Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
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Chapter 1
Exercise Solutions
EX1.1
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
GaAs: ni = ( 2.1× 1014 ) ( 300 )
⎛
⎞
−1.4
⎟ or ni = 1.8 × 106 cm −3
exp ⎜
−
6
⎜ 2 ( 86 × 10 ) ( 300 ) ⎟
⎝
⎠
3/ 2
⎛
⎞
− 0.66
⎟ or ni = 2.40 × 1013 cm −3
exp⎜⎜
−6
⎟
(
)
2
86
10
300
×
⎝
⎠
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(
)
Ge: n i = 1.66 × 1015 (300)
3/ 2
(
)
EX1.2
(a) (i)
n o = N d = 2×1016 cm
(
) = 1.125 ×10 cm
2
n i2
1.5 × 1010
=
no
2 × 1016
po =
(ii) p o = N a = 1015 cm
4
(
(ii)
) = 2.25 ×10 cm
2
no = N d = 2 × 1016 cm
(
ni2
1.8 × 10 6
=
no
2 × 1016
po =
p o = N a = 1015 cm
(
−3
−3
n2
1.5 × 10 10
no = i =
po
1015
(b) (i)
−3
5
−3
−4
cm −3
−3
) = 1.62 × 10
2
−3
)
2
n i2
1.8 × 10 6
=
= 3.24 × 10 − 3 cm −3
po
10 15
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no =
EX1.3
(a) For n-type;
ρ=
(b) J =
1
ρ
1
1
=
= 0.046 ohm-cm
−19
eμ n N d
1.6 × 10 (6800) 2 × 1016
(
)
(
)
⋅ Ε ⇒ Ε = ρ J = (0.046)(175) = 8.04 V/cm
--------------------------------------------------------------------------------------------------------------------------------EX1.4
Diffusion current density due to holes:
dp
J p = −eD p
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟
⎜L ⎟
⎜L ⎟
⎝ p⎠
⎝ p⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(a) At x = 0
(1.6 ×10 ) (10) (10 ) = 16 A / cm
J =
−19
16
2
10−3
−3
(b) At x = 10 cm
p
⎛ −10−3 ⎞
J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2
⎝ 10 ⎠
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EX1.5
( )( )
(
)
⎡ (10 )(10 ) ⎤
(b) V = (0.026 ) ln ⎢
⎥ = 0.374 V
⎢⎣ (2.4 × 10 ) ⎥⎦
⎡ 10 16 10 17 ⎤
(a) Vbi = (0.026 ) ln ⎢
⎥ = 1.23 V
6 2
⎣⎢ 1.8 × 10 ⎦⎥
16
bi
17
13 2
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EX1.6
−1/ 2
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
and
⎡N N ⎤
Vbi = VT ln ⎢ a 2 d ⎥
⎣ ni ⎦
⎡ (1017 )(1016 ) ⎤
⎥ = 0.757 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
5 ⎞
⎛
Then 0.8 = C jo ⎜ 1 +
⎟
⎝ 0.757 ⎠
or
C jo = 2.21 pF
−1/ 2
= C jo ( 7.61)
−1/ 2
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EX1.7
⎛I ⎞
(a) V D = VT ln⎜⎜ D ⎟⎟
⎝ IS ⎠
⎛ 50 × 10 −6 ⎞
⎟ = 0.563 V
(i) V D = (0.026) ln⎜⎜
−14 ⎟
⎝ 2 × 10 ⎠
⎛ 10 −3 ⎞
⎟ = 0.641 V
(ii) V D = (0.026 ) ln⎜⎜
−14 ⎟
⎝ 2 × 10 ⎠
⎛ 50 × 10 −6 ⎞
⎟ = 0.443 V
(b) (i) V D = (0.026) ln⎜⎜
−12 ⎟
⎝ 2 × 10 ⎠
⎛ 10 −3 ⎞
⎟ = 0.521 V
(ii) V D = (0.026 ) ln⎜⎜
−12 ⎟
⎝ 2 × 10 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX1.8
⎛V ⎞
VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
( 4 − VD )
so 4 = I D ( 4 × 103 ) + VD ⇒ I D =
4 ×103
and
⎛ V ⎞
I D = (10 −12 ) exp ⎜ D ⎟
⎝ 0.026 ⎠
By trial and error, we find I D ≅ 0.866 mA and V D ≅ 0.535 V.
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EX1.9
V PS − Vγ
8 − 0.7
⇒R=
= 6.08 k Ω
R
1.20
4 − 0. 7
(b) I D =
= 0.9429 mA
3. 5
PD = I DV D = (0.9429 )(0.7 ) = 0.66 mW
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ID =
(a)
EX1.10
PSpice Analysis
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EX1.11
8 − 0.7
= 0.365 mA
20
V
0.026
⇒ 71.2 Ω
rd = T =
I D 0.365
(a) I D =
0.25 sin ω t
⇒ 12.5 sin ω t ( μ A)
20 + 0.0712
8 − 0 .7
(b) I D =
= 0.73 mA
10
0.026
rd =
⇒ 35.6 Ω
0.73
0.25 sin ω t
id =
⇒ 24.9 sin ω t ( μ A)
10 + 0.0356
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id =
EX1.12
⎛I ⎞
⎛ 1.2 × 10−3 ⎞
or VD = 0.6871 V
For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 4 × 10 ⎠
⎝ IS ⎠
The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V
⎛V ⎞
Now I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or
1.2 × 10−3
⇒ I S = 1.07 × 10−10 A
⎛ 0.4221 ⎞
exp ⎜
⎟
⎝ 0.026 ⎠
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IS =
EX1.13
P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA
10 − 5.6
= 1.79 ⇒ R = 2.46 kΩ
R
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Also I =
Test Your Understanding Solutions
TYU1.1
(a) T = 400K
⎛ − Eg ⎞
Si: ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
ni = ( 5.23 × 1015 ) ( 400 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
or
ni = 4.76 × 1012 cm −3
Ge: ni = (1.66 × 1015 ) ( 400 )
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 400 ) ⎥⎦
or
ni = 9.06 × 1014 cm −3
GaAs:
ni = ( 2.1× 1014 ) ( 400 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
or
ni = 2.44 × 109 cm −3
(b) T = 250 K
Si: ni = ( 5.23 × 1015 ) ( 250 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 1.61× 108 cm −3
Ge: ni = (1.66 × 1015 ) ( 250 )
or
ni = 1.42 × 1012 cm −3
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
GaAs: ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 6.02 × 103 cm −3
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TYU1.2
(
)
(
)
(a) σ = eμ p N a = 1.6 × 10 −19 (480) 2 × 10 15 = 0.154 (ohm-cm)
ρ=
1
σ
=
1
= 6.51 Ω -cm
0.1536
(
)
(
)
(b) σ = eμ n N d = 1.6 × 10 −19 (1350) 2 × 1017 = 43.2 (ohm-cm)
ρ=
1
σ
=
−1
−1
1
= 0.0231 Ω -cm
43.2
________________________________________________________________________
TYU1.3
(a) J = σ Ε = (0.154)(4) = 0.616 A/cm 2
(b) J = σ Ε = (43.2)(4) = 172.8 A/cm 2
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TYU1.4
(a)
J n = eDn
⎛ 1015 − 1016 ⎞
dn
Δn
so J n = 1.6 × 10−19 ( 35 ) ⎜
= eDn
−4 ⎟
dx
Δx
⎝ 0 − 2.5 × 10 ⎠
(
)
or
J n = 202 A / cm 2
(b)
J p = −eD p
⎛ 1014 − 5 × 1015 ⎞
dp
Δp
so J p = − 1.6 × 10−19 (12.5 ) ⎜
= −eD p
−4 ⎟
dx
Δx
⎝ 0 − 4 × 10 ⎠
(
)
or
J p = −24.5 A / cm2
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TYU1.5
(a) no = N d = 8 × 1015 cm −3
n 2 (1.5 × 10 )
po = i =
= 2.81× 10 4 cm −3
no
8 × 1015
10 2
(b) n = no + δ n = 8 × 1015 + 0.1× 1015
or
n = 8.1×1015 cm−3
p = po + δ p = 2.81 × 10 4 + 1014
or
p ≅ 1014 cm −3
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU1.6
( )(
(
)
⎡ 10 15 5 × 10 16 ⎤
⎛N N ⎞
(a) Vbi = VT ln⎜⎜ a 2 d ⎟⎟ = (0.026 ) ln ⎢
⎥ = 0.679 V
2
⎢⎣ 1.5 × 10 10
⎥⎦
⎝ ni ⎠
( )(
(
( )(
(
)
)
⎡ 10 15 5 × 10 16 ⎤
(b) Vbi = (0.026 ) ln ⎢
⎥ = 1.15 V
2
⎢⎣ 1.8 × 10 6
⎥⎦
⎡ 10 15 5 × 10 16 ⎤
(c) Vbi = (0.026 ) ln ⎢
⎥ = 0.296 V
2
⎢⎣ 2.4 × 10 13
⎥⎦
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)
)
)
TYU1.7
⎛V ⎞
⎛ 0.55 ⎞
(a) (i) I D = I S exp⎜⎜ D ⎟⎟ = 10 −16 exp⎜
⎟ ⇒ 0.154 μ A
V
⎝ 0.026 ⎠
⎝ T ⎠
⎛ 0.65 ⎞
(ii) I D = 10 −16 exp⎜
⎟ ⇒ 7.20 μ A
⎝ 0.026 ⎠
(
(
)
(
)
)
⎛ 0.75 ⎞
(ii) I D = 10 −16 exp⎜
⎟ ⇒ 0.337 mA
⎝ 0.026 ⎠
(b) (i) I D = −10 −16 A
(ii) I D = −10 −16 A
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TYU1.8
ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV
Then VD = 0.650 − 0.20 = 0.450 V
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TYU1.9
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU1.10
(a) I D = 0
(b)
(c)
2 − 0.7
= 0.325 mA
4
5 − 0.7
ID =
= 1.075 mA
4
ID = 0
ID =
(d)
(e) I D = 0
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TYU1.11
P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA
VPS − Vγ
10 − 0.7
=
⇒ R = 6.2 kΩ
1.5
ID
______________________________________________________________________________________
Now R =
TYU1.12
ID
0.8
=
= 30.8 mS
VT 0.026
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gd =
TYU1.13
rd =
VT 0.026
=
= 2.6 k Ω
I D 0.010
0.026
⇒ 260 Ω
0.10
0.026
rd =
⇒ 26 Ω
1
----------------------------------------------------------------------------------------------------------------------------rd =
TYU1.14
rd =
VT
0.026
0.026
⇒ 50 =
⇒ ID =
50
ID
ID
or
I D = 0.52 mA
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TYU1.15
For the pn junction diode,
4 − 0.7
ID =
= 0.825 mA
4
4 − 0.3
= 0.925 mA
4
______________________________________________________________________________________
For the Schottky diode, I D =
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU1.16
Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10−3 ) ( 20 ) = 5.18 V
Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V
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TYU1.17
P = I Z VZ ⇒ I Z =
P 6.5
=
= 1.81 mA
V Z 3.6
V PS = I Z R + V Z = (1.81)(4 ) + 3.6 = 10.8 V
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