⎟ −1
⎣ ⎝ VT ⎠ ⎦ ⎝ VT ⎠
⎛V ⎞
exp ⎜ D ⎟ = 1 − 0.90 = 0.10
⎝ VT ⎠
VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V
b.
⎡ ⎛ VF ⎞ ⎤ ⎛ 0.2 ⎞
⎢ exp ⎜ ⎟ − 1⎥ exp ⎜ ⎟ −1
⎝ VT ⎠ ⎦
= S ⋅⎣ ⎝ 0.026 ⎠
IF I
=
IR IS ⎡ ⎛ VR ⎞ ⎤ ⎛ −0.2 ⎞
⎢exp ⎜ ⎟ − 1⎥ exp ⎜ 0.026 ⎟ − 1
⎝ ⎠
⎣ ⎝ VT ⎠ ⎦
2190
=
−1
IF
= 2190
IR
1.26
a.
⎛ 0.5 ⎞
I ≅ (10−11 ) exp ⎜ ⎟ ⇒ I = 2.25 mA
⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−11 ) exp ⎜ ⎟ ⇒ I = 0.105 A
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−11 ) exp ⎜ ⎟ ⇒ I = 4.93 A
⎝ 0.026 ⎠
b.
⎛ 0.5 ⎞
I ≅ (10−13 ) exp ⎜ ⎟ ⇒ I = 22.5 μ A
⎝ 0.026 ⎠
⎛ 0.6 ⎞
I = (10−13 ) exp ⎜ ⎟ ⇒ I = 1.05 mA
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I = (10−13 ) exp ⎜ ⎟ ⇒ I = 49.3 mA
⎝ 0.026 ⎠
1.27
(a) (
I = I S eVD / VT − 1 )
150 × 10 −6
= 10 −11
(e VD / VT
)
− 1 ≅ 10−11 eVD / VT

9. ⎛ 150 × 10−6 ⎞ ⎛ 150 × 10−6 ⎞
Then VD = VT ln ⎜ −11 ⎟ = (0.026) ln ⎜ −11 ⎟
⎝ 10 ⎠ ⎝ 10 ⎠
Or VD = 0.430 V
(b)
⎛ 150 × 10−6 ⎞
VD = VT ln ⎜ −13 ⎟
⎝ 10 ⎠
Or VD = 0.549 V
1.28
⎛ 0.7 ⎞
(a) 10−3 = I S exp ⎜ ⎟
⎝ 0.026 ⎠
I S = 2.03 × 10 −15 A
(b)
VD I D ( A ) ( n = 1) I D ( A )( n = 2 )
0.1 9.50 ×10 −14 1.39 ×10 −14
0.2 4.45 ×10 −12 9.50 ×10 −14
0.3 2.08 ×10 −10 6.50 ×10 −13
0.4 9.75 ×10 −9 4.45 ×10 −12
0.5 4.56 ×10 −7 3.04 ×10 −11
0.6 2.14 ×10 −5 2.08 ×10 −10
0.7 10 −3 1.42 ×10 −9
1.29
(a)
I S = 10 −12 A
VD(v) ID(A) log10ID
0.10 4.68 ×10−11 −10.3
0.20 2.19 ×10−9 −8.66
0.30 1.03 ×10−7 −6.99
0.40 4.80 ×10−6 −5.32
0.50 2.25 ×10−4 −3.65
0.60 1.05 ×10−2 −1.98
0.70 4.93 ×10−1 −0.307
(b)
I S = 10 −14 A
VD(v) ID(A) log10ID
0.10 4.68 ×10−13 −12.3
0.20 2.19 ×10−11 −10.66
0.30 1.03 ×10−9 −8.99
0.40 4.80 ×10−8 −7.32
0.50 2.25 ×10−6 −5.65
0.60 1.05 ×10−4 −3.98
0.70 4.93 ×10−3 −2.31
1.30
a.
ID2 ⎛ V − VD1 ⎞
= 10 = exp ⎜ D 2 ⎟
I D1 ⎝ VT ⎠
ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV

10. b. ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV
1.31
⎛I ⎞ ⎛ 150 × 10−6 ⎞
(a) (i) VD = Vt ln ⎜ D ⎟ = ( 0.026 ) ln ⎜ −15 ⎟
⎝ IS ⎠ ⎝ 10 ⎠
VD = 0.669 V
⎛ 25 × 10−6 ⎞
(ii) VD = ( 0.026)ln ⎜ −15 ⎟
⎝ 10 ⎠
VD = 0.622 V
⎛ 0.2 ⎞
(b) (i) I D = (10−15 )exp ⎜ −12
⎟ = 2.19 × 10 A
⎝ 0.026 ⎠
(ii) ID = 0
(iii) I D = −10 −15 A
(iv) I D = −10 −15 A
1.32
⎛I ⎞ ⎛ 2 × 10−3 ⎞
VD = Vt ln ⎜ D ⎟ = (0.026) ln ⎜ −14 ⎟
= 0.6347 V
⎝ IS ⎠ ⎝ 5 × 10 ⎠
⎛ 2 × 10−3 ⎞
VD = (0.026) ln ⎜ −12 ⎟
= 0.5150 V
⎝ 5 × 10 ⎠
0.5150 ≤ VD ≤ 0.6347 V
1.33
⎛V ⎞
(a) I D = I S exp ⎜ D ⎟
⎝ Vt ⎠
⎛ 1.10 ⎞
12 ×10−3 = I S exp ⎜ −21
⎟ ⇒ I S = 5.07 × 10 A
⎝ 0.026 ⎠
⎛ 1.0 ⎞
I D = ( 5.07 × 10−21 ) exp ⎜ ⎟
(b) ⎝ 0.026 ⎠
I D = 2.56 × 10−4 A = 0.256 mA
1.34
⎛ 1.0 ⎞
(a) I D = 10−23 exp ⎜ −7
⎟ = 5.05 × 10 A
⎝ 0.026 ⎠
⎛ 1.1 ⎞
(b) I D = 10−23 exp ⎜ −5
⎟ = 2.37 × 10 A
⎝ 0.026 ⎠
⎛ 1.2 ⎞
(c) I D = 10−23 exp ⎜ −3
⎟ = 1.11× 10 A
⎝ 0.026 ⎠
1.35
IS doubles for every 5C increase in temperature.
I S = 10 −12 A at T = 300K
For I S = 0.5 × 10 −12 A ⇒ T = 295 K
For I S = 50 × 10 −12 A, (2) n = 50 ⇒ n = 5.64
Where n equals number of 5C increases.
Then ΔT = ( 5.64 )( 5 ) = 28.2 K
So 295 ≤ T ≤ 328.2 K

11. 1.36
I S (T )
= 2ΔT / 5 , ΔT = 155° C
I S (−55)
I S (100)
= 2155 / 5 = 2.147 × 109
I S (−55)
VT @100°C ⇒ 373°K ⇒ VT = 0.03220
VT @− 55°C ⇒ 216°K ⇒ VT = 0.01865
⎛ 0.6 ⎞
exp ⎜ ⎟
I D (100) ⎝ 0.0322 ⎠
= (2.147 × 109 ) ×
I D (−55) ⎛ 0.6 ⎞
exp ⎜ ⎟
⎝ 0.01865 ⎠
=
( 2.147 ×10 )(1.237 ×10 )
9 8
( 9.374 ×10 ) 13
I D (100)
= 2.83 × 103
I D (−55)
1.37
3.5 = ID (105) + VD
⎛ V ⎞ ⎛ ID ⎞
(a) I D = 5 ×10−9 exp ⎜ D ⎟ ⇒ VD = 0.026 ln ⎜ −9 ⎟
⎝ 0.026 ⎠ ⎝ 5 × 10 ⎠
Trial and error.
VD ID VD
0.50 3 ×10 −5 0.226
0.40 3.1×10−5 0.227
0.250 3.25 ×10 −5 0.228
0.229 3.271×10 −5 0.2284
0.2285 3.2715 ×10 −5 0.2284
So VD ≅ 0.2285 V
I D ≅ 3.272 × 10−5 A
(b) I D = I S = 5 × 10−9 A
VR = ( 5 × 10−9 )(105 ) = 5 × 10−4 V
VD = 3.4995 V
1.38
⎛ I ⎞
10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D12 ⎟
−
⎝ 10 ⎠
Trial and error.
VD(v) ID(A) VD(v)
0.50 4.75 ×10−4 0.5194
0.517 4.7415 ×10 −4 0.5194
0.5194 4.740 ×10−4 0.5194
VD = 0.5194 V
I D = 0.4740 mA
1.39

12. I s = 5 × 10 −13 A
R1 ϭ 50 K
ϩ ϩ
1.2 V R2 ϭ 30 K VD
Ϫ ID Ϫ
RTH ϭ R1 ͉͉ R2 ϭ 18.75 K
ϩ ϩ
VTH VD
Ϫ Ϫ
ID
⎛ R2 ⎞ ⎛ 30 ⎞
VTH = ⎜ ⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V
⎝ R1 + R2 ⎠ ⎝ 80 ⎠
⎛I ⎞
0.45 = I D RTH + VD , VD = VT ln ⎜ D ⎟
⎝ IS ⎠
By trial and error:
I D = 2.56 μ A, VD = 0.402 V
1.40
ϩ VDϪ ϩ VDϪ
ϩ ϩ
I1
VI 1K V0
Ϫ IR I2 Ϫ
I S = 2 × 10 −13 A
V0 = 0.60 V
⎛V ⎞ ⎛ 0.60 ⎞
I 2 = I S exp ⎜ 0 ⎟ = ( 2 × 10−13 ) exp ⎜ ⎟
⎝ VT ⎠ ⎝ 0.026 ⎠
= 2.105 mA
0.6
IR = = 0.60 mA
1K
I1 = I 2 + I R = 2.705 mA
⎛I ⎞ ⎛ 2.705 × 10−3 ⎞
VD = VT ln ⎜ 1 ⎟ = (0.026) ln ⎜ −13 ⎟
⎝ IS ⎠ ⎝ 2 × 10 ⎠
= 0.6065
VI = 2VD + V0 ⇒ VI = 1.81 V
1.41
(a) Assume diode is conducting.
Then, VD = Vγ = 0.7 V
0. 7
So that I R 2 = ⇒ 23.3 μ A
30

13. 1.2 − 0.7
I R1 = ⇒ 50 μ A
10
Then I D = I R1 − I R 2 = 50 − 23.3
Or I D = 26.7 μ A
(b) Let R1 = 50 k Ω Diode is cutoff.
30
VD = ⋅ (1.2) = 0.45 V
30 + 50
Since VD < Vγ , I D = 0
1.42
ϩ5 V
3 k⍀ 2 k⍀
ID
VB ϭVAϪVr VA
2 k⍀ 2 k⍀
A&VA:
5 − VA V
(1) = ID + A
2 2
A& VA − Vr
5 − (VA − Vr ) (VA − Vr )
(2) + ID =
2 2
5 − (VA − Vr ) ⎡ 5 − VA VA ⎤ VA − Vr
So +⎢ − ⎥=
3 ⎣ 2 2⎦ 2
Multiply by 6:
10 − 2 (VA − Vr ) + 15 − 6VA = 3 (VA − Vr )
25 + 2Vr + 3Vr = 11VA
(a) Vr = 0.6 V
11VA = 25 + 5 ( 0.6 ) = 28 ⇒ VA = 2.545 V
5 − VA VA
From (1) I D = − = 2.5 − VA ⇒ I D Neg. ⇒ I D = 0
2 2
Both (a), (b) I D = 0
2
VA = 2.5, VB = ⋅ 5 = 2 V ⇒ VD = 0.50 V
5
1.43
Minimum diode current for VPS (min)
I D (min) = 2 mA, VD = 0.7 V
0.7 5 − 0.7 4.3
I2 = , I1 = =
R2 R1 R1
We have I1 = I 2 + I D

14. 4.3 0.7
so (1) = +2
R1 R2
Maximum diode current for VPS (max)
P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA
I1 = I 2 + I D
or
9.3 0.7
(2) = + 14.3
R1 R2
9.3 4.3
Using Eq. (1), = − 2 + 14.3 ⇒ R1 = 0.41 kΩ
R1 R1
Then R2 = 82.5Ω 82.5Ω
1.44
(a) Vo = 0.7 V
5 − 0.7
I= ⇒ I = 0.215 mA
20
10 − 0.7
(b) I= ⇒ I = 0.2325 mA
20 + 20
Vo = I (20 K) − 5 ⇒ Vo = −0.35 V
10 − 0.7
(c) I= ⇒ I = 0.372 mA
5 + 20
Vo = 0.7 + I (20) − 8 ⇒ Vo = +0.14 V
(d) I =0
Vo = I (20) − 5 ⇒ Vo = −5 V
1.45
⎛ ⎞
5 = I ( 2 × 109 ) + VD
I
(a) VD = ( 0.026 ) ln ⎜ −12 ⎟
⎝ 2 ×10 ⎠
VD → ID → VD Vo = VD = 0.482 V
0.6 2.2 ×10−4 0.481
0.482 2.259 ×10−4 0.482 I = 0.226 mA
⎛ ⎞
10 = I ( 4 × 10 4 ) + VD
I
(b) VD = ( 0.026 ) ln ⎜ −12 ⎟
⎝ 2 ×10 ⎠
Vo → I → VD VD = 0.483 V
0.5 2.375 ×10−4 0.4834 I = 0.238 mA
0.484 2.379 ×10−4 0.4834 Vo = −0.24 V
⎛ ⎞
10 = I ( 2.5 × 10 4 ) + VD
I
(c) VD = ( 0.026 ) ln ⎜ ⎟
⎝ 2 ×10−12 ⎠
Vo → I → VD VD = 0.496 V
0.480 3.808 ×10 −4 0.496 I = 0.380 mA
0.496 3.802 ×10 −4 0.496 Vo = −0.10 V
(d) I = − I S ⇒ I = 2 × 10−12 A
Vo ≅ −5 V
1.46
(a) Diode forward biased VD = 0.7 V

15. 5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V
(b) P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω
1.47
0.65
(a) I R 2 = I D1 = = 0.65 mA = I D1
1
ID2 = 2(0.65) = 1.30 mA
VI − 2Vr − V0 5 − 3(0.65)
ID2 = = = 1.30 ⇒ R1 = 2.35 K
R1 R1
0.65
(b) IR2 = = 0.65 mA
1
8 − 3(0.65)
ID2 = ⇒ I D 2 = 3.025 mA
2
I D1 = I D 2 − I R 2 = 3.025 − 0.65
I D1 = 2.375 mA
1.48
VT (0.026)
a. τd = = = 0.026 kΩ = 26Ω
I DQ 1
id = 0.05 I DQ = 50 μ A peak-to-peak
vd = idτ d = (26)(50) μ A ⇒ vd = 1.30 mV peak-to-peak
(0.026)
b. For I DQ = 0.1 mA ⇒ τ d = = 260Ω
0. 1
id = 0.05 I DQ = 5 μ A peak-to-peak
vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak
1.49
RS
ϩ
␯S ϩ ␯d
Ϫ Ϫ
a. diode resistance rd = VT /I
⎛ ⎞
⎛ rd ⎞ ⎜ VT /I ⎟
vd = ⎜ ⎟ vS = ⎜ V ⎟ vS
⎝ rd + RS ⎠ ⎜ T + RS
⎜ ⎟
⎟
⎝ I ⎠
⎛ VT ⎞
vd = ⎜ ⎟ vs = vo
⎝ VT + IRS ⎠
b. RS = 260Ω

16. v0 ⎛ VT ⎞ 0.026 v
I = 1 mA, =⎜ ⎟= ⇒ 0 = 0.0909
vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26) vS
v 0.026 v
I = 0.1 mA, 0 = ⇒ 0 = 0.50
vs 0.026 + ( 0.1)( 0.26 ) vS
v0 0.026 v
I = 0.01 mA. = ⇒ 0 = 0.909
vS 0.026 + (0.01)(0.26) vS
1.50
⎛V ⎞ ⎛ I ⎞
I ≅ I S exp ⎜ a ⎟ , Va = VT ln ⎜ ⎟
⎝ VT ⎠ ⎝ IS ⎠
⎛ 100 × 10−6 ⎞
pn junction, Va = (0.026) ln ⎜ −14 ⎟
⎝ 10 ⎠
Va = 0.599 V
⎛ 100 × 10−6 ⎞
Schottky diode, Va = (0.026) ln ⎜ −9 ⎟
⎝ 10 ⎠
Va = 0.299 V
1.51
Schottky
pn junction
I
⎛V ⎞
Schottky: I ≅ I S exp ⎜ a ⎟
⎝ VT ⎠
⎛ I ⎞ ⎛ 0.5 × 10−3 ⎞
Va = VT ln ⎜ ⎟ = (0.026) ln ⎜ −7 ⎟
⎝ IS ⎠ ⎝ 5 × 10 ⎠
= 0.1796 V
Then
Va of pn junction = 0.1796 + 0.30
= 0.4796
I 0.5 × 10−3
IS = =
⎛V ⎞ ⎛ 0.4796 ⎞
exp ⎜ a ⎟ exp ⎜ ⎟
⎝ VT ⎠ ⎝ 0.026 ⎠
I S = 4.87 × 10 −12 A
1.52
(a)
ϩ VD Ϫ
I1
0.5 mA
I2
I1 + I 2 = 0.5 × 10 −3

17. ⎛V ⎞ ⎛ VD ⎞
5 × 10−8 exp ⎜ D −12
⎟ + 10 exp ⎜ ⎟ = 0.5 × 10
−3
⎝ VT ⎠ ⎝ VT ⎠
⎛V ⎞
5.0001× 10 −8 exp ⎜ D ⎟ = 0.5 × 10 −3
⎝ VT ⎠
⎛ 0.5 × 10−3 ⎞
VD = (0.026) ln ⎜ −8 ⎟
⇒ VD = 0.2395
⎝ 5.0001× 10 ⎠
Schottky diode, I 2 = 0.49999 mA
pn junction, I1 = 0.00001 mA
(b)
ϩ VD1 Ϫ ϩ VD2 Ϫ
I
ϩ 0.90 V Ϫ
⎛V ⎞ ⎛V ⎞
I = 10 −12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ D 2 ⎟
⎝ VT ⎠ ⎝ VT ⎠
VD1 + VD 2 = 0.9
⎛V ⎞ ⎛ 0.9 − VD1 ⎞
10−12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ ⎟
⎝ VT ⎠ ⎝ VT ⎠
⎛ 0.9 ⎞ ⎛ −VD1 ⎞
= 5 ×10−8 exp ⎜ ⎟ exp ⎜ ⎟
⎝ VT ⎠ ⎝ VT ⎠
⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞ ⎛ 0.9 ⎞
exp ⎜ D1 ⎟ = ⎜ −12 ⎟
exp ⎜ ⎟
⎝ VT ⎠ ⎝ 10 ⎠ ⎝ 0.026 ⎠
⎛ 5 × 10−8 ⎞
2VD1 = VT ln ⎜ −12 ⎟
+ 0.9 = 1.1813
⎝ 10 ⎠
VD1 = 0.5907 pn junction
VD 2 = 0.3093 Schottky diode
⎛ 0.5907 ⎞
I = 10−12 exp ⎜ ⎟ ⇒ I = 7.35 mA
⎝ 0.026 ⎠
1.53
R ϭ 0.5 K
V0
ϩ I
ϩ
VPS ϭ 10 V VZ RL
Ϫ Ϫ
IL
IZ
VZ = VZ 0 = 5.6 V at I Z = 0.1 mA
rZ = 10Ω
I Z rZ = ( 0.1)(10 ) = 1 mV
VZ0 = 5.599
a. RL → ∞ ⇒
10 − 5.599 4.401
IZ = = = 8.63 mA
R + rZ 0.50 + 0.01
VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 )
VZ = V0 = 5.685 V

18. 11 − 5.599
b. VPS = 11 V ⇒ I Z = = 10.59 mA
0.51
VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V
9 − 5.599
VPS = 9 V ⇒ I Z = = 6.669 mA
0.51
VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V
ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V
c. I = IZ + IL
V0 V − V0 V − VZ 0
IL = , I = PS , IZ = 0
RL R rZ
10 − V0 V0 − 5.599 V0
= +
0.50 0.010 2
10 5.599 ⎡ 1 1 1⎤
+ = V0 ⎢ + + ⎥
0.50 0.010 ⎣ 0.50 0.010 2 ⎦
20.0 + 559.9 = V0 (102.5)
V0 = 5.658 V
1.54
9 − 6.8
a. IZ = ⇒ I Z = 11 mA
0.2
PZ = (11)( 6.8 ) ⇒ PZ = 74.8 mW
12 − 6.8
IZ = ⇒ I Z = 26 mA
0.2
b.
26 − 11
%= × 100 ⇒ 136%
11
PZ = ( 26 )( 6.8 ) = 176.8 mW
176.8 − 74.8
%= × 100 ⇒ 136%
74.8
1.55
I Z rZ = ( 0.1)( 20 ) = 2 mV
VZ 0 = 6.8 − 0.002 = 6.798 V
a. RL = ∞
10 − 6.798
IZ = ⇒ I Z = 6.158 mA
0.5 + 0.02
V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 )
V0 = 6.921 V
b. I = IZ + IL
10 − V0 V0 − 6.798 V0
= +
0.50 0.020 1
10 6.798 ⎡ 1 1 1⎤
+ = V0 ⎢ + +
0.30 0.020 ⎣ 0.50 0.020 1⎥⎦
359.9 = V0 (53)
V0 = 6.791 V
ΔV0 = 6.791 − 6.921
ΔV0 = −0.13 V
1.56

19. For VD = 0, I SC = 0.1 A
⎛ 0.2 ⎞
For ID = 0 VD = VT ln ⎜ −14
+ 1⎟
⎝ 5 × 10 ⎠
VD = VDC = 0.754 V