Microelectronics, Circuit Analysis and Design by Donald A. Neamen, 4th edition-solutions

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

Chapter 1
1.1

ni = BT 3 / 2 e
(a) Silicon

− Eg / 2 kT

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
= 2.067 × 1019 exp [ −25.58]
ni = 1.61× 108 cm −3

(i)

ni = ( 5.23 × 1015 ) ( 250 )

(ii)

ni = ( 5.23 × 1015 ) ( 350 )

(b)

GaAs

(i)

ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= 3.425 × 1019 exp [ −18.27 ]
ni = 3.97 ×1011 cm −3
3/ 2

3/ 2

⎡
⎤
−1.4
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

= ( 8.301× 1017 ) exp [ −32.56]
ni = 6.02 × 103 cm −3

(ii)

ni = ( 2.10 × 1014 ) ( 350 )

3/ 2

⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦

= (1.375 × 1018 ) exp [ −23.26]
ni = 1.09 × 108 cm −3

______________________________________________________________________________________
1.2

a.

⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
⎛
⎞

−1.1
1012 = 5.23 × 1015 T 3 / 2 exp ⎜
⎟
−6
⎝ 2(86 × 10 )(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

______________________________________________________________________________________


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.3

Silicon
(a)

ni = ( 5.23 × 1015 ) (100 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦

= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3

(b)

ni = ( 5.23 × 1015 ) ( 300 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3

(c)

ni = ( 5.23 × 1015 ) ( 500 )

3/ 2

⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦

= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3

Germanium.
(a)

ni = (1.66 × 1015 ) (100 )

3/ 2

⎡
⎤
−0.66
⎥ = (1.66 × 1018 ) exp [ −38.37 ]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦

ni = 35.9 cm −3
(b)

ni = (1.66 × 1015 ) ( 300 )

3/ 2

⎡
⎤
−0.66
⎥ = ( 8.626 × 1018 ) exp [ −12.79]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

ni = 2.40 × 1013 cm −3
(c)

ni = (1.66 × 1015 ) ( 500 )

3/ 2

⎡
⎤
−0.66
⎥ = (1.856 × 1019 ) exp [ −7.674]
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦

ni = 8.62 ×1015 cm −3
______________________________________________________________________________________
1.4

(a) n-type; no = 10

15

(

)

(

)

n2
2.4 × 1013
cm ; po = i =
no
1015
−3

2

2

= 5.76 × 1011 cm −3

ni2
1.5 × 1010
=
= 2.25 × 10 5 cm −3
no
1015
______________________________________________________________________________________

(b) n-type; no = 1015 cm −3 ; po =


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

1.5

(a) p-type; p o = 1016 cm −3 ; no =

(

ni2
1.8 × 10 6
=
po
1016

)

(

2

= 3.24 × 10 − 4 cm −3

)

2

ni2
2.4 × 1013
=
= 5.76 × 1010 cm −3
po
1016

______________________________________________________________________________________

(b) p-type; p o = 1016 cm −3 ; no =

1.6

(a)
(b)

n-type
no = N d = 5 × 1016 cm −3
10
ni2 (1.5 × 10 )
po =
=
= 4.5 × 103 cm −3
no
5 × 1016
2

(c)

no = N d = 5 × 1016 cm −3

From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3

( 3.97 × 10 )
=

11 2

= 3.15 × 106 cm −3
5 × 1016
______________________________________________________________________________________
po

1.7

(a) p-type; p o = 5 × 1016 cm −3 ; no =

(

)

(

)

ni2
1.5 × 1010
=
po
5 × 1016

2

= 4.5 × 10 3 cm −3

2

ni2
1.8 × 10 6
=
= 6.48 × 10 −5 cm −3
po
5 × 1016
______________________________________________________________________________________

(b) p-type; p o = 5 × 1016 cm −3 ; no =

1.8
(a) Add boron atoms
(b) N a = po = 2 × 1017 cm −3

(

)

2

ni2
1.5 × 1010
=
= 1.125 × 10 3 cm −3
po
2 × 1017
______________________________________________________________________________________

(c) no =

1.9

(a)

no = 5 × 1015 cm −3
10
n 2 (1.5 × 10 )
po = i =
⇒ po = 4.5 × 104 cm −3
no
5 × 1015
2

(b)

n o > p o ⇒ n-type

(c) no ≅ N d = 5 × 1015 cm −3
______________________________________________________________________________________


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.10
a.

b.

Add Donors
N d = 7 × 1015 cm −3
Want po = 106 cm −3 = ni2 / N d

So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21
⎛ − Eg ⎞
= B 2T 3 exp ⎜
⎟
⎝ kT ⎠

⎛
⎞
2
−1.1
⎟
7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜
6
−
⎜ ( 86 × 10 ) (T ) ⎟
⎝
⎠
By trial and error, T ≈ 324° K
______________________________________________________________________________________

1.11
(a) I = Aσ Ε = 10 −5 (1.5)(10) ⇒ I = 0.15 mA

( )

(

)

Iρ 1.2 × 10 −3 (0.4)
= 2.4 V/cm
=
A
ρ
2 × 10 − 4
______________________________________________________________________________________

(b) I =

AΕ

⇒Ε=

(

)

1.12

J 120
−1
=
= 6.67 (Ω − cm)
Ε 18
σ

(6.67)
σ ≅ eμ n N d ⇒ N d =
=
= 3.33 × 1016 cm −3
eμ n
1.6 × 10 −19 (1250)
______________________________________________________________________________________
J =σΕ ⇒σ =

(

)

1.13

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

(

)

1.14

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

(

(

)

)

1.15
(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
______________________________________________________________________________________


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.16
Dn = (0.026)(1250) = 32.5 cm 2 /s; D p = (0.026 )(450 ) = 11.7 cm 2 /s
J n = eDn

⎛ 1016 − 1012
dn
= 1.6 × 10 −19 (32.5)⎜⎜
dx
⎝ 0 − 0.001

(

J p = −eD p

)

⎞
⎟⎟ = −52 A/cm 2
⎠

⎛ 1012 − 1016
dp
= − 1.6 × 10 −19 (11.7 )⎜⎜
dx
⎝ 0 − 0.001

(

)

⎞
⎟⎟ = −18.72 A/cm 2
⎠

Total diffusion current density
J = −52 − 18.72 = −70.7 A/cm 2
______________________________________________________________________________________
1.17
J p = −eD p

dp
dx

⎛ −1 ⎞

⎛ −x ⎞
= −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟
⎜ Lp ⎟
⎜ Lp ⎟
⎝ ⎠
⎝ ⎠

(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞
−19

Jp =

15

⎜⎜ ⎟⎟
⎝ Lp ⎠

10 × 10 −4

J p = 2.4 e

− x / Lp

J p = 2.4 A/cm2

(a)

x=0

(b)

x = 10 μ m

J p = 2.4 e−1 = 0.883 A/cm 2

x = 30 μ m
J p = 2.4 e−3 = 0.119 A/cm 2
______________________________________________________________________________________
(c)

1.18
a.

N a = 1017 cm −3 ⇒ po = 1017 cm −3
n 2 (1.8 × 10
no = i =
1017
po

b.

)

6 2

⇒ no = 3.24 × 10−5 cm −3

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

______________________________________________________________________________________
⎛N N
1.19 Vbi = VT ln⎜⎜ a 2 d
⎝ ni

(a) (i)

⎞
⎟
⎟
⎠

(

)(

⎡ 5 × 1015 5 × 1015
Vbi = (0.026 ) ln ⎢
2
⎢⎣ 1.5 × 1010

(

(

)

)⎤⎥ = 0.661 V
⎥⎦

)( )
(
)
⎡ (10 )(10 ) ⎤
= (0.026 ) ln ⎢
⎥ = 0.937 V
⎣⎢ (1.5 × 10 ) ⎦⎥

(ii)

⎡ 5 × 1017 1015 ⎤
Vbi = (0.026 ) ln ⎢
⎥ = 0.739 V
10 2
⎥⎦
⎣⎢ 1.5 × 10

(iii)

Vbi

18

18

10 2


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1

By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

(

)(

⎡ 5 × 1015 5 × 1015
Vbi = (0.026 ) ln ⎢
2
⎢⎣
1.8 × 10 6

(b) (i)

(

(

)

)⎤⎥ = 1.13 V
⎥⎦

)( )⎤⎥ = 1.21 V
(
) ⎥⎦
⎡ (10 )(10 ) ⎤
= (0.026 ) ln ⎢

⎥ = 1.41 V
⎢⎣ (1.8 × 10 ) ⎥⎦

(ii)

⎡ 5 × 1017 1015
Vbi = (0.026 ) ln ⎢
2
⎢⎣ 1.8 × 10 6

(iii)

Vbi

18

18

6 2

______________________________________________________________________________________
1.20
⎛N N
Vbi = VT ln⎜⎜ a 2 d
⎝ ni

or

⎞
⎟

⎟
⎠

(n ) exp⎛⎜ V

(1.5 × 10) exp⎛ 0.712 ⎞ = 1.76 × 1016 cm −3
bi ⎞
⎜
⎟
⎜ V ⎟⎟ = 1016
Nd
⎝ 0.026 ⎠
⎝ T ⎠
______________________________________________________________________________________
Na =

1.21

2
i

2

⎡ N a (1016 ) ⎤
⎛N N ⎞
⎥
Vbi = VT ln ⎜ a 2 d ⎟ = ( 0.026 ) ln ⎢
10 2
⎢⎣ (1.5 × 10 ) ⎥⎦
⎝ ni ⎠

For N a = 1015 cm −3 , Vbi = 0.637 V
For N a = 1018 cm −3 , Vbi = 0.817 V

______________________________________________________________________________________
1.22

⎛ T ⎞
kT = (0.026) ⎜
⎟
⎝ 300 ⎠
200
250
300
350
400
450
500

kT
0.01733
0.02167
0.026
0.03033
0.03467
0.0390
0.04333

(T)3/2
2828.4

3952.8
5196.2
6547.9
8000.0
9545.9
11,180.3


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
⎛
⎞
−1.4
⎟
ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜
⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝
⎠
⎛N N ⎞
Vbi = VT ln ⎜ a 2 d ⎟
⎝ ni ⎠
T
ni
200
1.256
250
6.02 × 103

300
1.80 × 106
350
1.09 × 108
400
2.44 × 109
450
2.80 × 1010
500
2.00 × 1011

Vbi
1.405
1.389
1.370
1.349
1.327
1.302
1.277

______________________________________________________________________________________
1.23

⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠

−1/ 2

⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤

⎥ = 0.684 V
Vbi = ( 0.026 ) ln ⎢
⎢⎣
(1.5 ×10 10 ) 2 ⎥⎦
1 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
0.684
⎝
⎠

−1/ 2

(a)

3 ⎞
⎛
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠

−1/ 2

(b)

= 0.255 pF

= 0.172 pF
−1/ 2

5 ⎞
⎛
= 0.139 pF
C j = ( 0.4 ) ⎜ 1 +
⎟
⎝ 0.684 ⎠
______________________________________________________________________________________

(c)

1.24

(a)

⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠

−1 / 2

5 ⎞
⎛
For VR = 5 V, C j = (0.02) ⎜ 1 +
⎟
⎝ 0. 8 ⎠

−1 / 2

⎛ 1. 5 ⎞

For VR = 1.5 V, C j = (0.02) ⎜1 +
⎟
⎝ 0. 8 ⎠

= 0.00743 pF
−1 / 2

= 0.0118 pF


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.00743 + 0.0118
= 0.00962 pF
2
vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

C j (avg ) =

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

so that t2 = 8.09 × 10

−10

)

s

______________________________________________________________________________________
1.25

⎛ V
C j = C jo ⎜⎜1 + R
⎝ Vbi

⎞
⎟⎟
⎠

−1 / 2

(

)( )
)

⎡ 5 × 1015 1017 ⎤
; Vbi = (0.026 ) ln ⎢
⎥ = 0.739 V
10 2
⎥⎦
⎣⎢ 1.5 × 10

(

For V R = 1 V,
Cj =

0.60
1
1+
0.739

= 0.391 pF

For VR = 3 V,
Cj =

0.60
3
1+
0.739

= 0.267 pF

For V R = 5 V,
0.60

Cj =

(a)

fo =

(b) f o =

1
2π LC

(

5
1+
0.739

=

(

= 0.215 pF

2π 1.5 × 10

2π 1.5 × 10

1
−3

)(0.391× 10 )
−12

1
−3

)(0.267 × 10 )
−12

⇒ f o = 6.57 MHz

⇒ f o = 7.95 MHz


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

fo =

1

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

(

−3

)(

)

1.26

a.

⎡
⎛V ⎞ ⎤
⎛V ⎞
I = I S ⎢ exp ⎜ D ⎟ − 1⎥ − 0.90 = exp ⎜ D ⎟ − 1
⎝ VT ⎠ ⎦
⎝ VT ⎠
⎣
⎛V ⎞
exp ⎜ D ⎟ = 1 − 0.90 = 0.10
⎝ VT ⎠

VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V

b.
IF
IR

⎡
⎛ VF
⎢ exp ⎜
I
⎝ VT
= S ⋅⎣
IS ⎡

⎛ VR
⎢exp ⎜
⎝ VT
⎣
=

⎞ ⎤
⎛ 0.2 ⎞
⎟ − 1⎥ exp ⎜
⎟ −1
0.026 ⎠
⎠ ⎦
⎝
=
⎞ ⎤ exp ⎛ −0.2 ⎞ − 1
⎜
⎟
⎟ − 1⎥
⎝ 0.026 ⎠
⎠ ⎦

2190
−1

IF
= 2190
IR

______________________________________________________________________________________

⎡ ⎛V
1.27 I D = I S ⎢exp⎜⎜ D
⎣⎢ ⎝ VT
(a) (i)
(ii)
(iii)
(iv)

⎞ ⎤
⎟⎟ − 1⎥
⎠ ⎦⎥

(

)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 1.03 μ A
⎝ 0.026 ⎠
⎛ 0.5 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 2.25 mA
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I D = 10 −11 exp⎜
⎟ ⇒ 4.93 A
⎝ 0.026 ⎠
⎡ ⎛ − 0.02 ⎞ ⎤
−12
A
I D = 10 −11 ⎢exp⎜
⎟ − 1⎥ = −5.37 × 10
⎣ ⎝ 0.026 ⎠ ⎦

(v)

⎡ ⎛ − 0.20 ⎞ ⎤
−11
I D = 10 −11 ⎢exp⎜
⎟ − 1⎥ ≅ −10 A
0
.
026

⎠ ⎦
⎣ ⎝

(vi)

I D = − 10 −11 A

(b) (i)
(ii)
(iii)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 0.0103 μ A

⎝ 0.026 ⎠
⎛ 0.5 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 22.5 μ A
⎝ 0.026 ⎠
⎛ 0.7 ⎞
I D = 10 −13 exp⎜
⎟ ⇒ 49.3 mA
⎝ 0.026 ⎠


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

(

)

(iv)

⎡ ⎛ − 0.02 ⎞ ⎤
−14
I D = 10 −13 ⎢exp⎜
A
⎟ − 1⎥ = −5.37 × 10
0
.

026
⎝
⎠
⎣
⎦

(v)

I D ≅ −10 −13 A

I D ≅ −10 −13 A
(vi)
______________________________________________________________________________________
⎛I
1.28 V D = VT ln⎜⎜ D
⎝ IS

⎞
⎟
⎟
⎠

⎛ 10 × 10 −6
(a) (i) V D = (0.026 ) ln⎜⎜
−11
⎝ 10

⎞
⎟⎟ = 0.359 V
⎠

⎛ 100 × 10 −6
V D = (0.026 ) ln⎜⎜
−11
⎝ 10

⎞
⎟⎟ = 0.419 V
⎠

⎛ 10 −3 ⎞
V D = (0.026) ln⎜⎜ −11 ⎟⎟ = 0.479 V
⎝ 10 ⎠
⎡ ⎛ V ⎞ ⎤
(ii) − 5 × 10 −12 = 10 −11 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.018 V
⎣ ⎝ 0.026 ⎠ ⎦
⎛ 10 × 10 −6
(b) (i) V D = (0.026 ) ln⎜⎜
−13
⎝ 10

⎞
⎟⎟ = 0.479 V
⎠

⎛ 100 × 10 −6
V D = (0.026 ) ln⎜⎜
−13
⎝ 10
⎛ 10 −3

V D = (0.026) ln⎜⎜ −13
⎝ 10

⎞
⎟⎟ = 0.539 V
⎠

⎞
⎟⎟ = 0.599 V
⎠

⎡ ⎛ V ⎞ ⎤
(ii) − 10 −14 = 10 −13 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.00274 V
⎣ ⎝ 0.026 ⎠ ⎦
______________________________________________________________________________________

1.29

(a)

(b)

VD

⎛ 0.7 ⎞
10−3 = I S exp ⎜
⎟
⎝ 0.026 ⎠
I S = 2.03 × 10 −15 A
I D ( A ) ( n = 1)

I D ( A )( n = 2 )

0.1
9.50 ×10
1.39 ×10 −14
0.2
4.45 ×10 −12
9.50 ×10 −14
0.3
2.08 ×10 −10
6.50 ×10 −13
−
9
0.4
9.75 ×10
4.45 ×10 −12
0.5
4.56 ×10 −7
3.04 ×10 −11
−
5
0.6
2.14 ×10
2.08 ×10 −10
0.7
10 −3
1.42 ×10 −9
______________________________________________________________________________________
−14

Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.30
(a)
I S = 10 −12 A

VD(v)
0.10
0.20
0.30
0.40
0.50
0.60
0.70
(b)

ID(A)
4.68 ×10−11
2.19 ×10−9
1.03 ×10−7
4.80 ×10−6
2.25 ×10−4
1.05 ×10−2
4.93 ×10−1

log10ID
−10.3
−8.66
−6.99
−5.32
−3.65
−1.98
−0.307

I S = 10 −14 A

VD(v)
ID(A)
log10ID
−13
−12.3
0.10
4.68 ×10
−
11
−10.66
0.20
2.19 ×10
−
9
−8.99
0.30
1.03 ×10
−8
−7.32

0.40
4.80 ×10
−
6
−5.65
0.50
2.25 ×10
−4
−3.98
0.60
1.05 ×10
−
3
−2.31
0.70
4.93 ×10
______________________________________________________________________________________
1.31
a.
⎛ V − VD1 ⎞
ID2
= 10 = exp ⎜ D 2
⎟
I D1
⎝ VT
⎠
ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV

b.

ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV

______________________________________________________________________________________
1.32

⎛ 2 ⎞
(a) (i) V D = (0.026) ln⎜
⎟ = 0.539 V
−9
⎝ 2 × 10 ⎠
⎛ 20 ⎞
(ii) V D = (0.026) ln⎜
⎟ = 0.599 V
−9
⎝ 2 × 10 ⎠
⎛ 0.4 ⎞
(b) (i) I D = 2 × 10 −9 exp⎜
⎟ ⇒ 9.60 mA
⎝ 0.026 ⎠
⎛ 0.65 ⎞
(ii) I D = 2 × 10 −9 exp⎜
⎟ ⇒ 144 A
⎝ 0.026 ⎠
______________________________________________________________________________________

(

)

(

)


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.33
⎛I ⎞
⎛ 2 × 10−3 ⎞
= 0.6347 V
VD = Vt ln ⎜ D ⎟ = (0.026) ln ⎜
−14 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
⎛ 2 × 10−3 ⎞
= 0.5150 V
VD = (0.026) ln ⎜
−12 ⎟
⎝ 5 × 10 ⎠
0.5150 ≤ VD ≤ 0.6347 V

______________________________________________________________________________________
1.34

⎛ 0.30 ⎞
−8
(a) 1.5 × 10 −3 = I S exp⎜

⎟ ⇒ I S = 1.46 × 10 A
⎝ 0.026 ⎠
⎛ 0.35 ⎞
(b) (i) I D = 1.462 × 10 −8 exp⎜
⎟ ⇒ I D = 10.3 mA
⎝ 0.026 ⎠
⎛ 0.25 ⎞
(ii) I D = 1.462 × 10 −8 exp⎜
⎟ ⇒ I D = 0.219 mA
⎝ 0.026 ⎠
______________________________________________________________________________________

(

)

(

)

1.35

(

)

(

)

(

)

(

)

⎛ 0.8 ⎞
(a) I D = 10 − 22 exp⎜
⎟ ⇒ 2.31 nA
⎝ 0.026 ⎠
⎛ 1.0 ⎞
I D = 10 − 22 exp⎜
⎟ ⇒ 5.05 μ A
⎝ 0.026 ⎠
⎛ 1.2 ⎞
I D = 10 − 22 exp⎜
⎟ ⇒ 11.1 mA
⎝ 0.026 ⎠
⎡ ⎛ − 0.02 ⎞ ⎤
− 23
I D = 10 − 22 ⎢exp⎜
⎟ − 1⎥ = −5.37 × 10 A
0
.
026
⎠ ⎦
⎣ ⎝

For V D = −0.20 V, I D = −10 −22 A
For V D = −2 V, I D = −10 −22 A
(b)
⎛ 0.8 ⎞
I D = 5 × 10 −24 exp⎜
⎟ ⇒ 115 pA
⎝ 0.026 ⎠
⎛ 1.0 ⎞
I D = 5 × 10 − 24 exp⎜
⎟ ⇒ 0.253 μ A
⎝ 0.026 ⎠
⎛ 1.2 ⎞
I D = 5 × 10 − 24 exp⎜
⎟ ⇒ 0.554 mA
⎝ 0.026 ⎠

(

)

(

)

(

)

(

)

⎡ ⎛ − 0.02 ⎞ ⎤
− 24
A
I D = 5 × 10 − 24 ⎢exp⎜
⎟ − 1⎥ = −2.68 × 10
0
.
026
⎠ ⎦
⎣ ⎝

For V D = −0.20 V, I D = −5 × 10 −24 A
For V D = −2 V, I D = −5 × 10 −24 A
______________________________________________________________________________________


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

1.36

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
______________________________________________________________________________________
1.37
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

I D (100)
= (2.147 × 109 ) ×
I D (−55)

⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.0322 ⎠
⎛ 0.6 ⎞
exp ⎜
⎟
⎝ 0.01865 ⎠

( 2.147 ×10 )(1.237 ×10 )
( 9.374 ×10 )
9

=

8

13

I D (100)
= 2.83 × 103
I D (−55)
______________________________________________________________________________________
1.38
(a) V PS = I D R + V D

( )

2.8 = I D 10 6 + VD ;

(

)

⎛ V ⎞
I D = 5 × 10 −11 exp⎜ D ⎟
⎝ 0.026 ⎠

By trial and error,
V D = 0.282 V, I D = 2.52 μ A
(b)
I D ≅ −5 × 10 −11 A, VD = −2.8 V

______________________________________________________________________________________


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

1.39
⎛ I ⎞
10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D−12 ⎟
⎝ 10 ⎠
Trial and error.
VD(v)
ID(A)
VD(v)
−4
0.50
0.5194
4.75 ×10
−
4
0.517
0.5194
4.7415 ×10
−4
0.5194
0.5194
4.740 ×10

VD = 0.5194 V
I D = 0.4740 mA
______________________________________________________________________________________
1.40
I s = 5 × 10 −13 A

⎛ R2
VTH = ⎜
⎝ R1 + R2

⎞
⎛ 30 ⎞
⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V
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

______________________________________________________________________________________


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

Problem Solutions
______________________________________________________________________________________

1.41
(a) I D1 = I D 2 = 1 mA

(i)

⎛ 10 −3
V D1 = V D 2 = (0.026 ) ln⎜⎜ −13
⎝ 10

⎞
⎟⎟ = 0.599 V
⎠

(ii)

⎛ 10 −3
V D1 = (0.026 ) ln⎜⎜
−14
⎝ 5 × 10

⎞
⎟⎟ = 0.617 V
⎠

⎛ 10 −3
V D 2 = (0.026 ) ln⎜⎜
−13

⎝ 5 × 10

⎞
⎟⎟ = 0.557 V
⎠

(b) V D1 = V D 2

V D1 = V D 2

(ii)

Ii
= 0.5 mA
2
⎛ 0.5 × 10 −3
= (0.026 ) ln⎜⎜
−13
⎝ 10

I D1 = I D 2 =

(i)

⎞
⎟⎟ = 0.581 V
⎠

I
I D1

5 × 10 −14
= S1 =
= 0.10
I D 2 I S 2 5 × 10 −13
So I D1 = 0.10 I D 2

I D1 + I D 2 = 1.1I D 2 = 1 mA

So I D 2 = 0.909 mA, I D1 = 0.0909 mA
Now
⎛ 0.0909 × 10 −3 ⎞
⎟⎟ = 0.554 V
V D1 = (0.026 ) ln⎜⎜
−14
⎝ 5 × 10
⎠
⎛ 0.909 × 10 −3 ⎞
⎟⎟ = 0.554 V
V D 2 = (0.026 ) ln⎜⎜
−13
⎝ 5 × 10
⎠
______________________________________________________________________________________

1.42

(

)

⎛ 0.635 ⎞
(a) I D 3 = 6 × 10 −14 exp⎜
⎟ ⇒ 2.426 mA
⎝ 0.026 ⎠
0.635
= 0.635 mA
IR =
1
I D1 = I D 2 = 2.426 + 0.635 = 3.061 mA
⎛ 3.061 × 10 −3
V D1 = V D 2 = (0.026 ) ln⎜⎜
−14
⎝ 6 × 10
V I = 2(0.641) + 0.635 = 1.917 V

⎞
⎟⎟ = 0.641 V
⎠

(b) I D 3 = 2.426 mA
0.635
= 1.27 mA
IR =
0.5
I D1 = I D 2 = 2.426 + 1.27 = 3.696 mA
⎛ 3.696 × 10 −3
V D1 = V D 2 = (0.026 ) ln⎜⎜
−14
⎝ 6 × 10
V I = 2(0.6459 ) + 0.635 = 1.927 V

⎞
⎟⎟ = 0.6459 V
⎠


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
______________________________________________________________________________________
1.43
(a) Assume diode is conducting.
Then, VD = Vγ = 0.7 V
0. 7
⇒ 23.3 μ A
30
1.2 − 0.7
⇒ 50 μ A
I R1 =
10
Then I D = I R1 − I R 2 = 50 − 23.3

So that I R 2 =

Or I D = 26.7 μ A
(b) Let R1 = 50 k Ω Diode is cutoff.
30
⋅ (1.2) = 0.45 V

30 + 50
Since VD < Vγ , I D = 0
VD =

______________________________________________________________________________________
1.44

At node VA:
5 − VA
V
= ID + A
(1)
2
2
At node V B = V A − Vγ
(2)

5 − (VA − Vr )

+ ID =

(VA − Vr )

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


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5 − VA VA
−
= 2.5 − VA ⇒ I D Neg. ⇒ I D = 0
2
2
Both (a), (b) I D = 0

From (1) I D =

2

⋅ 5 = 2 V ⇒ VD = 0.50 V
5
______________________________________________________________________________________

VA = 2.5, VB =

1.45
(a) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 0.7 mA
VO = 0.7 V; I D = (I i − 0.7 ) mA; for I i ≥ 0.7 mA

(b) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 1.7 mA
VO = 1.7 V; I D = (I i − 1.7 ) mA; for I i ≥ 1.7 mA
(c) VO = 0.7 V; I D1 = I i ; I D 2 = 0 ; for 0 ≤ I i ≤ 2 mA
______________________________________________________________________________________
1.46
Minimum diode current for VPS (min)
I D (min) = 2 mA, VD = 0.7 V
I2 =

0.7
5 − 0.7 4.3
, I1 =
=
R2
R1
R1

We have I1 = I 2 + I D
4.3 0.7
=

+2
R1
R2
Maximum diode current for VPS (max)
P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA

so (1)

I1 = I 2 + I D

or
(2)

9.3 0.7
=
+ 14.3
R1
R2

Using Eq. (1),

9.3 4.3
=
− 2 + 14.3 ⇒
R1
R1

R1 = 0.41 kΩ

Then R2 = 82.5Ω 82.5Ω

______________________________________________________________________________________
1.47

5 − 0.7
= 0.215 mA, VO = 0.7 V
20
5 − 0.6
= 0.220 mA, VO = 0.6 V
(ii) I =
20
5 − 0.7 − (− 5)
= 0.2325 mA, VO = (0.2325)(20 ) − 5 = −0.35 V
(b) (i) I =
40
5 − 0.6 − (− 5)
I=
= 0.235 mA, VO = (0.235)(20 ) − 5 = −0.30 V
(iii)
40
(a) (i) I =


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

2 − 0.7 − (− 8)
= 0.372 mA, VO = 2 − (0.372 )(5) = 0.14 V

25
2 − 0.6 − (− 8)
= 0.376 mA, VO = 2 − (0.376 )(5) = 0.12 V
(ii) I =
25
(d) (i) I = 0 , VO = −5 V
(ii) I = 0 , VO = −5 V
______________________________________________________________________________________
(c) (i) I =

1.48

(a) I =

5 − VO
⎛ V ⎞
, I = 5 × 10 −14 exp⎜ D ⎟
20
⎝ 0.026 ⎠
By trial and error, V D = VO = 0.5775 V, I = 0.221 mA

(

)

10 − V D
, VO = 5 − I (20 ) − V D
40
I = 0.2355 mA, V D = 0.579 V, VO = −0.289

(b) I =

10 − V D
, VO = 2 − I (5)
25
I = 0.3763 mA, V D = 0.5913 V, VO = 0.1185

(c) I =

(d) I = −5 × 10 −14 A, VO ≅ −5 V
______________________________________________________________________________________
1.49
(a)

Diode forward biased VD = 0.7 V
5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V
(b) P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω
______________________________________________________________________________________
1.50

(a)

0.65
= 0.65 mA = I D1
1
= 2(0.65) = 1.30 mA

I R 2 = I D1 =
ID2

ID2 =

(b)

VI − 2Vr − V0 5 − 3(0.65)
=
= 1.30 ⇒ R1 = 2.35 K
R1
R1

0.65
= 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
IR2 =

______________________________________________________________________________________
1.51


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

V
(0.026)
τd = T =
= 0.026 kΩ = 26Ω
a.
1
I DQ

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
b.

(0.026)
= 260Ω
0. 1
= 5 μ A peak-to-peak

For I DQ = 0.1 mA ⇒ τ d =
id = 0.05 I DQ

vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak
______________________________________________________________________________________
1.52

(a) rd =

VT
0.026
=
=1 kΩ

I DQ 0.026

0.026
⇒ 100 Ω
0.26
0.026
⇒ 10 Ω
(c) rd =
2.6
______________________________________________________________________________________
(b) rd =

1.53

a.

diode resistance rd = VT /I
⎛
⎜ VT /I
⎛ rd ⎞
vd = ⎜
⎟ vS = ⎜ V
⎝ rd + RS ⎠
⎜⎜ T + RS
⎝ I
⎛ VT
⎞
vd = ⎜
⎟ vs = vo
⎝ VT + IRS ⎠

b.

⎞
⎟
⎟ vS
⎟⎟
⎠

RS = 260Ω

⎞
v0 ⎛ VT
v
0.026
=⎜
⇒ 0 = 0.0909
⎟=
vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26)
vS
v
v
0.026
I = 0.1 mA, 0 =
⇒ 0 = 0.50
vs 0.026 + ( 0.1)( 0.26 )
vS
I = 1 mA,

I = 0.01 mA.

v0
v
0.026
=
⇒ 0 = 0.909
vS 0.026 + (0.01)(0.26)
vS

______________________________________________________________________________________


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

1.54
pn junction diode
⎛ 0.72 × 10 −3 ⎞
⎟ = 0.548 V
V D = (0.026 ) ln⎜⎜
−13 ⎟
⎝ 5 × 10
⎠
Schottky diode
⎛ 0.72 × 10 −3 ⎞
⎟⎟ = 0.249 V
V D = (0.026 ) ln⎜⎜

−8
⎝ 5 × 10
⎠
______________________________________________________________________________________

1.55

⎛V ⎞
Schottky: I ≅ I S exp ⎜ a ⎟
⎝ VT ⎠

⎛ I ⎞
⎛ 0.5 × 10−3 ⎞
Va = VT ln ⎜ ⎟ = (0.026) ln ⎜
−7 ⎟
⎝ 5 × 10 ⎠
⎝ IS ⎠
= 0.1796 V
Then
Va of pn junction = 0.1796 + 0.30

= 0.4796
I
0.5 × 10−3
=
⎛V ⎞
⎛ 0.4796 ⎞
exp ⎜ a ⎟ exp ⎜
⎟
⎝ 0.026 ⎠

⎝ VT ⎠
I S = 4.87 × 10 −12 A

IS =

______________________________________________________________________________________
1.56

(a)


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I1 + I 2 = 0.5 × 10 −3

⎛V
5 × 10 −8 exp ⎜ D
⎝ VT

⎞
⎛ VD ⎞
−12
−3
⎟ + 10 exp ⎜ ⎟ = 0.5 × 10
V
⎠
⎝ T ⎠

⎛V
5.0001× 10 −8 exp ⎜ D
⎝ VT

⎞
−3
⎟ = 0.5 × 10
⎠

⎛ 0.5 × 10−3 ⎞
⇒ VD = 0.2395
VD = (0.026) ln ⎜
−8 ⎟
⎝ 5.0001 × 10 ⎠
Schottky diode, I 2 = 0.49999 mA

pn junction, I1 = 0.00001 mA
(b)

⎛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 ⎜
V
⎝ T ⎠
⎝ VT ⎠
⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞
⎛ 0. 9 ⎞
exp ⎜ D1 ⎟ = ⎜
exp ⎜
⎟
−12 ⎟
V
10
⎝ 0.026 ⎠
⎠
⎝ T ⎠ ⎝
⎛ 5 × 10−8 ⎞
+ 0.9 = 1.1813
2VD1 = VT ln ⎜
−12 ⎟
⎝ 10
⎠
VD1 = 0.5907 pn junction

VD 2 = 0.3093 Schottky diode
⎛ 0.5907 ⎞
I = 10−12 exp ⎜
⎟ ⇒ I = 7.35 mA
⎝ 0.026 ⎠
______________________________________________________________________________________
1.57

VZ = VZ 0 = 5.6 V at I Z = 0.1 mA
rZ = 10Ω


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I Z rZ = ( 0.1)(10 ) = 1 mV
VZ0 = 5.599

a.

RL → ∞ ⇒
IZ =

10 − 5.599
4.401
=
= 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
b.

11 − 5.599
= 10.59 mA
0.51
VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V
VPS = 11 V ⇒ I Z =

9 − 5.599
= 6.669 mA
0.51
VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V

VPS = 9 V ⇒ I Z =

ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V
c.

I = IZ + IL
V
V − V0
V − VZ 0
, IZ = 0
I L = 0 , I = PS
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

______________________________________________________________________________________
1.58

10 − 6.8
= 6.4 mA
0.5
P = I Z VZ = (6.4)(6.8) = 43.5 mW
(b) I Z = (0.1)(6.4) = 0.64 mA
I L = 6.4 − 0.64 = 5.76 mA
(a) I Z =

VZ
V
6.8
⇒ RL = Z =
= 1.18 k Ω
RL
I Z 5.76
______________________________________________________________________________________
IL =

1.59

I Z rZ = ( 0.1)( 20 ) = 2 mV
VZ 0 = 6.8 − 0.002 = 6.798 V

a.

RL = ∞
IZ =

10 − 6.798

⇒ I Z = 6.158 mA
0.5 + 0.02


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
______________________________________________________________________________________
1.60
For VD = 0, I SC = 0.1 A

⎛ 0.2
⎞
+ 1⎟
For ID = 0 VD = VT ln ⎜
−14
⎝ 5 × 10
⎠
VD = VDC = 0.754 V
______________________________________________________________________________________
1.61 V D = 0, I D = 0.2 A
V D = 0.60 V, I D = 0.1995 A
V D = 0.65 V, I D = 0.1964 A
V D = 0.70 V, I D = 0.1754 A
V D = 0.72 V, I D = 0.1468 A
V D = 0.74 V, I D = 0.0853 A
V D = 0.7545 V, I D = 0
______________________________________________________________________________________
1.62
⎡ ⎛ V ⎞ ⎤

(a) 0.16 = 0.20 − 5 × 10 −14 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = 0.7126 V
⎣ ⎝ 0.026 ⎠ ⎦
(b) P = (0.16 )(0.7126 ) = 0.114 W
______________________________________________________________________________________

(

)


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

Chapter 2
2.1
⎛ 1000 ⎞
(a) For υ I > 0.6 V, υ O = ⎜
⎟(υ I − 0.6 )
⎝ 1020 ⎠
For υ I < 0.6 V, υ O = 0

⎛ 1000 ⎞
(b) (ii) υ O = 0 = ⎜
⎟[10 sin (ω t )1 − 0.6]
⎝ 1020 ⎠
0. 6
Then sin (ω t )1 =

= 0.06 ⇒ (ω t )1 = 3.44° ⇒ 0.01911π rad
10
Also (ω t )2 = 180 − 3.44 = 176.56° ⇒ 0.9809π rad
Now

υ O (avg ) =
=

T

1
1
υ O (t )dt =
T 0
2π

∫

1
2π

2π

∫ [10 sin x − 0.6]dx
0

0.9809π ⎤
⎡
− 0.6 x
⎢− 10 cos x

⎥
0.01911π
0.01911π ⎦
⎣
0.9809π

1
[(− 10 )(− 0.9982 − 0.9982 ) − 0.6(0.9809π − 0.01911π )]
2π
υ O (avg ) = 2.89 V

=

⎤
⎛ 1000 ⎞ ⎡
⎛π ⎞
(iii) υ O ( peak ) = ⎜
⎟ ⎢10 sin ⎜ ⎟ − 0.6⎥ = 9.2157 V; i d (max ) = 9.2157 mA
1020
2
⎝
⎠⎣
⎝ ⎠
⎦
(iv) PIV = 10 V
______________________________________________________________________________________

2.2

v0 = vI − vD

⎛i ⎞
v
vD = VT ln ⎜ D ⎟ and iD = 0
I
R
⎝ S⎠
⎛ v ⎞
v0 = vI − VT ln ⎜ 0 ⎟
⎝ IS R ⎠

______________________________________________________________________________________


Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.3
⎛ 1⎞
(a) υ S = 120 2 ⎜ ⎟ = 16.97 V (peak)
⎝ 10 ⎠
υ O ( peak ) = 16.27 V

16.27
= 8.14 mA
2
(c) υ O = 16.97 sin ω t − 0.7

(b) i D ( peak ) =

sin (ω t )1 =

0. 7
= 0.04125 ⇒ (ω t )1 = 2.364°
16.97
(ω t )2 = 180 − 2.364 = 177.64°

⎛ 177.64 − 2.364 ⎞
%=⎜
⎟ × 100% = 48.7%
360
⎝
⎠

(d)

υ O (avg ) =

1
2π

0.9869π

∫ [16.97 sin x − 0.7]dx

0.01313π

0.9869π
0.9869π

⎤
1 ⎡
=
− 0. 7 x
(− 16.97 ) cos x
⎢
⎥
2π ⎣
0.01313π
0.01313π ⎦
1
[(− 16.97 )(− 0.99915 − 0.99915) − 0.7(0.9738π )]
=
2π
υ O (avg ) = 5.06 V

υ O (avg )

5.06
= 2.53 mA
2
2
______________________________________________________________________________________

(e) i D (avg ) =

=

2.4
(a) υ R (t ) = 15 sin ω t − 0.7 − 9 = 15 sin ω t − 9.7

(ω t )1 = sin −1 ⎛⎜ 9.7 ⎞⎟ = 40.29° ⇒ 0.2238π

(ω t )2

rad
⎝ 15 ⎠
= 180 − 40.29 = 139.71° ⇒ 0.7762π rad

υ R (avg ) =

1
2π

0.7762π

∫ [15 sin x − 9.7]dx

0.2238π

0.7762π
0.7762π
⎤
1 ⎡
1
(− 15) cos x
− 9.7 x
⎢
⎥ = 2π [(− 15)(− 0.7628 − 0.7628 ) − 9.7(0.5523π )]
2π ⎣

0.2238π
0.2238π ⎦
υ R (avg ) = 0.9628 V

=

i D (avg ) = 0.8 =

0.9628
⇒ R = 1.20 Ω
R

(b)
⎛ 139.71 − 40.29 ⎞
%=⎜
⎟ × 100% = 27.6%
360
⎝
⎠
______________________________________________________________________________________