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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
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
×
⎝
⎠
______________________________________________________________________________________
(
)
Ge: n i = 1.66 × 1015 (300)
3/ 2
(
)
EX1.2
(a) (i)
n o = N d = 2×1016 cm
(
)
2
n i2
1.5 × 1010
=
no
2 × 1016
po =
(ii) p o = N a = 1015 cm
(
)
2
no = N d = 2 × 1016 cm
(
ni2
1.8 × 10 6
=
no
2 × 1016
po =
(ii)
= 1.125 × 10 4 cm −3
−3
n2
1.5 × 10 10
no = i =
po
1015
(b) (i)
p o = N a = 1015 cm
(
−3
)
= 2.25 × 10 5 cm −3
−3
2
= 1.62 × 10 − 4 cm −3
−3
)
2
n i2
1.8 × 10 6
=
= 3.24 × 10 − 3 cm −3
po
10 15
______________________________________________________________________________________
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⎠
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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
=
−19
Jp
(b) At x = 10−3
16
2
10−3
cm
⎛ −10−3 ⎞
J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2
⎝ 10 ⎠
______________________________________________________________________________________
EX1.5
( )( )
(
)
⎡ (10 )(10 ) ⎤
= (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
(b) Vbi
17
13 2
______________________________________________________________________________________
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
______________________________________________________________________________________
EX1.7
⎛I ⎞
(a) V D = VT ln⎜⎜ D ⎟⎟
⎝ IS ⎠
⎛ 50 × 10 −6
(i) V D = (0.026) ln⎜⎜
−14
⎝ 2 × 10
⎞
⎟ = 0.563 V
⎟
⎠
⎛ 10 −3
(ii) V D = (0.026 ) ln⎜⎜
−14
⎝ 2 × 10
⎞
⎟ = 0.641 V
⎟
⎠
⎛ 50 × 10 −6
(b) (i) V D = (0.026) ln⎜⎜
−12
⎝ 2 × 10
⎞
⎟ = 0.443 V
⎟
⎠
⎛ 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.
______________________________________________________________________________________
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
______________________________________________________________________________________
ID =
(a)
EX1.10
PSpice Analysis
______________________________________________________________________________________
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
______________________________________________________________________________________
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 ⎠
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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 ⎠
______________________________________________________________________________________
IS =
EX1.13
P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA
10 − 5.6
= 1.79 ⇒ R = 2.46 kΩ
R
______________________________________________________________________________________
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 )
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
or
ni = 1.42 × 1012 cm −3
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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
______________________________________________________________________________________
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
______________________________________________________________________________________
TYU1.4
(a)
J n = eDn
⎛ 1015 − 1016 ⎞
Δn
dn
so J n = 1.6 × 10−19 ( 35 ) ⎜
= eDn
−4 ⎟
Δx
dx
⎝ 0 − 2.5 × 10 ⎠
(
)
or
J n = 202 A / cm 2
(b)
J p = −eD p
⎛ 1014 − 5 × 1015 ⎞
Δp
dp
so J p = − 1.6 × 10−19 (12.5 ) ⎜
= −eD p
−4 ⎟
Δx
dx
⎝ 0 − 4 × 10 ⎠
(
)
or
J p = −24.5 A / cm2
______________________________________________________________________________________
TYU1.5
(a) no = N d = 8 × 1015 cm −3
10
ni2 (1.5 × 10 )
po =
=
= 2.81× 10 4 cm −3
no
8 × 1015
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 ⎤
⎞
⎟ = (0.026 ) ln ⎢
⎥ = 0.679 V
2
⎟
⎢⎣ 1.5 × 10 10
⎥⎦
⎠
⎡ 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
⎥⎦
______________________________________________________________________________________
⎛N N
(a) Vbi = VT ln⎜⎜ a 2 d
⎝ ni
( )(
(
( )(
(
)
)
)
)
)
TYU1.7
⎛V
(a) (i) I D = I S exp⎜⎜ D
⎝ VT
(
)
(
)
(ii) I D = 10 −16
⎞
⎛ 0.55 ⎞
⎟⎟ = 10 −16 exp⎜
⎟ ⇒ 0.154 μ A
⎝ 0.026 ⎠
⎠
⎛ 0.65 ⎞
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
______________________________________________________________________________________
TYU1.8
ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV
Then VD = 0.650 − 0.20 = 0.450 V
______________________________________________________________________________________
TYU1.9
______________________________________________________________________________________
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lOMoARcPSD|14951455
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
______________________________________________________________________________________
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
______________________________________________________________________________________
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 =
ID
ID
50
or
I D = 0.52 mA
______________________________________________________________________________________
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 =
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lOMoARcPSD|14951455
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
______________________________________________________________________________________
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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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 2
Exercise Solutions
EX2.1
V S − V B − Vγ
12 − 4.5 − 0.6
= 27.6 mA
0.25
R
υ R (max ) = V S + V B = 12 + 4.5 = 16.5 V
Conduction cycle:
υ I = 12 sin ω t1 = 4.5 + 0.6 = 5.1 V
or
⎛ 5.1 ⎞
ω t1 = sin −1 ⎜ ⎟ = 25.15°
⎝ 12 ⎠
ωt 2 = 180 − 25.15 = 154.85°
154.85 − 25.15
× 100% = 36.0%
Percent time =
360
______________________________________________________________________________________
i D ( peak ) =
=
EX2.2
(a) vO = 12sin θ1 − 1.4 = 0
1.4
= 0.1166
12
which yields
θ1 = 6.7°
By symmetry, θ 2 = 180 − 6.7 = 173.3°
or sin θ 1 =
Then
173.3 − 6.7
× 100% = 46.3%
360
1.4
sinθ1 =
= 0.35
4
(b)
which yields
θ1 = 20.5°
% time =
By symmetry, θ 2 = 180 − 20.5 = 159.5°
159.5 − 20.5
× 100% = 38.6%
% time =
360
Then
______________________________________________________________________________________
EX2.3
(a) C =
VM
12
=
⇒ 125 μ F
2 fRVr 2(60 ) 2 × 10 3 (0.4)
(
)
VM
12
=
⇒ 250 μ F
fRVr (60 ) 2 × 10 3 (0.4 )
______________________________________________________________________________________
(b) C =
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX2.4
Vr =
VM
VM
⇒R=
f RC
f CVr
R=
or
75
( 60 ) ( 50 ×10−6 ) ( 4 )
Then R = 6.25 k Ω
______________________________________________________________________________________
EX2.5
10 ≤ VPS ≤ 14 V , VZ = 5.6 V , 20 ≤ RL ≤ 100 Ω
5.6
= 0.28 A,
20
5.6
I L ( min ) =
= 0.056 A
100
⎡VPS ( max ) − VZ ⎦⎤ ⋅ I L ( max )
⎣⎡VPS ( min ) − VZ ⎦⎤ ⋅ I L ( min )
−
I Z (max) = ⎣
VPS ( min ) − 0.9VZ − 0.1VPS ( max ) VPS ( min ) − 0.9VZ − 0.1VPS ( max )
I L ( max ) =
I L ( max ) =
or
(14 − 5.6 )( 280 ) − (10 − 5.6 )( 56 )
10 − ( 0.9 )( 5.6 ) − ( 0.1)(14 )
I L ( max ) = 591.5 mA
or
Power(min) = I Z ( max ) ⋅ VZ = ( 0.5915)( 5.6 )
So Power(min) = 3.31 W
VPS ( max ) − VZ
14 − 5.6
=
Ri =
I Z ( max ) + I L ( min ) 0.5915 + 0.056
or Ri ≅ 13 Ω
Now
______________________________________________________________________________________
EX2.6
13.6 − 9
= 0.2383 A
15.3 + 4
= 9 + ( 4 )( 0.2383) = 9.9532 V
For vPS = 13.6 V , I Z =
vL ,max
11 − 9
= 0.1036 A
15.3 + 4
= 9 + ( 4 )( 0.1036 ) = 9.4144 V
For vPS = 11 V , I Z =
vL ,min
ΔvL
9.9532 − 9.4144
× 100% =
× 100%
ΔvPS
13.6 − 11
Source Reg = 20.7%
Source Reg =
or
13.6 − 9
= 0.2383 A
15.3 + 4
= 9 + ( 4 )( 0.2383) = 9.9532 V
For I L = 0, I Z =
vL , noload
For I L = 100 mA,
which yields
IZ =
13.6 − ⎡⎣9 + I Z ( 4 ) ⎤⎦
15.3
− 0.10
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
I Z = 0.1591 A
vL , full load = 9 + ( 4 )( 0.1591A ) = 9.6363 V
Load Reg =
vL , noload − vL , full load
vL , full load
× 100%
9.9532 − 9.6363
× 100%
9.6363
or Load Reg = 3.29%
______________________________________________________________________________________
=
EX2.7
For vI < 5 V , D2 on ⇒ VO = −5 V
Then, V2 = 4.3 V.
D1 turns on when v1 = 2.5 V,
Then, V1 = 1.8 V.
Δv
R2
1
1
=
vI > 2.5 V , O = ⇒
ΔvI 3
R1 + R2 3
For
So that R1 = 2R2
______________________________________________________________________________________
EX2.8
For υ O = +2 V, D is on. Δυ I = 10 V, so Δυ O = 10 V.
Output = Square wave between +2 and −8 V.
______________________________________________________________________________________
EX2.9
10 − 4.4
= 0.5895 mA
9.5
vI = 4.4 − 0.6 − ( 0.5895 )( 0.5) = 3.505 V
vO = 4.4 V , I =
Set I = ID1, then
Summary: For 0 ≤ vI ≤ 3.5 V , vO = 4.4 V
For vI > 3.5 V , D2 turns on and when vI ≥ 9.4 V , vO = 10 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
______________________________________________________________________________________
EX2.10
(a) If D1 is on, υ O = υ I − Vγ − V B = 5 − 0.7 − 1 = 3.3 V.
3.3 − 0.7
= 0.65 mA
4
5 − 3.3
= 1.0 mA, but I D 2 < I R1 is impossible.
Now I R1 =
1.7
D1 is cutoff and I D1 = 0
5 − 0.7
= 0.754 mA
Then I R1 = I D 2 =
1.7 + 4
υ O = 0.7 + (0.754 )(4 ) = 3.72 V
(b) υ I = 10 V, Both D1 and D2 are on.
υ O = 10 − 0.7 − 1 ⇒ υ O = 8.3 V
8.3 − 0.7
I D2 =
= 1.9 mA
4
1.7
I R1 =
= 1.0 mA
1.7
I D1 = 1.9 − 1.0 = 0.9 mA
______________________________________________________________________________________
Then I D 2 =
EX2.11
D2 cutoff, I D 2 = 0
− 0.7 − (− 5)
= 2.15 mA
2
5 − 0.7 − (− 10) 14.3
=
=
= 1.19 mA
8+4
R1 + R2
V B = −0.7 V, I D 3 =
I D1
V A = 5 − (1.19 )(8) = −4.53 V
V A < V B so that D2 is cutoff.
______________________________________________________________________________________
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX2.12
(a)
I ph = η eΦ
⎡ 6.4 × 10−2 ⎤
⎥ ( 0.5 )
I = ( 0.8 ) (1.6 × 10 −19 ) ⎢
⎢⎣ ( 2 ) (1.6 × 10−19 ) ⎥⎦
so
I = 12.8 mA
or ph
v = (12.8)(1) = 12.8 V .
(b) We have O
The diode must be reverse biased so that VPS > 12.8 V
______________________________________________________________________________________
EX2.13
The equivalent circuit is
I=
So
5 − 1.7 − 0.2
= 15 mA
rf + R
15 − 1.7 − 0.2 3.1
=
= 0.207 kΩ
15
15
Or
Then R = 207 − 15 ⇒ R = 192 Ω
______________________________________________________________________________________
rf + R =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU2.1
(a) i D ( peak ) for V B = 4 V.
15 − 0.7 − 4
i D ( peak ) = 18 =
⇒ R = 572 Ω
R
15 − 0.7 − 8
= 11.0 mA
(b) i D =
0.572
Then
11.0 ≤ i D ( peak ) ≤ 18 mA
For V B = 4 V,
⎛ 4.7 ⎞
15 sin ω t1 = 4.7 ⇒ ω t1 = sin −1 ⎜
⎟ = 18.26°
⎝ 15 ⎠
ω t 2 = 180 − 18.26 = 161.74°
161.74 − 18.26
× 100% = 39.9%
duty cycle =
360
For V B = 8 V,
⎛ 8.7 ⎞
15 sin ω t1 = 8.7 ⇒ ω t1 = sin −1 ⎜
⎟ = 35.45°
⎝ 15 ⎠
ω t 2 = 180 − 35.45 = 144.55°
144.55 − 35.45
× 100% = 30.3%
duty cycle =
360
Then
30.3 ≤ duty cycle ≤ 39.9%
______________________________________________________________________________________
TYU2.2
vI = 120sin ( 2π 60t ) , Vγ = 0.7 V ,
and R = 2.5 kΩ
Full-wave rectifier: Turns ratio 1:2 so that
vS = v I
VM = 120 − 0.7 = 119.3 V
Vr = 119.3 − 100 = 19.3 V
C=
VM
119.3
=
2 f RVr 2 ( 60 ) ( 2.5 x103 ) (19.3)
C = 20.6 μ F
So
or
_____________________________________________________________________________
TYU2.3
vI = 50sin ( 2π 60t ) , Vγ = 0.7 V ,
C=
and R = 10 kΩ. Full-wave rectifier
( 50 − 1.4 )
VM
=
2 f RVr 2 ( 60 ) (10 × 103 ) ( 2 )
C = 20.3 μ F
or
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU2.4
Using Equation (2.16)
ω Δt =
2Vr
=
VM
2 ( 4)
75
= 0.327
⎛ 0.327 ⎞
Percent time = ⎜
⎟ × 100% = 5.2%
⎝ 2π ⎠
(a)
ω Δt =
2Vr
=
VM
2 (19.3)
119.3
= 0.569
⎛ 0.569 ⎞
Percent time = ⎜
⎟ × 100% = 18.1%
⎝ π ⎠
(b)
ω Δt =
2Vr
=
VM
2 ( 2)
48.6
= 0.287
⎛ 0.287 ⎞
Percent time = ⎜
⎟ × 100% = 9.14%
⎝ π ⎠
(c)
______________________________________________________________________________________
TYU2.5
(a) P = I Z V Z
1 = I Z (6.2 + 3I Z ) = 3I Z2 + 6.2 I Z
3I Z2 + 6.2 I Z − 1 = 0 ⇒ I Z (max) = 150 mA
VZ = 6.2 + 3(0.15) = 6.65 V
12 − 6.65
⇒ Ri = 35.7 Ω
0.15
(b) For I Z = (0.1)(150 ) = 15 mA
Ri =
VZ = VO = 6.2 + 3(0.015) = 6.245 V
12 − 6.245
= 161.2 mA
I L = I i − I Z ⇒ Ii =
0.0357
I L = 161.2 − 15 = 146.2 mA
6.245
= 42.7 Ω
RL =
0.1462
υ (no load ) − υ L ( full load )
Load Regulation = L
× 100%
υ L (no load )
6.65 − 6.245
=
× 100% = 6.09%
6.65
______________________________________________________________________________________
TYU2.6
IZ =
VPS − VZ
− IL
Ri
11 − 9
− 0.1 = 0
20
For VPS (min) and IL (max), then
(Minimum Zener current is zero.)
13.6 − 9
I Z ( max ) =
− 0 ⇒ I Z ( max ) = 230 mA
20
For VPS (max) and IL (min), then
I Z ( min ) =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
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Exercise Solutions
______________________________________________________________________________________
The characteristic of the minimum Zener current being one-tenth of the maximum value is violated. The
proper circuit operation is questionable.
______________________________________________________________________________________
TYU2.7
I Z ( min ) =
VPS ( min ) − VZ
Ri
− I L ( max )
10 − 9
− I L ( max )
I ( max ) = 35.4 mA
0.0153
which yields L
so
______________________________________________________________________________________
30 =
TYU2.8
For υ I ≤ −3.7 V, D2 on ⇒ υ O = −3.7 V
D1 turns on when υ I = 1.7 V so
−3.7 ≤ υ I ≤ 1.7 V ⇒ υ O = υ I
For υ I > 1.7 V,
i1 =
υ I − 1.7
R1 + R2
=
υ I − 1.7
7
⎛ υ I − 1.7 ⎞
⎟(2 ) + 1.7
⎝ 7 ⎠
υ O = i1 R2 + 1.7 = ⎜
Or
υ O = 0.286υ I + 1.21
______________________________________________________________________________________
TYU2.9
As vS goes negative, D turns on and vO = +5 V . As vS goes positive, D turns off. Output is a square
wave oscillating between +5 and +35 volts.
______________________________________________________________________________________
TYU2.10
V1 = 3 − 0.7 = 2.3 V
V2 = 2 − 0.7 = 1.3 V
1
⇒ R1 = R2
2
Put resistor in series with D2 ,
For υ I > 3 V, slope =
R3
1
=
⇒ R1 = 2 R3
3 R1 + R3
______________________________________________________________________________________
For υ I < 2 V, slope =
TYU2.11
D2 and D3 cutoff so that I D 2 = I D 3 = 0
14 − 0.7 − (− 5)
18.3
I D1 =
=
= 1.22 mA
5+5+5
R1 + R2 + R3
V A = 14 − 0.7 − (1.22 )(5) = 7.2 V ⇒ D2 cutoff
V B = (1.22 )(5) − 5 = 1.1 V ⇒ D3 cutoff
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU2.12
D2 cutoff, I D 2 = 0
V B = −0.7 V
14 − 0.7 − (− 0.7 )
14
I D1 =
=
= 0.7 mA
8 + 12
R1 + R2
V A = 14 − 0.7 − (0.7 )(8) = 7.7 V ⇒ D2 cutoff
− 0.7 − (− 5)
I R3 =
= 1.72 mA
2.5
I D1 + I D 3 = I R 3
I D 3 = I R 3 − I D1 = 1.72 − 0.7 = 1.02 mA
______________________________________________________________________________________
TYU2.13
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU2.14
(a) VO = 0.6 V for all V1.
(b)
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 3
Exercise Solutions
EX3.1
VTN = 1 V , VGS = 3 V , VDS = 4.5 V
VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V
Transistor biased in the saturation region
I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2
2
2
(a) VGS = 2 V, VDS = 4.5 V
Saturation region:
I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA
2
(b) VGS = 3 V, VDS = 1 V
Nonsaturation region:
2
I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA
⎣
⎦
______________________________________________________________________________________
EX3.2
0.5 = K p (3 − 1.2) ⇒ K p = 0.154 mA/V
2
2
(a) i D = K p (υ SG + VTP ) = 0.154(2 − 1.2 ) = 0.0986 mA
2
2
[
]
[
]
2
= 0.154 2(5 − 1.2 )(2) − (2 ) = 1.72 mA
(b) i D = K p 2(υ SG + VTP )υ SD − υ SD
2
______________________________________________________________________________________
EX3.3
⎛ R2
VGS = ⎜⎜
⎝ R1 + R2
⎞
⎛ 245 ⎞
⎟⎟ ⋅ V DD = ⎜
⎟(2.2) = 0.8983 V
⎝ 245 + 355 ⎠
⎠
I D = (25)(0.8983 − 0.35) = 7.52 μ A
V DS = 2.2 − (0.00752)(100 ) = 1.45 V
______________________________________________________________________________________
2
EX3.4
I DQ = 0.5 mA, V SDQ = 2.0 V
3.3 − 2.0
= 2.6 k Ω
0.5
2
I D = K p VSGQ + VTP
RD =
(
0.5 = 0.2(V
)
− 0.6 )
2
SGQ
⇒ VSGQ = 2.18 V
VG = 3.3 − 2.18 = 1.12 V
⎛ R2 ⎞
1
⎟⎟ ⋅ V DD =
(R1 R2 )(3.3)
VG = ⎜⎜
R1
⎝ R1 + R2 ⎠
1
1.12 =
(300)(3.3) ⇒ R1 = 884 k Ω
R1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
884 R2
= 300 ⇒ R2 = 454 k Ω
884 + R2
______________________________________________________________________________________
EX3.5
⎛ 30 ⎞
(a) VGSQ = ⎜
⎟(5) = 1.667 V
⎝ 30 + 60 ⎠
2
I DQ = (0.5)(1.667 − 0.6) = 0.5689 mA
V DSQ = 5 − (0.5689 )(4 ) = 2.724 V
(b) K n (+ 5% ) = 0.525 mA/V 2 , VTN (− 5% ) = 0.57 V
I D = 0.525(1.667 − 0.57) = 0.6314 mA
V DS = 5 − (0.6314 )(4 ) = 2.474 V
2
K n (− 5% ) = 0.475 mA/V 2 , VTN (+ 5% ) = 0.63 V
I D = 0.475(1.667 − 0.63) = 0.5105 mA
V DS = 5 − (0.5105)(4) = 2.958 V
Then
0.5105 ≤ I D ≤ 0.6314 mA
2.474 ≤ V DS ≤ 2.958 V
______________________________________________________________________________________
2
EX3.6
⎛ 345 ⎞
(a) VG = ⎜
⎟(4.4) − 2.2 = 0.330 V
⎝ 345 + 255 ⎠
2.2 = I DQ R S + V SGQ + VG
(
2.2 = K p RS VSGQ + VTP
(
)
2
+ VSGQ + 0.330
)
2
1.87 = (0.035)(6 ) VSGQ
− 0.6VSGQ + 0.09 + VSGQ
0.21V
2
SGQ
+ 0.874VSGQ − 1.8511 = 0 ⇒ VSGQ = 1.545 V
We find
2
I DQ = 35(1.545 − 0.3) = 54.22 μ A
V SDQ = 4.4 − (0.05422 )(6 + 42 ) = 1.797 V
(b)
For VTP = −0.315 V
We have
2
1.87 = (0.035)(6) VSGQ
− 0.63VSGQ + 0.099225 + VSGQ
(
)
2
0.21VSGQ
+ 0.8677VSGQ − 1.849 = 0 ⇒ VSGQ = 1.550 V
Then
2
I DQ = 35(1.550 − 0.315) = 53.36 μ A
For VTP = −0.285 V
We have
2
1.87 = (0.035)(6 ) VSGQ
− 0.57VSGQ + 0.081225 + VSGQ
(
)
2
0.21VSGQ
+ 0.8803VSGQ − 1.8529 = 0 ⇒ VSGQ = 1.5395 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
2
I DQ = 35(1.5395 − 0.285) = 55.08 μ A
Therefore
53.36 ≤ I DQ ≤ 55.08 μ A
______________________________________________________________________________________
EX3.7
4.4 = VSD (sat ) + I D (6 + 42 )
4.4 = VSG + VTP + (0.035)(48)(VSG + VTP )
2
(
2
4.4 = VSG − 0.3 + 1.68 VSG
− 0.6VSG + 0.09
)
1.68V − 0.008VSG − 4.5488 = 0 ⇒ VSG = 1.648 V
We find
2
I D = 35(1.648 − 0.3) = 63.59 μ A
2
SG
VSD = 4.4 − (0.0636)(48) = 1.348 V
Note: VSD (sat ) = 1.648 − 0.3 = 1.348 V
______________________________________________________________________________________
EX3.8
2
(a) I DQ = 60 = 30 VSGQ − 0.4 ⇒ VSGQ = 1.814 V
(
I DQ =
)
3 − 1.814
= 0.060 ⇒ RS = 19.77 k Ω
RS
V D = 1.814 − 2.5 = −0.686 V
− 0.686 − (− 3)
RD =
= 38.57 k Ω
0.060
(b) VTP (+ 5% ) = 0.42 V, K p (− 5% ) = 28.5 μ A/V 2
(
)
2
3 = I D RS + VSG = (0.0285)(19.77) VSG
− 0.84VSG + 0.1764 + VSG
which yields
V SG = 1.849 V
I D = (28.5)(1.849 − 0.42) = 58.2 μ A
V SD = 6 − (0.0582)(19.77 + 38.57 ) = 2.605 V
2
VTP (− 5%) = 0.38 V, K p (+ 5% ) = 31.5 μ A/V
(
2
)
2
3 = (0.0315)(19.77 ) VSG
− 0.76VSG + 0.1444 + VSG
which yields
VSG = 1.780 V
I D = 31.5(1.780 − 0.38) = 61.72 μ A
VSD = 6 − (0.06172)(19.77 + 38.57 ) = 2.399 V
Then
58.2 ≤ I D ≤ 61.72 μ A
2.399 ≤ V SD ≤ 2.605 V
______________________________________________________________________________________
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX3.9
(a)
VI = 4 V, Driver in Non ⋅ Sat.
2
K nD ⎣⎡ 2 (VI − VTND ) VO − VO2 ⎦⎤ = K nL [VDD − VO − VTNL ]
5 ⎣⎡ 2 ( 4 − 1) VD − VD2 ⎦⎤ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2
2
2
6VD2 − 38VO + 16 = 0
VD =
38 ± 1444 − 384
2 ( 6)
VD = 0.454 V
(b) VI = 2 V Driver: Sat
K nD [VI − VTND ] = K nL [VDD − VO − VTNL ]
2
2
5 [ 2 − 1] = [5 − VO − 1]
2
2
5 = 4 − VO ⇒ VO = 1.76 V
______________________________________________________________________________________
EX3.10
(a) For VI = 5 V, Load in saturation and driver in nonsaturation.
I DD = I DL
K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL ( −VTNL )
2
K nD ⎡
K
2
2 ( 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06
⎦
K nL ⎣
K nL
(b)
I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡⎣ − ( −2 ) ⎤⎦
2
2
K nL = 50 μ A / V 2 and K nD = 103 μ A / V 2
______________________________________________________________________________________
EX3.11
For M N
I DN = I DP
K n (VGSN − VTN ) = K p (Vscop + VTP )
2
2
VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI
Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V
For M P : VI = 1.75 V
VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V
So Vot = 5 − 0.75 ⇒ Vot = 4.25 V
______________________________________________________________________________________
EX3.12
Transistor in nonsaturation
(a) V DS = 0.2 V, VGS = 5 V
[
2
VO = VDD − I D RD = VDD − K n RD 2(VGS − VTN )VDS − VDS
[
]
]
0.2 = 5 − K n (0.5) 2(5 − 1)(0.2) − (0.2) ⇒ K n = 6.154 mA/V 2
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
[
]
(b) I D = (6.154) 2(5 − 1)(0.2) − (0.2) = 9.60 mA
P = I DV DS = (9.60 )(0.2 ) = 1.92 mW
______________________________________________________________________________________
EX3.13
a.
2
V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0
I D = K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ =
5 − VO
RD
( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦ = 5 − V0
1.5V02 − 13V0 + 5 = 0
V0 =
(13) − 4 (1.5)( 5)
⇒ V0 = 0.40 V
2 (1.5 )
13 ±
2
5 − 0.40
⇒ I R = I D1 = 0.153 mA
30
V1 = V2 = 5 V
I R = I D1 =
b.
5 − VO
= 2 K n ⎣⎡ 2 (VI − VTN ) VO − VO2 ⎤⎦
RD
{
}
5 − V0 = 2 ( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦
3V02 − 25V0 + 5 = 0
V0 =
25 ±
( 25) − 4 ( 3)( 5 )
⇒ V0 = 0.205 V
2 ( 3)
2
5 − 0.205
⇒ I R = 0.160 mA
30
= I D 2 = 0.080 mA
IR =
I D1
______________________________________________________________________________________
EX3.14
VGS 3 =
I REF1
120
+ VTN =
+ 0.4 = 1.814 V
K n3
60
VGS 2 = VGS 3 = 1.814 V
I Q1 = K n 2 (VGS 2 − VTN ) = 30(1.814 − 0.4) = 60 μ A
2
VGS1 =
I Q1
K n1
+ VTN =
2
60
+ 0.4 = 1.495 V
50
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX3.15
2
⎛ 0.04 ⎞
0.1 = ⎜
⎟ (15 )(VSGC − 0.6 )
⎝ 2 ⎠
VSGC = 1.177 V = VSGB
2
⎛ 0.04 ⎞ ⎛ W ⎞
0.2 = ⎜
⎟ ⎜ ⎟ (1.177 − 0.6 )
⎝ 2 ⎠ ⎝ L ⎠B
⎛W ⎞
⎜ ⎟ = 30
⎝ L ⎠B
2
⎛ 0.04 ⎞
0.2 = ⎜
⎟ ( 25 )(VSGA − 0.6 )
⎝ 2 ⎠
VSGA = 1.23 V
______________________________________________________________________________________
EX3.16
I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN )
VGS 3 = 2 V ⇒ VGS 4 = 3 V
2
( 2 − 1)
(a)
2
=
2
Kn4
K
1
2
( 3 − 1) ⇒ n 4 =
K n3
K n3 4
I Q = K n 2 (VGS 2 − VTN )
But VGS 2 = VGS 3 = 2 V
2
0.1 = K n 2 ( 2 − 1) ⇒ K n 2 = 0.1 mA / V 2
2
(b)
0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2
2
0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2
(c)
______________________________________________________________________________________
2
EX3.17
VS 2 = 5 − 5 = 0
RS 2 =
I D 2 = K n 2 (VGS 2 − VTN 2 )
5
= 16.7 K
0.3
2
0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V
2
5 − 2.425
= 25.8 K
0.1
VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V
RD1 =
RS 1 =
−2.575 − ( −5 )
0.1
⇒ RS 1 = 24.3 K
I D1 = K n1 (VGS 1 − VTN 1 )
2
0.1 = 0.5 (VGS 1 − 1.2 ) ⇒ VGS 1 = 1.647 V
VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V
2
⎛ R2 ⎞
1
VG1 = ⎜
⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5
+
R
R
R
⎝ 1
2 ⎠
1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
1
−0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K
R1
491 R2
= 200 ⇒ R2 = 337 K
491 + R2
______________________________________________________________________________________
EX3.18
VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V
I DQ = K n (VGS 1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V
VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V
⎛
⎞
R3
R3
VG1 = ⎜
(5) ⇒ R3 = 50.7 K
⎟ (5) ⇒ 0.507 =
500
⎝ R1 + R2 + R3 ⎠
VS 2 = VS 1 + VDS 1 = −1 + 2.5 = 1.5 V
VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V
⎛ R2 + R3 ⎞
⎛ R2 + R3 ⎞
VG 2 = ⎜
⎟ (5) ⇒ 3.007 = ⎜
⎟ (5)
+
+
R
R
R
⎝ 500 ⎠
2
3 ⎠
⎝ 1
R2 + R3 = 300.7
R2 = 300.7 − 50.7 ⇒ R2 = 250 K
R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K
VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V
5−4
⇒ RD = 4 K
0.25
______________________________________________________________________________________
RD =
EX3.19
VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V
⎛ ( −1.2 ) ⎞
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 −
⎟⎟ ⇒ I D = 6.45 mA
⎜
VP ⎠
⎝
⎝ ( −4.5 ) ⎠
______________________________________________________________________________________
2
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX3.20
I D = 2.5 mA
⎛ V ⎞
I D = I DSS ⎜ 1 − GS ⎟
VP ⎠
⎝
2
2
⎛
V ⎞
2.5 = 6 ⎜ 1 − GS ⎟ ⇒ VGS = −1.42 V
⎜ ( −4 ) ⎟
⎝
⎠
VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5
VS = −4.375
VDS = 6 ⇒ VD = 6 − 4.375 = 1.625
5 − 1625
⇒ RD = 1.35 kΩ
2.5
RD =
( 20 )
2
R1 + R2
= 2 ⇒ R1 + R2 = 200 kΩ
VG = VGS + VS = −1.42 − 4.375 = −5.795
⎛ R2 ⎞
VG = ⎜
⎟ ( 20 ) − 10
⎝ R1 + R2 ⎠
⎛ R ⎞
−5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ
⎝ 200 ⎠
R1 = 157.95 kΩ → 158 kΩ
______________________________________________________________________________________
EX3.21
VS = −VGS . I D =
0 − VS VGS
=
RS
RS
⎛ V ⎞
I D = I DSS ⎜1 − GS ⎟
VP ⎠
⎝
2
⎛ V
VGS
V2 ⎞
⎛ V ⎞
= 6 ⎜1 − GS ⎟ = 6 ⎜1 − GS + GS ⎟
16 ⎠
1
4 ⎠
2
⎝
⎝
2
0.375VGS − 4VGS + 6 = 0
2
VGS =
4 ± 16 − 4 ( 0.375 )( 6 )
2 ( 0.375 )
VGS = 8.86 or VGS = 1.806 V
14243
impossible
ID =
VGS
= 1.806 mA
RS
VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278
VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V
VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19
So VSD > VSD ( sat )
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX3.22
RR
Rib = R1 R 2 = 1 2 = 100 k Ω
R1 + R 2
I DQ = 5 mA, V S = − I DQ R S = −(5)(1.2 ) = −6 V
V SDQ = 12 V, V D = V S − V SDQ
= −6 − 12 = −18 V
RD =
−18 − ( −20 )
5
⇒ RD = 0.4 kΩ
2
I DQ
VGS
⎛ V ⎞
⎛ V ⎞
= I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟
4 ⎠
VP ⎠
⎝
⎝
= 0.838 V
2
VG = VGS + VS = 0.838 − 6 = −5.162
⎛ R2 ⎞
VG = ⎜
⎟ ( −20 )
⎝ R1 + R2 ⎠
1
−5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ
R1
R1 R2
= 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2
R1 + R2
( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ
______________________________________________________________________________________
Test Your Understanding Solutions
TYU3.1
(a)
VTN = 1.2 V , VGS = 2 V
V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V
(i)
VDS = 0.4 ⇒ Nonsaturation
(ii)
VDS = 1 ⇒ Saturation
(iii) VDS = 5 ⇒ Saturation
VTN = −1.2 V , VGS = 2 V
V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V
(b)
(i) VDS = 0.4 ⇒ Nonsaturation
(ii)
VDS = 1 ⇒ Nonsaturation
(iii) VDS = 5 ⇒ Saturation
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU3.2
Wμ n ∈ox
20 × 10 −4 (500 )(3.9 ) 8.85 × 10 −14
(a) K n =
=
⇒ 1.08 mA/V 2
2 Lt ox
2 0.8 × 10 − 4 200 × 10 −8
(b)
2
i D = (1.08) 2(2 − 1.2)(0.4) − (0.4) = 0.518 mA
(i)
(
)
(
(
)(
)
[
(ii)
)
i D = (1.08)(2 − 1.2) = 0.691 mA
]
2
(iii)
i D = (1.08)(2 − 1.2) = 0.691 mA
(i)
i D = (1.08) 2(2 + 1.2)(0.4) − (0.4) = 2.59 mA
(ii)
iD
2
[
]
= (1.08)[2(2 + 1.2)(1) − (1) ] = 5.83 mA
2
2
i D = (1.08)(2 + 1.2) = 11.1 mA
(iii)
______________________________________________________________________________________
2
TYU3.3
(a)
VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V
(b)
(i) Non Sat
(ii) Sat (iii) Sat
VSD (sat) = 2 + 1.2 = 3.2 V
(i) Non Sat
(ii) Non Sat
(iii) Sat
______________________________________________________________________________________
TYU3.4
(a) K p =
(b)
(i)
(ii)
Wμ p ∈ox
2 Lt ox
=
(10 × 10 )(300)(3.9)(8.85 × 10 ) ⇒ 0.324 mA/V
2(0.8 × 10 )(200 × 10 )
−4
−14
−4
[
]
i D = (0.324) 2(2 − 1.2)(0.4) − (0.4) = 0.156 mA
2
i D = (0.324)(2 − 1.2) = 0.207 mA
2
(iii)
i D = (0.324)(2 − 1.2) = 0.207 mA
(i)
i D = (0.324) 2(2 + 1.2)(0.4) − (0.4) = 0.778 mA
(ii)
iD
(iii)
2
−8
2
[
]
= (0.324)[2(2 + 1.2)(1) − (1) ] = 1.75 mA
2
2
i D = (0.324)(2 + 1.2) = 3.32 mA
2
______________________________________________________________________________________
TYU3.5
(a), (i) (ii)
i D = (10)(0.5 − 0.25) = 0.625 μ A
2
(b)
(i)
i D = K n (υ GS − VTN ) (1 + λυ DS )
2
i D = (10)(0.5 − 0.25) [1 + (0.03)(0.5)] = 0.6344 μ A
2
(ii)
i D = (10)(0.5 − 0.25) [1 + (0.03)(1.2)] = 0.6475 μ A
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c) For (a),
ro = ∞
1
1
For (b), ro =
=
= 53.3 M Ω
λI DQ (0.03)(0.625)
______________________________________________________________________________________
TYU3.6
(a) VTN = VTNO = 0.4 V
[ 2φ + υ − 2φ ]
= 0.4 + 0.15[ 2(0.35 ) + 0.5 − 2(0.35) ] ⇒ V
= 0.4 + 0.15[ 2(0.35 ) + 1.5 − 2(0.35) ] ⇒ V
(b) VTN = VTNO + γ
f
SB
f
TN
(c)
VTN
TN
= 0.439 V
= 0.497 V
______________________________________________________________________________________
TYU3.7
V DS = 2.2 − (0.07 )R D = 1.2 ⇒ RD = 14.3 k Ω
VGS =
I DQ
Kn
70
+ 0.25 = 1.778 V
30
+ VTN =
⎛ R2 ⎞
⎛ R ⎞
⎟⎟ ⋅ V DD = ⎜ 2 ⎟(2.2)
VGS = 1.778 = ⎜⎜
R
+
R
⎝ 500 ⎠
2 ⎠
⎝ 1
We find
R2 = 404 k Ω , R1 = 96 k Ω
______________________________________________________________________________________
TYU3.8
3.3 − 1.6
= 0.17 mA
10
W
⎛ 0.1 ⎞⎛ W ⎞
2
0.17 = ⎜
= 2.36
⎟⎜ ⎟(1.6 − 0.4) ⇒
L
⎝ 2 ⎠⎝ L ⎠
______________________________________________________________________________________
ID =
TYU3.9
(a) The transition point is
VIt =
=
(
VDD − VTNL + VTND 1 + K nD / K nL
(
)
1 + K nD /K nL
5 − 1 + 1 1 + 0.05/ 0.01
)
1 + 0.05/ 0.01
7.236
=
⇒ VIt = 2.236 V
3.236
VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V
(b)
We may write
I D = K n D (VGSD − VTND ) = ( 0.05)( 2.236 − 1) ⇒ I D = 76.4 μ A
2
2
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU3.10
(a) V DS > [0 − (− 1.2)] ⇒ Saturation
3.3 − 1.8
= 0.1875 mA
8
W
⎛ 0.08 ⎞⎛ W ⎞
2
0.1875 = ⎜
= 3.26
⎟⎜ ⎟[0 − (− 1.2 )] ⇒
L
L
2
⎝
⎠⎝ ⎠
ID =
(b) V DS < [0 − (− 1.2)] ⇒ Nonsaturation
3.3 − 0.8
ID =
= 0.3125 mA
8
W
⎛ 0.08 ⎞⎛ W ⎞
2
0.3125 = ⎜
= 6.10
⎟⎜ ⎟ 2(0 − (− 1.2))(0.8) − (0.8) ⇒
2
L
L
⎠⎝ ⎠
⎝
______________________________________________________________________________________
[
]
TYU3.11
(a) Transition point for the load transistor – Driver is in the saturation region.
I DD = I DL
K nD (VGSD − VTND ) = K nL (VGSL − VTNL )
VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V
2
2
Then VOt = 5 − 2 = 3 V , VOt = 3 V
K nD
(VIt − 1) = ( −VTNL )
K nL
0.08
(VIt − 1) = 2 ⇒ VIt = 1.89 V
0.01
(b) For the driver:
VOt = VIt − VTND
VIt = 1.89 V , VOt = 0.89 V
______________________________________________________________________________________
TYU3.12
Transistor biased in nonsaturation
2
2
I D = K n 2(VGS − VTN )VDS − VDS
= (4) 2(10 − 0.7 )(0.2) − (0.2) = 14.72 mA
10 − 0.2
RD =
= 0.666 k Ω
14.72
______________________________________________________________________________________
[
]
[
]
TYU3.13
(a) Transistor biased in the nonsaturation region
5 − 1.5 − VDS
= 12
ID =
R
2
⎤⎦
I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS
2
⎤⎦
12 = 4 ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS
2
4VDS
− 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V
5 − 1.5 − 0.374
⇒ R = 261 Ω
12
Then
______________________________________________________________________________________
R=
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU3.14
5 − VO
= K n ⎡⎣ 2 (V2 − VTN ) VO − VO2 ⎤⎦
ID =
RD
5 − ( 0.10 )
25
a.
2
= K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2
⎣
⎦
5 − V0
= 2 ( 0.248 ) ⎡⎣ 2 ( 5 − 1) V0 − V02 ⎤⎦
25
5 − V0 = 12.4 ⎡⎣8V0 − V02 ⎤⎦
12.4V02 − 100.2V0 + 5 = 0
100.2 ±
(100.2 ) − 4 (12.4 )( 5 )
⇒ V0 = 0.0502 V
2 (12.4 )
2
V0 =
b.
______________________________________________________________________________________
TYU3.15
VSGC =
I REF 2
− VTP =
K pC
40
+ 0.3 ⇒ VSGC = VSGB = 1.30 V
40
I Q 2 = K pB (VSGB + VTP ) = 60(1.30 − 0.3) = 60 μ A
2
V SGA =
I Q2
K pA
2
60
+ 0.3 = 1.19 V
75
− VTP =
______________________________________________________________________________________
TYU3.16
2
I Q = K n1 (VGS1 − VTN )
120 = K n1 (1.5 − 0.7 ) ⇒ K n1 = 187.5 μ A/V 2
VGS 2 = VGS 3 = 2 V
2
120 = K n 2 (2 − 0.7 ) ⇒ K n 2 = 71.0 μ A/V 2
2
I REF = K n3 (VGS 3 − VTN )
2
80 = K n3 (2 − 0.7 ) ⇒ K n3 = 47.3 μ A/V 2
VGS 4 = 5 − 2 = 3 V
2
80 = K n 4 (3 − 0.7 ) ⇒ K n 4 = 15.12 μ A/V 2
______________________________________________________________________________________
2
TYU3.17
2
2
I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V
VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V
VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V
VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU3.18
2
⎤⎦
I D = K ⎡⎣ 2 (VGS − VTN ) VDS − VDS
2
= 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤
⎣
⎦
ID = 9 μA
2.5 − 0.1
⇒ RD = 267 kΩ
0.009
______________________________________________________________________________________
RD =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 4
Exercise Solutions
EX4.1
⎞⎛ W ⎞
⎟⎜ ⎟ I DQ
⎟⎝ L ⎠
⎠
⎛ k'
g m = 2 ⎜⎜ n
⎝ 2
W
⎛ 0.1 ⎞⎛ W ⎞
1.8 = 2 ⎜
= 20.25
⎟⎜ ⎟(0.8) ⇒
L
⎝ 2 ⎠⎝ L ⎠
______________________________________________________________________________________
EX4.2
⎛ k ' ⎞⎛ W ⎞ ⎛ 0.1 ⎞
2
(a) K n = ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜
⎟(50) = 2.5 mA/V
2
2
L
⎠
⎝ ⎠⎝ ⎠ ⎝
2
I DQ = K n VGSQ − VTN
(
0.25 = 2.5(V
GSQ
)
− 0.4)
2
⇒ VGSQ = 0.716 V
V DSQ = V DD − I DQ R D = 3.3 − (0.25 )(10 ) = 0.8 V
V DS (sat ) = VGS − VTN = 0.717 − 0.4 = 0.316 V ⇒ V DS > V DS (sat )
(b)
g m = 2 K n I DQ = 2 (2.5)(0.25) = 1.58 mA/V
ro =
(c) Aυ
1
=
1
= 160 k Ω
(0.025)(0.25)
= − g m (ro RD ) = −(1.58)(160 10) = −14.9
λI DQ
______________________________________________________________________________________
EX4.3
⎛ R2 ⎞
⎛ 320 ⎞
VGS = ⎜
⎟ VDD = ⎜
⎟ ( 5 ) = 1.905 V
⎝ 520 + 320 ⎠
⎝ R1 + R2 ⎠
I DQ = 0.20 (1.905 − 0.8 ) = 0.244 mA
2
g m = 2 K n I DQ = 2
(a)
(b)
(c)
( 0.2 )( 0.244 ) = 0.442 mA/V
ro = ∞
Av = − g m RD = − ( 0.422 )(10 ) = −4.22
Ri = R1 R2 = 520 320 = 198 k Ω
(d) RO = RD = 10 K
______________________________________________________________________________________
EX4.4
At transition point, I D = 1 mA
I D = K n (VGSt − VTN ) = K n (VDS ( sat ) )
2
2
1 = 0.2 (VDS ( sat ) ) ⇒ VDS ( sat ) = 2.236 V
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
5 − 2.236
+ 2.236 = 3.62 V
Want V DSQ =
2
5 − 3.62
RD =
= 2.76 k Ω
0.5
(
)
0.5 = 0.2 VGSQ − 0.8 ⇒ VGSQ = 2.38 V
2
⎛ R2 ⎞
1
⎟⎟V DD =
(R1 R2 )V DD
VGSQ = ⎜⎜
+
R
R
R
2 ⎠
1
⎝ 1
1
So 2.38 = ( 200 )( 5 ) ⇒ R1 = 420 K and R2 = 382 K
R1
Av = − g m RD
( 0.2 )( 0.5) = 0.6325 mA/V
Av = − ( 0.6325 )( 2.76 )
g m = 2 K n I DQ = 2
= −1.75
______________________________________________________________________________________
EX4.5
⎛ 35 ⎞
(a) VG = ⎜
⎟(10) − 5 = −3.25 V
⎝ 35 + 165 ⎠
VG = VGS + I D RS − 5
So
2
5 − 3.25 = VGS + K n R S (VGS − VTN )
(
2
1.75 = VGS + (1)(0.5) VGS
− 1.6VGS + 0.64
)
or
2
0.5VGS
+ 0.2VGS − 1.43 = 0 ⇒ VGS = 1.503 V
Then
I DQ = (1)(1.503 − 0.8) = 0.4942 mA
2
V DSQ = 10 − (0.4942 )(7 + 0.5) = 6.29 V
(b)
g m = 2 K n I DQ = 2 (1)(0.4942) = 1.406 mA/V
− g m RD
− (1.406)(7 )
=
= −5.78
1 + g m RS 1 + (1.406 )(0.5)
______________________________________________________________________________________
Aυ =
EX4.6
⎛ k p'
(a) K p = ⎜
⎜ 2
⎝
⎞⎛ W ⎞ ⎛ 0.04 ⎞
2
⎟⎜ ⎟ = ⎜
⎟(40 ) = 0.80 mA/V
⎟⎝ L ⎠ ⎝ 2 ⎠
⎠
(
3 = K p RS VSGQ + VTP
(
3 = (0.8)(1.2) V
2
SGQ
)
2
+ VSGQ
)
− 0.8VSGQ + 0.16 + VSGQ
or
2
0.96V SGQ
+ 0.232V SGQ − 2.846 = 0 ⇒ V SGQ = 1.605 V
I DQ = (0.8)(1.605 − 0.4) = 1.162 mA
2
V SDQ = 6 − (1.162 )(1.2 + 2 ) = 2.283 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)
g m = 2 K p I DQ = 2 (0.8)(1.162) = 1.928 mA/V
ro =
Aυ
1
1
=
= 43.03 k Ω
(0.02)(1.162)
= − g m (ro RD ) = −(1.928)(43.03 2) = −3.68
λI DQ
______________________________________________________________________________________
EX4.7
VDSQ = VDD − I DQ RS
5 = 10 − (1.5 ) RS ⇒ RS = 3.33 kΩ
I DQ = K n (VGS − VTN ) ⇒ 1.5 = (1)(VGS − 0.8 )
2
2
⎛ R2
VGS = 2.025 V = VG − VS = VG − 5 ⇒ VG = 7.025 V = ⎜
⎝ R1 + R2
So
⎞
R2
⋅ 10
⎟ VDD =
400
⎠
R2 = 281 kΩ, R1 = 119 kΩ
Neglecting R Si , Aυ =
[
ro = λI DQ
]
−1
g m (R S ro )
1 + g m (R S ro )
= [(0.015)(1.5)] = 44.4 k Ω
−1
R S ro = 3.33 44.4 = 3.1 k Ω
g m = 2 K n I DQ = 2 (1)(1.5) = 2.45 mA/V
Aτ =
(2.45)(3.1) ⇒ A
υ
1 + (2.45)(3.1)
= 0.884
______________________________________________________________________________________
EX4.8
(a) VSG = V DD − I D RS = 5 − (1.5)(2 ) = 2 V
⎛ k 'p ⎞⎛ W ⎞
2
I D = ⎜ ⎟⎜ ⎟(V SG + VTP )
⎜ 2 ⎟⎝ L ⎠
⎝ ⎠
W
⎛ 0.04 ⎞⎛ W ⎞
2
1.5 = ⎜
= 117
⎟⎜ ⎟(2 − 1.2) ⇒
L
⎝ 2 ⎠⎝ L ⎠
(b)
⎛ k p'
Kp = ⎜
⎜ 2
⎝
⎞⎛ W ⎞ ⎛ 0.04 ⎞
2
⎟⎜ ⎟ = ⎜
⎟(117 ) = 2.344 mA/V
⎟⎝ L ⎠ ⎝ 2 ⎠
⎠
g m = 2 K p I D = 2 (2.344)(1.5) = 3.75 mA/V
Aυ =
g m RS
(3.75)(2) = 0.882
=
1 + g m RS 1 + (3.75)(2)
g m (RS R L )
(c)
Aυ =
Then
(0.794)[1 + (3.75)(RS
1 + g m (RS R L )
= (0.9 )(0.882) = 0.794
]
RL ) = (3.75)(RS RL ) ⇒ RS RL = 2 RL = 1.028
So
R L = 2.12 k Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX4.9
⎛ R2 ⎞
9.3 ⎞
⎛
VG = ⎜
⎟ VDD = ⎜
⎟ (5)
⎝ 70.7 + 9.3 ⎠
⎝ R1 + R2 ⎠
= 0.581 V
I DQ = K p (VSG − VTP
)
= K P (VS − VG − VTP
2
=
)
2
5 − VS
RS
Then ( 0.4 )( 5 )(VS − 0.581 − 0.8 ) = 5 − VS
2
2 (VS − 1.381) = 5 − VS
2
2 (VS2 − 2.762VS + 1.907 ) = 5 − VS
2VS2 − 4.52VS − 1.19 = 0
VS =
4.52 ±
(4.52 )2 + 4(2)(1.19 )
2(2 )
V S = 2.50 V ⇒ I DQ =
5 − 2.5
= 0.5 mA
5
g m = 2 K p I DQ = 2 (0.4)(0.5) = 0.894 mA/V
Aυ =
=
R1 R 2
g m RS
⋅
1 + g m R S R1 R 2 + R Si
(0.894)(5) ⋅ 70.7 9.3 ⇒ A
υ
1 + (0.894 )(5) 70.7 9.3 + 0.5
= 0.770
Neglecting RSi , Av = 0.817
1
1
Ro = R S
=5
= 5 1.12 ⇒ R o = 0.915 k Ω
0.894
gm
______________________________________________________________________________________
EX4.10
VO = g mV sg (R D R L ) and V sg = V i
Aυ = g m (R D R L )
I DQ =
5 − V SG
= K p (V SG − VTP
RS
5 − V SG = (1)(4)(V SG − 0.8)
(
4V
2
2
2
5 − V SG = 4 V SG
− 1.6V SG + 0.64
2
SG
)
)
− 5.4V SG − 2.44 = 0
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
V SG =
5. 4 ±
(5.4)2 + (4)(4)(2.44)
2(4 )
V SG = 1.71 V
I DQ =
5 − 1.71
= 0.822 mA
4
g m = 2 K p I DQ = 2 (1)(0.822) = 1.81 mA/V
Aυ = (1.81)(2 4) = (1.81)(1.33) ⇒ Aυ = 2.41
1
1
=4
= 4 0.552 ⇒ Rin = 0.485 k Ω
1.81
gm
Rin = R S
______________________________________________________________________________________
EX4.11
(W L )D
(W L )L
Aυ = 8 =
(a)
(W L )D
=
1. 2
⎛
(b) VGSDt =
(V DD − VTNL ) + VTND ⎜⎜1 +
⎝
1+
⎛W ⎞
⇒ ⎜ ⎟ = 76.8
⎝ L ⎠D
K nD
K nL
K nD
K nL
⎞
⎟
⎟ (3.3 − 0.4) + (0.4)(1 + 8)
⎠ =
1+ 8
or VGSDt = 0.7222 V
So
0.7222 − 0.4
VGSDQ =
+ 0.4 = 0.561 V
2
______________________________________________________________________________________
EX4.12
Aυ = − g mD (roD roL )
roD = roL =
1
1
=
= 500 k Ω
λI DQ (0.02 )(0.1)
g mD = 2 K nD I DQ = 2 (0.25)(0.1) = 0.3162 mA/V
Then
Aυ = −(0.3162) 500 500 = −79.1
(
)
______________________________________________________________________________________
EX4.13
Aυ = − g m ron rop
(
ron = rop =
)
1
(0.015)(0.1)
= 666.7 k Ω
− 250 = − g m (666.7 666.7 )
g m = 0.75 mA/V = 2 K n I DQ = 2 K n (0.1)
k n′ ⎛ W ⎞ ⎛ 0.080 ⎞⎛ W ⎞
⎛W ⎞
⎜ ⎟=⎜
⎟⎜ ⎟ ⇒ ⎜ ⎟ = 35.2
2 ⎝ L ⎠ ⎝ 2 ⎠⎝ L ⎠
⎝ L ⎠1
______________________________________________________________________________________
K n = 1.406 mA/V 2 =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX4.14
(a)
1
1
ro 2 ro1 ≈
g m1
g m1
RO =
1 1
= = 0.5 mA/V
R0 2
So g m1 =
g m1 = 2 K n I D
( 0.2 ) I D
0.5 = 2
(b)
⇒ I D = 0.3125 mA
g m1 (ro1 ro 2 )
Aυ =
1 + g m1 (ro1 ro 2 )
1
= 320 k Ω
(0.01)(0.3125)
(0.5)(320 320)
ro1 = ro 2 =
Aυ =
1 + (0.5)(320 320 )
Aυ = 0.988
______________________________________________________________________________________
EX4.15
(a)
1
ro1 2 K n I D + λ1 I D
Av =
=
1
1
λ2 I D + λ1 I D
+
ro 2 ro1
g m1 +
120 =
2 0.2 I D + 0.01I D
0.01I D + 0.01I D
2.4 I D − 0.01I D = 2 0.2 I D
2.39 I D = 2 0.2 ⇒ I D = 0.140 mA
g m1 = 2
( 0.2 )( 0.14 ) ⇒ g m1 = 0.335 mA/V
(b)
Ro = ro1 ro 2
ro1 = ro 2 =
1
= 714 k Ω
(0.01)(0.14 )
Ro = 714 714 = 357 k Ω
______________________________________________________________________________________
EX4.16
Ro = R S 2
1
g m2
g m 2 = 0.632 mA/V, R S 2 = 8 k Ω
1
= 8 1.58 ⇒ R o = 1.32 k Ω
0.632
______________________________________________________________________________________
Ro = 8
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX4.17
54.6
⎛ 54.6 ⎞
⎞
⎛
(a) VG1 = ⎜
⎟(5) = 0.91 V
⎟(5) = ⎜
⎝ 300 ⎠
⎝ 54.6 + 150 + 95.4 ⎠
⎛ 54.6 + 150 ⎞
VG 2 = ⎜
⎟(5) = 3.41 V
⎝ 300 ⎠
VG1 = VGS1 + K n1 R S (VGS1 − VTN ) − 5
2
(
5.91 = VGS1 + (3)(10) VGS2 1 − 1.6VGS1 + 0.64
)
or
2
30VGS
1 − 47VGS 1 + 13.29 = 0 ⇒ VGS 1 = 1.196 V
Then
2
I DQ = (3)(1.196 − 0.8) = 0.471 mA
V D1 = VG 2 − VGS 2 = 3.41 − 1.196 = 2.214 V
VS1 = VG1 − VGS1 = 0.91 − 1.196 = −0.286 V
Then
V DSQ1 = 2.214 − (− 0.286 ) = 2.5 V
V D 2 = 5 − (0.471)(2.5) = 3.8225 V
V DSQ 2 = 3.8225 − 2.214 = 1.61 V
(b)
g m1 = 2 K n1 I DQ = 2 (3)(0.471) = 2.377 mA/V
Aυ = − g m1 R D = −(2.377 )(2.5) = −5.94
______________________________________________________________________________________
EX4.18
V S = I DQ R S = (1.2 )(2.7 ) = 3.24 V
V D = V S + V DSQ = 3.24 + 12 = 15.24 V
RD =
20 − 15.24
⇒ R D = 3.97 k Ω
1.2
⎛ V
I D = I DSS ⎜⎜1 − GS
⎝ VP
⎞
⎟⎟
⎠
2
2
⎛ V ⎞
V
1.2 = 4⎜⎜1 − GS ⎟⎟ ⇒ GS = 0.4523
V
VP
P ⎠
⎝
VGS = (0.4523)(− 3) = −1.357 V
VG = V S + VGS = 3.24 − 1.357 = 1.883 V
⎛ R2 ⎞
⎛ R ⎞
⎟⎟(20 ) = ⎜⎜ 2 ⎟⎟(20 ) = 1.88 ⇒ R 2 = 47 k Ω , R1 = 453 k Ω
VG = ⎜⎜
⎝ 500 ⎠
⎝ R1 + R 2 ⎠
1
1
ro =
=
= 167 k Ω
λI DQ (0.005)(1.2)
gm =
2 I DSS ⎛ VGS
⎜1 −
(− V P ) ⎜⎝ V P
(
⎞ 2(4 ) ⎛ 1.357 ⎞
⎟⎟ =
⎟ = 1.46 mA/V
⎜1 −
3 ⎠
⎠ (3) ⎝
)
(
)
Aυ = − g m ro R D R L = −(1.46 ) 167 3.97 4 ⇒ Aυ = −2.87
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX4.19
a.
2
2
⎛ V ⎞
⎛ V ⎞
V
I DQ = I DSS ⎜ 1 − GS ⎟ 2 = 8 ⎜ 1 − GS ⎟ ⇒ GS = 0.5
V
V
VP
⎝
⎝
P ⎠
P ⎠
VGS = ( 0.5 )( −3.5 ) ⇒ VGS = −1.75
Also I DQ =
−VGS − ( −10 )
RS
2=
1.75 + 10
⇒ RS = 5.88 kΩ
RS
b.
2 I DSS ⎛ VGS ⎞ 2(8) ⎛ 1.75 ⎞
⎜1 −
⎟=
⎟ = 2.29 mA/V
⎜1 −
3.5 ⎠
V P ⎟⎠ 3.5 ⎝
V P ⎜⎝
1
= 50 k Ω
ro =
(0.01)(2)
Vi
Vi = V gs + g m R S V gs ⇒ V gs =
1 + g m RS
gm =
Aυ =
(2.29)(5.88 50)
g m (R S ro )
Vo
=
=
⇒ Aυ = 0.9234
Vi 1 + g m (R S ro ) 1 + (2.29 )(5.88 50 )
c.
Aυ =
(
g m R S R L ro
)
) = (0.80)(0.9234) = 0.7387
(2.29)(R R r )
(
1 + g m R S R L ro
0.7387 =
S
L
o
1 + (2.29)(R S R L ro )
⇒ R S R L ro = 1.235 k Ω
R S ro = 5.261 k Ω
(5.261)R L
= 1.235 ⇒ R L = 1.61 k Ω
5.261 + R L
______________________________________________________________________________________
Then
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU4.1
(a)
⎛ k'
g m = 2 ⎜⎜ n
⎝ 2
⎞⎛ W ⎞
⎟⎜ ⎟ I DQ
⎟⎝ L ⎠
⎠
⎤ W
⎡⎛ 0.1 ⎞⎛ W ⎞
= 4 ⎢⎜
= 26.0
⎟⎜ ⎟(1.2 )⎥ ⇒
L
L
2
⎠⎝ ⎠
⎦
⎣⎝
1
1
=
= 55.6 k Ω
(b) ro =
λI DQ (0.015)(1.2)
(2.5)2
______________________________________________________________________________________
TYU4.2
(a) I DQ = K n VGSQ − VTN
(
0.15 = 0.5(V
)
− 0.4)
2
2
GSQ
⇒ VGSQ = 0.948 V
V DSQ = 3.3 − (0.15 )(8) = 2.1 V
(b)
g m = 2 K n I DQ = 2 (0.5)(0.15) = 0.548 mA/V
ro =
Aυ
1
=
1
= 333 k Ω
(0.02)(0.15)
= − g m (ro RD ) = −(0.548)(333 8) = −4.28
λI DQ
______________________________________________________________________________________
TYU4.3
i D = I DQ + id = I DQ + g mυ gs
i D = 0.15 + (0.548)(0.025)sin ω t
or
i D = 0.15 + 0.0137 sin ω t (mA)
Also
υ DS = V DSQ + υ d = 2.1 − (0.0137 )(8)sin ω t
or
υ DS = 2.1 − (0.11)sin t (V)
______________________________________________________________________________________
TYU4.4
(a) V SDQ = V DD − I DQ R D
3 = 5 − I DQ (5)⇒ I DQ = 0.4 mA
(
I DQ = K p VSGQ + VTP
(
)
2
)
0.4 = 0.4 VSGQ − 0.4 ⇒ VSGQ = 1.4 V
(b)
2
g m = 2 K p I DQ = 2 (0.4)(0.4) = 0.8 mA/V
Aυ = − g m R D = −(0.8)(5) = −4
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU4.5
η=
η=
(a)
γ
2 2φ f + vSB
0.40
2 2 ( 0.35 ) + 1
η=
0.40
2 2 ( 0.35 ) + 3
g m = 2 K n I DQ = 2
(b)
For (a),
⇒ η = 0.153
⇒ η = 0.104
( 0.5)( 0.75) = 1.22 mA / V
g mb = g m η = (1.22 )( 0.153) ⇒ g mb = 0.187 mA / V
g = (1.22 )( 0.104 ) ⇒ g mb = 0.127 mA / V
For (b), mb
______________________________________________________________________________________
TYU4.6
a.
With RG ⇒ VGS = VDS ⇒ transistor biased in sat. region
2
2
I D = K n (VGS − VTN ) = K n (VDS − VTN )
VDS = VDD − I D RD = VDD − K n RD (VDS − VTN )
VDS = 15 − ( 0.15 )(10 )(VDS − 1.8 )
2
2
2
= 15 − 1.5 (VDS
− 3.6VDS + 3.24 )
2
− 4.4VDS − 10.14 = 0
1.5VDS
VDS =
4.4 ±
( 4.4 )
2
+ ( 4 )(1.5 )(10.14 )
2 (1.5 )
⇒ VDSQ = 4.45 V
I DQ = ( 0.15 )( 4.45 − 1.8 ) ⇒ I DQ = 1.05 mA
2
b.
Neglecting effect of RG:
Aυ = − g m (R D R L )
g m = 2 K n (VGS − VTN ) = 2(0.15)(4.45 − 1.8) ⇒ g m = 0.795 mA/V
Aυ = −(0.795)(10 5) ⇒ Aυ = −2.65
RG ⇒ establishes VGS = VDS ⇒ essentially no effect on small-signal voltage gain.
c.
______________________________________________________________________________________
TYU4.7
a.
5 = I DQ RS + VSG and I DQ = K p (VSG + VTP ) 2
0.8 = 0.5(VSG + 0.8) 2 ⇒ VSG = 0.465 V
5 = ( 0.8 ) RS + 0.465 ⇒ RS = 5.67 kΩ
VSDQ = 10 − I DQ ( RS + RD )
3 = 10 − ( 0.8 )( 5.67 + RD )
RD =
10 − ( 0.8 )( 5.67 ) − 3
0.8
⇒ RD = 3.08 kΩ
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
b.
Vo = g mV sg (R D ro ) = − g mVi (R D ro )
Vo
= − g m (R D ro )
Vi
g m = 2 K p (V SG + VTP ) = 2(0.5)(0.465 + 0.8) = 1.265 mA/V
Aυ =
ro =
Aυ
1
1
=
= 62.5 k Ω
(0.02)(0.8)
= −(1.265)(3.08 62.5) ⇒ Aυ
λI O
= −3.71
______________________________________________________________________________________
TYU4.8
(a)
V0 = g mVgs r0
Vi = Vgs + V0 ⇒ Vgs = Vi − V0
So V0 = g m r0 (Vi − V0 )
Av =
( 4 )( 50 )
V0
g m r0
=
=
⇒ Av = 0.995
Vi 1 + g m r0 1 + ( 4 )( 50 )
I x + g mVgs =
I x = g mVx +
(b)
Vx
and Vgs = −Vx
r0
Vx
1
1
⇒ R0 = r0
= 50 ⇒ R0 ≅ 0.25 kΩ
r0
gm
4
With R S = 4 k Ω ⇒ Aυ =
g m (ro R S )
1 + g m (ro R S )
r0 || Rs = 50 || 4 = 3.7 kΩ ⇒ Av =
( 4 )( 3.7 )
⇒ Av = 0.937
1 + ( 4 )( 3.7 )
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU4.9
(a)
g m = 2 K n I DQ ⇒ 2 = 2 K n ( 0.8 ) ⇒ K n = 1.25 mA / V 2
μ n Cox W
Kn =
⋅
2
W
So
= 62.5
L
⎛W ⎞
⇒ 1.25 = ( 0.020 ) ⎜ ⎟
L
⎝L⎠
I DQ = K n (VGS − VTN ) ⇒ 0.8 = 1.25 (VGS − 2 ) ⇒ VGS = 2.8 V
2
b.
[
ro = λI DQ
Aυ =
]
−1
2
= [(0.01)(0.8)] = 125 k Ω
−1
g m (ro R L )
1 + g m (ro R L )
ro R L = 125 4 = 3.88 k Ω
(2)(3.88) ⇒ A
υ
1 + (2)(3.88)
Aυ =
= 0.886
1
1
ro = 125 ⇒ R o ≅ 0.5 k Ω
2
gm
______________________________________________________________________________________
Ro =
TYU4.10
Rin =
1
= 0.35 k Ω ⇒ g m = 2.86 mA/V
gm
Vo
= R D R L = 2.4 = R D 4 ⇒ R D = 6 k Ω
Ii
g m = 2 K n I DQ
2.86 = 2 K n (0.5) ⇒ K n = 4.09 mA/V 2
I DQ = K n (VGS − VTN )
2
0.5 = 4.09(VGS − 1) ⇒ VGS = 1.35 V ⇒ V S = −1.35 V
V D = 5 − (0.5)(6 ) = 2 V
V DS = V D − V S = 2 − (− 1.35) = 3.35 V
We have VDS = 3.35 > VGS − VTN = 1.35 − 1 = 0.35 V ⇒ Biased in the saturation region
2
______________________________________________________________________________________
TYU4.11
K n1 =
Kn2
μ n Cox ⎛ W ⎞
⋅⎜
⎝L
μ n Cox ⎛ W
=
⋅⎜
2 ⎝L
Av = −
2
2
⎟ = ( 0.020 )( 80 ) = 1.6 mA / V
⎠1
⎞
2
⎟ = ( 0.020 )(1) = 0.020 mA / V
⎠2
K n1
1.6
=−
⇒ Av = −8.94
Kn2
0.020
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
The transition point is determined from vGSt − VTND = VDD − VTNL −
vGSt − 0.8 = ( 5 − 0.8 ) − ( 8.94 )( vGSt − 0.8 )
vGSt =
K n1
( vGSt − VTND )
Kn2
( 5 − 0.8) + (8.94 )( 0.8) + 0.8
1 + 8.94
vGSt = 1.22 V
1.22 − 0.8
+ 0.8 ⇒ VGS = 1.01 V
2
For Q-point in middle of saturation region
______________________________________________________________________________________
VGS =
TYU4.12
(a)
⎛ R2
VG1 = ⎜⎜
⎝ R1 + R2
⎞
⎛ 135 ⎞
⎟⎟(10 ) − 5 = ⎜
⎟(10 ) − 5 = −2.394 V
⎝ 135 + 383 ⎠
⎠
VG1 = VGS1 + K n1 RS1 (VGS1 − VTN ) − 5
2
or
(
2
5 − 2.394 = VGS1 + (1.5)(3.9) VGS
1 − 1.2VGS 1 + 0.36
)
so
2
5.85VGS
1 − 6.02VGS 1 − 0.5 = 0 ⇒ VGS 1 = 1.106 V
Then
2
I DQ1 = (1.5)(1.106 − 0.6 ) = 0.3845 mA
V DSQ1 = 10 − (0.3845 )(3.9 + 16.1) = 2.31 V
VG 2 = 5 − (0.3845)(16.1) = −1.190 V
VG 2 = VGS 2 + K n 2 RS 2 (VGS 2 − VTN ) − 5
2
or
(
2
5 − 1.19 = VGS 2 + (2)(8) VGS
2 − 1.2VGS 2 + 0.36
)
so
2
16VGS
2 − 18.2VGS 2 + 1.95 = 0 ⇒ VGS 2 = 1.018 V
Then
I DQ 2 = (2 )(1.018 − 0.6) = 0.349 mA
2
V DSQ 2 = 10 − (0.349 )(8) = 7.208 V
(b)
g m1 = 2 K n1 I DQ1 = 2 (1.5)(0.3845) = 1.519 mA/V
g m 2 = 2 K n 2 I DQ 2 = 2 (2)(0.349) = 1.671 mA/V
From Example 4.16
− g m1 g m 2 R D1 (RS 2 R L )
Ri
Aυ =
⋅
1 + g m 2 (RS 2 RL )
Ri + RSi
=
− (1.519 )(1.671)(16.1)(8 4)
1 + (1.671)(8 4)
⋅
99.8
99.8 + 4
or
Aυ = −19.2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c)
1
1
RS 2 =
8 = 0.5984 8 ⇒ Ro = 557 Ω
g m2
1.671
______________________________________________________________________________________
Ro =
TYU4.13
From Example 6.18
g m = 3.0 mA/V, ro = 41.7 k Ω
R1 R2 = 420 180 = 126 k Ω
V gs =
Aυ =
⎛ 126 ⎞
⋅ Vi = ⎜
⎟ ⋅ Vi = 0.863 ⋅ Vi
R1 R 2 + Ri
⎝ 126 + 20 ⎠
R1 R 2
(
− g mV gs ro R D R L
Vi
)
(
)
= −(3.0 )(0.863) 41.7 2.7 4 = −(2.589 )(1.55) ⇒ Aυ = −4.01
_____________________________________________________________________________
TYU4.14
a.
⎛ R2 ⎞
VG1 = ⎜
⎟ (VDD )
⎝ R1 + R2 ⎠
⎛ 430 ⎞
VG1 = ⎜
⎟ ( 20 ) = 17.2 V
⎝ 430 + 70 ⎠
2
⎛ V ⎞
⎛ V −V ⎞
I DQ1 = I DSS ⎜1 − GS ⎟ = 6 ⎜1 − G1 S 1 ⎟
V
2
⎝
⎠
⎝
P ⎠
2
20 − VS1
⎛ 17.2 VS1 ⎞
⎛V
⎞
= 6 ⎜1 −
+
= 6 ⎜ S 1 − 7.6 ⎟ and I DQ1 =
⎟
2
2 ⎠
4
⎝
⎝ 2
⎠
2
2
20 − VS 1
⎛V
⎞
= 6 ⎜ S 1 − 7.6 ⎟
4
⎝ 2
⎠
2
⎛V
⎞
20 − VS1 = 24 ⎜ S 1 − 7.6VS1 + 57.76 ⎟
⎝ 4
⎠
2
Then
= 6VS21 − 182.4VS 1 + 1386.24
6VS21 − 181.4VS 1 + 1366.24 = 0
VS1 =
181.4 ±
(181.4 ) − 4 ( 6 )(1366.24 )
2 (6)
2
VS1 = 14.2 V ⇒ VGS 1 = 17.2 − 14.2 = 3 V > VP
So VS1 = 16.0 ⇒ VGS1 = 17.2 − 16 = 1.2 < VP − Q
20 − 16
on I DQ1 =
⇒ I DQ1 = 1 mA
4
VSDQ1 = 20 − I DQ1 ( RS 1 + RD1 )
= 20 − (1)( 8 ) ⇒ VSDQ1 = 12 V
VD1 = I DQ1 RD1 = (1)( 4 ) = 4 V = VG 2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ V − VS 2 ⎞
⎛ V ⎞
= I DSS ⎜ 1 − GS ⎟ = 6 ⎜⎜1 − G 2
⎟
VP ⎠
( −2 ) ⎟⎠
⎝
⎝
2
I DQ 2
2
2
2
V
V
⎛ 4 V ⎞
⎛ V ⎞
= 6 ⎜ 1 + − S 2 ⎟ = 6 ⎜ 3 − S 2 ⎟ and I DQ 2 = S 2 = S 2
R
2
2
2
4
⎝
⎠
⎝
⎠
S2
V
⎛ V ⎞
Then S 2 = 6 ⎜ 3 − S 2 ⎟
4
2 ⎠
⎝
2
⎛
V2 ⎞
VS 2 = 24 ⎜ 9 − 3VS 2 + S 2 ⎟
4 ⎠
⎝
2
6VS 2 − 73VS 2 + 216 = 0
73 ±
VS 2 =
( 73)
2
− 4 ( 6 )( 216 )
2 (6)
⇒ VS 2 = 7.09 V or = 5.08 V
For VS 2 = 5.08 V ⇒ VGS 2 = 4 − 5.08 = −1.08 transistor biased ON
5.08
⇒ I DQ 2 = 1.27 mA
4
= 20 − VS 2 = 20 − 5.08 ⇒ VDS 2 = 14.9 V
I DQ 2 =
VDS 2
b.
V g 2 = g m1V sg1 R D1 = − g m1Vi R D1
Vo = g m 2V gs 2 (R S 2 R L )
V g 2 = V gs 2 + V o ⇒ V gs 2 =
Aυ =
Vg 2
1+ g m 2 (R S 2 R L )
Vo
− g m1 R D1
=
Vi 1+ g m 2 (R S 2 R L )
2 I DSS ⎛ VGS ⎞
⎜1 −
⎟
VP ⎝
VP ⎠
2 ( 6 ) ⎛ 1.2 ⎞
=
⎜1 −
⎟ = 2.4 mA/V
2 ⎝
2 ⎠
2 ( 6 ) ⎛ 1.08 ⎞
gm2 =
⎜1 −
⎟ = 2.76 mA/V
2 ⎝
2 ⎠
− (2.4 )(4 )
Then Aυ =
= −2.05
1 + (2.76 )(4 2 )
g m1 =
_____________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 5
Exercise Solutions
EX5.1
I E = (1 + β )I B
IE
1.20
=
= 141.2 ⇒ β = 140.2
I B 0.0085
β
140.2
α=
=
= 0.9929
1 + β 141.2
I C = I E − I B = 1.20 − 0.0085 = 1.1915 mA
______________________________________________________________________________________
1+ β =
EX5.2
BVCEO =
BVCBO
β
=
200
or BVCEO = 40.5 V
______________________________________________________________________________________
EX5.3
IB =
n
3
120
V BB − V BE (on ) 2 − 0.7
=
⇒ I B = 3.02 μ A
430
RB
I C = βI B = (150)(3.02 ) μ A ⇒ I C = 0.453 mA
VCE = VCC − I C RC = 3.3 − (0.453)(3.2 ) = 1.85 V
P ≅ I C VCE = (0.453)(1.85) = 0.838 mW
______________________________________________________________________________________
EX5.4
IB =
V + − V EB (on ) − V BB 3.3 − 0.7 − 1.2
=
⇒ I B = 3. 5 μ A
400
RB
I C = β I B = (80 )(3.5) μ A ⇒ I C = 0.28 mA
VEC = V + − I C RC = 3.3 − (0.28)(5.25) = 1.83 V
______________________________________________________________________________________
EX5.5
(a) I B =
V + − V EB (on ) − V BB 3.3 − 0.7 − 2
=
⇒ IB = 4μ A
RB
150
I C = βI B = (110)(4 ) μ A ⇒ I C = 0.44 mA
VEC = V + − I C RC = 3.3 − (0.44)(5) = 1.1 V
(b)
3.3 − 0.7 − 1
⇒ I B = 10.7 μ A
150
V + − V EC (sat ) 3.3 − 0.2
=
= 0.62 mA
IC =
5
RC
IB =
V EC = 0.2 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX5.6
For 0 ≤ VI < 0.7 V , Q is cutoff, VO = 9 V
n
0.2 = 9 −
(100 )(VI − 0.7 )( 4 )
When Qn is biased in saturation, we have
So, for VI ≥ 5.1 V , V = 0.2 V
200
⇒ VI = 5.1 V
O
______________________________________________________________________________________
EX5.7
VBB = I B RB + VBE (on) + I E RE + V −
I E = (1 + β )I B
So
3.3 − 0.7
⇒ I B = 3.116 μ A
640 + (81)(2.4 )
I C = βI B = (80 )(3.116) μ A ⇒ I C = 0.249 mA
IB =
⎛1+ β ⎞
⎛ 81 ⎞
⎟⎟ I C = ⎜ ⎟(0.249 ) = 0.252 mA
I E = ⎜⎜
β
⎝ 80 ⎠
⎝
⎠
VCE = [3.3 − (− 3.3)] − (0.249)(10) − (0.252)(2.4 ) = 3.51 V
______________________________________________________________________________________
EX5.8
I EQ =
V + − V EB (on )
3 − 0. 7
⇒ RE =
= 18.4 k Ω
0.125
RE
VC = V EB (on ) − V ECQ = 0.7 − 2.2 = −1.5 V
⎛ β ⎞
⎛ 110 ⎞
⎟⎟ I EQ = ⎜
I CQ = ⎜⎜
⎟(0.125) = 0.1239 mA
+
β
1
⎝ 111 ⎠
⎝
⎠
V −V −
− 1.5 − (− 3)
RC = C
=
= 12.1 k Ω
I CQ
0.1239
______________________________________________________________________________________
EX5.9
(a)
(b)
(c)
5 = I E RE + VEB ( on ) + I B RB − 2
180 ⎞
⎛
5 + 2 − 0.7 = I E ⎜ 2 +
⎟ I E = 0.9859 mA
41 ⎠
⎝
I C = 0.962 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.2725 mA
61 ⎠
⎝
I C = 1.25 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.6657 mA
101 ⎠
⎝
I C = 1.64 mA
180 ⎞
⎛
6.3 = I E ⎜ 2 +
⎟ I E = 1.97365 mA
151 ⎠
⎝
I C = 1.94 mA
(d)
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX5.10
IE =
VBB − VEB ( on )
RE
4 − 0.7
1.0
⇒ RE =
or RE = 3.3 k Ω
I C = α I E = ( 0.992 )(1) = 0.992 mA
I B = I E − I C = 1.0 − 0.992 or I B = 8 μ A
VCB = I C RC − VCC = ( 0.992 )(1) − 5
or VCB = −4.01 V
______________________________________________________________________________________
EX5.11
(a) R1 =
V + − Vγ − VCE (sat )
I C1
=
5 − 1.5 − 0.2
⇒ R1 = 220 Ω
15
I C1 15
=
= 0.30 mA
50 50
5 − 0.7
=
= 14.3 k Ω
0.3
I B1 =
R B1
(b)
IC2
2
=
= 0.08 A
25 25
12 − 0.7 − 0
= 141 Ω
RB2 =
0.08
______________________________________________________________________________________
I B2 =
EX5.12
(a) For V1 = V2 = 0, All currents are zero and VO = 5 V.
(b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0,
5 − 0.7
= 4.53 mA
I B1 =
0.95
5 − 0.2
⇒ I C 1 = I R = 8 mA
I C1 =
0.6
VO = 0.2 V
(c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V
______________________________________________________________________________________
EX5.13
In active region,
υ O = mυ I + b ⇒ m = −6.5
At υ I = 0.7 V, υ O = 5 V
5 = −6.5(0.7 ) + b ⇒ b = 9.55
Then
υ O = −6.5υ I + 9.55
When
υ O = 0.2 = −6.5υ I + 9.55 ⇒ υ I = 1.438 V
Q-point
1.438 − 0.7
+ 0.7 = 1.069 V
υ IQ =
2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
υ OQ =
5 − 0.2
+ 0.2 = 2.6 V
2
Now
1.069 − 0.7
⇒ 4.61 μ A
80
I CQ = β I BQ = (120 )(4.61) μ A ⇒ I CQ = 0.5535 mA
At Q-point
υ OQ = 5 − I CQ RC
I BQ =
2.6 = 5 − (0.5535)RC ⇒ RC = 4.34 k Ω
______________________________________________________________________________________
EX5.14
RC =
VCC − VCEQ
I CQ
I CQ
=
2.8 − 1.4
= 11.7 k Ω
0.12
0.12
⇒ I BQ = 0.80 μ A
150
β
VCC − V BEQ 2.8 − 0.7
=
= 2.625 M Ω
RB =
0.80
I BQ
I BQ =
=
______________________________________________________________________________________
EX5.15
(a) RTH = R1 R2 = 85 35 = 24.8 k Ω
⎛ R2 ⎞
⎛ 35 ⎞
⎟⎟(VCC ) = ⎜
VTH = ⎜⎜
⎟(3.3) = 0.9625 V
⎝ 35 + 85 ⎠
⎝ R1 + R2 ⎠
(b) VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E
so
VTH − V BE (on )
0.9625 − 0.7
=
⇒ I BQ = 2.617 μ A
RTH + (1 + β )R E 24.8 + (151)(0.5)
= β I BQ = (150 )(0.002617 ) = 0.3926 mA
I BQ =
I CQ
⎛1+ β ⎞
⎛ 151 ⎞
⎟⎟ I CQ = ⎜
I EQ = ⎜⎜
⎟(0.3926 ) = 0.3952 mA
⎝ 150 ⎠
⎝ β ⎠
VCEQ = 3.3 − (0.3926 )(4 ) − (0.3952 )(0.5) = 1.53 V
0.9625 − 0.7
⇒ I BQ = 4.18 μ A
24.8 + (76 )(0.5)
Then I CQ = (75 )(0.00418 ) = 0.3135 mA
(c) I BQ =
I EQ = (76 )(0.00418 ) = 0.3177 mA
VCEQ = 3.3 − (0.3135 )(4 ) − (0.3177 )(0.5) = 1.89 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX5.16
VCEQ ≅ VCC − I CQ ( RC + RE )
or 2.5 ≅ 5 − I CQ (1 + 0.2 )
which yields
I CQ = 2.08 mA,
RTH
I CQ
2.08
= 0.0139 mA
150
β
= ( 0.1)(1 + β ) RE = ( 0.1)(151)( 0.2 )
I BQ =
=
or RTH = 3.02 k Ω
⎛ R2 ⎞
1
Now VTH = ⎜
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
⎝ 1
2 ⎠
1
1
so VTH = ( 3.02 )( 5 )
R1
V = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
We can write TH
1
( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 )
R1
or
We obtain R1 = 13 k Ω and then R2 = 3.93 k Ω
______________________________________________________________________________________
EX5.17
I CQ =
V + − VO 5 − 0
=
= 0.5 mA
10
RC
⎛1+ β ⎞
⎛ 151 ⎞
⎟⎟ I CQ = ⎜
I EQ = ⎜⎜
⎟(0.5) = 0.5033 mA
β
⎝ 150 ⎠
⎠
⎝
VCEQ = V + − V − − I CQ RC − I EQ R E
(
)
= 10 − (0.5)(10 ) − (0.5033)(2 ) = 3.99 V
Now
RTH
I CQ
0.5
⇒ I BQ = 3.33 μ A
150
= (0.1)(1 + β )RE = (0.1)(151)(2 ) = 30.2 k Ω
I BQ =
β
=
⎛ R2
VTH = ⎜⎜
⎝ R1 + R2
Also
⎞
1
⎟⎟(10 ) − 5 =
(RTH )(10) − 5 = 1 (30.2)(10) − 5
R1
R1
⎠
VTH = I BQ RTH + V BE (on ) + I EQ R E − 5
= (0.00333)(30.2 ) + 0.7 + (0.5033)(2) − 5 = −3.193 V
Then
1
(30.2)(10) − 5 = −3.193
R1
or
R1 = 167 k Ω and 167 R2 = 30.2 ⇒ R2 = 36.9 k Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX5.18
V EO = −0.7 V, VCO = −0.7 + 1.6 = 0.9 V
RC =
V + − VCO 3.3 − 0.9
=
= 20 k Ω
I CQO
0.12
⎛1+ β ⎞
⎛ 61 ⎞
⎟⎟ I CQO = ⎜ ⎟(0.12 ) = 0.122 mA
I Q = ⎜⎜
β
⎝ 60 ⎠
⎝
⎠
⎛
2⎞
2 ⎞
⎛
I 1 = ⎜⎜1 + ⎟⎟ I Q = ⎜1 + ⎟(0.122 ) = 0.126 mA
60
β
⎠
⎝
⎠
⎝
0 − V BE (on ) − (− 3.3) 3.3 − 0.7
=
⇒ R1 = 20.6 k Ω
I 1 = 0.126 =
R1
R1
______________________________________________________________________________________
EX5.19
RTH 50 || 100 = 33.3 k Ω
⎛ 50 ⎞
VTH = VTH = ⎜
⎟ (10 ) − 5 = −1.67 V
⎝ 50 + 100 ⎠
−1.67 − 0.7 − ( −5 )
I B1 =
⇒ 11.2 μ A
33.3 + (101)( 2 )
I C1 = 1.12 mA, I E1 = 1.13 mA
VE1 = I E1 RE1 − 5 = (1.13)( 2 ) − 5 = −2.74 V
VCE1 = 3.25 V ⇒ VC1 = 0.51 V
Now VE 2 = 0.51 + 0.7 = 1.21 V
5 − 1.21
= 1.90 mA ⇒ I B 2 = 18.8 μ A
2
= 1.88 mA
IE2 =
IC 2
I R1 = I C1 − I B 2 = 1.12 − 0.0188 = 1.10 mA
5 − 0.51
RC1 =
= 4.08 k Ω
1.10
VEC 2 = 2.5 ⇒ VC 2 = VE 2 − VEC 2
= 1.21 − 2.5 = −1.29 V
RC 2 =
−1.29 − ( −5 )
= 1.97 k Ω
1.88
______________________________________________________________________________________
EX5.20
12
= 240 k Ω = R1 + R2 + R3
0.05
Then VB1 = ( 0.5 )( 2 ) + 0.7 = 1.7 V
We find
1.7
= 34 k Ω
0.05
= ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 V
R3 =
Also VB 2
ΔVR 2 = 5.7 − 1.7 = 4 V
so R2 =
4
= 80 k Ω
0.05
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
and R1 = 240 − 80 − 34 = 126 k Ω
VC 2 = 1 + 4 + 4 = 9 V
Then RC =
V + − VC 2 12 − 9
=
= 6 kΩ
0.5
I CQ
______________________________________________________________________________________
Test Your Understanding Solutions
TYU5.1
(a) α =
β
=
60
= 0.9836
61
1+ β
150
= 0.9934
α=
151
0.982
α
=
= 54.6
(b) β =
1 − α 1 − 0.982
0.9925
= 132.3
β=
1 − 0.9925
______________________________________________________________________________________
TYU5.2
I E = I C + I B = 0.620 + 0.005 = 0.625 mA
I C 0.620
=
= 124
I B 0.005
β
124
α=
=
= 0.992
1 + β 125
______________________________________________________________________________________
β=
TYU5.3
I C = αI E = (0.9915)(1.20 ) = 1.19 mA
0.9915
α
β=
=
= 116.6
1 − α 1 − 0.9915
I
1.20
IB = E =
⇒ I B = 10.2 μ A
1 + β 117.6
______________________________________________________________________________________
TYU5.4
VA
⇒ V A = (225)(0.8) = 180 V
IC
180
⇒ 2.25 M Ω
ro =
(b) (i)
0.08
180
= 22.5 k Ω
ro =
(ii)
8
______________________________________________________________________________________
(a)
ro =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU5.5
⎛ V ⎞
I C = I O ⎜1 + CE ⎟
VA ⎠
⎝
At VCE = 1 V , I C = 1 mA
1 ⎞
⎛
I C = 1 = I O ⎜1 + ⎟ ⇒ I O = 0.9868 mA
⎝ 75 ⎠
(a) For VA = 75 V ,
Then, at VCE = 10 V
⎛ 10 ⎞
I C = ( 0.9868 ) ⎜1 + ⎟ = 1.12 mA
⎝ 75 ⎠
1 ⎞
⎛
I C = 1 = I O ⎜1 +
⎟ ⇒ I O = 0.9934 mA
V
=
150
V
,
150
⎝
⎠
A
(b) For
10 ⎞
⎛
I C = ( 0.9934 ) ⎜1 +
⎟ = 1.06 mA
V
=
10
V
,
150
⎝
⎠
At CE
______________________________________________________________________________________
TYU5.6
BVCEO =
BVCBO
BVCBO = 3 100 ( 30 ) = 139 V
so
______________________________________________________________________________________
TYU5.7
(a)
(c)
and
For
n β
VI = 0.2 V < VBE ( on ) ⇒ I B = I C = 0, VO = 5 V and P = 0
For VI = 3.6 V , transistor is driven into saturation, so
V + − VCE ( sat ) 5 − 0.2
IC =
=
= 10.9 mA
RC
0.440
IB =
VI − VBE ( on )
RB
=
3.6 − 0.7
= 4.53 mA
0.64
I C 10.9
=
= 2.41 < β
I B 4.53
Note that
which shows that the transistor is indeed driven into saturation. Now,
P = I BVBE ( on ) + I CVCE ( sat )
= ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mW
______________________________________________________________________________________
TYU5.8
For VBC = 0 ⇒ VO = 0.7 V
I
9.77
5 − 0.7
IB = C =
= 0.195 mA
= 9.77 mA
50
β
0.44
Then
and
V = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7
or VI = 0.825 V
Now I
Also P = I BVBE ( on ) + I CVCE
IC =
= ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mW
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU5.9
IC =
IE =
3.3 − VC 3.3 − 2.27
=
= 0.2575 mA
4
RC
− V BE (on ) − (− 3.3) 3.3 − 0.7
=
= 0.260 mA
10
RE
I B = I E − I C = 0.260 − 0.2575 ⇒ I B = 2.5 μ A
I C 0.2575
=
= 103
I B 0.0025
β
103
α=
=
= 0.99038
1 + β 104
______________________________________________________________________________________
β=
TYU5.10
IE =
5 − V EB (on ) 5 − 0.7
=
= 0.5375 mA
8
RE
⎛ β ⎞
⎛ 85 ⎞
⎟⎟ I E = ⎜ ⎟(0.5375) = 0.531 mA
I C = ⎜⎜
⎝ 86 ⎠
⎝1+ β ⎠
I
0.5375
IB = E =
⇒ I B = 6.25 μ A
1+ β
86
VEC = 10 − (0.531)(4) − (0.5375)(8) = 3.575 V
______________________________________________________________________________________
TYU5.11
VBB = I B RB + VBE ( on ) + I E RE
or
VBB = I B RB + VBE ( on ) + (1 + β ) I B RE
IB =
VBB − VBE ( on )
RB + (1 + β ) RE
Then
or I B = 15.1 μ A
=
2 − 0.7
10 + ( 76 )(1)
I = ( 75 )(15.1 μ A) = 1.13 mA
I = ( 76 )(15.1 μ A) = 1.15 mA
Also C
and E
Now VCE = VCC + VBB − I C RC − I E RE
= 8 + 2 − (1.13)( 2.5 ) − (1.15 )(1) = 6.03 V
______________________________________________________________________________________
TYU5.12
VE = 5 − VCE = 5 − 2.2 = 2.8 V
IB =
VBB − VBE (on ) − VE 5 − 0.7 − 2.8
=
= 0.15 mA
10
RB
I E = (1 + β )I B = (121)(0.15) = 18.15 mA
VE
2 .8
=
= 0.154 k Ω
I E 18.15
______________________________________________________________________________________
RE =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU5.13
IE
1.2
=
= 0.01319 mA
1+ β
91
= I E R E + V EB (on ) + I B RB
= (1.2 )(1) + 0.7 + (0.01319)(50 ) = 2.56 V
I E = 1.2 mA, I B =
(a)
VBB
⎛ β ⎞
⎛ 90 ⎞
⎟⎟ I E = ⎜ ⎟(1.2 ) = 1.187 mA
I C = ⎜⎜
1
+
β
⎝ 91 ⎠
⎠
⎝
VEC = 5 − I E RE = 5 − (1.2 )(1) = 3.8 V
______________________________________________________________________________________
(b)
TYU5.14
For vI = 0, iB = iC = 0, vO = 12 V , P = 0
v − VBE ( on ) 12 − 0.7
iB = I
=
= 47.1 mA
0.24
RB
(b) For vI = 12 V ,
V − VCE ( sat ) 12 − 0.1
iC = CC
=
= 2.38 A
5
RC
(a)
vO = 0.1 V
and
P = iBVBE ( on ) + iCVCE ( sat )
= ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 W
______________________________________________________________________________________
TYU5.15
I CQ =
(a)
5 − VCEQ
RC
I CQ
=
5 − 2.5
= 1.25 mA
2
1.25
⇒ I BQ = 10.42 μ A
120
5 − V BE (on ) 5 − 0.7
=
= 413 k Ω
RB =
0.01042
I BQ
I BQ =
β
=
I BQ = 10.42 μ A
(b)
For β = 80 ⇒ I CQ = 0.8336 mA
For β = 160 ⇒ I CQ = 1.667 mA
Now
VCEQ = 5 − I CQ (2 )
For β = 80 ⇒ VCEQ = 5 − (0.8336 )(2 ) = 3.33 V
For β = 160 ⇒ VCEQ = 5 − (1.667 )(2 ) = 1.67 V
So
1.67 ≤ VCEQ ≤ 3.33 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU5.16
5 − 0.7
= 0.005375 mA
800
β = 75, I CQ = β I BQ = ( 75)( 0.005375)
I BQ =
For
Or
For
Or
I CQ = 0.403 mA
β = 150, I CQ = (150 )( 0.005375)
I CQ = 0.806 mA
Largest
I CQ ⇒ Smallest VCEQ
β = 150, RC =
For
β = 75, RC =
For
5 −1
= 4.96 k Ω
0.806
5−4
= 2.48 k Ω
0.403
I CQ = 0.604 mA
VCEQ = 2.5 V , RC =
5 − 2.5
= 4.14 k Ω
0.604
For a nominal
and
I = 0.403 mA, VCEQ = 5 − ( 0.403)( 4.14 ) = 3.33 V
Now for CQ
I = 0.806 mA, VCEQ = 5 − ( 0.806 )( 4.14 ) = 1.66 V
For CQ
1.66 ≤ VCEQ ≤ 3.33 V
So, for RC = 4.14 k Ω,
______________________________________________________________________________________
TYU5.17
RTH = R1 R2 = 440 230 = 151 k Ω
(a)
(b)
⎛ R2 ⎞
⎛ 230 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(5) = 1.716 V
⎝ 440 + 230 ⎠
⎝ R1 + R2 ⎠
V − V BE (on )
1.716 − 0.7
I BQ = TH
⇒ I BQ = 3.364 μ A
=
RTH + (1 + β )RE 151 + (151)(1)
I CQ = β I BQ = 0.5046 mA; I EQ = (1 + β )I BQ = 0.508 mA
VCEQ = VCC − I CQ RC − I EQ R E
(c)
= 5 − (0.5046 )(4 ) − (0.508)(1) = 2.47 V
RTH = 151 k Ω ; VTH = 1.716 V
1.716 − 0.7
⇒ I BQ = 4.2 μ A
I BQ =
151 + (91)(1)
I CQ = β I BQ = 0.378 mA; I EQ = (1 + β )I BQ = 0.382 mA
VCEQ = 5 − (0.378 )(4 ) − (0.382 )(1) = 3.11 V
______________________________________________________________________________________
TYU5.18
(a) RTH = (0.1)(1 + β )RE = (0.1)(151)(1) = 15.1 k Ω
I BQ =
I EQ
I CQ
β
=
⎛1+ β
= ⎜⎜
⎝ β
0.4
⇒ I BQ = 2.667 μ A
150
⎞
⎛ 151 ⎞
⎟⎟ I CQ = ⎜
⎟(0.4 ) = 0.4027 mA
⎝ 150 ⎠
⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
VCEQ = VCC − I CQ RC − I EQ R E
2.7 = 5 − (0.4 )RC − (0.4027 )(1) ⇒ RC = 4.74 k Ω
Now
1
(RTH )VCC = I BQ RTH + VBE (on) + I EQ RE
VTH =
R1
1
(15.1)(5) = (0.002667)(15.1) + 0.7 + (0.4027)(1)
R1
yields R1 = 66 k Ω ; 66 R2 = 15.1 ⇒ R2 = 19.6 k Ω
VTH = 1.143 V; RTH = 15.1 k Ω
(b)
VTH − VBE (on )
1.143 − 0.7
⇒ I BQ = 4.175 μ A
=
RTH + (1 + β )R E 15.1 + (91)(1)
= β I BQ = 0.376 mA; I EQ = (1 + β )I BQ = 0.380 mA
I BQ =
I CQ
V CEQ = 5 − (0.376 )(4.74 ) − (0.380 )(1) = 2.84 V
______________________________________________________________________________________
TYU5.19
V ECQ = 5 ≅ 10 − I CQ (RC + R E ) = 10 − I CQ (4.5 + 0.5) ⇒ I CQ ≅ 1 mA
I BQ =
I CQ
β
=
1
⇒ I BQ = 8.33 μ A
120
⎛1+ β ⎞
⎛ 121 ⎞
⎟⎟ I CQ = ⎜
I EQ = ⎜⎜
⎟(1) = 1.008 mA
⎝ 120 ⎠
⎝ β ⎠
RTH = (0.1)(1 + β )RE = (0.1)(121)(0.5) = 6.05 k Ω
V + = I EQ RE + VEB (on ) + I BQ RTH + VTH
5 = (1.008)(0.5) + 0.7 + (0.00833)(6.05) + VTH ⇒ VTH = 3.746 V
⎛ R2 ⎞
1
⎟⎟(10) − 5 =
(RTH )(10) − 5
VTH = ⎜⎜
R1
⎝ R1 + R2 ⎠
So
1
(6.05)(10) − 5 = 3.746 ⇒ R1 = 6.92 k Ω
R1
6.92 R2 = 6.05 ⇒ R2 = 48.1 k Ω
______________________________________________________________________________________
TYU5.20
⎛ β ⎞
⎛ 120 ⎞
⎟⎟ I Q = ⎜
(a) I CQ = ⎜⎜
⎟(0.25) = 0.2479 mA
⎝ 121 ⎠
⎝1+ β ⎠
I CQ = I S eVBE VT
or
⎛ I CQ ⎞
⎛ 0.2479 × 10 −3 ⎞
⎟ = (0.026) ln⎜
V BE = VT ln⎜⎜
⎜ 3 × 10 −14 ⎟⎟ = 0.5937 V
⎟
⎠
⎝
⎝ IS ⎠
I CQ 0.2479
I BQ =
=
⇒ I BQ = 2.066 μ A
β
120
VB = −(0.002066)(75) = −0.155 V
VE = VB − VBE = −0.155 − 0.5937 = −0.7487 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
VC = V + − I CQ RC = 2.5 − (0.2479 )(4) = 1.508 V
VCEQ = VC − V E = 1.508 − (− 0.7487 ) = 2.26 V
⎛ 60 ⎞
(b) I CQ = ⎜ ⎟(0.25) = 0.2459 mA
⎝ 61 ⎠
⎛ 0.2459 × 10 −3 ⎞
⎟⎟ = 0.5935 V
V BE = (0.026) ln⎜⎜
−14
⎠
⎝ 3 × 10
I CQ 0.2459
I BQ =
=
⇒ I BQ = 4.098 μ A
β
60
VB = −(0.004098)(75) = −0.307 V
VE = −0.307 − 0.5935 = −0.901 V
VC = 2.5 − (0.2459 )(4) = 1.516 V
VCEQ = 1.516 − (− 0.901) = 2.42 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 6
Exercise Solutions
EX6.1
V BB − V BE (on ) 0.85 − 0.7
=
⇒ I BQ = 0.833 μ A
RB
180
= β I BQ = (120 )(0.000833 ) = 0.10 mA
(a) I BQ =
I CQ
VCEQ = VCC − I CQ RC = 3.3 − (0.1)(15 ) = 1.8 V
I CQ
(b) g m =
VT
βVT
rπ =
I CQ
=
=
0.1
= 3.846 mA/V
0.026
(120)(0.026 ) = 31.2
0.1
kΩ
⎛ rπ ⎞
31.2 ⎞
⎟⎟ = −(3.846)(15)⎛⎜
(c) Aυ = − g m RC ⎜⎜
⎟ = −8.52
⎝ 31.2 + 180 ⎠
⎝ rπ + R B ⎠
______________________________________________________________________________________
EX6.2
V BB − V BE (on ) 1.025 − 0.7
=
= 0.00325 mA
RB
100
= β I BQ = (150 )(0.00325 ) = 0.4875 mA
(a) I BQ =
I CQ
gm =
rπ =
ro =
I CQ
VT
β VT
I CQ
=
=
0.4875
= 18.75 mA/V
0.026
(150 )(0.026) = 8 k Ω
0.4875
VA
150
=
= 308 k Ω
I CQ 0.4875
⎛ rπ ⎞
8 ⎞
⎟⎟ = −(18.75)(308 6)⎛⎜
(b) Aυ = − g m (ro RC )⎜⎜
⎟ = −8.17
+
8
r
R
+
⎝ 100 ⎠
B ⎠
⎝ π
______________________________________________________________________________________
EX6.3
(a)
I BQ =
VBB − VEB ( on )
RB
=
1.145 − 0.7
50
I BQ = 0.0089 mA
or
Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA
Now
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
gm =
rπ =
ro =
I CQ
=
VT
β VT
I CQ
0.801
= 30.8 mA / V
0.026
=
( 90 )( 0.026 )
0.801
= 2.92 k Ω
VA
120
=
= 150 k Ω
I CQ 0.801
(
We have Vo = g mVπ ro RC
(b)
)
⎛ r
⎞
Vπ = − ⎜ π ⎟ Vs
⎝ rπ + RB ⎠
and
so
⎛ rπ
V
Aυ = o = − g m ⎜⎜
Vs
⎝ rπ + R B
⎞
⎟(ro RC )
⎟
⎠
⎛ 2.92 ⎞
= −(30.8)⎜
⎟(150 2.5)
⎝ 2.92 + 50 ⎠
which yields Aυ = −4.18
______________________________________________________________________________________
EX6.4
Using Figure 6.23
I = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10 −4 < hre < 50 × 10 −4 ,
(a) For CQ
5 < hoe < 13 μ mhos
I = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10 −4 < hre < 1.6 × 10 −4 ,
(b) For CQ
22 < hoe < 35 μ mhos
______________________________________________________________________________________
EX6.5
RTH = R1 R2 = 250 75 = 57.7 k Ω
⎛ R2
VTH = ⎜⎜
⎝ R1 + R 2
⎞
⎛ 75 ⎞
⎟⎟ ⋅ VCC = ⎜
⎟(5)
⎝ 75 + 250 ⎠
⎠
or
VTH = 1.154 V
I BQ =
or
VTH − VBE ( on )
RTH + (1 + β ) RE
=
1.154 − 0.7
57.7 + (121)( 0.6 )
I BQ = 3.48 μ A
I CQ = β I BQ = (120 )( 3.38 μ A ) = 0.418 mA
(a)
Now
gm =
rπ =
I CQ
VT
=
0.418
= 16.08 mA / V
0.026
βVT (120 )( 0.026 )
=
= 7.46 kΩ
I CQ
0.418
We have
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Vo = − g mVπ RC
We find
Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 )
or
Rib = 80.1 k Ω
Also
R1 R2 = 250 75 = 57.7 k Ω
R1 R2 Rib = 57.7 80.1 = 33.54 k Ω
We find
⎛ R1 R 2 Rib
V s′ = ⎜
⎜R R R +R
S
⎝ 1 2 ib
or
⎞
⎟ ⋅ V = ⎛⎜ 33.54 ⎞⎟ ⋅ V
⎟ s ⎝ 33.54 + 0.5 ⎠ s
⎠
Vs′ = ( 0.985) Vs
Now
⎡ ⎛1+ β
Vs′ = Vπ ⎢1 + ⎜
⎣⎢ ⎝ rπ
or
⎞ ⎤
⎟ RE ⎥ = Vπ
⎠ ⎦⎥
⎡ ⎛ 121 ⎞
⎤
⎢1 + ⎜ 7.46 ⎟ ( 0.6 ) ⎥
⎝
⎠
⎣
⎦
Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 )Vs
So
Av =
Vo
= − (16.08 )( 0.0932 )( 0.985 )( 5.6 )
Vs
or
Av = −8.27
______________________________________________________________________________________
EX6.6
(a) RTH = R1 R2 = 14.4 110 = 12.73 k Ω
⎛ R2 ⎞
⎛ 110 ⎞
⎟⎟VCC = ⎜
VTH = ⎜⎜
⎟(12 ) = 10.61 V
⎝ 110 + 14.4 ⎠
⎝ R1 + R2 ⎠
12 = (101)I BQ (0.3) + 0.7 + I BQ (12.73) + 10.61
so
I BQ = 0.0160 mA
I CQ = β I BQ = 1.60 mA; I EQ = (1 + β )I BQ = 1.62 mA
V ECQ = 12 − (1.6 )(4 ) − (1.62 )(0.3) = 5.11 V
(b) g m =
rπ =
ro =
I CQ
VT
β VT
I CQ
=
1.60
= 61.54 mA/V
0.026
=
(100 )(0.026 ) = 1.625 k Ω
1.60
VA
=∞
I CQ
− β (RC RL )
− (100)(4 10)
= −8.95
rπ + (1 + β )RE 1.625 + (101)(0.3)
______________________________________________________________________________________
(c) Aυ =
=
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX6.7
(a) Ri = RS + RB rπ
I BQ RB + 0.7 + (1 + β )I BQ R E + V − = 0
5 − 0.7 = I BQ [100 + (121)(4 )] ⇒ I BQ = 0.007363 mA
I CQ = β I BQ = 0.8836 mA
I CQ
gm =
rπ =
VT
β VT
I CQ
=
0.8836
= 33.98 mA/V
0.026
=
(120 )(0.026 ) = 3.53
0.8836
kΩ
Ri = 0.5 + 100 3.53 = 3.91 k Ω
(b)
ro =
VA
80
=
= 90.5 k Ω
I CQ 0.8836
⎛ 100 3.53 ⎞
⎛ RB rπ ⎞
⎟ ⋅ υ = (0.872)υ s
⎟ ⋅υ s = ⎜
Vπ = ⎜
⎜ 100 3.53 + 0.5 ⎟ s
⎜ RB rπ + RS ⎟
⎠
⎝
⎠
⎝
V
Aυ = − g m (RC ro ) π = −(33.98)(4 90.5)(0.872 ) = −114
υs
______________________________________________________________________________________
EX6.8
(a)
gm =
I CQ
VT
=
0.25
= 9.615 mA/V
0.026
V
100
= 400 k Ω
ro = A =
I CQ 0.25
Aυ = − g m (ro rc ) = −(9.615)(400 100) = −769
(b)
(
)
(
Aυ = − g m ro rc rL = −(9.615 ) 400 100 100
)
Aυ = −427
______________________________________________________________________________________
EX6.9
I BQ =
5 − 0.7
= 0.00672 mA
10 + (126 )( 5 )
I CQ = 0.84 mA, I EQ = 0.847 mA
VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 )
or
VCEQ = 3.83 V
dc load line
VCE ≅ (V + − V − ) − I C ( RC + RE )
or
VCE = 10 − I C ( 7.3)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
ac load line (neglecting ro)
υ ce = −i c (RC R L ) = −i c (2.3 5) = −i c (1.58)
______________________________________________________________________________________
EX6.10
(b) I BQ R B + 0.7 + (1 + β )R E − 5 = 0
⇒ I BQ = 0.007363 mA; I CQ = β I BQ = 0.884 mA; I EQ = 0.8909 mA
VCEQ = 10 − (0.8836 )(4 ) − (0.8909 )(4 ) = 2.90 V
(c) ΔVCE = −ΔI C (RC ro ) = −ΔI C (4 90.5) = −ΔI C (3.831)
For ΔVCE = 2.9 − 0.5 = 2.4 V
2.4
= 0.626 mA; Δυ ce = 4.8 V, peak-to-peak
3.831
______________________________________________________________________________________
Then ΔI C =
EX6.11
⎛ R2 ⎞
1
VTH = ⎜
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
2 ⎠
1
⎝ 1
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1)
(a)
so
RTH = 12.1 k Ω, VTH =
1
(12.1)(12 )
R1
We can write
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
We have
I CQ = 1.6 mA, I BQ =
1.6
= 0.0133 mA
120
Then
1
(12.1)(12 )
R1
12.1 + (121)(1)
12 − 0.7 −
I BQ = 0.0133 =
which yields
R1 = 15.24 k Ω
Since RTH = R1 R 2 = 12.1 k Ω , we find R2 = 58.7 k Ω
Also
VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)
ac load line Δυ ec = −i c (RC R L )
Want
Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA
Also Δvec = 3.99 − 0.5 = 3.49 V
Δυ ec 3.49
Now
=
= 2.327 k Ω = RC R L
1.5
Δi c
So 4 R L = 2.327 k Ω which yields RL = 5.56 k Ω
______________________________________________________________________________________
EX6.12
(a) RTH = R1 R2 = 1.3 4.2 = 0.9927 k Ω
⎛ R2 ⎞
⎛ 4.2 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(12 ) = 9.1636 V
⎝ 1.3 + 4.2 ⎠
⎝ R1 + R2 ⎠
9.1636 − 0.7
= 2.473 mA
I BQ =
0.9927 + (81)(0.03)
I EQ = (1 + β )I BQ = 0.20 A, I CQ = β I BQ = 0.1978 A
VCEQ = 12 − (0.20 )(30 ) = 6.0 V
(b) g m =
I CQ
VT
=
0.1978
= 7.608 A/V
0.026
rπ = 10.52 Ω , ro = 379.2 Ω
(1 + β )(ro R E )
(81)(379.2 30)
=
= 0.9953
rπ + (1 + β )(ro R E ) 10.52 + (81)(379.2 30)
Rib = rπ + (1 + β )(ro RE ) = 10.52 + (81)(379.2 30)
Aυ =
(c)
or
Rib = 2.26 k Ω
______________________________________________________________________________________
EX6.13
R o = R E ro
rπ
10.52
= 30 379.2
= 0.129 Ω
81
1+ β
______________________________________________________________________________________
EX6.14
(a)
I CQ = 1.25 mA and β = 100, we find
I = 1.26 mA and I BQ = 0.0125 mA
For EQ
VCEQ = 10 − I EQ RE
Now
Or
4 = 10 − (1.26 ) RE
which yields
RE = 4.76 k Ω
Then
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 )
or
RTH = 48.1 k Ω
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
We have
⎛ R2 ⎞
1
VTH = ⎜
⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5
R1
⎝ R1 + R2 ⎠
or
VTH =
1
( 481) − 5
R1
I BQ =
We can write
Or
VTH − 0.7 − ( −5 )
RTH + (1 + β ) RE
1
( 481) − 5 − 0.7 + 5
R1
0.0125 =
48.1 + (101)( 4.76 )
which yields
R1 = 65.8 k Ω
Since R1 R 2 = 48.1 k Ω , we obtain
R2 = 178.8 k Ω
(b)
rπ =
ro =
β VT
I CQ
=
(100 )( 0.026 )
1.25
= 2.08 k Ω
VA
125
=
= 100 k Ω
I CQ 1.25
We may note that
g mVπ = g m ( I b rπ ) = β I b
Also
Rib = rπ + (1 + β ) R E R L ro
(
)
= 2.08 + (101)(4.76 1 100 )
or
Rib = 84.9 k Ω
Now
⎛ RE ro ⎞
Io = ⎜
1 + β ) Ib
⎜ R r + R ⎟⎟ (
L ⎠
⎝ E o
where
⎛ R1 R2 ⎞
Ib = ⎜
⋅I
⎜ R R + R ⎟⎟ s
ib ⎠
⎝ 1 2
We can then write
⎛ RE ro ⎞
⎛ R1 R2
I
AI = o = ⎜
1+ β )⎜
(
⎟
⎜R R +R
I s ⎜⎝ RE ro + RL ⎟⎠
ib
⎝ 1 2
We have
RE ro = 4.76 100 = 4.54 k Ω
⎞
⎟⎟
⎠
so
48.1 ⎞
⎛ 4.54 ⎞
⎛
AI = ⎜
⎟ (101) ⎜
⎟
+
+ 84.9 ⎠
4.54
1
48.1
⎝
⎠
⎝
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or
AI = 29.9
(c)
R o = R E ro
rπ
2.08
= 4.76 100
101
1+ β
Ro = 20.5 Ω
or
_____________________________________________________________________________
EX6.15
(a)
RTH = R1 R2 = 70 6 = 5.53 k Ω
⎛ R2 ⎞
⎛ 6 ⎞
VTH = ⎜
⎟ (10 ) − 5 = ⎜
⎟ (10 ) − 5
+
R
R
⎝ 70 + 6 ⎠
⎝ 1
2 ⎠
or
VTH = −4.2105 V
We find
−4.2105 − 0.7 − ( −5 )
⇒ 2.91 μ A
I BQ1 =
5.53 + (126 )( 0.2 )
and
I CQ1 = β I BQ1 = (125 )( 2.91 μ A) = 0.364 mA
I EQ1 = (1 + β ) I BQ1 = 0.368 mA
At the collector of Q1,
V − 0.7 − ( −5)
5 − VC1
= I CQ1 + C1
RC1
(1 + β )( RE 2 )
or
V − 0.7 − ( −5 )
5 − VC1
= 0.364 + C1
5
(126 )(1.5)
which yields
VC1 = 2.99 V
also
VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5
or
VE1 = −4.93 V
Then
VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V
We find
I EQ 2 =
VC1 − 0.7 − ( −5 )
1.5
= 4.86 mA
and
⎛ β ⎞
⎛ 125 ⎞
I CQ 2 = ⎜
⎟ ⋅ I EQ1 = ⎜
⎟ ( 4.86 ) = 4.82 mA
⎝ 126 ⎠
⎝1+ β ⎠
We find
VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V
and
VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)
The small-signal transistor parameters are:
rπ 1 =
β VT
I CQ1
g m1 =
rπ 2 =
gm2 =
=
I CQ1
I CQ 2
0.364
=
I CQ 2
VT
= 8.93 k Ω
0.364
= 14.0 mA / V
0.026
=
VT
β VT
(125 )( 0.026 )
(125)( 0.026 )
=
4.82
= 0.674 k Ω
4.82
= 185 mA / V
0.026
R = r + (1 + β ) RE1 = 8.93 + (126 )( 0.2 )
We find ib1 π 1
Or
Rib1 = 34.1 k Ω
and
Rib = rπ 2 + (1 + β )(R E 2 R L )
= 0.674 + (126)(1.5 10) = 165 k Ω
The small-signal equivalent circuit is:
We can write
Vo = (1 + β )I b 2 (R E 2 R L )
where
⎛ RC1 ⎞
Ib2 = ⎜
⎟ ( − g m1Vπ 1 )
⎝ RC1 + Rib 2 ⎠
V
Vπ 1 = s ⋅ rπ 1
Rib1
Then
⎛
R C1
V
Aυ = o = (1 + β )(R E 2 R L )⎜⎜
Vi
⎝ RC1 + Rib 2
⎞⎛ − g m1 rπ 1
⎟⎜
⎟⎜ R
ib1
⎠⎝
⎞
⎟
⎟
⎠
so
⎛ 5 ⎞⎛ 125 ⎞
Aυ = −(126 )(1.5 10 )⎜
⎟⎜
⎟
⎝ 5 + 165 ⎠⎝ 34.1 ⎠
or
Av = −17.7
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c)
Ri = R1 R 2 Rib1 = 70 6 34.1 = 4.76 k Ω
⎛ r + RC1 ⎞
⎛ 0.676 + 5 ⎞
Ro = RE 2 ⎜ π 2
⎟ = 1.5 ⎜
⎟
⎝ 126 ⎠
⎝ 1+ β ⎠
and
or
Ro = 43.7 Ω
______________________________________________________________________________________
Test Your Understanding Solutions
TYU6.1
ib =
υs
R B + rπ
=
υs
180 + 31.2
=
υs
211.2
0.065 sin ω t
⇒ ib = 0.308 sin ω t ( μ A)
211.2
i B = I BQ + ib = 0.833 + 0.308 sin ω t ( μ A)
ib =
⎛
⎞
⎞
π
⎟⎟ ⋅ υ s = ⎛⎜
υ be = ⎜⎜
⎟(0.065 sin ω t ) = 0.00960 sin ω t (V)
r
R
31
.
2
180
+
+
⎠
⎝
B ⎠
⎝ π
r
31.2
υ BE = V BE (on ) + υ be = 0.7 + 0.00969 sin ω t (V)
υ ce = Aυ ⋅ υ s = (8.52)(0.065 sin ω t ) = 0.554 sin ω t (V)
υ CE = VCEQ + υ ce = 1.8 − 0.554 sin ω t (V)
______________________________________________________________________________________
TYU6.2
(a)
V + − V EB (on ) − V BB 3.3 − 0.7 − 2.455
=
⇒ I BQ = 1.8125 μ A
RB
80
= β I BQ = 0.20 mA
I BQ =
I CQ
V ECQ = 3.3 − (0.2 )(7 ) = 1.9 V
(b)
(c)
(d)
gm =
I CQ
VT
=
0. 2
= 7.692 mA/V
0.026
rπ =
β VT (110 )(0.026 )
=
= 14.3 k Ω
0.20
I CQ
ro =
VA
80
=
= 400 k Ω
I CQ 0.20
⎛ rπ ⎞
14.3 ⎞
⎟⎟ = −(7.692 )(7 400 )⎛⎜
Aυ = − g m (RC ro )⎜⎜
⎟ = −8.02
⎝ 14.3 + 80 ⎠
⎝ rπ + R B ⎠
Ri = R B + rπ = 80 + 14.3 = 94.3 k Ω
Ro = RC ro = 7 400 = 6.88 k Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU6.3
RTH = R1 R2 = 100 25 = 20 k Ω
⎛ R2 ⎞
⎛ 25 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(5) = 1.0 V
R
R
+
⎝ 100 + 25 ⎠
2 ⎠
⎝ 1
V − V BE (on )
1.0 − 0.7
I BQ = TH
=
= 0.00597 mA
RTH + (1 + β )R E 20 + (121)(0.25)
I CQ = β I BQ = 0.7164 mA
Now
gm =
I CQ
=
0.7164
= 27.55 mA/V
0.026
VT
β VT (120 )(0.026 )
rπ =
=
= 4.355 k Ω
0.7164
I CQ
We find
RTH [rπ + (1 + β )R E ] = 20 [4.355 + (121)(0.25)] = 20 34.605 = 12.67 k Ω
Then
− β RC
− (120 )(4 )
⎛ 12.67
⎞
Aυ =
⋅⎜
⋅ (0.9807 )
⎟=
rπ + (1 + β )R E ⎝ 12.67 + 0.25 ⎠ 4.355 + (121)(0.25)
or
Aυ = −13.6
______________________________________________________________________________________
TYU6.4
Av ≅ −
RC
RE
As a first approximation,
Resulting gain is always smaller than this value. The effect of RS is very small.
RC
= 10
R
Set E
5 ≅ I C ( RC + RE ) + VCEQ
Now
5 = ( 0.5 )( RC + RE ) + 2.5
or
which yields RC + RE = 5 k Ω
We have RC = 10 R E
so RE = 0.454 k Ω and RC = 4.54 k Ω
We have
I BQ =
and
I CQ
β
=
0.5
= 0.005 mA
100
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 )
or RTH = 4.59 k Ω
Also
⎛ R2
VTH = ⎜
⎝ R1 + R2
⎞
1
⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
R
1
⎠
or
VTH =
1
23
( 4.59 )( 5 ) =
R1
R1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Also
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
so that
23
= ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 )
R1
R1 = 24.1 k Ω
which yields
Since R1 R2 = 4.59 k Ω , then R2 = 5.67 k Ω
______________________________________________________________________________________
TYU6.5
As a first approximation
R
Av ≅ − C
RE
Set
RC
=9
RE
Now
VCC ≅ I CQ ( RC + RE ) + VECQ
7.5 = ( 0.6 )( 9 RE + RE ) + 3.75
which yields
RE = 0.625 k Ω and RC = 5.62 k Ω
We have
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 )
or RTH = 6.31 k Ω
Also
VTH =
1
1
⋅ RTH ⋅ VCC = ( 6.31)( 7.5 )
R1
R1
We have
I BQ =
and
I CQ
β
=
0.6
= 0.006 mA
100
V CC = (1 + β )I BQ R E + V EB (on ) + I BQ RTH + VTH
7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) +
1
( 6.31)( 7.5)
R1
which yields
R1 = 7.41 k Ω
Since RTH = R1 R 2 = 6.31 k Ω , then R 2 = 42.5 k Ω
_____________________________________________________________________________
TYU6.6
We have
Av =
⎛R ⎞
− β RC
= − ( 0.95 ) ⎜ C ⎟
rπ + (1 + β ) RE
⎝ RE ⎠
or
⎛ 2 ⎞
Av = − ( 0.95) ⎜
⎟ = −4.75
⎝ 0.4 ⎠
Assume, from Example 6.5 that, rπ = 1.2 k Ω
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
−β ( 2)
1.2 + (1 + β )( 0.4 )
or
= −4.75
β = 76
_____________________________________________________________________________
TYU6.7
Dc analysis: by symmetry, VTH = 0
RTH = R1 R 2 = 20 20 = 10 k Ω
We can write
0 − 0.7 − ( −5 )
= 0.00672 mA
I BQ =
10 + (126 )( 5 )
I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA
Small-signal transistor parameters:
β VT (125 )( 0.026 )
rπ =
=
= 3.87 k Ω
I CQ
0.84
gm =
ro =
I CQ
VT
=
0.84
= 32.3 mA / V
0.026
VA
200
=
= 238 k Ω
I CQ 0.84
(a) We can write
V o = − g mVπ ro RC R L
(
so
Aυ =
or
(b)
(
)
and Vπ = V s
)
(
Vo
= − g m ro RC R L = −(32.3) 238 2.3 5
Vs
)
Aυ = −50.5
Ro = ro RC = 238 2.3 = 2.28 k Ω
______________________________________________________________________________________
TYU6.8
We find
I CQ = 0.418 mA,
⎛ 121 ⎞
VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜
⎟ ( 0.418 )( 0.6 )
⎝ 120 ⎠
or V CEQ = 2.41 V
So
ΔvCE = ( 2.41 − 0.5 ) × 2
, or ΔvCE = 3.82 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU6.9
For I CQ ≅ I EQ ,
VCEQ = 10 − I CQ (4 + 4 ) = 10 − I CQ (8)
ΔI C = I CQ − 0.1
ΔVCE = VCEQ − 0.7
(
)
ΔVCE = ΔI C (4 ) = I CQ − 0.1 (4 ) = VCEQ − 0.7
So
(
)
VCEQ = I CQ − 0.1 (4 ) + 0.7
Then
(I
CQ
)
− 0.1 (4 ) + 0.7 = 10 − I CQ (8)⇒ I CQ = 0.8083 mA
VCEQ = 10 − (0.8083 )(8) ⇒ VCEQ = 3.533 V
Then peak-to-peak values are
ΔVCE = (3.533 − 0.7 )(2 ) = 5.67 V
ΔI C = (0.8083 − 0.1)(2) = 1.42 mA
______________________________________________________________________________________
TYU6.10
We can write
0 − 0.7 − ( −10 )
⇒ 6.60 μ A
I BQ =
100 + (131)(10 )
I CQ = (130 )( 6.60 μ A) = 0.857 mA
Assume nominal small-signal parameters of:
hie = 4 k Ω, h fe = 134
hre = 0, hoe = 12 μ S ⇒
1
= 83.3 k Ω
hoe
We find
⎛
1 ⎞
Rib = hie + (1 + h fe ) ⎜ RE RL
⎟
h
oe ⎠
⎝
= 4 + (135 )(10 || 10 || 83.3) = 641 k Ω
To find the voltage gain:
100 641
R B Rib
⋅Vs =
⋅ V s = (0.896 )V s
V s′ =
100 641 + 10
R B Rib + R S
Also
1 + h fe R ′
Vo
=
V s′ hie + 1 + h fe R ′
(
(
)
)
where
R′ = RE RL
Then
Av =
1
= 10 10 83.3 = 4.72 k Ω
hoe
Vo ( 0.896 )(135 )( 4.72 )
=
= 0.891
4 + (135 )( 4.72 )
Vs
To find the current gain
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛
⎜ RE 1
⎜
hoe
I
Ai = o = ⎜
Ii ⎜
1
⎜ RE h + RL
oe
⎝
⎞
⎟
⎟
⎟ 1 + h fe
⎟
⎟
⎠
(
)⎛⎜⎜ R
RB
⎝ B + Rib
⎞
⎟
⎟
⎠
⎛ 10 83.3 ⎞
⎟(135)⎛⎜ 100 ⎞⎟
=⎜
⎜ 10 83.3 + 10 ⎟
⎝ 100 + 641 ⎠
⎝
⎠
or
Ai = 8.59
To find the output resistance:
1 hie + R S R B
Ro = R E
hoe
1 + h fe
= 10 83.3
4 + 10 100
135
⇒ R o = 96.0 Ω
_____________________________________________________________________________
TYU6.11
RTH = R1 R2 = 50 50 = 25 k Ω
⎛ R2 ⎞
⎛1⎞
⎟⎟ ⋅ VCC = ⎜ ⎟(5) = 2.5 V
VTH = ⎜⎜
⎝2⎠
⎝ R1 + R 2 ⎠
Now
V − VEB ( on ) − VTH
I BQ = CC
RTH + (1 + β ) RE
=
5 − 0.7 − 2.5
= 0.00793 mA
25 + (101)( 2 )
and
I CQ = (100 )( 0.00793) = 0.793 mA
Small-signal transistor parameters:
I CQ 0.793
=
= 30.5 mA / V
gm =
0.026
VT
rπ =
ro =
β VT
I CQ
=
(100 )( 0.026 )
0.793
= 3.28 k Ω
VA
125
=
= 158 k Ω
I CQ 0.793
(a) Define
R′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω
Av =
(1 + β ) R′
(101)( 0.4 )
=
rπ + (1 + β ) R′ 3.28 + (101)( 0.4 )
or Aυ = 0.925
(b)
Rib = rπ + (1 + β ) R′ = 3.28 + (101)( 0.4 )
Rib = 43.7 k Ω
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
R o = R E ro
rπ
3.28
= 2 158
101
1+ β
or
Ro = 32.0 Ω
_____________________________________________________________________________
TYU6.12
(a) I BQ RB + VBE (on ) + (1 + β )I BQ R E + V − = 0
3. 3 − 0 . 7
= 0.001358 mA
100 + (121)(15)
= 0.1643 mA; I CQ = 0.1629 mA
I BQ =
I EQ
VCEQ = 6.6 − (0.1643 )(15 ) = 4.14 V
(b)
Aυ =
(1 + β )(RE RL ) ⎛ Ri ⎞
⎟
⋅⎜
rπ + (1 + β )(R E R L ) ⎜⎝ Ri + RS ⎟⎠
We find
0.1629
= 6.265 mA/V
0.026
(120 )(0.026 ) = 19.15 k Ω
rπ =
0.1629
Now
Rib = rπ + (1 + β )(R E R L ) = 19.15 + (121)(15 2) = 232.7 k Ω
gm =
Ri = Rib RB = 232.7 100 = 69.94 k Ω
Then
(121)(15 2)
⎛ 69.94 ⎞
Aυ =
⋅⎜
⎟ = 0.892
19.15 + (121)(15 2) ⎝ 69.94 + 2 ⎠
Also
⎛ RB ⎞⎛ R E ⎞
100
⎞⎛ 15 ⎞
⎛
⎟⎟⎜⎜
⎟⎟ = (121)⎜
Ai = (1 + β )⎜⎜
⎟
⎟⎜
R
R
R
R
+
+
+
100
232
.
7
⎠⎝ 15 + 2 ⎠
⎝
ib ⎠⎝ E
L ⎠
⎝ B
Ai = 32.1
(c) We found
Rib = 232.7 k Ω
Now
⎛ 19.15 + 100 2 ⎞
⎛ rπ + R B RS ⎞
⎟ 15 = 0.1745 15
⎟ RE = ⎜
Ro = ⎜
⎜
⎟
⎟
⎜ 1+ β
121
⎝
⎠
⎠
⎝
R o = 172 Ω
___________________________________________________________________________________
TYU6.13
(a) dc analysis:
V − VEB ( on ) 10 − 0.7
=
= 0.93 mA
I EQ = EE
10
RE
⎛ β ⎞
⎛ 100 ⎞
I CQ = ⎜
⎟ I EQ = ⎜
⎟ ( 0.93) = 0.921 mA
⎝ 101 ⎠
⎝1+ β ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
VECQ = VEE − I EQ RE − I CQ RC − ( −VCC )
= 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 )
or
VECQ = 6.1 V
(b) Small-signal transistor parameters:
β VT (100 )( 0.026 )
rπ =
=
= 2.82 k Ω
I CQ
0.921
I CQ
0.921
=
= 35.42 mA / V
0.026
VT
Small-signal current gain:
I o = g mVπ , and Vπ = Vs
gm =
Also
Ii =
⎛ 1
⎞
+ g mVπ = Vs ⎜
+ gm ⎟
⎜
⎟
RE rπ
⎝ RE rπ
⎠
Vs
Then
Ai =
Io
=
Ii
=
g m Vπ
⎛ 1
⎞
+ gm ⎟
Vπ ⎜
⎜R r
⎟
⎝ E π
⎠
(35.42)(10 2.82)
=
g m (R E rπ )
1 + g m (R E rπ )
1 + (35.42 )(10 2.82 )
or AI = 0.987
(c) Small-signal voltage gain:
Vo = g mVπ RC = g mVs RC
Av =
Vo
= g m RC = ( 35.42 )( 5 )
Vs
Av = 177
or
___________________________________________________________________________
TYU6.14
(a) I BQ R B + V BE (on ) + (1 + β )I BQ R E = V EE
3.3 − 0.7
= 0.001675 mA
100 + (121)(12 )
= 0.201 mA
I BQ =
I CQ
gm =
(b)
0.201
(120)(0.026) = 15.52 k Ω ; r = ∞
= 7.73 mA/V; rπ =
o
0.026
0.201
⎛
⎜
RE
Ai = ⎜
⎜
rπ
⎜ RE +
+
1
β
⎝
Ai = 0.654
⎞
⎟
⎟⎛⎜ β
⎟⎜⎝ 1 + β
⎟
⎠
⎞⎛ RC
⎟⎟⎜⎜
⎠⎝ RC + R L
⎞
⎛
⎟ 120
⎞ ⎜
12
⎞⎛ 12 ⎞
⎟⎛⎜
⎟=⎜
⎟
⎟⎜
⎟ ⎜
⎠ ⎜ 12 + 15.52 ⎟⎟⎝ 121 ⎠⎝ 12 + 6 ⎠
121 ⎠
⎝
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Now
⎛ 12 6 ⎞ ⎡15.52
⎛ RC R L ⎞ ⎡ rπ
⎤
⎤
⎟
⎟⎢
R E RS ⎥ = (7.73)⎜
Aυ = g m ⎜
⎢ 121 12 0.5⎥
⎟
⎜
⎜ R ⎟ ⎣1 + β
0
.
5
⎦
⎦
S
⎠⎣
⎝
⎠
⎝
Aυ = (7.73)(8)(0.1012 ) = 6.26
(c)
Ri = R E
rπ
15.52
= 12
⇒ Ri = 127 Ω
121
1+ β
Ro = RC = 12 k Ω
______________________________________________________________________________________
TYU6.15
dc analysis
5 = I BQ RB + VBE ( on ) + I EQ RE
I BQ =
5 − 0.7
4.3
=
RB + (101) RE RB + (101) RE
I CQ =
(100 )( 4.3)
RB + (101) RE
Also
5 = I CQ RC + VCEQ + I EQ RE − 5
or
⎡
⎛ 101 ⎞ ⎤
VCEQ = 10 − I CQ ⎢ RC + ⎜
⎟ RE ⎥
⎝ 100 ⎠ ⎦
⎣
ac analysis :
Vo = − g mVπ ( RC RL )
and
Vs = −Vπ −
Vπ
⋅ RB = −Vπ
rπ
⎛ RB ⎞
⎜1 +
⎟
rπ ⎠
⎝
or
⎛ r
⎞
Vπ = − ⎜ π ⎟ ⋅ Vs
+
r
R
B ⎠
⎝ π
Then
V
β
Av = o =
( RC RL )
Vs rπ + RB
where
β = g m rπ
For
I CQ = 1 mA,
rπ =
Then
Av = 20 =
β VT
I CQ
=
(100 )( 0.026 )
1
= 2.6 k Ω
(100 ) ( 2 2 )
2.6 + RB
which yields
RB = 2.4 k Ω
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then from
I CQ = 1 =
(100)(4.3)
2.4 + (101)R E
we find
R E = 4.23 k Ω
______________________________________________________________________________________
TYU6.16
(a) dc analysis
I
= 1 mA,
For EQ 2
⎛ 100 ⎞
I CQ 2 = ⎜
⎟ (1) = 0.990 mA
⎝ 101 ⎠
I EQ 2
1
I EQ1 =
=
= 0.0099 mA
1 + β 101
I BQ1 =
I EQ1
1+ β
=
0.0099
= 0.000098 mA
101
I CQ1 = (100 )( 0.000098) = 0.0098 mA
VB1 = − I BQ1 RB = − ( 0.000098 )(10 )
VB1 = −0.00098 ≅ 0
VE1 = −0.7 V
VE 2 = −1.4 V
I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA
VO = 5 − (1)( 4 ) = 1 V
VCEQ 2 = 1 − ( −1.4 ) = 2.4 V
VCEQ1 = 1 − ( −0.7 ) = 1.7 V
(b) small-signal transistor parameters:
β VT (100 )(0.026 )
rπ 1 =
=
= 265 k Ω
I CQ1
0.0098
g m1 =
rπ 2 =
g m2 =
I CQ1
VT
=
0.0098
= 0.377 mA/V
0.026
β VT (100 )(0.026 )
=
= 2.63 k Ω
I CQ 2
0.990
I CQ 2
VT
=
0.990
= 38.1 mA/V
0.026
ro1 = ro 2 = ∞
(c) small-signal voltage gain
Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC
Vs = Vπ 1 + Vπ 2
⎛V
⎞
⎛1+ β ⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜
⎟ ⋅ Vπ 1rπ 2
⎝ rπ 1
⎠
⎝ rπ 1 ⎠
⎡
⎤
⎛1+ β ⎞
Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜
⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC
⎝ rπ 1 ⎠
⎣
⎦
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛1+ β ⎞
Vs = Vπ 1 + ⎜
⎟ ⋅ rπ 2Vπ 1
⎝ rπ 1 ⎠
⎡
⎛ r ⎞⎤
= Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥
⎝ rπ 1 ⎠ ⎦
⎣
Vs
Vπ 1 =
⎛r ⎞
1 + (1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
Now
⎡
⎛ rπ 2 ⎞ ⎤
⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC
V
⎝ rπ 1 ⎠ ⎦
Av = o = − ⎣
Vs
⎛r ⎞
1 + (1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
⎡
⎛ 2.63 ⎞ ⎤
⎢ 0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 )
⎝
⎠⎦
Av = − ⎣
⎛ 2.63 ⎞
1 + (101) ⎜
⎟
⎝ 265 ⎠
or
Av = −77.0
(d)
Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63)
or
Ri = 531 k Ω
_____________________________________________________________________________
TYU6.17
(a)
R1 + R2 + R3 =
RE =
9
= 90 k Ω
0.1
0.7
= 0.7 k Ω
1
⎛R ⎞
V B1 = 0.7 + 0.7 = 1.4 V ⇒ ⎜ 3 ⎟(9 ) = 1.4 ⇒ R3 = 14 k Ω
⎝ 90 ⎠
⎛ R + R3 ⎞
V B 2 = 0.7 + 2.5 + 0.7 = 3.9 V ⇒ ⎜ 2
⎟(9 ) = 3.9 ⇒ R2 = 25 k Ω
⎝ 90 ⎠
Then R1 = 51 k Ω
VC 2 = 0.7 + 2.5 + 2.5 = 5.7 V
So
9 − 5.7
= 3.3 k Ω
RC =
1
1
(100)(0.026) = 2.6 k Ω
= 38.46 mA/V; rπ =
(b) g m =
1
0.026
⎞
⎛ r
2 ⎛ 2.6 ⎞
(c) Aυ = − g m1 g m 2 ⎜⎜ π 2 ⎟⎟(RC R L ) = −(38.46 ) ⎜
⎟(3.3 10 ) = −94.5
⎝ 101 ⎠
⎝1+ β2 ⎠
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU6.18
(a) dc analysis
RTH = R1 R2 = 125 30 = 24.2 k Ω
⎛ R2 ⎞
⎛ 30 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ (12 )
+
R
R
⎝ 125 + 30 ⎠
⎝ 1
2 ⎠
or
VTH = 2.32 V
Now
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
2.32 − 0.7
= 0.0250 mA
24.2 + ( 81)( 0.5 )
I BQ =
I CQ = ( 80 )( 0.025) = 2.00 mA
⎡
⎛1+ β ⎞ ⎤
VCEQ = VCC − I CQ ⎢ RC + ⎜
⎟ RE ⎥
⎝ β ⎠ ⎦
⎣
⎡ ⎛ 81 ⎞
⎤
= 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥
80
⎣ ⎝ ⎠
⎦
or
VCEQ = 6.99 V
Power dissipated in RC :
2
PRC = I CQ
RC = ( 2.0 ) ( 2 ) = 8.0 mW
2
Power dissipated in R L :
I LQ = 0 ⇒ PRL = 0
Power dissipated in transistor:
PQ = I BQVBEQ + I CQVCEQ
= ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW
(b) With
vs = 18cos ω t ( mV )
β VT
rπ =
I CQ
=
(80 )( 0.026 )
2.0
= 1.04 k Ω
We can write
vce =
β
rπ
(R
C
RL ) VP cos ω t
Power dissipated in R L :
pRL =
=
vce ( rms )
2
RL
1
1
⋅
2 2 × 103
⎤
1 1 ⎡β
= ⋅
⋅ ⎢ ( RC RL ) VP ⎥
2 RL ⎣ rπ
⎦
⎡ 80
⋅⎢
( 2 2 ) ( 0.018)⎤⎥
⎣1.04
⎦
2
2
or
pRL = 0.479 mW
Power dissipated in RC :
Since RC = RL = 2 k Ω, , we find
pRC = 8.0 + 0.479 = 8.48 mW
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
PQ ≅ I CQ VCEQ
⎛β
− ⎜⎜
⎝ rπ
⎞
⎟
⎟
⎠
2
2
⎛ VP ⎞
⎟ (RC R L )
⎜
⎟
⎜
⎝ 2⎠
2
(
80
⎞ ⎛⎜ 0.018 ⎞⎟
⎛
2 ×10 3 2 ×10 3
= 2 × 10 −3 (6.99) − ⎜
⎟ ⎜
3
⎟
⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠
(
)
2
)
or
pQ = 13.0 mW
_____________________________________________________________________________
TYU6.19
(a) dc analysis
RTH = R1 R2 = 53.8 10 = 8.43 k Ω
⎛ R2 ⎞
⎛ 10 ⎞
VTH = ⎜
⎟ ⋅ VCC = ⎜
⎟ ( 5)
⎝ 53.8 + 10 ⎠
⎝ R1 + R2 ⎠
or
VTH = 0.7837 V
Now
0.7837 − 0.7
I BQ =
= 0.00993 mA
8.43
I CQ = (100 )( 0.00993) = 0.993 mA
VCEQ = VCC − I CQ RC
2.5 = 5 − ( 0.993) RC
which yields
RC = 2.52 k Ω
( b)
Power dissipated in RC :
2
PRC = I CQ
RC = ( 0.993 ) ( 2.52 )
2
or
PRC = 2.48 mW
Power dissipated in transistor:
PQ ≅ I CQVCEQ = ( 0.993)( 2.5)
or
PQ = 2.48 mW
(c) ac analysis
Maximum ac collector current:
ic = ( 0.993) cos ω t ( mA)
Power dissipated in RC :
pRC =
1
1
2
2
( 0.993) RC = ( 0.993) ( 2.52 )
2
2
or
pRC = 1.24 mW
Now
pRC
1.24
Fraction =
=
= 0.25
PRC + PQ 2.48 + 2.48
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 7
Exercise Solutions
EX7.1
(a) (i) f L =
1
⇒τS =
2πτ S
τ S = (RS + RP )C S
1
1
=
⇒ 3.183 ms
2π f L 2π (50 )
3.183 × 10 −3 = (2 + 8) × 10 3 (C S ) ⇒ C S = 0.318 μ F
⎡
f
⎢
⎛ RP ⎞⎢
fL
⎟
(ii) T = ⎜⎜
⎟
⎢
2
⎝ R P + RS ⎠ ⎢
⎞
1 + ⎛⎜ f
⎟
⎢⎣
⎝ fL ⎠
⎡
⎤
f
⎢
⎥
fL
⎥ = ⎛ 8 ⎞⎢
⎥ ⎜⎝ 8 + 2 ⎟⎠ ⎢
2
⎞
⎢ 1 + ⎛⎜ f
⎥
⎟
⎢⎣
⎥⎦
⎝ fL ⎠
⎤
⎥
⎥
⎥
⎥
⎥⎦
⎡
⎤
20
f
0.4
⎥ = 0.297
=
= 0.4 ; T = (0.8)⎢
⎢ 1 + (0.4)2 ⎥
f L 50
⎣
⎦
1)
50
(
f
= 0.566
=
= 1 ; T = (0.8)
For f = 50,
f L 50
2
100
f
(2) = 0.716
=
= 2 ; T = (0.8)
For f = 100,
2
50
fL
1 + (2)
(b)
(i) τ P = (RS R P )C P = (4.7 25)× 10 3 × 120 × 10 −12 = 4.747 × 10 −7 s
For f = 20,
(
fH =
1
2πτ P
)
1
=
⇒ 335 kHz
2π 4.747 × 10 −7
(
⎛ RP
(ii) T = ⎜⎜
⎝ R P + RS
)
⎞
⎟⎟ ⋅
⎠
1
2
=
(0.84175)
⎞
⎞
1 + ⎛⎜ f
1 + ⎛⎜ f
⎟
⎟
⎝ fH ⎠
⎝ fH ⎠
(0.84175) = 0.825
For f = 0.2 f H ; T =
2
1 + (0.2 )
For f = f H ; T =
For f = 8 f H ; T
2
(0.84175) = 0.595
2
(0.84175) = 0.104
=
2
1 + (8)
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX7.2
(a) Aυ (mid ) =
RP
R P + RS
⎛ RP ⎞
RP
⎟⎟ ⇒
= 0.7943
− 2 = 20 log10 ⎜⎜
R
+
R
R
S ⎠
P + RS
⎝ P
7.5
= 0.7943 ⇒ RS = 1.942 k Ω
7.5 + RS
1
1
1
⇒τ S =
=
= 0.7958 × 10 −3 s
2πτ S
2π f L 2π (200)
fL =
τ S = (RS + RP )C S
0.7958 × 10 −3 = (1.942 + 7.5) × 10 3 (C S ) ⇒ C S = 0.0843 μ F
τ P = (RS RP )C P = (1.942 7.5)× 10 3 × (80 × 10 −12 ) = 1.234 × 10 −7 s
1
1
fH =
=
⇒ 1.29 MHz
2πτ P 2π 1.234 × 10 −7
(
)
(b) τ S = 0.796 ms
τ P = 0.123 μ s
________________________________________________________________________
EX7.3
(a) RTH = R1 R2 = 110 42 = 30.39 k Ω
⎛ R2 ⎞
⎛ 42 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(3) = 0.8289 V
⎝ 110 + 42 ⎠
⎝ R1 + R2 ⎠
V − V BE (on )
0.8289 − 0.7
I BQ = TH
=
= 0.001217 mA
RTH + (1 + β )R E 30.39 + (151)(0.5)
I CQ = β I BQ = 0.1826 mA
τ S = (R1 R2 Rib )CC
where Rib = rπ + (1 + β )RE
Now
rπ =
(150)(0.026) = 21.36 k Ω
0.1826
Then Rib = 21.36 + (151)(0.5) = 96.86 k Ω
(
)
τ S = 110 42 96.86 × 10 3 × (0.47 × 10 −6 ) ⇒ 10.87 ms
(b) f L =
Aυ =
1
2πτ S
=
1
= 14.6 Hz
2π 10.87 × 10 −3
(
)
− βRC
− (150)(7 )
=
= −10.84
rπ + (1 + β )R E 21.36 + (151)(0.5)
________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX7.4
(a) I DQ = K n (VGSQ − VTN )2
(
)
⇒ VGSQ = 1.981 V
=
3 − 1.981
= 4.08 k Ω
0.25
250 = 100 VGSQ − 0.4
− VGSQ − (− 3)
RS =
I DQ
2
V D = −VGSQ + V DSQ = −1.981 + 1.7 = −0.281 V
RD =
3 − V D 3 − (− 0.281)
=
= 13.1 k Ω
I DQ
0.25
(
)
(b) τ S = (RD + RL )CC = (13.1 + 20) × 103 × 0.7 × 10 −6 ⇒ τ S = 23.17 ms
1
1
=
= 6.87 Hz
fL =
2πτ S 2π 23.17 × 10 −3
______________________________________________________________________________________
(
EX7.5
)
τ S = (R L + Ro )C C 2
f =
1
2π τ S
⇒ CC 2 =
1
2π (R L + R o )
⎧⎪ rπ + (R S R B )⎫⎪
R o = R E ro ⎨
⎬
1+ β
⎪⎩
⎪⎭
From Example 7-5, R0 = 35.5 Ω
CC 2 =
1
2π (10 ) ⎡⎣10 × 10 3 + 35.5⎤⎦
CC 2 = 1.59 μ F
______________________________________________________________________________________
EX7.6
(a)
RTH = 5 K
VTH = −3.7527
I BQ =
−3.7527 − 0.7 − ( −5) 0.54726
=
5 + (101)( 0.5)
55.5
= 0.00986
I CQ = 0.986 mA
gm = 37.925 rπ = 2.637 K
5 − Vo Vo
− = 1 − Vo ( 0.4 )
5
5
Vo = 0.035
0.986 =
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)
Vo = − g mVπ (RC R L )
Rib = rπ + (1 + β )R E = 2.64 + (101)(0.5) = 53.14 k Ω
Vπ =
Vb =
Vb
⎛1+ β ⎞
⎟RE
1 + ⎜⎜
⎟
⎝ rπ ⎠
RTH Rib
RTH Rib + R S
Aυ =
=
Vb
Vb
=
14
.885
101
⎞
⎛
1+ ⎜
⎟(0.5)
⎝ 2.637 ⎠
⋅ Vi =
5 53.14
5 53.14 + 0.1
⋅ Vi = (0.9786 )Vi
− (37.925 )(2.5)
(0.9786 ) ⇒ Aυ = −6.23
14.885
(c)
______________________________________________________________________________________
EX7.7
a.
I BQ =
0 − 0.7 − ( −10 )
0.5 + (101)( 4 )
= 0.0230 mA
I CQ = 2.30 mA
rπ =
gm =
τB =
β VT
I CQ
I CQ
VT
=
=
(100 )( 0.026 )
2.30
= 1.13 kΩ
2.30
= 88.46 mA / V
0.026
R E (R S + rπ )C E
R S + rπ + (1 + β )R E
(4 ×10 )(0.5 + 1.13)C
3
=
E
0.5 + 1.13 + (101)(4)
1
1
=
⇒ τ B = 0.7958 ms
τB =
2π f B 2π (200 )
τ B = 16.07C E ⇒ C E =
b.
0.796 × 10 −3
⇒ C E = 49.5 μ F
16.07
τ A = R E C E = (4 ×10 3 )(49.5 ×10 −6 ) ⇒ τ A = 0.198 s
1
⇒ f A = 0.804 Hz
2π τ A 2π (0.198)
_____________________________________________________________________________
fA =
1
=
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX7.8
(a)
fβ =
rπ =
1
2π rπ (Cπ + C μ )
(120)(0.026) = 15.6
0.2
Cπ + C μ =
kΩ
1
1
=
⇒ 0.113 pF
2π rπ f β 2π 15.6 × 10 3 90 × 10 6
(
)(
)
Cπ = 0.113 − 0.02 = 0.093 pF
(b) h fe =
βo
⎛ f ⎞
⎟
1+ ⎜
⎜ fβ ⎟
⎠
⎝
2
120
For f = 50 MHz, h fe =
(i)
For f = 125 MHz, h fe =
(ii)
⎛ 50 ⎞
1+ ⎜ ⎟
⎝ 90 ⎠
120
For f = 500 MHz, h fe =
2
⎛ 125 ⎞
1+ ⎜
⎟
⎝ 90 ⎠
120
= 105
2
= 70.1
= 21.3
2
⎛ 500 ⎞
1+ ⎜
⎟
⎝ 90 ⎠
______________________________________________________________________________________
(iii)
EX7.9
rπ =
(150)(0.026) = 26 k Ω
0.15
1
1
=
fβ =
2π rπ (Cπ + C μ ) 2π 26 × 10 3 (0.8 + 0.012 ) × 10 −12
(
)
or
f β = 7.54 MHz
f T = β o f β = (150 )(7.54 ) ⇒ f T = 1.13 GHz
______________________________________________________________________________________
EX7.10
(a) RTH = R1 R2 = 200 220 = 104.8 k Ω
⎛ R2 ⎞
⎛ 220 ⎞
⎟⎟ ⋅ VCC = ⎜
VTH = ⎜⎜
⎟(5) = 2.619 V
R
R
+
⎝ 200 + 220 ⎠
2 ⎠
⎝ 1
V − V BE (on )
2.619 − 0.7
=
I BQ = TH
= 0.009325 mA
RTH + (1 + β )R E 104.8 + (101)(1)
I CQ = β I BQ = 0.9325 mA
Now
rπ =
(100)(0.026) = 2.788 k Ω
0.9325
; gm =
0.9325
= 35.87 mA/V
0.026
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ rs R1 R2
Ib = ⎜
⎜r R R +r
⎝ s 1 2 π
⎞
⎞
100 104.8
⎟ ⋅ I = ⎛⎜
⎟ ⋅ I = 0.9483I s
⎟ s ⎜ 100 104.8 + 2.788 ⎟ s
⎠
⎝
⎠
⎛ RC
I o = − βI b ⎜⎜
⎝ RC + R L
⎞
⎛ 2.2 ⎞
⎟⎟ = −(100 )(0.9483)I s ⎜
⎟
⎝ 2.2 + 4.7 ⎠
⎠
or
Io
= −30.24
Is
(b) (i) For C μ = 0 ⇒ C M = 0
Ai =
[
)]
= (0.08)[1 + (35.87)(2.2 4.7 )] = 4.38 pF
(
(ii) For C μ = 0.08 pF ⇒ C M = C μ 1 + g m RC RL
CM
(c) Req = rs RTH rπ = 100 104.8 2.788 = 2.644 k Ω
(i)
f 3dB =
1
1
=
⇒ 60.2 MHz
2πReq Cπ
2π 2.644 × 10 3 10 −12
(
)(
)
1
⇒ 11.2 MHz
2π 2.644 × 10 (1 + 4.38) × 10 −12
______________________________________________________________________________________
(ii)
f 3dB =
(
3
)
EX7.11
g m = 2π f T C gs + C gd = 2π 3 × 10 9 (60 + 8) × 10 −15 ⇒ 1.282 mA/V
(
(
)
)
2
g m = 2 K n I DQ ⇒ I DQ =
1 ⎛ gm ⎞
1 ⎛ 1.282 ⎞
⎜
⎟ =
⎟
⎜
1.2 ⎝ 2 ⎠
Kn ⎝ 2 ⎠
2
or
I DQ = 0.342 mA
______________________________________________________________________________________
EX7.12
⎛ 166 ⎞
(a) VG = ⎜
⎟(10) = 4.15 V
⎝ 166 + 234 ⎠
2
VG = VGS + K n RS (VGS − VTN )
(
2
4.15 = VGS + (0.8)(0.5) VGS
− 4VGS + 4
)
or
2
0.4VGS
− 0.6VGS − 2.55 = 0
which yields
VGS = 3.384 V
I D = (0.8)(3.384 − 2) = 1.532 mA
Now
g m = 2 K n I D = 2 (0.8)(1.532 ) = 2.214 mA/V
2
RTH = R1 R2 = 166 234 = 97.11 k Ω
So
⎛ RTH
Aυ = − g m (R D R L )⎜⎜
⎝ RTH + Ri
⎞
⎛ 97.11 ⎞
⎟⎟ = −(2.214 )(4 20 )⎜
⎟
⎝ 97.11 + 10 ⎠
⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or
Aυ = −6.69
[
)]
(
[
(
)]
(b) C M = C gd 1 + g m RD RL = 20 1 + (2.214) 4 20 = 167.6 fF
(c)
f 3−dB =
−3
gm
2.214 × 10
=
2π C gs + C M
2π (100 + 167.6) × 10 −15
(
)
or
f 3− dB = 1.32 GHz
______________________________________________________________________________________
EX7.13
dc analysis
VTH = 0, RTH = 10 kΩ
I BQ =
0 − 0.7 − ( −5 )
10 + (126 )( 5 )
= 0.00672 mA
I CQ = 0.840 mA
rπ =
gm =
r0 =
β VT
I CQ
I CQ
VT
=
=
(125)( 0.026 )
0.840
= 3.87 kΩ
0.840
= 32.3 mA/V
0.026
VA
200
=
= 238 kΩ
I CQ 0.84
High-frequency equivalent circuit
a.
Miller Capacitance
C M = C μ (1 + g m R L′ )
R L′ = ro RC R L = 238 2.3 5 = 1.565 k Ω
C M = (3)[1 + (32.3)(1.565)] ⇒ C M = 155 pF
b.
Req = R S R B rπ = R S R1 R 2 rπ
= 1 20 20 3.87 = 0.736 k Ω
τ p = R eq (C π + C M ) = (0.736 × 10 3 )(24 + 155)× 10 −12
fH
= 1.314 × 10 −7 s
1
=
⇒ f H = 1.21 MHz
2π 1.314 × 10 −7
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
c.
or
( Aυ )M
⎡ 10 3.87 ⎤
⎡ R B rπ ⎤
= − g m R L′ ⎢
⎥
⎥ = −(32.3)(1.565)⎢
⎣⎢10 3.87 + 1 ⎦⎥
⎣⎢ R B rπ + R S ⎥⎦
( Aυ )M
= −37.2
______________________________________________________________________________________
EX7.14
The dc analysis
10 − 0.7
= 0.00838 mA
I BQ =
100 + (101)(10 )
I CQ = 0.838 mA
rπ =
gm =
β VT
I CQ
=
(100 )( 0.026 )
0.838
= 3.10 kΩ
I CQ
= 32.22 mA/V
VT
For the input
⎤
⎡⎛ r ⎞
⎤
⎡⎛ 3.10 ⎞
−12
3
τ pπ = ⎢⎜⎜ π ⎟⎟ R E R S ⎥C π = ⎢⎜
⎟ 10 1⎥ × 10 × 24 × 10
1
101
β
+
⎠
⎝
⎠
⎝
⎦
⎣
⎦
⎣
= 7.13 × 10 −10 s
1
1
f Hπ =
=
⇒ f Hπ = 223 MHz
2π τ pπ
2π 7.13 × 10 −10
For the output
τ pμ = (RC R L )C μ = (10 1)× 10 3 × 3 × 10 −12
(
f Hμ
)
= 2.73 × 10 −9 s
1
1
=
=
⇒ f Hμ = 58.4 MHz
2π τ pμ 2π 2.73 × 10 −9
( Aυ )M
(
)
⎡
⎛ rπ ⎞
⎟⎟
⎢ R E ⎜⎜
⎝1+ β ⎠
⎢
= g m (RC R L )⎢
⎢ R ⎛⎜ rπ ⎞⎟ + R
S
⎢ E ⎜⎝ 1 + β ⎟⎠
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎛ 3.1 ⎞ ⎤
⎟ ⎥
⎢ 10 ⎜
⎝ 101 ⎠ ⎥
⎢
(
)
= (32.22) 10 1
⇒ ( Aυ ) M = 0.870
⎢ ⎛ 3.1 ⎞ ⎥
⎥
⎢10 ⎜
+
1
⎟
⎣⎢ ⎝ 101 ⎠ ⎦⎥
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX7.15
⎛
⎞
R3
7.92
⎛
⎞
VB1 = ⎜
⎟ (12 ) = ⎜
⎟ (12 ) = 0.9502 V
+
+
+
+
R
R
R
58.8
33.3
7.92
⎝
⎠
2
3 ⎠
⎝ 1
Neglecting base currents
0.9502 − 0.7
= 0.50 mA
IC =
0.5
β VT (100 )( 0.026 )
=
= 5.2 K
rπ =
0.5
IC
IC
0.5
=
= 19.23 mA/V
VT 0.026
From Eq (7.119(a)),
τ pπ = R S R B1 rπ (C π 1 + C M 1 )
gm =
(
)
R B1 = R2 R3 = 33.3 7.92 = 6.398 k Ω
C M 1 = 2C μ 1 = 6 pF
Then
τ pπ = 1 6.398 5.2 × 10 3 × (24 + 6 )× 10 −12 ⇒ τ pπ = 22.24 ns
(
f Hπ =
)
1
2π τ pπ
=
1
⇒ f Hπ = 7.15 MHz
2π 22.24 × 10 −9
(
)
From Eq (7.120(a)),
τ pμ = (RC R L )C μ 2 = (7.5 2 )× 10 3 × 3 × 10 −12 ⇒ τ pμ = 4.737 ns
f Hμ =
1
2π τ pμ
=
1
⇒ f Hμ = 33.6 MHz
2π 4.737 × 10 −9
(
)
From Eq. (7.125),
Aυ
M
Aυ
M
⎡ R B1 rπ 1 ⎤
⎡ 6.40 5.2 ⎤
= g m 2 (RC R L )⎢
⎥ = (19.23)(7.5 2)⎢
⎥
⎣⎢ R B1 rπ 1 + R S ⎦⎥
⎣⎢ 6.40 5.2 + 1 ⎦⎥
= 22.5
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU7.1
a.
V0 = − ( gmVπ ) RL
rπ
× Vi
Vπ =
1
+ RS
rπ +
sCC
T (s) =
=
V0 ( s )
Vi ( s )
=
− gm rπ RL
rπ + RS + (1 / sCC )
− gm rπ RL ( sCC )
1 + s ( rπ + RS ) CC
gm rπ = β
T (s) =
− β RL
rπ + RS
⎛ s ( rπ + RS ) CC
×⎜
⎜ 1 + s (r + R ) C
S
C
π
⎝
⎞
⎟⎟
⎠
Then τ = ( rπ + RS ) CC
b.
f3− dB =
1
2π ( rπ + RS ) CC
1
⇒ f3 dB = 53.1 Hz
2π ⎣⎡ 2 × 10 3 + 1 × 10 3 ⎦⎤ ⎣⎡10 −6 ⎦⎤
( 2 )( 50 )( 4 )
rg R
T ( jω ) max = π m L =
2 +1
rπ + RS
f3− dB =
T ( jω ) max = 133
c.
______________________________________________________________________________________
TYU7.2
(a) τ = RL C L = 10 × 10 3 2 × 10 −12 ⇒ 0.02 μ s
1
1
=
⇒ 7.96 MHz
(b) f 3− dB =
2πτ 2π 0.02 × 10 −6
(
)(
(
)
)
rπ
⎛ 2.4 ⎞
⋅ (g m R L ) = ⎜
⎟(50)(10 ) = 480
rπ + RS
⎝ 2.4 + 0.1 ⎠
______________________________________________________________________________________
Aυ
max
=
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU7.3
(a) τ S = (RS + rπ )CC = (0.1 + 2.4) × 10 3 × 5 × 10 −6 ⇒ 12.5 ms
τ P = RL C L = (10 × 10 )(4 × 10
3
(
−12
) ⇒ 0.04 μ s
)
⎛ rπ ⎞
2.4 ⎞
⎟⎟(g m RL ) = −⎛⎜
(b) Aυ = −⎜⎜
⎟(50 )(10 ) = −480
⎝ 2.4 + 0.1 ⎠
⎝ rπ + RS ⎠
1
1
=
= 12.7 Hz
(c) f L =
2πτ S 2π 12.5 × 10 −3
1
1
=
⇒ 3.98 MHz
fH =
2πτ P 2π 0.04 × 10 −6
______________________________________________________________________________________
(
)
(
)
TYU7.4 Computer Analysis
______________________________________________________________________________________
TYU7.5 Computer Analysis
______________________________________________________________________________________
TYU7.6
rπ =
(120)(0.026) = 26 k Ω
0.12
1
1
fβ =
⇒ (Cπ + C μ ) =
2π rπ (Cπ + C μ )
2π rπ f β
or
Cπ + C μ =
1
⇒ 0.408 pF
2π 26 × 10 3 15 × 10 6
(
)(
)
Then
Cπ = 0.408 − 0.08 = 0.328 pF
______________________________________________________________________________________
TYU7.7
h fe =
βo
2
⎛ f ⎞
⎟
⎜ f ⎟
⎝ β ⎠
φ = − tan −1 ⎜
;
⎛ f ⎞
⎟
1+ ⎜
⎜ fβ ⎟
⎝
⎠
f β = 167 MHz ; β o = 120
Then
(a) For f = 150 MHz; h fe =
120
= 89.3
2
⎛ 150 ⎞
1+ ⎜
⎟
⎝ 167 ⎠
⎛ 150 ⎞
φ = − tan −1 ⎜
⎟ = −41.9°
⎝ 167 ⎠
120
= 38.0
For f = 500 MHz; h fe =
2
500
⎛
⎞
1+ ⎜
⎟
⎝ 167 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ 500 ⎞
⎟ = −71.5°
⎝ 167 ⎠
120
h fe =
= 5.0
For f = 4 GHz;
2
⎛ 4000 ⎞
1+ ⎜
⎟
⎝ 167 ⎠
⎛ 4000 ⎞
φ = − tan −1 ⎜
⎟ = −87.6°
⎝ 167 ⎠
______________________________________________________________________________________
φ = − tan −1 ⎜
TYU7.8
(a)
10 9
⇒ f β = 6.67 MHz
β o 150
1
1
fβ =
⇒ Cπ + C μ =
2π rπ Cπ + C μ
2π rπ f β
fβ =
fT
=
) (
(
)
or
(C
π
+ Cμ ) =
1
⇒ 1.989 pF
2π 12 × 10 6.667 × 10 6
(
3
)(
)
Then
Cπ = 1.989 − 0.15 = 1.84 pF
(b) rπ =
β VT
I CQ
⇒ I CQ =
βVT
rπ
=
(150 )(0.026 ) = 0.325 mA
12
______________________________________________________________________________________
TYU7.9
(a)
gm = 2 K n (VGS − VTN ) = 2 ( 0.4 )( 3 − 1) ⇒ gm = 1.6 mA/V
gm′ = 80% of gm = 1.28 mA/V
gm′ =
gm
1 + gm rS
1 + gm rS =
gm
gm′
⎞ 1 ⎛ 1.6
1 ⎛ gm
⎞
− 1⎟ =
− 1⎟
⎜
⎜
′
gm ⎝ gm
⎠ 1.6 ⎝ 1.28 ⎠
rS = 0.156 kΩ ⇒ rS = 156 ohms
rS =
(b)
gm = 2 K n (VGS − VTN ) = 2 ( 0.4 )( 5 − 1) ⇒ gm = 3.2 mA/V
gm
3.2
gm′ =
=
= 2.134
1 + gm rS 1 + ( 3.2 )( 0.156 )
Δgm 3.2 − 2.134
=
⇒ A 33.3% reduction
3.2
gm
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU7.10
⎛ 0.1 ⎞
gm = 2 ⎜
⎟(15)(0.1) = 0.5477 mA/V
⎝ 2 ⎠
gm
gm
fT =
⇒ C gs + C gsp + C gdp =
2π C gs + C gsp + C gdp
2π f T
) (
(
)
=
(0.5477 × 10 ) ⇒ 72.64 fF
2π (1.2 × 10 )
−3
9
Then
C gs = 72.64 − 3 − 3 = 66.6 fF
______________________________________________________________________________________
TYU7.11
fT =
) (
)
gm
gm
⇒ C gs + C gsp + C gdp =
2π C gs + C gsp + C gdp
2π f T
(
=
1.2 × 10 −3
⇒ 76.39 fF
2π 2.5 × 10 9
(
)
C gsp + C gdp = 76.39 − 60 = 16.39
or
C gsp = C gdp = 8.2 fF
______________________________________________________________________________________
TYU7.12
dc analysis
⎛ 50 ⎞
VG = ⎜
⎟ (10 ) − 5 = −2.5
⎝ 50 + 150 ⎠
VS − ( −5 )
VS = VG − VGS . I D =
RS
K n (VGS − VTN ) =
2
VG − VGS + 5
RS
(1)( 2 ) ⎡⎣VGS2 − 1.6VGS + 0.64 ⎤⎦ = −2.5 − VGS + 5
2VGS2 − 2.2VGS − 1.22 = 0
VGS =
2.2 ±
( 2.2 )
2
+ 4 ( 2 )(1.22 )
2 (2)
⇒ VGS = 1.505 V
gm = 2K n (VGS − VTN ) = 2 (1)(1.505 − 0.8 )
= 1.41 mA/V
Equivalent circuit
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(a)
(b)
C M = C gd (1 + gm RD ) = ( 0.2 ) ⎡⎣1 + (1.42 )( 5 ) ⎤⎦ ⇒ C M = 1.61 pF
τ P = (Ri RG )(C gs + C M )
(
)
= 20 50 150 × 10 3 × (2 + 1.61)× 10 −12 = 4.71 × 10 −8 s
fH =
( Av )M
1
2π τ P
=
1
⇒ f H = 3.38 MHz
2π 4.71× 10 −8
(
)
⎛ RG ⎞
= − gm RD ⎜
⎟
⎝ RG + RS ⎠
⎛ 37.5 ⎞
= − (1.41)( 5 ) ⎜
⎟ ⇒ ( Av ) M = −4.60
⎝ 37.5 + 20 ⎠
( Av )M
c.
______________________________________________________________________________________
TYU7.13 Computer Analysis
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 8
Exercise Solutions
EX8.1
1
VCC = 12 V
2
⎛ 25 ⎞
= 2 I CQ = 2⎜ ⎟ = 4.17 A
⎝ 12 ⎠
(a) PT = I CVCE ; At VCEQ =
25 = I CQ (12) ⇒ I C ,max
24
= 5.76 Ω
4.167
PQ , max = 25 W
RL =
⎞
⎛1
(b) 25 = I CQ ⎜ ⋅ VCC ⎟ = I CQ (6) ⇒ I CQ = 4.17 A, ⇒ I C , max = 5 A
⎠
⎝2
12
= 2.4 Ω
RL =
5
At I CQ = 2.5 A, VCEQ = 6 V
PQ ,max = (2.5)(6 ) = 15 W
______________________________________________________________________________________
EX8.2
PQ = (2 )(8) = 16 W
(a) Tdev = 25 + (16 )(3 + 1 + 4) = 153 ° C
(b) Tcase = 25 + (16 )(1 + 4) = 105° C
(c) Tsnk = 25 + (16 )(4 ) = 89° C
______________________________________________________________________________________
EX8.3
θ dev − case =
PD ,max =
TJ ,max − Tamb
PD,rated
=
200 − 25
= 3.5°C/W
50
TJ ,max − Tamb
θ dev − case + θ case − snk + θ snk − amb
200 − 25
⇒ PD ,max = 29.2 W
3.5 + 0.5 + 2
= Tamb + PD ,max (θ case −snk + θ snk − amb )
=
Tcase
= 25 + ( 29.2 )( 0.5 + 2 ) ⇒ Tcase = 98°C
______________________________________________________________________________________
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX8.4
a.
b.
I DQ =
10 − 4
⇒ I DQ = 60 mA
0.1
⎛9⎞
vds = − ⎜ ⎟ ( 60 )( 0.050 ) = −2.7 V ⇒ vDS ( min ) = 4 − 2.7 = 1.3 V
⎝ 10 ⎠
So maximum swing is determined by drain-to-source voltage.
VPP = 2 × ( 2.5) = 5.0 V
c.
1 VP2 1 ( 2.5 )
⋅
= ⋅
⇒ PL = 31.25 mW
2 RL 2 0.1
2
PL =
PS = VDD ⋅ I DQ = (10 )( 60 ) = 600 mW
η=
PL
PS
=
31.25
⇒ η = 5.2%
600
______________________________________________________________________________________
EX8.5 Computer Analysis
______________________________________________________________________________________
EX8.6 No Exercise Problem
______________________________________________________________________________________
EX8.7
8
= 0.32 A
25
= (0.2 )(0.32 ) ⇒ 64 mA
(a) For υ O = 8 V, i L =
I DQ
I DQ = K (VGS − VTN )
2
64 = 250(VGS − 1.2 ) ⇒ VGS =
2
Then V BB = 3.412 V
V BB
= 1.706 V
2
V BB
V
− υ GSn ⇒ υ I = υ O − BB + υ GSn
2
2
dυ GSn
dυ I
= 1+
dυ O
dυ O
We have
dυ GSn dυ GSn di dn
=
⋅
dυ O
didn dυ O
(b) υ O = υ I +
υ GSn =
idn
dυ GSn 1 1
1
+ VTN ⇒
= ⋅
⋅
2 K
K
di dn
idn
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(i) For υ O ≈ 0 (very small), then
idn = i L + idp ⇒ − Δi dp ≅ Δidn
so Δi dn ≅
then
Δidn
Δυ O
1
Δi L
2
1 Δi
1 1
1
= ⋅ L = ⋅
=
= 0.02
2 Δυ O 2 R L 50
For υ O ≈ 0 , idn = 0.064 A
dυ GSn 1
1
1
= ⋅
⋅
= 3.953
didn
2 0.25 0.064
Then
dυ I
= 1 + (3.953)(0.02 ) = 1.079
dυ O
Or
dυ O
= 0.927
dυ I
(ii) For υ O ≅ 8 V, i dn = i L
didn
di
1
= L =
= 0.04
dυ O dυ O R L
then
dυ GSn 1
1
1
= ⋅
⋅
= 1.768
di dn
2 0.25
0.32
dυ I
= 1 + (1.768)(0.04 ) = 1.0707
dυ O
Or
dυ O
= 0.934
dυ I
______________________________________________________________________________________
EX8.8
a.
Rb = rπ + (1 + β ) RE′ and RE′ = a 2 RL = (10 ) ( 8 ) = 800 Ω
Ri = 1.5 kΩ = RTH Rb
2
IQ =
rπ =
VCC
18
=
= 22.5 mA
a 2 RL (10 )2 ( 8 )
(100 )( 0.026 )
= 0.116 kΩ
22.5
Rb = 0.116 + (101)( 0.8 ) = 80.9 kΩ
1.5 = RTH 80.9 =
RTH ( 80.9 )
RTH + ( 80.9 )
⇒ ( 80.9 − 1.5 ) RTH = (1.5 )( 80.9 ) ⇒ RTH = 1.53 kΩ
⎛ R2 ⎞
1
VTH = ⎜
⎟ VCC = ⋅ RTH ⋅ VCC
+
R
R
R
⎝ 1
2 ⎠
1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
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Exercise Solutions
______________________________________________________________________________________
I BQ
IQ
22.5
= 0.225 mA
100
V − 0.7
1
= TH
⇒ (1.53)(18 ) = ( 0.225 )(1.53) + 0.7 ⇒ R1 = 26.4 kΩ
RTH
R1
I BQ =
β
=
26.4 R2
= 1.53
26.4 + R2
( 26.4 − 1.53) R2 = (1.53)( 26.4 ) ⇒ R2 = 1.62 kΩ
b.
vE = 0.9VCC = ( 0.9 )(18 ) = 16.2 V
iE = 0.9 I CQ = ( 0.9 )( 22.5 ) = 20.25 mA
v
16.2
v0 = E =
⇒ VP = 1.62 V
10
a
i0 = aiE = (10 )( 20.25 ) ⇒ I P = 203 mA
PL =
1
(1.62 )( 0.203) ⇒ PL = 0.164 W
2
______________________________________________________________________________________
EX8.9
⎛ 10 −3
(a) For I Q = 1 mA, υ BE = (0.026) ln⎜⎜
−14
⎝ 2 × 10
Then V BB = 1.281 V
(
⎞
⎟⎟ = 0.6405 V
⎠
)
⎛ 0.6405 ⎞
For D1 , D2 ; I Bias = 1.2 × 10 −14 exp⎜
⎟ = 0.60 mA
⎝ 0.026 ⎠
1.2
= 1.2 mA
(b) υ O = 1.2 V, i L =
1
1st approximation:
iCn = 1.6 mA, i Bn = 0.016 mA
⎛ 1.6 × 10 −3 ⎞
⎟ = 0.65274 V
−14 ⎟
⎝ 2 × 10 ⎠
I D = 0.60 − 0.016 = 0.584 mA
υ BEn = (0.026 ) ln⎜⎜
⎛ 0.584 × 10 −3 ⎞
⎟ = 1.27963 V
V BB = 2(0.026 ) ln⎜⎜
−14 ⎟
⎠
⎝ 1.2 × 10
then υ EBp = 1.27963 − 0.65274 = 0.62689 V
(
)
⎛ 0.62689 ⎞
iCp = 2 × 10 −14 exp⎜
⎟ = 0.59206 mA
⎝ 0.026 ⎠
2nd approximation:
i En = 1.2 + 0.59206 = 1.792 mA; iCn = 1.7743 mA
After 4 iterations:
iCn = 1.73 mA; iCp = 0.547 mA
υ BEn = 0.6547 V; υ EBp = 0.6248 V
I D = 0.5827 mA
(c) υ O = 3 V; i L =
3
= 3 mA
1
1st approximation;
iCn = 3.3 mA, i Bn = 0.033 mA
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
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Exercise Solutions
______________________________________________________________________________________
⎛ 3.3 × 10 −3 ⎞
⎟ = 0.67156 V
−14 ⎟
⎝ 2 × 10 ⎠
I D = 0.6 − 0.033 = 0.567 mA
υ BEn = (0.026) ln⎜⎜
⎛ 0.567 × 10 −3 ⎞
⎟ = 1.2781 V
V BB = 2(0.026) ln⎜⎜
−14 ⎟
⎠
⎝ 1.2 × 10
υ EBp = 1.2781 − 0.67156 = 0.6065 V
(
)
⎛ 0.6065 ⎞
iCp = 2 × 10 −14 exp⎜
⎟ = 0.2706 mA
⎝ 0.026 ⎠
Then i En = 3 + 0.2706 = 3.2706 mA; iCn = 3.2382 mA
After 4 iterations:
iCn = 3.24 mA; i Cp = 0.276 mA
υ BEn = 0.671 V, υ EBp
= 0.607 V
I D = 0.5676 mA
______________________________________________________________________________________
EX8.10 No Exercise EX8.10
______________________________________________________________________________________
EX8.11
a.
vI = 0 = v0 , vB 3 = 0.7 V
12 − 0.7 11.3
=
⇒ I R1 = 45.2 mA
I R1 =
0.25
R1
If transistors are matched, then
iE1 = iE 3
iR1 = iE1 + iB 3 = iE1 +
iE 3
1+ β
⎛
1 ⎞
1⎞
⎛
iR1 = iE1 ⎜1 +
⎟ = iE1 ⎜1 + ⎟
1
41
β
+
⎝
⎠
⎝
⎠
45.2
⇒ iE1 = iE 2 = 44.1 mA
iE1 =
1.024
i
44.1
⇒ iB1 = iB 2 = 1.08 mA
iB1 = iB 2 = E1 =
1+ β
41
b.
For vI = 5 V ⇒ v0 = 5 V
5
⇒ i0 = 0.625 A
8
0.625
iE 3 ≅ 0.625 A, iB 3 =
⇒ iB 3 = 15.2 mA
41
12 − 5.7
vB 3 = 5.7 V ⇒ iR1 =
= 25.2 mA
0.25
10
iE1 = 25.2 − 15.2 ⇒ iE1 = 10.0 mA ⇒ iB1 =
= 0.244 mA
41
vB 4 = 5 − 0.7 = 4.3 V
i0 =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
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Exercise Solutions
______________________________________________________________________________________
IR2 =
4.3 − ( −12 )
0.25
= 65.2 mA ≅ iE 2
65.2
= 1.59 mA
41
iI = iB 2 − iB1 = 1.59 − 0.244 ⇒ iI = 1.35 mA
iB 2 =
AI =
i0 625
=
⇒ AI = 463
iI 1.35
c.
From Equation (8.55)
(1 + β ) R ( 41)( 250 )
AI =
2 RL
=
2 (8)
= 641
______________________________________________________________________________________
Test Your Understanding Solutions
TYU8.1
For
For
VDS = 0, I D ( max ) =
24
= 1.2 A = ID ( max )
20
I D = 0 ⇒ VDS ( max ) = 24 V
Maximum power when
VDS =
ID =
VDS ( max )
2
I D ( max )
2
= 12 V and
= 0.6 A ⇒ PD ( max ) = (12 )( 0.6 ) = 7.2 Watts
______________________________________________________________________________________
TYU8.2
(a) RE =
VCC − (− VCC ) 12 − (− 12 )
=
= 96 Ω
I C ,max
0.25
(b) For I CQ = 0.125 A, VCEQ = 12 V
PQ , max = (0.125 )(12 ) = 1.5 W
______________________________________________________________________________________
TYU8.3
(a) ΔT = P ⋅ θ = (6 )(1.8) = 10.8° C
ΔT 100
=
= 40 W
(b) P =
2.5
θ
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU8.4
(a) VCEQ = 6 = VCC − I CQ R L = 12 − I CQ (1) ⇒ I CQ = 6 mA
PQ = I CQVCEQ = (6 )(6 ) = 36 mW
1 V P2 1 (4.5)
⋅
=
= 10.1 mW
2 RL 2 1
10.1
10.1
× 100% =
× 100% = 14.1%
(ii) η =
I CQVCC
(6)(12)
2
(b) (i) PL =
(iii) PQ = 36 − 10.1 = 25.9 mW
______________________________________________________________________________________
TYU8.5
PL =
1 VP2
20
⋅
⇒ VP = 2 RL PL = 2 ( 8 )( 25 ) ⇒ VP = 20 V ⇒ VCC =
⇒ VCC = 25 V
2 RL
0.8
IP =
VP 20
=
⇒ I P = 2.5 A
8
RL
PQ =
VCCVP VP2
−
π RL 4 RL
PQ =
( 25)( 20 ) ( 20 )
−
π (8)
4 (8)
a.
b.
2
c.
η=
= 19.9 − 12.5 ⇒ PQ = 7.4 W
π VP π 20
= ⋅
⇒ η = 62.8%
4VCC 4 25
d.
______________________________________________________________________________________
TYU8.6
PL =
( 4)
1 VP2
⋅
=
⇒ PL = 80 mW
2 RL 2 ( 0.1)
IP =
VP
4
=
⇒ I P = 40 mA
RL 0.1
PQ =
VCCVP VP2
−
π RL 4 RL
PQ =
( 5 )( 4 ) ( 4 )
−
= 63.7 − 40 ⇒ PQ = 23.7 mW
π ( 0.1) 4 ( 0.1)
2
a.
b.
2
c.
η=
π VP π 4
= ⋅ ⇒ η = 62.8%
4VCC 4 5
d.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU8.7
a.
I CQ ≅
1 ⎛ 2VCC
⋅⎜
2 ⎝ RL
⎞ VCC 12
=
= 8 mA
⎟=
⎠ RL 1.5
RTH = R1 R2
⎛ R2 ⎞
1
VTH = ⎜
⎟ VCC = ⋅ RTH ⋅ VCC
R1
⎝ R1 + R2 ⎠
I CQ
VTH − VBE
8
=
= 0.107 mA =
I BQ =
β
75
RTH + (1 + β ) RE
Let RTH = (1 + β ) RE = ( 76 )( 0.1) = 7.6 kΩ
1
⋅ ( 7.6 )(12 ) − 0.7
R1
0.107 =
7.6 + 7.6
1
⋅ ( 91.2 ) = 2.33 ⇒ R1 = 39.1 kΩ
R1
39.1R2
= 7.6 ⇒ ( 39.1 − 7.6 ) R2 = ( 7.6 )( 39.1) ⇒ R2 = 9.43 kΩ
39.1 + R2
b.
2
2
1
1
⋅ ( 0.9 I CQ ) RL = ⎡⎣( 0.9 )( 8 ) ⎤⎦ (1.5 ) ⇒ PL = 38.9 mW
2
2
PS = VCC I CQ = (12 )( 8 ) = 96 mW
PL =
PQ = PS − PL = 96 − 38.9 ⇒ PQ = 57.1 mW
η=
PL
PS
=
38.9
⇒ η = 40.5%
96
______________________________________________________________________________________
TYU8.8
⎛ 10 −3 ⎞
⎟ = 0.73643 V
(a) υ BEn = (0.026) ln⎜⎜
−16 ⎟
⎝ 5 × 10 ⎠
⎛ 10 −3 ⎞
⎟ = 0.72421 V
υ EBp = (0.026) ln⎜⎜
−16 ⎟
⎝ 8 × 10 ⎠
V BB = υ BEn + υ EBp = 1.4606 V
(b) See above
(c) υ I = υ BEn −
V BB
1.4606
= 0.73643 −
2
2
or
υ I = 6.1 mV
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU8.9
I E = I E 3 + IC 4 + IC 5
= I E 3 + IC 4 + β5 I B5
= I E 3 + β 4 I B 4 + β 5 (1 + β 4 ) I B 4
I E = (1 + β 3 ) I B 3 + β 4 β 3 I B 3 + β 5 (1 + β 4 ) β 3 I B 3
If β 4 and β 5 are large, then I E ≅ β 3 β 4 β 5 I B 3
β ≅ β3 β 4 β5
So that composite current gain is
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 9
Exercise Solutions
EX9.1
(a) i1 =
υI
R1
Aυ = −
=
25 × 10 −3
= 10 × 10 −6 ⇒ R1 = 2.5 k Ω
R1
R2
− R2
⇒ −25 =
⇒ R2 = 62.5 k Ω
R1
2.5
(
)
(b) υ O = Aυ ⋅ υ I = (25) 25 × 10 −3 = 0.625 V
−0.625 ≤ υ O ≤ 0.625 V
______________________________________________________________________________________
EX9.2
⎛
R
R ⎞
⎜⎜1 + 3 + 3 ⎟⎟
R
R
4
2 ⎠
⎝
Aυ = −75 , Let R1 = 20 k Ω
Aυ = −
R2
R1
Aυ = −75 = −
Let
R2
R1
⎛
R ⎞ R
⎜⎜1 + 3 ⎟⎟ − 3
R4 ⎠ R1
⎝
R 2 R3
=
=8
R1
R1
Then R2 = R3 = 160 k Ω
⎛
R ⎞
75 = 8⎜⎜1 + 3 ⎟⎟ + 8
R4 ⎠
⎝
R3
or
= 7.375
R4
160
= 21.7 k Ω
So R4 =
7.375
______________________________________________________________________________________
EX9.3
(a) Aυ = −
R2
1
⋅
R1 ⎡
R ⎞⎤
1 ⎛
⎜⎜1 + 2 ⎟⎟⎥
⎢1 +
R1 ⎠⎥⎦
⎣⎢ Aod ⎝
R1 = 25 k Ω , Aυ = −15.0 , Aod = 10 4
Then
R
1
− 15.0 = − 2 ⋅
R1 ⎡
R2 ⎞⎤
1 ⎛
⎟⎥
⎢1 + 4 ⎜⎜1 +
R1 ⎟⎠⎦⎥
⎣⎢ 10 ⎝
which yields
R2
= 15.024 ⇒ R2 = 375.6 k Ω
R1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
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Exercise Solutions
______________________________________________________________________________________
(b) (i) Aod = 10 5
Then Aυ = −(15.024) ⋅
1
1
⎤
⎡
⎢1 + 10 5 (16.024)⎥
⎦
⎣
= −15.0216
(ii) Aod = 10 3
Aυ = −(15.024) ⋅
1
= −14.787
1
⎡
⎤
(
)
+
1
16
.
024
⎢ 10 3
⎥
⎣
⎦
______________________________________________________________________________________
EX9.4
(a) υ O = −3(υ I 1 + 2υ I 2 + 0.3υ I 3 + 4υ I 4 )
υ O = −3υ I 1 − 6υ I 2 − 0.9υ I 3 − 12υ I 4
Then
R
R
RF
R
= 3 , F = 6 , F = 0.9 , F = 12
R1
R2
R3
R4
R3 will be the maximum resistance.
Let R3 = 400 k Ω ⇒ R F = 360 k Ω , R1 = 120 k Ω , R2 = 60 k Ω , R4 = 30 k Ω
(b) (i) υ O = −3(0.1) − 6(− 0.2 ) − 0.9(− 1) − 12(0.05) = +1.2 V
(ii) υ O = −3(− 0.2) − 6(0.3) − 0.9(1.5) − 12(− 0.1) = −1.35 V
______________________________________________________________________________________
EX9.5
We may note that
R3
3
=
=2
R2 1.5
and
RF 20
=
=2
R1 10
so that
R3 RF
=
R2 R1
Then
iL =
− ( −3 )
− vI
=
⇒ iL = 2 mA
R2 1.5 kΩ
vL = iL Z L = ( 2 × 10−3 ) ( 200 ) = 0.4 V
i4 =
vL
0.4
=
= 0.267 mA
R2 1.5 kΩ
i3 = i4 + iL = 0.267 + 2 = 2.267 mA
v0 = i3 R3 + vL = ( 2.267 × 10−3 )( 3 × 103 ) − 0.4 ⇒ v0 = 7.2 V
______________________________________________________________________________________
EX9.6
(a) Ad =
R2
= 50
R1
For υ I 2 = 50 mV and υ I 1 = −50 mV
υ O = 50(0.05 − (− 0.05)) = 5 V
5 − 0.05
iR2 ≅
= 50 μ A
R2
Set R2 = R4 = 100 k Ω
R1 = R3 = 2 k Ω
0.05
0.05
=
⇒ 0.49 μ A
(b) i R 3 =
R3 + R4 100 + 2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
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Exercise Solutions
______________________________________________________________________________________
______________________________________________________________________________________
EX9.7
We have the general relation that
⎛ R ⎞ ⎛ [ R4 / R3 ] ⎞
R2
v0 = ⎜1 + 2 ⎟ ⎜
⎟⎟ vI 2 − vI 1
⎜
R1 ⎠ ⎝ 1 + [ R4 / R3 ] ⎠
R1
⎝
R1 = R3 = 10 kΩ, R2 = 20 kΩ, R4 = 21 kΩ
⎛ 20 ⎞ ⎛ [ 21/10] ⎞
⎛ 20 ⎞
v0 = ⎜1 + ⎟ ⎜
⎟ vI 2 − ⎜ ⎟ vI 1
⎝ 10 ⎠ ⎜⎝ 1 + [ 21/10] ⎟⎠
⎝ 10 ⎠
v0 = 2.0323vI 2 − 2.0vI
vI 1 = 1, vI 2 = −1
a.
b.
c.
v0 = −2.0323 − 2.0 ⇒ v0 = −4.032 V
vI 1 = vI 2 = 1 V
v0 = 2.0323 − 2.0 ⇒ v0 = 0.0323 V
vcm = vI 1 = vI 2
so common-mode gain
v
Acm = 0 = 0.0323
vcm
d.
⎛ A ⎞
C M R RdB = 20 log10 ⎜ d ⎟
⎝ Acm ⎠
2.0323
⎛ 1⎞
Ad =
− ( 2.0 ) ⎜ − ⎟ = 2.016
2
⎝ 2⎠
⎛ 2.016 ⎞
C M R RdB = 20 log10 ⎜
⎟ = 35.9 d B
⎝ 0.0323 ⎠
______________________________________________________________________________________
EX9.8
⎛ 2 R2 ⎞
⎟
⎜⎜1 +
R1 ⎟⎠
⎝
90 ⎛ 2(50) ⎞
Ad (max ) =
⎟ = 153
⎜1 +
30 ⎝
2 ⎠
(a) Ad =
R4
R3
90 ⎛
2(50) ⎞
⎟ = 5.94
⎜1 +
30 ⎝ 2 + 100 ⎠
5.94 ≤ Ad ≤ 153
Ad (min ) =
0.025 − (− 0.025)
⇒ i1 = 25 μ A
R1
2
______________________________________________________________________________________
(b) i1 =
υ I1 − υ I 2
=
EX9.9
(a) R1C 2 = 10 4 0.1 × 10 −6 ⇒ 1 ms
(i)
0 < t <1
1 1
υ O = 0 − ⋅ t ' = −1 V
1 0
(ii)
1< t < 2
(− 1) ⋅ t ' 2 = −1 + 1(2 − 1) = 0
υ O = −1 −
1
1
( )(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
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Exercise Solutions
______________________________________________________________________________________
(iii)
2<t<3
3
1
1
υ O = 0 − ⋅ t ' = −1 V
(iv)
3<t <4
υ O = −1 −
2
(− 1) ⋅ t ' 4 = 0
1
3
( )( ) ⇒ 10 ms
(b) R1C 2 = 10 10
(i)
0 < t <1
4
−6
υO = 0 −
(ii)
(1) ⋅ t ' 1 = −0.1 V
10
1< t < 2
υ O = −0.1 −
(iii)
2<t<3
υO = 0 −
(iv)
0
(− 1) ⋅ t ' 2 = 0
10
1
(1) ⋅ t ' 3 = −0.1 V
10
3<t <4
υ O = −0.1 −
2
(− 1) ⋅ t ' 4 = 0
10
3
______________________________________________________________________________________
EX9.10
(a) RN = R1 R2 = 40 20 = 13.33 k Ω
R P = R A R B RC = 50 50 100 = 20 k Ω
⎛
⎤
RF
R
R ⎞⎡ R
R
υ I 1 − F υ I 2 + ⎜⎜1 + F ⎟⎟ ⎢ P υ I 3 + P υ I 4 ⎥
R1
R2
R
R
R
B
N ⎠⎣ A
⎦
⎝
80
80
80 ⎞ ⎡ 20
20
⎛
⎤
υ O = − υ I 1 − υ I 2 + ⎜1 +
⎟⎢ υ I 3 + υ I 4 ⎥
40
20
50
⎝ 13.33 ⎠ ⎣ 50
⎦
υ O = −2υ I 1 − 4υ I 2 + 2.8υ I 3 + 2.8υ I 4
υO = −
(b) (i) υ O = −2(0.1) − 4(0.15) + 2.8(0.2) + 2.8(0.3) = 0.6 V
(ii) υ O = −2(− 0.2) − 4(0.25) + 2.8(− 0.1) + 2.8(0.2) = −0.32 V
______________________________________________________________________________________
EX9.11 Computer Analysis
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU9.1
(a) Aυ = −
R2
240
⇒ −12 = −
⇒ R1 = 20 k Ω
R1
R1
υI
−0.15
=
⇒ i1 = −7.5 μ A
R1
20
0.25
⇒ i1 = 12.5 μ A
(ii) i1 =
20
______________________________________________________________________________________
(b) (i) i1 =
TYU9.2
(a)
Av =
− R2
R1 + RS
−100
= −4.926
19 + 1.3
−100
Av ( max ) =
= −5.076
19 + 0.7
so 4.926 ≤ Av ≤ 5.076
Av ( min ) =
(b)
0.1
= 5.076 μ A
19 + 0.7
0.1
i1 ( min ) =
= 4.926 μ A
19 + 1.3
so 4.926 ≤ i1 ≤ 5.076 μ A
i1 ( max ) =
(c) Maximum current specification is violated.
______________________________________________________________________________________
TYU9.3
Aυ = −
R2
1
200
=−
⋅
⋅
R1 ⎡
20 ⎡
⎤
1
R2 ⎞
1 ⎛
⎟⎟⎥
⎜⎜1 +
⎢1 + 4
⎢1 +
10
A
R
⎣
1 ⎠⎥
od ⎝
⎣⎢
⎦
1
200 ⎞⎤
⎛
⎟⎥
⎜1 +
20 ⎠⎦
⎝
or
Aυ = −9.989
(a) υ O = (− 9.989 )(50 ) ⇒ υ O = −0.49945 V
υ1 = −
(b) υ I =
υO
Aod
υO
Aυ
υ1 = −
=
υO
Aod
=−
(− 0.49945) ⇒ υ
10 4
1
= 49.945 μ V
+5
= −0.50055 V
− 9.989
=−
+5
⇒ υ1 = −0.5 mV
10 4
( )
(c) υ O = − Aodυ1 = − 10 4 (0.2) ⇒ υ O = −2 V
υO
−2
=
= 0.20022 V
Aυ − 9.989
______________________________________________________________________________________
υI =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU9.4
RF
R
R
υ I1 − F υ I 2 − F υ I 3
R1
R2
R3
200
200
200
=−
υ I1 −
υI2 −
υI3
20
40
50
= −10υ I 1 − 5υ I 2 − 4υ I 3
υO = −
(a) υ O = −10(− 0.25) − 5(0.30 ) − 4(− 0.50) ⇒ υ O = 3 mV
(b) υ O = −10(10 ) − 5(− 40 ) − 4(25) = 0
______________________________________________________________________________________
TYU9.5
vI 1 + vI 2 + vI 3 RF
=
( vI 1 + vI 2 + vI 3 )
3
R
RF 1
= ⇒ R1 = R2 = R3 ≡ R = 1 M Ω
R 3
1
Then RF = M Ω = 333 k Ω
3
vO =
______________________________________________________________________________________
TYU9.6
(a)
⎛
R ⎞
R
Aυ = 10 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 9
R1 ⎠
R1
⎝
Set R2 = 180 k Ω , R1 = 20 k Ω
⎛
R ⎞
R
(b) Aυ = 5 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 4
R1 ⎠
R1
⎝
For υ O = 5 V, υ1 = 1 V
5 −1
⇒ R2 = 40 k Ω , then R1 = 10 k Ω
i R 2 = 100 μ A =
R2
______________________________________________________________________________________
TYU9.7
v0 = Aod ( v2 − v1 ) = Aod ( vI − v1 )
v0
v
− vI = −v1 or v1 = vI − 0
Aod
Aod
i1 =
v −v
v1
= i2 and i2 = 0 1
R1
R2
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ 1 ⎞ v −v
Then v1 ⎜ ⎟ = 0 1
R2
⎝ R1 ⎠
⎛ 1
1 ⎞ v
v1 ⎜ + ⎟ = 0
⎝ R1 R2 ⎠ R2
⎛ R ⎞
⎛ R ⎞⎛
v ⎞
v0 ⎜ 1 + 2 ⎟ v1 = ⎜1 + 2 ⎟ ⎜ vI − 0 ⎟
Aod ⎠
R1 ⎠
R1 ⎠ ⎝
⎝
⎝
⎛ R2 ⎞
⎜1 + ⎟
R1 ⎠
v0
⎝
So Av = =
vI
1 ⎛ R2 ⎞
1+
⎜1 + ⎟
Aod ⎝
R1 ⎠
______________________________________________________________________________________
TYU9.8
⎛ Rb ⎞
For vI 2 = 0, v2 = ⎜
⎟ vI 1 and
⎝ Rb + Ra ⎠
⎛ R ⎞ ⎛ Rb ⎞
v0 ( vI 1 ) = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 1
R1 ⎠ ⎝ Rb + Ra ⎠
⎝
⎛ 70 ⎞ ⎛ 50 ⎞
= ⎜1 + ⎟ ⎜
⎟ vI 1
5 ⎠ ⎝ 50 + 25 ⎠
⎝
= 10vI 1
⎛ Ra ⎞
For vI 1 = 0, v2 = ⎜
⎟ vI 2
⎝ Rb + Ra ⎠
⎛ R ⎞ ⎛ Ra ⎞
v0 ( vI 2 ) = ⎜ 1 + 2 ⎟ ⎜
⎟ vI 2
R1 ⎠ ⎝ Rb + Ra ⎠
⎝
⎛ 70 ⎞ ⎛ 25 ⎞
= ⎜1 + ⎟ ⎜
⎟ vI 2
5 ⎠ ⎝ 25 + 50 ⎠
⎝
= 5vI 2
Then
v0 = v0 ( vI 1 ) + v0 ( vI 2 )
v0 = 10vI 1 + 5vI 2
______________________________________________________________________________________
TYU9.9
R S >> Ri so i1 = i2 = iS = 100 μ A
v0 = −iS RF
−10 = − (100 × 10−6 ) R ⇒ R = 100 kΩ
F
F
We want
______________________________________________________________________________________
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU9.10
We want iL = 1 mA when vI = −5 V
iL =
− ( −5 )
−VI
−v
⇒ R2 = I =
⇒ R2 = 5 kΩ
R2
i2
10−3
vL = iL Z L = (10−3 ) ( 500 ) ⇒ vL = 0.5 V
i4 =
vL
0.5
=
⇒ i4 = 0.1 mA
R2 5 kΩ
i3 = i4 + iL = 0.1 + 1 = 1.1 mA
If op-amp is biased at ±10 V, output must be limited to ≈ 8 V.
So v0 = i3 R3 + vL
8 = (1.1× 10−3 ) R3 + 0.5 ⇒ R3 = 6.82 kΩ
Let
R3 = 7.0 kΩ
Then we want
R3 RF 7
=
= = 1.4
R2 R1 5
R = 10 kΩ
R = 14 kΩ
Can choose 1
and F
______________________________________________________________________________________
TYU9.11
⎛υ −υI 2
(a) υ O1 = υ I 1 + i1 R' 2 = υ I 1 + ⎜⎜ I 1
⎝ R1
⎛
R' ⎞
R'
⎛
R ⎞
R
⎞
⎟⎟ R' 2
⎠
υ O1 = ⎜⎜1 + 2 ⎟⎟υ I 1 − 2 υ I 2
R1 ⎠
R1
⎝
υ O 2 = ⎜⎜1 + 2 ⎟⎟υ I 2 − 2 υ I 1
R1 ⎠
R1
⎝
Now
υO =
R4
(υ O 2 − υ O1 )
R3
We can write υ I 1 = υ cm −
υd
and υ I 2 = υ cm +
2
υd
2
Then
⎛
⎛ 2 R' 2 ⎞⎛ υ d ⎞
υ ⎞ R' ⎛
υ ⎞
R ' ⎞⎛
⎟⎜ ⎟
υ O1 = ⎜⎜1 + 2 ⎟⎟⎜υ cm − d ⎟ − 2 ⎜υ cm + d ⎟ = υ cm − ⎜⎜1 +
2
2
R
R
R1 ⎟⎠⎝ 2 ⎠
⎠
⎠
1 ⎠⎝
1 ⎝
⎝
⎝
⎛
⎛ 2 R ⎞⎛ υ ⎞
υ ⎞ R ⎛
υ ⎞
R ⎞⎛
υ O 2 = ⎜⎜1 + 2 ⎟⎟⎜υ cm + d ⎟ − 2 ⎜υ cm − d ⎟ = υ cm + ⎜⎜1 + 2 ⎟⎟⎜ d ⎟
⎝
R1 ⎠⎝
υO =
R4
R3
⎡⎛ 2 R2
⎢⎜⎜1 +
R1
⎣⎢⎝
υO =
R4
R3
⎡ R2 + R ' 2 ⎤
⎥ ⋅υ d
⎢1 +
R1 ⎦
⎣
2 ⎠
R1 ⎝
2 ⎠
⎝
R1 ⎠⎝ 2 ⎠
Then
⎛ 2 R' 2
⎞⎛ υ d ⎞
⎟⎟⎜ ⎟ + υ cm + ⎜⎜1 +
R1
⎝
⎠⎝ 2 ⎠
⎤
⎞⎛ υ d ⎞
⎟⎟⎜ ⎟ − υ cm ⎥
⎠⎝ 2 ⎠
⎦⎥
or
So
Acm = 0
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b) For Ad (max ) , let R1 = 2 k Ω , R ' 2 = 50 k Ω +5% = 52.5 k Ω
90 ⎡ 50 + 52.5 ⎤
1+
⎥ = 156.75
30 ⎢⎣
2
⎦
For Ad (min ) , let R1 = 102 k Ω , R ' 2 = 50 k Ω −5% = 47.5 k Ω
Then Ad (max ) =
Then Ad (min ) =
90 ⎡ 50 + 47.5 ⎤
= 5.87
1+
30 ⎢⎣
102 ⎥⎦
(c) CMRR = ∞
______________________________________________________________________________________
TYU9.12
i1 =
υ I1 − υ I 2
R1
⇒ R1 ( fixed ) =
[2 − (− 2)]× 10 −3
2 × 10 − 6
⇒ 2 kΩ
2 R2 ⎞
⎛
Ad (max ) = (2.5)⎜1 +
⎟ = 500 ⇒ R2 = 199 k Ω
2 ⎠
⎝
⎛
2(199) ⎞
⎟⎟ = 5 ⇒ R1 (var ) = 396 k Ω
Ad (min ) = (2.5)⎜⎜1 +
⎝ 2 + R1 (var ) ⎠
______________________________________________________________________________________
TYU9.13
End of 1st pulse: υ o =
−1
τ
×t
After 10 pulses: υ o = −5 =
So τ =
10 μs
0
=
− 10 × 10 −6
τ
(
− (10) 10 ×10 −6
)
τ
100 × 10 −6
⇒ τ = 20 μ s
5
τ = 20 ×10 −6 = R1C 2
C = 0.01× 10−6 = 0.01 μ F ⇒ R1 = 2 kΩ
For example, 2
______________________________________________________________________________________
TYU9.14
(a)
R − ΔR
⎞ + ⎛ R − ΔR ⎞ +
⎛
⎟ ⋅V
⎟ ⋅V = ⎜
R
R
R
R
−
Δ
+
+
Δ
⎝ 2R ⎠
⎠
⎝
R + ΔR
⎞ + ⎛ R + ΔR ⎞ +
⎛
υB = ⎜
⎟ ⋅V
⎟ ⋅V = ⎜
⎝ 2R ⎠
⎝ R + ΔR + R − ΔR ⎠
⎡⎛ R − ΔR ⎞ ⎛ R + ΔR ⎞⎤ +
υ O1 = υ A − υ B = ⎢⎜
⎟⎥ ⋅ V
⎟−⎜
⎣⎝ 2 R ⎠ ⎝ 2 R ⎠⎦
υA = ⎜
so
(
)
⎛ − ΔR ⎞ + ⎛ − 5 ⎞
−4
⎟ ⋅V = ⎜
⎟ΔR = − 2.5 × 10 ΔR
3
R
20
10
×
⎠
⎝
⎠
⎝
−4
υ
=
−
2
.
5
×
10
(
−
100
) ⇒ υO1 = 25 mV
We have O1
υ O1 = ⎜
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)
For the instrumentation amplifier
2R ⎞
R ⎛
υ O = 4 ⎜⎜1 + 2 ⎟⎟(υ O1 )
R3 ⎝
R1 ⎠
3=
R4
R3
⎛
2 R2
⎜⎜1 +
R1
⎝
⎞
⎟⎟(0.025)
⎠
⎛
2 R2 ⎞
R
R
⎟⎟ ; For example, set 4 = 10 and 2 = 5.5
⎜⎜1 +
R1 ⎠
R3
R1
⎝
______________________________________________________________________________________
or 120 =
R4
R3
TYU9.15
(a)
⎛ R ⎞ + 1
⎟ ⋅ V = (3) = 1.5 V
2
⎝R+R⎠
⎛
⎞ + ⎛ 1 ⎞
R
⎟⎟ ⋅ V = ⎜
υ B = ⎜⎜
⎟(3)
⎝ 2+δ ⎠
⎝ R + R(1 + δ ) ⎠
υA = ⎜
⎛ 1
⎝ 2+δ
υ O1 = υ A − υ B = 1.5 − ⎜
(2 + δ )(1.5) − 3 = 1.5δ ≅ 0.75δ V
⎞
⎟(3) =
2+δ
2 +δ
⎠
(b)
For an instrumentation amplifier
R ⎛ 2R ⎞
υ O = 4 ⎜⎜1 + 2 ⎟⎟(υ O1 )
R3 ⎝
R1 ⎠
For δ = 0.025 , want υ O = 3 V
3=
R4
R3
⎛ 2 R2
⎜⎜1 +
R1
⎝
⎞
⎟⎟(0.75)(0.025)
⎠
or
⎛
2 R2 ⎞
⎟
⎜⎜1 +
R1 ⎟⎠
⎝
R
R
For example, set 4 = 10 and 2 = 7.5
R3
R1
______________________________________________________________________________________
160 =
R4
R3
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 10
Exercise Solutions
EX10.1
I REF =
V + − V BE (on ) − V − 3 − 0.7 − (− 3)
=
= 0.1128 mA
47
R1
I REF
0.112766
= 0.1109 mA
=
2 ⎞
⎛
⎞
⎛
2
⎟
⎜⎜1 + ⎟⎟ ⎜1 +
β ⎠ ⎝ 120 ⎠
⎝
I
= I B 2 = O ⇒ 0.9243 μ A
IO =
I B1
β
______________________________________________________________________________________
EX10.2
I REF =
V + − VBE ( on ) − V −
R1
=
5 − 0.7 − ( −5 )
12
I REF = 0.775 mA
I0 =
I REF
0.775
=
= 0.7549 mA
2
2
1+
1+
75
β
ΔI 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and ΔI 0 =
ΔVCE 2
1
ΔVCE 2 ⇒ r0 =
ΔI 0
r0
V
4
= 265 kΩ = A ⇒ VA = ( 265 )( 0.7549 ) ⇒ VA ≅ 200 V
0.0151
I0
______________________________________________________________________________________
r0 =
EX10.3
3 − 0.6 − 0.7 − (− 3)
= 0.15667 mA
30
I REF
0.15667
IO =
=
= 0.15663 mA
2
2
1+
1+
(120)(81)
β (1 + β 3 )
I C1 = I C 2 = I O
I REF =
I B1 = I B 2 =
IO
β
⇒ 1.3053 μ A
I E 3 = I B1 + I B 2 = 2.6106 μ A
I E3
= 0.03223 μ A
1+ β3
______________________________________________________________________________________
I B3 =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX10.4
R1 =
V + − V BE1 − V − 3 − 0.6 − (− 3)
=
= 54 k Ω
0.10
I REF
VT ⎛ I REF ⎞ 0.026 ⎛ 100 ⎞
⎟=
ln⎜
ln⎜
⎟ = 2.09 k Ω
I O ⎜⎝ I O ⎟⎠ 0.02 ⎝ 20 ⎠
V BE 2 = V BE1 − I O R E = 0.6 − (0.02 )(2.09 ) = 0.558 V
______________________________________________________________________________________
RE =
EX10.5
⎛ 120 × 10 −6
⎞
⎟⎟ = (0.026) ln⎜⎜
−16
⎝ 2 × 10
⎠
⎛I
(a) V BE1 = VT ln⎜⎜ REF
⎝ I S1
⎛I
(b) V BE 2 = VT ln⎜⎜ O
⎝ IS2
RE =
⎛ 50 × 10 −6
⎞
⎟⎟ = (0.026) ln⎜⎜
−16
⎝ 2 × 10
⎠
VT ⎛ I REF
ln⎜
I O ⎜⎝ I O
⎞
⎟⎟ = 0.7051 V
⎠
⎞
⎟⎟ = 0.6824 V
⎠
⎞ 0.026 ⎛ 120 ⎞
⎟=
⎟ 0.05 ln⎜⎝ 50 ⎟⎠ ⇒ 455 Ω
⎠
⎛I
(c) I O RE = VT ln⎜⎜ REF
⎝ IO
⎞
⎟⎟
⎠
⎛ 0.120 ⎞
⎟
I O (0.7 ) = (0.026) ln⎜⎜
⎟
⎝ IO ⎠
By trial and error, I O = 40.4 μ A
Now,
V BE 2 = V BE1 − I O RE = 0.7051 − (0.0404 )(0.7 ) = 0.6768 V
______________________________________________________________________________________
EX10.6
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
0.026 ⎛ 0.70 ⎞
ln ⎜
RE =
⎟ ⇒ RE = 3.465 kΩ
0.025 ⎝ 0.025 ⎠
I
0.025
⇒ g m 2 = 0.9615 mA/V
gm2 = 0 =
VT 0.026
rπ 2 =
r02 =
β VT
I0
=
(150 )( 0.026 )
0.025
= 156 kΩ
VA
100
=
= 4000 kΩ
I 0 0.025
RE′ = RE || rπ 2 = 3.47 || 156 = 3.39 kΩ
R0 = r02 (1 + g m 2 RE′ ) = 4000 ⎡⎣1 + ( 0.962 )( 3.39 ) ⎤⎦
R0 = 17.04 MΩ
1
3
⋅ dVC 2 =
⇒ dI 0 = 0.176 μ A
R0
17, 040
______________________________________________________________________________________
dI 0 =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX10.7
I REF = I R + I BR + I B1 + ... + I BN
I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN =
I 01
β
⎛I ⎞
⎛ N +1⎞
I REF = I 01 + ( N + 1) ⎜ 01 ⎟ = I 01 ⎜ 1 +
⎟
β ⎠
⎝ β ⎠
⎝
I REF
So I 01 = I 02 = ... = I 0 N =
N +1
1+
β
I 01
1
= 0.90 =
+1
N
I REF
1+
50
N +1 1
1+
=
50
0.9
⎛ 1
⎞
− 1⎟ ( 50 )
N +1 = ⎜
⎝ 0.9 ⎠
⎛ 1
⎞
− 1⎟ ( 50 ) − 1
N =⎜
⎝ 0.9 ⎠
N = 4.55 ⇒ N = 4
______________________________________________________________________________________
EX10.8
V DS 2 (sat ) = 0.4 = VGS 2 − 0.4 ⇒ VGS 2 = 0.8 V
⎛ k ' ⎞⎛ W ⎞
2
I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN )
2
L
⎝
⎠
2
⎝ ⎠
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
0. 1 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 12.5
⎝ 2 ⎠⎝ L ⎠ 2
⎝ L ⎠2
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.5 = ⎜
⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 62.5
L
2
⎝
⎠⎝ ⎠1
⎝ L ⎠1
(
)
VGS 3 = V + − V − − VGS1 = 1.8 − (− 1.8) − 0.8 = 2.8 V
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
I REF = 0.5 = ⎜
⎟⎜ ⎟ (2.8 − 0.4 ) ⇒ ⎜ ⎟ = 1.74
⎝ 2 ⎠⎝ L ⎠ 3
⎝ L ⎠3
______________________________________________________________________________________
EX10.9
a.
b.
I REF = K n (VGS − VTN )
2
0.020 = 0.080 (VGS − 1)
2
VGS = 1.5 V all transistors
VG 4 = VGS 3 + VGS1 + V − = 1.5 + 1.5 − 5 = −2 V
VS 4 = VG 4 − VGS 4 = −2 − 1.5 = −3.5 V
VD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 − VTN = 1.5 − 1 = 0.5 V
So VD 4 ( min ) = −3.5 + 0.5 ⇒ VD 4 ( min ) = −3.0 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
c.
R0 = r04 + r02 (1 + g m r04 )
1
1
=
= 2500 kΩ
r02 = r04 =
λ I 0 ( 0.02 )( 0.020 )
g m = 2 K n (VGS − VTN ) = 2 ( 0.080 )(1.5 − 1) ⇒ g m = 0.080 mA / V
R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) ⇒ R0 = 505 MΩ
______________________________________________________________________________________
EX10.10
For Q2 : vDS ( min ) = VP = 2 V ⇒ VS ( min ) = vDS ( min ) − 5 = 2 − 5 ⇒ VS ( min ) = −3 V
I 0 = I DSS 2 (1 + λ vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) ⇒ I 0 = 0.65 mA
⎛ v ⎞
I 0 = I DSS 1 ⎜1 − GS1 ⎟
⎝ VP1 ⎠
2
2
⎛ v ⎞
0.65 = 0.80 ⎜1 − GS 1 ⎟
−2 ⎠
⎝
vGS 1
= 0.0986 ⇒ vGS1 = −0.197 V
−2
vGS 1 = VI − VS − 0.197 = VI − ( −3) ⇒ VI ( min ) = −3.2 V
Vgs 2 = 0, Vgs1 = −VX
IX =
VX VX − V1
+
+ g m1VX
r02
r01
(1)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
V1 V1 − VX
+
= g m1VX
RD
r01
(2)
⎛ 1
⎞
VX ⎜ + g m1 ⎟
⎝ r01
⎠
V1 =
1
1
+
RD r01
⎞
1 ⎛ 1
⎜ + g m1 ⎟
r r
IX
1
1
1
⎠
=
=
+ + g m1 − 01 ⎝ 01
1
1
VX R0 r02 r01
+
RD r01
1
⎡
⎤
⎢
⎥
⎛
⎞
r
1
1
01
⎥
=
+ ⎜ + g m1 ⎟ ⎢1 −
1
1
r02 ⎝ r01
⎠⎢
+ ⎥
⎢⎣ RD r01 ⎥⎦
1
⎛
⎞
⎜
⎞
RD ⎟
1 ⎛ 1
⎟
=
+ ⎜ + g m1 ⎟ ⎜
r02 ⎝ r01
⎠⎜ 1 + 1 ⎟
⎜R
⎟
⎝ D r01 ⎠
⎞
1
1 ⎛ 1
For R D << ro1 ⇒
≅
+ ⎜⎜
+ g m1 ⎟⎟
Ro ro 2 ⎝ ro1
⎠
For Q1:
⎛ V ⎞ 2 ( 0.8 ) ⎛ −0.197 ⎞
2I
g m1 = DSS 1 ⎜1 − GS 1 ⎟ =
⎜1 −
⎟
VP ⎝
VP ⎠
2 ⎝
−2 ⎠
g m1 = 0.721 mA/V
1
1
=
= 10.3 kΩ
r0 =
λ I 0 ( 0.15 )( 0.65 )
1
1
1
=
+
+ 0.721 = 0.915 ⇒ R0 = 1.09 kΩ
R0 10.3 10.3
______________________________________________________________________________________
EX10.11
a.
⎛V ⎞
I REF = I S exp ⎜ EB 2 ⎟
⎝ VT ⎠
⎛I ⎞
⎛ 0.5 × 10−3 ⎞
VEB 2 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜
⎟ ⇒ VEB 2 = 0.521 V
−12
⎝ 10
⎠
⎝ IS ⎠
5 − 0.521
⇒ R1 = 8.96 kΩ
0.5
Combining Equations (10.79), (10.80), and (10.81), we find
R1 =
b.
(c)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ VEC 2 ⎞
⎜1 +
⎟
⎡
VAP ⎠
⎛ VI ⎞ ⎤ ⎛ VCEo ⎞
⎝
I S 0 ⎢exp ⎜ ⎟ ⎥ ⎜ 1 +
⎟ = I REF ×
⎛ VEB 2 ⎞
⎝ VT ⎠ ⎦ ⎝ VAN ⎠
⎣
⎜1 +
⎟
⎝ VAP ⎠
2.5 ⎞
⎛
⎜1 +
⎟
⎡
⎛ VI ⎞ ⎤ ⎛
2.5 ⎞
100 ⎠
⎝
−3
10 ⎢exp ⎜ ⎟ ⎥ ⎜ 1 +
⎟ = ( 0.5 × 10 )
⎛ 0.521 ⎞
⎝ VT ⎠ ⎦ ⎝ 100 ⎠
⎣
⎜1 +
⎟
100 ⎠
⎝
⎛V ⎞
⎛V ⎞
1.025 × 10−12 exp ⎜ I ⎟ = 5.098 × 10−4 exp ⎜ I ⎟ = 4.974 × 108 ⇒ VI = 0.521 V
⎝ VT ⎠
⎝ VT ⎠
−12
⎛ 1 ⎞
1
−⎜ ⎟
−
VT ⎠
⎝
0.026 = −38.46 ⇒ A = −1923
=
Av =
v
1
1
1
1
0.01 + 0.01
+
+
VAN VAP 100 100
d.
______________________________________________________________________________________
EX10.12
(a) Aυ = − g mo ron rop
(
ron =
)
V AN
100
=
= 400 k Ω
0.25
I Co
V AP
60
=
= 240 k Ω
0.25
I Co
0.25
g mo =
= 9.615 mA/V
0.026
Aυ = −(9.615) 400 240 = −1442
rop =
(
(b)
)
(
Aυ = −(0.6 )(1442 ) = −865 = − g mo ron rop R L
)
− 865 = −(9.615)(150 RL ) ⇒ RL = 225 k Ω
______________________________________________________________________________________
EX10.13
(a) I REF = I O = K n V IQ − VTN
(
(
)
)
2
0.20 = 0.10 V IQ − 0.5 ⇒ V IQ = 1.914 V
(
2
(b) Aυ = − g mo ron rop
)
g mo = 2 K n I Q = 2 (0.1)(0.2) = 0.2828 mA/V
ron = rop =
1
λI Q
=
1
(0.015)(0.2)
= 333 k Ω
Aυ = −(0.2828)(333 333) = −47.1
(
(c) Aυ = −(0.5)(47.1) = −23.55 = − g mo ron rop R L
(
)
)
23.55 = (0.2828 ) 333 333 R L ⇒ R L = 166.5 k Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU10.1
⎛
2⎞
2 ⎞
⎛
I REF = ⎜⎜1 + ⎟⎟ I O = ⎜1 +
⎟(0.20 ) = 0.2033 mA
β⎠
⎝ 120 ⎠
⎝
2.5 − 0.7 − (− 2.5)
R1 =
= 21.15 k Ω
0.2033
______________________________________________________________________________________
TYU10.2
Neglecting base currents
⎛ 150 × 10 −6
V BE1 = (0.026 ) ln⎜⎜
−15
⎝ 8 × 10
⎞
⎟⎟ = 0.6150 V
⎠
0
.
6150
⎛
⎞
I O = 5 × 10 −15 exp⎜
⎟ ⇒ 93.75 μ A
⎝ 0.026 ⎠
______________________________________________________________________________________
(
)
TYU10.3
I 0 = I REF ⋅
I B3 =
I0
β
1
⎛
⎞
2
⎜⎜ 1 +
⎟
+
β
β
1
(
) ⎟⎠
⎝
=
0.50
⎛
2 ⎞
⎜⎜1 +
⎟
50
( 51) ⎟⎠
⎝
⇒ I 0 = 0.4996 mA
⇒ I B 3 = 9.99 μ A
⎛ 1+ β ⎞
I E3 = ⎜
⎟ I C 3 = I E 3 = 0.5096 mA
⎝ β ⎠
I E3
0.5096
IC 2 =
=
⇒ I C 2 = 0.490 mA = I C1
2⎞
⎛
2⎞ ⎛
+
1
+
1
⎟
⎜
⎟ ⎜
⎝ β ⎠ ⎝ 50 ⎠
I B1 = I B 2 =
IC 2
β
⇒ I B1 = I B 2 = 9.80 μ A
______________________________________________________________________________________
TYU10.4
For circuit - Figure 10.2(b)
(a) I O ≅ I REF = 1 mA
(b) Ro =
V A 50
=
= 50 k Ω
1
IO
(c) dI O =
ΔVC 2
3
=
= 0.06 mA
50
Ro
dI O 0.06
=
⇒ 6%
1
IO
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
From circuit - Figure 10.9
⎞
⎛I
(a) I O RE = VT ln⎜⎜ REF ⎟⎟
⎝ IO ⎠
⎛ 1 ⎞
I O (2 ) = (0.026 ) ln⎜⎜ ⎟⎟
⎝ IO ⎠
By trial and error, I O = 41.4 μ A
50
⇒ 1.208 M Ω
(b) ro 2 =
0.0414
(200)(0.026) = 125.6 k Ω
0.0414
g m2 =
= 1.5923 mA/V ; rπ 2 =
0.026
0.0414
RE rπ 2 = 2 125.6 = 1.969 k Ω
Ro = (1.208)[1 + (1.5923)(1.969 )] ⇒ 5 M Ω
3
(c) dI O = = 0.6 μ A
5
dI O
0.6
=
⇒ 1.45%
41.4
IO
______________________________________________________________________________________
TYU10.5
⎛ k ' ⎞⎛ W ⎞
⎛ k ' ⎞⎛ W ⎞
2
2
(a) I REF = ⎜⎜ n1 ⎟⎟⎜ ⎟ (VGS1 − VTN 1 ) = ⎜⎜ n3 ⎟⎟⎜ ⎟ (VGS 3 − VTN 3 )
L
L
2
2
⎝
⎠
⎝
⎠
1
3
⎠
⎝
⎠
⎝
VGS 3 = V + − VGS1 = 2.5 − VGS1
Then
⎛ 95 ⎞
⎛ 100 ⎞
2
2
⎜
⎟(12.5)(VGS1 − 0.38) = ⎜ ⎟(1.18)(2.5 − VGS1 − 0.42)
⎝ 2⎠
⎝ 2 ⎠
We find 25(VGS1 − 0.38) = (7.4867 )(2.08 − VGS1 )
Or VGS1 = VGS 2 = 0.7718 V
⎛ 100 ⎞
2
I REF = ⎜
⎟(12.5)(0.7718 − 0.38) = 95.93 μ A
⎝ 2 ⎠
⎛ k ' ⎞⎛ W ⎞
⎛ 105 ⎞
2
2
I O = ⎜⎜ n 2 ⎟⎟⎜ ⎟ (VGS 2 − VTN 2 ) = ⎜
⎟(7.5)(0.7718 − 0.40 )
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠ 2
or
I O = 54.43 μ A
ΔI REF 95.93 − 100
(b)
=
× 100% = −4.07%
100
I REF
ΔI O 54.43 − 60
=
× 100% = −9.28%
60
IO
______________________________________________________________________________________
TYU10.6
⎛ k ' ⎞⎛ W ⎞
⎛ k ' ⎞⎛ W ⎞
2
2
(a) I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS1 − VTN ) = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 3 − VTN )
L
L
2
2
⎝
⎠
⎝
⎠
1
3
⎝ ⎠
⎝ ⎠
(
)
VGS 3 = V + − V − − VGS1 = 6 − VGS1
then
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
12 (VGS1 − 0.5) = 3 (6 − VGS1 − 0.5) ⇒ VGS1 = 2.167 V
⎛ 80 ⎞
2
I REF = ⎜ ⎟(12)(2.167 − 0.5) ⇒ 1.33 mA
⎝ 2⎠
⎛ k ' ⎞⎛ W ⎞
2
(b) I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN ) [1 + λV DS ]
L
2
⎝
⎠
2
⎝ ⎠
⎛ 0.08 ⎞
2
=⎜
⎟(6)(2.167 − 0.5) [1 + (0.02)(2)] = 0.6936 mA
⎝ 2 ⎠
⎛ 0.08 ⎞
2
(c) I O = ⎜
⎟(6)(2.167 − 0.5) [1 + (0.02)(4)] = 0.7203 mA
⎝ 2 ⎠
______________________________________________________________________________________
TYU10.7
2
2
I REF = K n1 (VGS1 − VTN ) = K n3 (VGS 3 − VTN )
We have
Then
(
)
VGS 3 = V + − V − − VGS1 = 6 − VGS1
0.35 (VGS1 − 0.7 ) = 0.10 (6 − VGS1 − 0.7 )
which yields
VGS1 = 2.302 V
then
2
I REF = 0.35(2.302 − 0.7 ) = 0.8986 mA
I O = 3(0.30)(2.302 − 0.7 ) = 2.31 mA
______________________________________________________________________________________
2
TYU10.8
All transistors are identical
⇒ I 0 = I REF = 250 μ A
I REF = K n (VGS − VTN )
2
0.25 = 0.20 (VGS − 1) ⇒ VGS = 2.12 V
2
______________________________________________________________________________________
TYU10.9
a.
b.
⎛ 0.1× 10−3 ⎞
VEB 2 = ( 0.026 ) ln ⎜
⇒ VEB 2 = 0.557 V
−14 ⎟
⎝ 5 × 10 ⎠
R1 =
5 − 0.557
⇒ R1 = 44.4 kΩ
0.1
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
c.
⎛ VEC 2
⎜ 1+ V
⎞ ⎤ ⎛ VCE 0 ⎞
AP
+
=
×
1
I
⎥
⎟ REF ⎜ V
⎟ ⎜
V
⎜
EB
2
⎠⎦ ⎝
AN ⎠
⎜ 1+ V
AP
⎝
⎡
⎛V
I S 0 ⎢exp ⎜ I
⎝ VT
⎣
⎞
⎟
⎟
⎟
⎟
⎠
2.5 ⎞
⎛
1+
⎜
⎞⎤ ⎛
2.5 ⎞
−3
100 ⎟
⎟ ⎥ ⎜1 +
⎟ = ( 0.1× 10 ) ⎜ 0.557 ⎟
⎠ ⎦ ⎝ 100 ⎠
⎜⎜ 1 +
⎟⎟
100 ⎠
⎝
⎡
⎛V
5 × 10−14 ⎢ exp ⎜ I
⎝ VT
⎣
⎛
I
⎝
T
( 5.125 ×10 ) exp ⎜ VV
−14
⎞
−4
⎟ = 1.019 × 10
⎠
⎛V ⎞
exp ⎜ I ⎟ = 1.988 × 109 ⇒ VI = 0.557 V
⎝ VT ⎠
1
0.026 ⇒ A = −1923
Av =
v
1
1
+
100 100
d.
______________________________________________________________________________________
−
TYU10.10
2
(a) I REF = K p (VSG + VTP )
0.15 = 0.12(VSG − 0.7 ) ⇒ VSG = 1.818 V
2
(b) VO =
[1 + λ (V
+
p
− VSG
λn + λ p
)] − K (V − V )
I (λ + λ )
2
n
I
REF
TN
n
p
Set VO = 2.5 V
[1 + (0.02)(5 − 1.818)] − 0.12(VI − 0.7 )2 ⇒ V = 1.798 V
I
0.04
0.15(0.04 )
− 2 K n (V I − VTN ) − 2(0.12 )(1.798 − 0.7 )
=
= −43.9
=
(0.15)(0.04)
I REF (λ n + λ p )
2. 5 =
(c)
Aυ
______________________________________________________________________________________
TYU10.11
2
(a) I REF = 80 = 50(VSG − 0.7 ) ⇒ VSG = 1.965 V
(b) VO =
[1 + λ (V
p
+
− VSG
λn + λ
)] − K (V − V )
I (λ + λ )
2
n
REF
I
TN
n
p
[1 + (0.02)(5 − 1.965)] − (0.05)(VI − 0.7 )2 ⇒ V = 1.940 V
I
(0.08)(0.04)
0.04
− 2 K n (V I − VTN ) − 2(0.05)(1.940 − 0.7 )
=
=
= −38.74
(0.08)(0.04)
I REF (λn + λ P )
2. 5 =
(c) Aυ
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU10.12
a.
I
0.5
⇒ g m = 19.2 mA/V
gm = C 0 =
VT
0.026
r0 =
VAN 120
=
⇒ r0 = 240 kΩ
I CQ 0.5
r02 =
VAP 80
=
⇒ r02 = 160 kΩ
I CQ 0.5
Av = − g m ( r0 || r02 || RL ) = − (19.2 ) [ 240 || 160 || 50] ⇒ Av = −631
b.
______________________________________________________________________________________
TYU10.13
1
= 38.46 mA/V
0.026
(100 )( 0.026 )
rπ 1 = rπ 2 =
= 2.6 K
1
80
rO1 = rO 2 =
= 80 K
1
120
rO =
= 120 K
1
1
RO1 = 2.6
80 = 0.0257 K
38.46
For R1 = 9.3 K
I C = 1mA,
gm =
( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K
RE′′ = 1 [ 2.6 + 0.924] = 0.779 K
RO 2 = 80 ⎡⎣1 + ( 38.46 )( 0.779 ) ⎤⎦ = 2476.7 K
Av = − g m ( rO || RO 2 ) = − ( 38.46 )(120 || 2476.7 ) = − ( 38.46 )(114.5 )
R ′ = R1
Av = −4404
For RL = 100 K
Av = −38.46 ⎡⎣114.5 100 ⎤⎦ = −2053
For RL = 10 K
Av = −38.46 [114.5 || 10] = −354
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU10.14
M 1 and M 2 identical ⇒ I o = I REF
a.
I O = K n (VI − VYN )
0.25 = 0.2 (VI − 1)
2
2
VI = 2.12 V
g m = 2 K n (VI − VTN ) = 2 ( 0.2 )( 2.12 − 1) ⇒ g m = 0.447 mA/V
r0 n =
1
1
=
⇒ r0 n = 400 kΩ
λn I 0 ( 0.01)( 0.25 )
r0 p =
1
1
=
⇒ r0 p = 200 kΩ
λ p I 0 ( 0.02 )( 0.25 )
Av = − g m ( r0 || r02 || RL )
Av = − ( 0.447 ) [ 400 || 200 || 100] ⇒ Av = −25.5
b.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 11
Exercise Solutions
EX11.1
(a) υ E = 0 − 0.7 = −0.7 V
υ C1 = υ C 2 = 5 − (0.15)(20 ) = 2 V
υ CE1 = υ CE 2 = 2 − (− 0.7 ) = 2.7 V
(b) υ E = −1 − 0.7 = −1.7 V
υ C1 = υ C 2 = 2 V
υ CE1 = υ CE 2 = 3.7 V
(c) υ E = +1 − 0.7 = +0.3 V
υ C1 = υ C 2 = 2 V
υ CE1= υ CE 2 = 2 − 0.3 = 1.7 V
______________________________________________________________________________________
EX11.2
i
(a) C1 = 0.25 =
IQ
1
⎛ − υd
1 + exp⎜⎜
⎝ VT
⎞
⎟⎟
⎠
⎛ −υd ⎞
⎟⎟ = 3 ⇒ −υ d = (0.026 ) ln (3)
We find, exp⎜⎜
⎝ VT ⎠
Or υ d = −28.56 mV
(b)
iC 2
=
IQ
1
⎛ + υd
1 + exp⎜⎜
⎝ VT
⎞
⎟⎟
⎠
= 0.9
⎛υ ⎞
exp⎜⎜ d ⎟⎟ = 0.1111 ⇒ υ d = (0.026 ) ln (0.1111)
⎝ VT ⎠
Or υ d = −57.13 mV
______________________________________________________________________________________
We find,
EX11.3
(a) CMRR dB = 75 dB ⇒ CMRR = 5623.4
5623.4 =
1 ⎡ (101)(0.8)Ro ⎤
⇒ Ro = 362 k Ω
1+
2 ⎢⎣ (0.026 )(100 ) ⎥⎦
(b) CMRR dB = 95 dB ⇒ CMRR = 56,234
1 ⎡ (101)(0.8)Ro ⎤
⎥ ⇒ Ro = 3.62 M Ω
⎢1 +
2 ⎣ (0.026 )(100 ) ⎦
______________________________________________________________________________________
56,234 =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX11.4
(a) υ C1 = − g m ⋅
υd
2
⋅ RC1
g R
υ C1
= −150 = − m C1
υd
2
0.1
= 3.846 mA/V
We find g m =
0.026
2(150)
= 78.0 k Ω
Then RC1 =
3.846
υ
υ C 2 = + g m ⋅ d ⋅ RC 2
2
υC 2
g R
= 100 = m C 2
υd
2
2(100)
= 52.0 k Ω
Then RC 2 =
3.846
(b) For VCB = 0 , υ C1 = υ C 2 = 1.5 V for υ cm = 1.5 V
Then 1.5 = V + − I CQ RC = V + − (0.1)(78) ⇒ V + = +9.3 V
So V + = −V − = 9.3 V
______________________________________________________________________________________
EX11.5
(a) υ o = Ad υ d + Acmυ cm
υ d = υ1 − υ 2 = −20 μ V
υ cm =
υ1 + υ 2
=0
2
Then υ o = (150)(− 20 ) ⇒ υ o = −3 mV
υ d = υ1 − υ 2 = −20 μ V
υ + υ2
= 200 μ V
υ cm = 1
(b)
2
Now CMRR dB = 50 dB ⇒ CMRR = 316.2 =
Ad
150
=
⇒ Acm = 0.474
Acm
Acm
Then υ o = Ad υ d + Acmυ cm = (150)(− 20 ) + (0.474 )(200)
Or υ o = −2.905 mV
______________________________________________________________________________________
EX11.6
Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦
β VT (100 )( 0.026 )
rπ =
=
= 10.4 K
I CQ
0.25
Rid = 2 ⎡⎣10.4 + (101)( 0.5 ) ⎤⎦ = 122 K
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX11.7
Ad =
10 =
g m RC
2 (1 + g m RE )
( 9.62 )(10 )
2 ⎡⎣1 + ( 9.62 ) RE ⎤⎦
1 + ( 9.62 ) RE = 4.81 ⇒ RE = 0.396 K
Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦ = 2 ⎡⎣10.4 + (101)( 0.396 ) ⎤⎦
Rid = 100.8 K
______________________________________________________________________________________
EX11.8
I1 =
10 − VGS 4
2
= K n 3 (VGS 4 − VTN )
R1
10 − VGS 4 = ( 0.1)( 80 )(VGS 4 − 0.8 )
2
10 − VGS 4 = 8 (VGS2 4 − 1.6VGS 4 + 0.64 )
8VGS2 4 − 11.8VGS 4 − 4.88 = 0
VGS 4 =
11.8 ±
(11.8) + 4 (8 )( 4.88 )
= 1.81 V
2 (8)
2
10 − 1.81
= 0.102 mA
80
0.102
= ID2 =
= 0.0512 mA
2
I1 = I Q =
I D1
= K n1 (VGS 1 − VTN )
2
0.0512 = 0.050 (VGS 1 − 0.8 ) ⇒ VGS 1 = 1.81 V
v01 = v02 = 5 − ( 0.0512 )( 40 ) = 2.95 V
2
Max vcm : VDS 1 ( sat ) = VGS 1 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( max ) = v01 − VDS1 ( sat ) + VGS 1
= 2.95 − 1.01 + 1.81
vcm ( max )
= 3.75 V
vcm ( min )
= 1.81 + 1.01 − 5
= −2.18 V
Min vcm : VDS 4 ( sat ) = VGS 4 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( min ) = VGS 1 + VDS 4 ( sat ) − 5
−2.18 ≤ vcm ≤ 3.75 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX11.9
⎛ k1
(a) Ad = ⎜⎜ n
⎝ 2
⎞⎛ W
⎟⎜
⎟⎝ L
⎠
⎞⎛⎜ I Q
⎟⎜
⎠⎝ 2
⎞
⎟ ⋅ RD
⎟
⎠
⎛ 0.1 ⎞⎛ W ⎞⎛ 0.2 ⎞
⎛W
15 = ⎜
⎟⎜ ⎟⎜
⎟ ⋅ (15) ⇒ ⎜
L
2
2
⎝
⎠⎝ ⎠⎝
⎠
⎝L
⎞
⎟ = 200
⎠
⎛ k ' ⎞⎛ W ⎞⎛ I Q ⎞
⎛ 0. 2 ⎞
⎛ 0.1 ⎞
(b) g f (max ) = ⎜⎜ n ⎟⎟⎜ ⎟⎜⎜ ⎟⎟ = ⎜
⎟ = 1 mA/V
⎟(200 )⎜
L
2
2
2
⎝ 2 ⎠
⎠
⎝
⎠
⎝
⎝ ⎠
⎝ ⎠
______________________________________________________________________________________
EX11.10
(a) ro 2 =
V A 2 150
=
= 750 k Ω
I CQ
0.2
ro 4 =
V A4
90
=
= 450 k Ω
I CQ 0.2
gm =
I CQ
VT
=
0. 2
= 7.692 mA/V
0.026
Ad = g m (ro 2 ro 4 ) = (7.692)(750 450) = 2163
(
)
(
)
(b) Ad = g m ro 2 ro 4 R L = (7.692 ) 750 450 250 = 1018
(c) Rid = 2rπ , rπ =
(120)(0.026) = 15.6
Rid = 31.2 k Ω
0.2
kΩ
(d) Ro = ro 2 ro 4 = 750 450 = 281 k Ω
______________________________________________________________________________________
EX11.11
⎛W ⎞
⎛W ⎞
⎛W ⎞
We have ⎜ ⎟ = 10 , ⎜ ⎟ = ⎜ ⎟ = 0.33
L
L
⎝ ⎠4
⎝ ⎠5 ⎝ L ⎠6
⎛ k ′ ⎞⎛ W ⎞
⎛ k ′ ⎞⎛ W ⎞
2
2
I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 4 − VTN ) = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 5 − VTN )
⎝ 2 ⎠⎝ L ⎠ 5
⎝ 2 ⎠⎝ L ⎠ 4
(10)(VGS 4 − 0.5)2 = (0.33)(VGS 5 − 0.5)2
VGS 5 =
Then
6 − VGS 4
2
(5.505)(VGS 4 − 0.5) = (V gs 5 − 0.5) =
Which yields VGS 4 = 0.8747 V
6 − VGS 4
− 0.5
2
⎛ 80 ⎞
Then I REF = I Q = ⎜ ⎟(10 )(0.8747 − 0.5)2 = 56.16 μ A
⎝ 2 ⎠
⎛ k′
Ad = 2 2⎜⎜ n
⎝ 2
1
1
⎞⎛ W ⎞ ⎛⎜ 1 ⎞⎟
⎛ 80 ⎞ ⎛ 1 ⎞
⋅
= 2 2⎜ ⎟(10 )⎜
⎟⎟⎜ ⎟
⎟⋅
⎟
⎜
2
+
λ
λ
56
.
16
0
.
02
0.02
+
L
I
⎠
⎝ ⎠ ⎝
⎠⎝ ⎠ n ⎝ Q ⎠ n
p
Ad = 188.7
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX11.12
Ad = g m ro 2 Ro
(
)
400 = g (500 101000) ⇒ g
m
m
= 0.8 mA/V
g m = 2 K n I DQ
⎛ 0.08 ⎞⎛ W ⎞
⎛W ⎞ ⎛W ⎞
0. 8 = 2 ⎜
⎟⎜ ⎟ (0.01) ⇒ ⎜ ⎟ = ⎜ ⎟ = 40
⎝ 2 ⎠⎝ L ⎠ 1
⎝ L ⎠1 ⎝ L ⎠ 2
______________________________________________________________________________________
EX11.13
I e 6 = (1 + β ) I b 6 = I b 7
I c 7 = β I b 7 = β (1 + β ) I b 6
Ic7
= β (1 + β ) = (100 )(101) = 1.01× 104
Ib6
______________________________________________________________________________________
EX11.14
I CQB = 0.5 mA, I BQB =
I CQA = 0.002747 mA
rπA =
0.5
= 0.00278 mA = I EQA
180
(90)(0.026) = 851.8 k Ω
0.002747
rπA + 300 851.8 + 300
=
= 12.66 k Ω
RoA =
91
91
(180)(0.026) = 9.36 k Ω
rπB =
0.5
r + RoA
⎛ 9.36 + 12.66 ⎞
Ro = 10 πB
= 10 ⎜
⎟ = 10 0.1217
1+ βB
181
⎠
⎝
or
R o = 120 Ω
______________________________________________________________________________________
EX11.15
I1 =
10 − 0.7 − ( −10 )
R1
⎛I
I Q R2 = VT ln ⎜ 1
⎜I
⎝ Q
= 0.6 ⇒ R1 = 32.2 K
⎞
⎟⎟
⎠
( 0.2 ) R2 = ( 0.026 ) ln ⎛⎜
I R 6 = I1 ⇒ R3 = 0
0.6 ⎞
⎟ ⇒ R2 = 143 Ω
⎝ 0.2 ⎠
.
10 = I C1 RC + VCE1 − 0.7
10.7 = ( 0.1) RC + 4 ⇒ RC = 67 K
vo 2 = −0.7 + 4 = 3.3 V
vE 4 = 3.3 − 1.4 = 1.9 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
IR4 =
vE 4
1.9
⇒ R4 =
⇒ R4 = 3.17 K
0.6
R4
vC 3 = vO 2 − 1.4 + vCE 4
= 3.3 − 1.4 + 3 = 4.9 V
I R 5 = I R 4 = 0.6 =
10 − 4.9
⇒ R5 = 8.5 K
R5
vE 5 = 4.9 − 0.7 = 4.2 V ⇒ R6 =
R7 =
0 − ( −10 )
5
4.2 − 0.7
= 5.83 K
0.6
⇒ R7 = 2 K
______________________________________________________________________________________
EX11.16
Ri 2 = rπ 3 + (1 + β ) rπ 4
rπ 4 =
rπ 3 ≈
(100 )( 0.026 )
0.6
β 2VT
I R4
= 4.333 K
(100 ) ( 0.026 )
2
=
0.6
= 433.3 K
Ri 2 = 433.3 + (101)( 4.333) ⇒ Ri 2 = 871 K
Ri 3 = rπ 5 + (1 + β ⎡⎣ R6 + rπ 6 + (1 + β ) R7 ⎤⎦
(100 )( 0.026 )
= 4.333 K
rπ 5 =
0.6
(100 )( 0.026 )
= 0.52 K
rπ 6 =
5
Ri 3 = 4.333 + (101) ⎡⎣5.83 + 0.52 + (101)( 2 ) ⎤⎦
Ri 3 = 21.0 MΩ
gm
0.1
( RC Ri 2 ) gm = 0.026 = 3.846 mA/V
2
3.846
Ad 1 =
( 67 871) = 119.64
2
⎛I ⎞
0.6
A2 = ⎜ R 4 ⎟ ( R5 Ri 3 ) =
(8.5 21000 ) = 98.037
2 ( 0.026 )
⎝ 2VT ⎠
Ad 1 =
A = Ad 1 ⋅ A2 = (119.64 )( 98.037 ) = 11, 729
______________________________________________________________________________________
EX11.17
1
1
=
6
2π R o C o 2π 10 × 10 0.2 × 10 −12
f z = 79.6 kHz
fz =
fp =
(
)(
)
1
2π R eq C o
From Example 11.17, R eq = 51.98 Ω
1
2π (51.98) 0.2 × 10 −12
f p = 15.3 GHz
fp =
(
)
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU11.1
(a) υ d = υ1 − υ 2 = 2.10 − 2.12 = −0.02 V
υ1 + υ 2
2.10 + 2.12
= 2.110 V
2
2
(b) υ d = υ1 − υ 2 = 0.25 − 0.002 sin ω t − [0.5 + 0.002 sin ω t ]
υ cm =
=
= −0.25 − 0.004 sin ω t (V)
0.25 − 0.002 sin ω t + 0.5 + 0.002 sin ω t
= 0.375 V
2
______________________________________________________________________________________
υ cm =
TYU11.2
(a) Need υ C1 = υ C 2 = 3 = 5 − (0.2)RC ⇒ RC = 10 k Ω
0.2
= 7.692 mA/V
(b) g m =
0.026
Ad = g m RC = (7.692 )(10 ) = 76.9
______________________________________________________________________________________
TYU11.3
(a) υ o = Ad υ d + Acmυ cm
υ d = υ1 − υ 2 = 0.995 sin ω t − 1.005 sin ω t = −0.01sin ω t (V)
υ cm =
υ1 + υ 2
2
=
0.995 sin ω t + 1.005 sin ω t
= 1.0 sin ω t (V)
2
Then
υ o = (80)(− 0.01sin ω t ) + (− 0.20)(1.0 sin ω t ) = −1.0 sin ω t (V)
(b) υ d = υ1 − υ 2 = −0.01sin ω t (V)
υ cm = 2 V
Then
υ o = (80)(− 0.01sin ω t ) + (− 0.20)(2) = −0.4 − 0.8 sin ω t (V)
______________________________________________________________________________________
TYU11.4
From Equation (11.41)
CMRR =
g m RO
⎛ ΔRC ⎞
⎜
⎟
⎝ RC ⎠
For CMRR |dB = 75 dB ⇒ CMRR = 5623.4
5623.4 =
Then
( 3.86 )(100 )
ΔRC
⋅ (10 )
Or ΔRC = 0.686 K
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU11.5
From Equation (11.49)
CMRR =
1 + 2 RO g m
⎛ Δg ⎞
2⎜ m ⎟
⎝ gm ⎠
For CMRR |dB = 90 dB ⇒ CMRR = 31622.8
⎛ 1 + 2 (100 )( 3.86 ) ⎞
31622.8 = ⎜
⎟ ( 3.86 )
2Δg m
⎝
⎠
Then
Δg m 0.0472
=
= 0.0122 ⇒ 1.22%
gm
3.86
Δg m = 0.0472 mA/V
Or
or
______________________________________________________________________________________
TYU11.6
(a) I EQ = 0.2 mA, I B1 = I B 2 =
0.2
⇒ 1.32 μ A
151
2(150)(0.026)
= 39 k Ω
0.2
10 sin ω t (mV )
Ib =
⇒ I b = 0.256 sin ω t ( μ A)
39 kΩ
39
1
= 19.5 k Ω
(c) Ricm = [rπ + (1 + β )(2 Ro )] ; rπ =
2
2
1
Ricm = [19.5 + (151)(2)(100)] ⇒ 15.11 M Ω
2
1 υ cm
1 3 sin ω t
= ⋅
⇒ 0.0993 sin ω t ( μ A)
icm = ⋅
2 Ricm 2 15.11
______________________________________________________________________________________
(b) Rid = 2rπ =
TYU11.7
⎛ k ⎞⎛ W
(a) Ad = ⎜⎜ n ⎟⎟⎜
⎝ 2 ⎠⎝ L
'
⎞⎛⎜ I Q
⎟⎜
⎠⎝ 2
⎞
⎟ ⋅ RD
⎟
⎠
⎛ 0.1 ⎞⎛ W ⎞⎛ 0.4 ⎞
⎛W ⎞
12 = ⎜
⎟⎜ ⎟⎜
⎟ ⋅ (7.5) ⇒ ⎜ ⎟ = 256
⎝ 2 ⎠⎝ L ⎠⎝ 2 ⎠
⎝L⎠
⎛ k ' ⎞⎛ W ⎞
2
(b) i D 2 = ⎜⎜ n ⎟⎟⎜ ⎟(υ GS 2 − VTN )
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
0.2 = ⎜
⎟(256)(υ GS 2 − 0.5) ⇒ υ GS 2 = 0.625 V
2
⎠
⎝
υ DS 2 (sat ) = 0.625 − 0.5 = 0.125 V
υ O = V + − i D 2 R D = 3 − (0.2)(7.5) = 1.5 V
υ CM (max ) = υ O − υ DS 2 (sat ) + υ GS 2 = 1.5 − 0.125 + 0.625
or
υ CM (max ) = 2 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU11.8
From Example 11-8,
Kn2 IQ
I Q = 0.587 mA
( 0.1)( 0.587 )
⋅ (16) ⇒ Ad = 2.74
2
1
1
For M 4 , R0 =
=
⇒ R0 = 85.2 kΩ
λ4 I Q ( 0.02)( 0.587 )
Ad =
2
⋅ RD =
g m = 2 K n (VGS 2 − VTN ) = 2 ( 0.1)( 2.71 − 1)
= 0.342 mA/V
Acm =
− ( 0.342 )(16 )
− g m RD
=
⇒ Acm = −0.0923
1 + 2 g m R0 1 + 2 ( 0.342 )(85.2 )
⎛ 2.74 ⎞
C M RRdB = 20 log10 ⎜
⎟ ⇒ C M RRdB = 29.4 dB
⎝ 0.0923 ⎠
______________________________________________________________________________________
TYU11.9
(a) CMRR =
⎛ k'
1 ⎡⎢
1 + 2 2⎜⎜ n
2⎢
⎝ 2
⎣
⎤ 1⎡
⎤
⎞⎛ W ⎞
0.1 ⎞
⎟⎜ ⎟ I Q ⋅ Ro ⎥ = ⎢1 + 2 2⎛⎜
⋅
(
)(
)
(
)
10
0
.
1
1000
⎥
⎟
⎟⎝ L ⎠
⎥ 2 ⎢⎣
⎝ 2 ⎠
⎥⎦
⎠
⎦
( )
Or
CMRR = 316.73 ⇒ CMRR dB = 50 dB
(b) CMRR dB = 80 dB ⇒ CMRR = 10 4
Then
⎤
1⎡
⎛ 0. 1 ⎞
10 4 = ⎢1 + 2 2⎜
⎟(10 )(0.1) ⋅ Ro ⎥ ⇒ Ro = 31.6 M Ω
2 ⎣⎢
⎝ 2 ⎠
⎦⎥
______________________________________________________________________________________
TYU11.10
Ro = ro 4 + ro 2 (1 + g m 4 ro 4 )
−1
Assume I REF = I O = 100 μ A and λ = 0.01 V
ro 2 = ro 4 =
Let
1
λ ID
=
1
( 0.01)( 0.1)
⇒1 MΩ
K n ( all devices ) = 0.1 mA / V 2
Then
gm4 = 2 Kn I D = 2
( 0.1)( 0.1) = 0.2 mA / V
Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) ⇒ 202 M Ω
VGS 1 = VGS 2 =
Now
ID
0.05
+ VTN =
+ 1 = 1.707 V
Kn
0.1
VDS1 ( sat ) = VGS1 − VTN = 1.707 − 1 = 0.707 V
So vo1 ( min ) = +4 − VGS1 + VDS1 ( sat ) = 4 − 1.707 + 0.707
vo1 ( min ) = 3 V = 10 − I D RD = 10 − ( 0.05 ) RD ⇒ RD = 140 kΩ
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
For a one-sided output, the differential gain is:
Ad =
1
g m1 RD where g m1 = 2 K n I D
2
( 0.1)( 0.05 ) = 0.1414 mA / V
=2
Ad =
1
( 0.1414 )(140 ) ⇒ Ad = 9.90
2
The common-mode gain is:
Acm =
2 K n I Q ⋅ RD
1 + 2 2 K n I Q ⋅ Ro
CMRRdB = 20 log10
=
2 ( 0.1)( 0.1) ⋅ (140 )
1 + 2 2 ( 0.1)( 0.1) ⋅ ( 202000 )
⇒ Acm = 0.0003465
Ad
⇒ CMRRdB = 89.1 dB
Acm
Then
______________________________________________________________________________________
TYU11.11
I B5 =
a.
So
b.
IQ
β (1 + β )
=
0.5
(180 )(181)
⇒ 15.3 nA
I 0 = 15.3 nA
For a balanced condition
VEC 4 = VEC 3 = VEB 3 + VEB 5 ⇒ VEC 4 = 1.4 V
VCE 2 = VC 2 − VE 2 = (10 − 1.4 ) − ( −0.7 ) ⇒ VCE 2 = 9.3 V
______________________________________________________________________________________
TYU11.12
ro 2 =
V AN
120
=
= 2400 k Ω
I CQ 0.05
ro 4 =
V AP
80
=
= 1600 k Ω
I CQ 0.05
gm =
I CQ
VT
=
0.05
= 1.923 mA/V
0.026
Ad = g m (ro 2 ro 4 ) = (1.923)(2400 1600) = 1846
______________________________________________________________________________________
TYU11.13
P = ( I Q + I REF ) ( 5 − ( −5 ) )
10 = ( 0.1 + I REF )(10 ) ⇒ I REF = 0.9 mA
R1 =
5 − 0.7 − ( −5 )
I REF
=
9.3
⇒ R1 = 10.3 kΩ
0.9
⎛I ⎞
I Q RE = VT ln ⎜ REF ⎟
⎜ I ⎟
⎝ Q ⎠
0.026 ⎛ 0.9 ⎞
RE =
ln ⎜
⎟ ⇒ RE = 0.571 kΩ
0.1
⎝ 0.1 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
ro 2 =
VA 2 120
=
⇒ 2.4 M Ω
I C 2 0.05
ro 4 =
VA 4
80
=
⇒ 1.6 M Ω
I C 4 0.05
gm =
0.05
= 1.923 mA / V
0.026
(
)
(
)
Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 ⇒ Ad = 158
______________________________________________________________________________________
TYU11.14
(a) Ro = ro 2 ro 4
ro 2 =
V AN
120
=
⇒ 2. 4 M Ω
0.05
I CQ
ro 4 =
V AP
80
=
⇒ 1.6 M Ω
I CQ 0.05
Ro = 2.4 1.6 = 0.96 M Ω
(b) RL = Ro = 0.96 M Ω
______________________________________________________________________________________
TYU11.15
g m = 2 K n I DQ = 2 (0.18)(0.1) = 0.2683 mA/V
ro 2 =
ro 4 =
1
1
=
= 666.7 k Ω
λ n I DQ (0.015)(0.1)
1
λ p I DQ
Ad = g m (ro 2
=
1
= 400 k Ω
(0.025)(0.1)
ro 4 ) = (0.2683)(666.7 400) = 67.1
______________________________________________________________________________________
TYU11.16
I E 2 = 75 μ A , I B 2 = 0.497 μ A , I C 2 = 74.50 μ A
I D1 = 25 + 0.497 = 25.497 μ A
g m1 = 2 K n I D1 = 2 (0.05)(0.025497 ) ⇒ 71.4 ( μ A/V)
I C 2 0.0745
=
= 2.865 mA/V
0.026
VT
(150)(0.026) = 52.3 k Ω
=
0.0745
g m2 =
rπ 2
Then
g m1 (1 + g m 2 rπ 2 ) (0.0714)[1 + (2.865)(52.3)]
=
= 2.275 mA/V
1 + g m1rπ 2
1 + (0.0714 )(52.3)
______________________________________________________________________________________
g mC =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU11.17
From Figure 11.41
80
= 160 kΩ
0.5
R0 ≅ β r04 = (150 )(160 ) kΩ ⇒ R0 = 24 MΩ
r04 =
From Figure 11.42
1
1
=
⇒ r06 = 160 kΩ
λ I D ( 0.0125 )( 0.5 )
r06 =
0.5 = 0.5 (VGS − 1) ⇒ VGS = 2 V
g m 6 = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2 − 1) = 1 mA / V
r04 = 160 kΩ
R0 = ( g m 6 )( r06 )( β r04 ) = (1)(160 )(150 )(160 ) ⇒ R0 = 3.840 MΩ
2
______________________________________________________________________________________
TYU11.18
From Equation (11.126)
2(1 + β )β VT 2(121)(120 )(0.026 )
Ri =
=
⇒ Ri = 1.51 M Ω
IQ
0. 5
β VT
rπ 11 =
IQ
=
(120 )(0.026) = 6.24 k Ω
0. 5
R E′ = rπ 11 R3 = 6.24 0.1 = 0.0984 k Ω
g m11 =
ro11 =
IQ
VT
=
0. 5
= 19.23 mA/V
0.026
V A 120
=
= 240 k Ω
0. 5
IQ
Then RC11 = r011 (1 + g m11 RE′ )
= 240 ⎡⎣1 + (19.23)( 0.0984 ) ⎤⎦
= 694 kΩ
rπ 8 =
βVT
I C8
=
(120)(0.026) = 1.56 k Ω
2
Rb8 = rπ 8 + (1 + β )R 4 = 1.56 + (121)(5) = 607 k Ω
Then R L 7 = RC11 Rb8 = 694 607 = 324 k Ω
⎛ IQ ⎞
⎡ 0.5 ⎤
⎟R L7 = ⎢
Then Aυ = ⎜⎜
⎥ (324) ⇒ Aυ = 3115
⎟
2
V
⎣ 2(0.026) ⎦
⎝ T ⎠
⎛r +Z ⎞
⎟⎟
R o = R 4 ⎜⎜ π 8
⎝ 1+ β ⎠
Z = RC11 RC 7
RC 7 =
V A 120
=
= 240 k Ω
0 .5
IQ
Z = 694 240 = 178 k Ω
⎛ 1.56 + 178 ⎞
Ro = 5 ⎜
⎟ = 5 1.48 ⇒ Ro = 1.14 k Ω
⎝ 121 ⎠
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU11.19
⎛ IQ ⎞
Av = ⎜
⎟ RL 7
⎝ 2VT ⎠
⎛ 0.5 ⎞
103 = ⎜⎜
⎟⎟ RL 7 ⇒ RL 7 = 104 kΩ
⎝ 2 ( 0.026 ) ⎠
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 12
Exercise Solutions
EX12.1
(a) (i) A f =
A
1 + Aβ
⇒ 50 =
5 × 10 4
⇒ β = 0.01998
1 + 5 × 10 4 β
(
)
Af
50
=
= 0.999
1
1
β
0.01998
100
(b) (i) 20 =
⇒ β = 0.04
1 + (100 )β
Af
20
(ii)
=
= 0.80
1
1
β
0.04
______________________________________________________________________________________
(ii)
EX12.2
(a) A f =
(
5 × 10 5
A
⇒ 50 =
⇒ β = 0.019998
1 + Aβ
1 + 5 × 10 5 β
(
)
)
A = 5 × 10 (0.85) = 4.25 × 10
Now
4.25 × 10 5
Af =
= 49.99912
1 + 4.25 × 10 5 (0.019998 )
49.99912 − 50
× 100% = −1.76 × 10 −3 %
Percent change =
50
100
⇒ β = 0.04
(b) 20 =
1 + (100)β
A = (100)(0.85) = 85
Now
85
= 19.318
Af =
1 + (85)(0.04 )
19.318 − 20
× 100% = −3.41%
Percent change =
20
______________________________________________________________________________________
5
5
(
)
EX12.3
(a) (i) 80 =
5 × 10 4
⇒ β = 0.01248
1 + 5 × 10 4 β
(
)
[
(
)]
(
)
(ii) ω fH = ω H (1 + βAo ) = (2π )(5) 1 + (0.01248) 5 × 10 4 = (2π ) 3.125 × 10 3 rad/s
5 × 10
= 159.7
⎛ 0.01248 ⎞
1 + 5 × 10 4 ⎜
⎟
2
⎠
⎝
Percent change = 100%
(b) (i) A f (0 ) =
4
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(
)
(
)
⎡ ⎛ 0.01248 ⎞
4 ⎤
3
(ii) ω fH = (2π )(5)⎢1 + ⎜
⎟ 5 × 10 ⎥ = (2π ) 1.565 × 10 rad/s
2
⎠
⎦
⎣ ⎝
Percent change ≅ −50%
______________________________________________________________________________________
EX12.4
(a)
vOA = A1 A2 vi + A2 vn
= (100 )(10 ) vi + (10 ) vn
So 1000vi
S
=
= 100 i
10vn
No
Ni
(b)
A1 A2
A2
vi +
vn
1 + β Α1 Α2
1 + β Α1 Α2
vOC =
105
10
vi +
vn
1 + ( 0.001)105
1 + ( 0.001) (105 )
=
S o 103 vi
S
=
= 10 4 i
N o 0.1vn
Ni
______________________________________________________________________________________
EX12.5
a.
Vε = VS − V fb = 100 − 99 = 1 m V
5
⇒ Av = 5000 V/V
0.001
V fb 0.099
=
⇒ β = 0.0198 V/V
V fb = β V0 ⇒ β =
V0
5
V0 = AvVε ⇒ Av =
Av
5000
=
⇒ Avf = 50 V/V
1 + β Av 1 + ( 0.0198 )( 5000 )
Avf =
Rif = Ri (1 + β Av ) = ( 5 ) ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ⇒ Rif = 500 kΩ
R0 f =
R0
4
=
⇒ R0 f ⇒ 40 Ω
1 + β Av 1 + ( 0.0198 )( 5000 )
b.
______________________________________________________________________________________
EX12.6
a.
I ε = I S − I fb = 100 − 99 = 1 μ A
Ai =
I0
5
=
⇒ Ai = 5000 A/A
I ε 0.001
β=
I fb
Aif =
I0
=
0.099
⇒ β = 0.0198 A/A
5
Ai
5000
=
⇒ Aif = 50 A/A
1 + Ai β 1 + ( 5000 )( 0.0198 )
Rif =
Ri
5
=
⇒ Rif ⇒ 50 Ω
1 + β Ai 1 + ( 0.0198 )( 5000 )
R0 f = (1 + β Ai ) R0 = ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ( 4 ) ⇒ R0 f = 400 kΩ
b.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX12.7
⎛
R ⎞ ⎛ 60 ⎞
(a) Aυ f = ⎜⎜1 + 2 ⎟⎟ = ⎜1 + ⎟ = 5
R1 ⎠ ⎝ 15 ⎠
⎝
Vo = Aυ f ⋅ Vi = (5)(0.1) = 0.5 V
Aυ
5 × 10 4
=
= 4.9995
Aυ
5 × 10 4
1+
1+
⎛1 + R 2 ⎞
5
⎟
⎜
R1 ⎠
⎝
Vo = (4.9995)(0.1) = 0.49995 V
(b) (i) Aυ f =
(ii) V∈ =
Vo
0.49995
=
⇒ V∈ = 9.999 μ V
Aυ
5 × 10 4
5 × 10 5
= 4.99995
5 × 10 5
1+
5
0.499995
⇒ V∈ = 0.99999 μ V
(ii) V∈ =
5 × 10 5
______________________________________________________________________________________
(c) (i) Aυ f =
EX12.8
Use a non inverting op-amp.
R2
R
= 15 ⇒ 2 = 14
R1
R1
R2 = 140 K
R1 = 10 K
1+
Let
β=
1
= 0.066667
R
1+ 2
R1
Input resistance.
Rif = 5 ( 0.06667 ) ( 5 × 103 ) ≅ 1.67 MΩ
Rof =
50
( 0.066667 ) ( 5 ×103 )
≡ 0.15Ω
______________________________________________________________________________________
EX12.9
⎞
RE
⎟⎟ ⋅
rπ
⎠ ⎛⎜
⎜ RE + 1 + h
FE
⎝
(
80 )(0.026 )
= 4.16 k Ω
rπ =
0.5
Then
rπ
4.16
=
= 0.0514 k Ω
1 + h FE
81
Then we want
⎛ h
i o = ⎜⎜ FE
⎝ 1 + h FE
⎞
⎟⎟
⎠
⋅ is
⎞
io
RE
⎛ 80 ⎞ ⎛
= 0.95 = ⎜ ⎟ ⎜
⎟
ii
⎝ 81 ⎠ ⎝ RE + 0.0514 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or
⎛
⎞
RE
⎜
⎟ = 0.9619
0.0514
R
+
⎝ E
⎠
which yields
RE ( min ) = 1.30 k Ω
and
⎛ 81 ⎞
V + = I E RE + 0.7 = ⎜ ⎟ ( 0.5 )(1.3) + 0.7 ⇒ V + ( min ) = 1.36 V
⎝ 80 ⎠
______________________________________________________________________________________
EX12.10
Use the configuration shown in figure 12.20.
RS = 500 Ω, RL = 200 Ω
1+
Let
RF
= 15
R1
R1 = 2 K
For example, let RF = 28 K
______________________________________________________________________________________
EX12.11
⎛ 20 ⎞
(a) (i) VG = ⎜
⎟(10) − 5 = −1 V
⎝ 20 + 30 ⎠
2
VG = VGS + I DQ RS − 5 = VGS + K n RS (VGS − VTN ) − 5
(
2
4 = VGS + (2)(0.4) VGS
− 4VGS + 4
2
GS
0.8V
So
)
− 2.2VGS − 0.8 = 0 ⇒ VGS = 3.075 V
I DQ = 2(3.075 − 2) = 2.312 mA
2
(ii) Vi = V gs + g mV gs RS ⇒ V gs =
⎛ RD
I o = −⎜⎜
⎝ RD + RL
Agf =
Vi
1 + g m RS
⎞
Vi
⎟⎟ g m ⋅
g
+
1
m RS
⎠
⎛ gm
Io
= −⎜⎜
Vi
⎝ 1 + g m RS
⎞⎛ R D
⎟⎟⎜⎜
⎠⎝ R D + R L
⎞
⎟⎟
⎠
Now
g m = 2 K n I DQ = 2 (2)(2.312) = 4.30 mA/V
⎡
⎤⎛ 2 ⎞
4.30
Agf = − ⎢
⎟ = −0.7904 mA/V
⎥⎜
⎣1 + (4.30 )(0.4 ) ⎦⎝ 2 + 2 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(
2
− 4VGS + 4
(b) (i) 4 = VGS + (1.8)(0.4) VGS
)
2
0.72VGS
− 1.88VGS − 1.12 = 0 ⇒ VGS = 3.111 V
So
2
I DQ = (1.8)(3.111 − 2 ) = 2.222 mA
(ii) g m = 2 (1.8)(2.222 ) = 4.0 mA/V
⎤⎛ 2 ⎞
⎡
4.0
Agf = − ⎢
⎥⎜ 2 + 2 ⎟ = −0.7692 mA/V
(
)(
)
1
4
.
0
0
.
4
+
⎠
⎦⎝
⎣
0.7692 − 0.7904
× 100% = −2.68%
Percent change =
0.7904
______________________________________________________________________________________
EX12.12
Use the circuit with the configuration shown in Figure 12.27.
The LED replaces RL .
Agf = 10 mS = 10 × 10−3 =
1
⇒ RE = 100 Ω
RE
______________________________________________________________________________________
EX12.13
⎛ 150 ⎞
(a) (i) VGS = ⎜
⎟(5) = 1.50 V
⎝ 150 + 350 ⎠
2
2
I DQ = K n (VGS − VTN ) = (1.5)(1.5 − 0.8) = 0.735 mA
g m = 2 K n I DQ = 2 (1.5)(0.735) = 2.1 mA/V
⎛ R1 R2
Vo = − g mV gs R D ; V gs = ⎜
⎜R R +R
S
⎝ 1 2
Aυ = −(0.84 )(2.1)(2 ) = −3.528
(ii)
V gs − Vo
RF
+
V gs
R1 R2
+
V gs − Vi
RS
⎞
⎟ ⋅ Vi = ⎛⎜ 105 ⎞⎟ ⋅ Vi = 0.84Vi
⎟
⎝ 105 + +20 ⎠
⎠
=0
⎛ 1
1
1 ⎞⎟ Vo Vi
+
+
V gs ⎜
=
+
⎜R
⎟
⎝ F R1 R2 RS ⎠ RF RS
V
1
1 ⎞ V
⎛ 1
+
+ ⎟= o + i
V gs ⎜
⎝ 47 105 20 ⎠ 47 20
V gs = Vo (0.26337 ) + Vi (0.61881)
Now
Vo − V gs
Vo
+ g mV gs +
=0
RD
RF
Vo Vo ⎛
1 ⎞
⎟[Vo (0.26337 ) + Vi (0.61881)] = 0
+
+ ⎜⎜ g m −
2 47 ⎝
R F ⎟⎠
or
Vo (1.0687 ) + Vi (1.2863) = 0
which yields
V
Aυ = o = −1.204
Vi
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b) K n = 1.275 mA/V 2
(i) I DQ = (1.275)(1.5 − 0.8) = 0.62475 mA
2
g m = 2 (1.275 )(0.62475 ) = 1.785 mA/V
Aυ = −(0.84 )(1.785)(2 ) = −2.9988
2.9988 − 3.528
× 100% = −15%
Percent change =
3.528
(ii) Vo (0.52128) + (1.7637 )[Vo (0.26337 ) + Vi (0.61881)] = 0
Vo (0.98579) + Vi (1.091395) = 0
Vo
= −1.107
Vi
1.107 − 1.204
× 100% = −8.06%
Percent change =
1.204
______________________________________________________________________________________
Aυ =
EX12.14
From EX12.13; I DQ = 0.735 mA, g m = 2.1 mA/V
Vx
Vx Vx
+
+ g mV gs +
R D ro
RF + R1 R2 RS
(a) I x =
We find
⎛ R1 R2 RS
V gs = ⎜
⎜R R R +R
F
⎝ 1 2 S
⎞
⎟ ⋅V
⎟ x
⎠
(
Ix
1
1
1 1 + g m R1 R2 RS
=
=
+ +
V x Rof
R D ro
R F + R1 R2 RS
)
Now
R1 R2 R S = 150 350 20 = 16.8 k Ω
For λ = 0 ⇒ ro = ∞
Then
1
1 1 + (2.1)(16.8)
= +
= 0.5 + 0.56865
2
47 + 16.8
Rof
or
Rof = 0.9358 k Ω
(b) For λ = 0.04 ⇒ ro =
1
= 34.01 k Ω
(0.04)(0.735)
Then
1
= 0.5 + 0.0294 + 0.56865
Rof
or
Rof = 0.9107 k Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX12.15
⎛ 5.5 ⎞
VTH = ⎜
⎟ (10 ) = 0.9735 V
⎝ 5.5 + 51 ⎠
RTH = 5.5 || 51 = 4.965 kΩ
I BQ =
0.973 − 0.7
= 0.00217 mA
4.96 + (121)(1)
I CQ = 0.2605 mA
rπ = 11.98 kΩ, g m = 10.02 mA / V
Req = RS || R1 || R2 || rπ
= (10 ) || 51 || 5.5 || 12
= 2.598 kΩ
From Equation (12.99(b)):
⎛
⎞
Req
T = ( g m RC ) ⎜
⎜ R + R + R ⎟⎟
F
eq ⎠
⎝ C
2.598
⎛
⎞
= (10 )(10 ) ⎜
⎟ ⇒ T = 2.75
⎝ 10 + 82 + 2.598 ⎠
______________________________________________________________________________________
EX12.16 Computer Analysis
______________________________________________________________________________________
EX12.17
Vε = −Vt , V0 = AvVε = − AvVt
⎛ R1 || Ri ⎞
⎛ R1 || Ri ⎞
Vr = ⎜
⎟ V0 = − ⎜
⎟ ( AvVt )
⎝ R1 || Ri + R2 ⎠
⎝ R1 || Ri + R2 ⎠
T =−
⎛ R1 || Ri ⎞
Vr
= Av ⎜
⎟
Vt
⎝ R1 || Ri + R2 ⎠
or
T=
Av
R
1+ 2
R1 Ri
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX12.18
⎛ f
3
⎝ 10
≅ 105 Hz,
φ = − tan −1 ⎜
At f180
T ( f180 ) = 1 =
or
1=
β (3000 )
(100 )(2)
⎞
−1 ⎛ f ⎞
⎟ − 2 tan ⎜ 5 ⎟
⎠
⎝ 10 ⎠
φ = −180°
β (3000)
⎛ 10 5
1 + ⎜⎜ 3
⎝ 10
⎞
⎟
⎟
⎠
2
⎡ ⎛ 10 5
⎢1 + ⎜ 5
⎢ ⎜⎝ 10
⎣
⎞
⎟
⎟
⎠
2
⎤
⎥
⎥
⎦
⇒ β = 0.0667
______________________________________________________________________________________
EX12.19
A f (0) =
T =1=
3000
3000
=
= 120
1 + β (3000 ) 1 + (0.008)(3000 )
(0.008)(3000)
2
2
⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤
1 + ⎜ 3 ⎟ ⎢1 + ⎜ 5 ⎟ ⎥
⎝ 10 ⎠ ⎣⎢ ⎝ 10 ⎠ ⎦⎥
By trial and error,
f ≅ 2.28 × 10 4 Hz
⎛ f ⎞
−1 ⎛ f ⎞
⎟ − 2 tan ⎜ 5 ⎟
3
⎝ 10 ⎠
⎝ 10 ⎠
4
⎛ 2.28 × 10 4
⎛ 2.28 × 10 ⎞
⎟⎟ − 2 tan −1 ⎜⎜
= − tan −1 ⎜⎜
3
5
⎝ 10
⎠
⎝ 10
φ = − tan −1 ⎜
⎞
⎟⎟ = −87.49 − 2(12.84 ) = −113.18
⎠
Then
Phase margin = −113.18 − (− 180 ) = 66.8°
______________________________________________________________________________________
EX12.20
⎛ f ⎞
⎛ f ⎞
(a) φ = −180 = − tan −1 ⎜ 1803 ⎟ − 2 tan −1 ⎜ 1805 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
f180 ≅ 105 Hz
T ( f180 ) =
250
⎛ 10
1 + ⎜⎜ 3
⎝ 10
5
⎞
⎟
⎟
⎠
2
⎡ ⎛ 10
⎢1 + ⎜ 5
⎢ ⎜⎝ 10
⎣
5
⎞
⎟
⎟
⎠
2
⎤
⎥
⎥
⎦
=
250
(100)(2)
= 1.25
T > 1 at f 180
(b) T =
250
⎛
f
⎜⎜1 + j
f PD
⎝
⎞⎛
f ⎞⎛
f ⎞
⎟⎟⎜1 + j 3 ⎟⎜1 + j 5 ⎟
10 ⎠⎝
10 ⎠
⎠⎝
⎛ f
⎝ f PD
φ = −120 = − tan −1 ⎜⎜
2
⎞
⎛ f ⎞
⎛ f ⎞
⎟⎟ − tan −1 ⎜ 3 ⎟ − 2 tan −1 ⎜ 5 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
f ≅ 0.577 × 10 3 Hz
Now
250
T =1=
⎛ 0.577 × 10 3 ⎞
⎟⎟
1 + ⎜⎜
f PD
⎠
⎝
2
⎛ 0.577 × 10 3
1 + ⎜⎜
10 3
⎝
2
⎞
⎟⎟ (1)
⎠
⎛ 0.577 × 10 3 ⎞
250
⎟⎟ =
⎜⎜
f PD
⎠ 1.155
⎝
which yields
f PD = 2.67 Hz
______________________________________________________________________________________
EX12.21
A f (0 ) =
10 5
= 40
1 + (0.025) 10 5
( )
f = (10)[1 + (0.025)(10 )] ≅ 25 kHz
5
______________________________________________________________________________________
EX12.22
⎛ f135
⎝ f PD
φ = −135 = − tan −1 ⎜⎜
T ( f135 ) = 1 =
⎞
⎛ f ⎞
⎟⎟ − 2 tan −1 ⎜ 1355 ⎟ ⇒ f135 ≅ 0.414 × 10 5 Hz
⎝ 10 ⎠
⎠
250
⎛ 0.414 × 10 5
1 + ⎜⎜
f PD
⎝
⎞
⎟
⎟
⎠
2
⎡ ⎛ 0.414 × 10 5
⎢1 + ⎜
10 5
⎢ ⎜⎝
⎣
⎞
⎟
⎟
⎠
2
⎤
⎥
⎥
⎦
0.414 × 10 5
250
=
⇒ f PD = 194 Hz
1.171
f PD
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU12.1
A
1 + Aβ
A
50 =
1 + A(0.019 )
(a) A f =
or
50 = A[1 − (50)(0.019)] ⇒ A = 103
5 × 10 5
= 52.63
1 + 5 × 10 5 (0.019 )
______________________________________________________________________________________
(b) A f =
(
)
TYU12.2
dAf
Af
=
1
(1 + β A)
dA dAf
=
A
Af
⋅
⎛ A
⋅⎜
⎜A
⎝ f
dA ⎛ Af
=⎜
A ⎝ A
⎞ dA
⎟.
⎠ A
⎞
⎛ 5 × 105 ⎞
dA
= ±5%
⎟⎟ = ( 0.001) ⎜
⎟⇒
A
⎝ 100 ⎠
⎠
______________________________________________________________________________________
TYU12.3
(a) 5 × 10 5 (6 ) = 200 × 10 3 A f (0 )
(
)
A f (0 ) = 15
(b)
(
)
(5 × 10 )(6) = (100 × 10 )A (0)
5
3
f
A f (0 ) = 30
______________________________________________________________________________________
TYU12.4
Vε = VS − V fb = 100 − 99 = 1 mV
Ag =
β=
I 0 5 mA
=
⇒ Ag = 5 A/V
Vε 1 mV
V fb
I0
=
99 mV
⇒ β = 19.8 V/A
5 mA
Ag
Agf =
=
5
⇒ Agf = 0.05 A/V = 50 mA/V
1 + (19.8 )( 5 )
1 + β Ag
______________________________________________________________________________________
TYU12.5
I ε = I S − I fb = 100 − 99 = 1 μ A
Az =
V0
5V
=
⇒ Az = 5 × 106 V/A
Iε 1 μ A
β=
I fb
Azf =
V0
=
99 μ A
⇒ β = 1.98 × 10 −5 A/V
5V
Az
1 + β Az
=
5 × 106
1 + (1.98 × 10−5 )( 5 × 106 )
⇒ Azf = 5 × 10 4 V/A = 50 V/mA
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU12.6
(a) rπ =
Aυ f
Ri f
hFEVT
(120)(0.026) = 2.6 k Ω
=
I CQ
1. 2
⎞
⎛1
⎜⎜ + g m ⎟⎟ R E
rπ
(1 + hFE )RE = (121)(1.5) = 0.985877
⎠
= ⎝
=
2.6 + (121)(1.5)
r
⎞
⎛1
π + (1 + hFE )R E
1 + ⎜⎜ + g m ⎟⎟ R E
⎠
⎝ rπ
= rπ + (1 + hFE )R E = 2.6 + (121)(1.5) = 184 .1 k Ω
RE
Ro f =
=
(RE )(rπ )
+ (1 + hFE )RE
rπ
⎛1
⎞
1 + ⎜⎜ + g m ⎟⎟ R E
⎝ rπ
⎠
(180)(0.026) = 3.9 k Ω
(b) rπ =
1.2
(181)(1.5) = 0.985839
Aυ f =
3.9 + (181)(1.5)
ΔAυ f
× 100% = −0.00385%
Aυ f
=
(1.5)(2.6) ⇒ R
of
2.6 + (121)(1.5)
= 21.18 Ω
Ri f = 3.9 + (181)(1.5) = 275.4 k Ω
ΔRi f
Ri f
× 100% = +49.6%
(1.5)(3.9) ⇒ R
of
3.9 + (181)(1.5)
Ro f =
ΔRo f
Ro f
= 21.24 Ω
× 100% = +0.283%
______________________________________________________________________________________
TYU12.7
(a)
g m RS
g m = 2 K n I DQ = 2 (0.5)(0.25) = 0.7071 mA/V
,
1 + g m RS
(0.7071)(3) = 0.67962
=
1 + (0.7071)(3)
RS
3
=
=
= 0.96114 k Ω
1 + g m RS 1 + (0.7071)(3)
Aυ f =
Aυ f
Ro f
(b) g m = 2 (0.5)(1) = 1.414 mA/V
(1.414 )(3) = 0.80923
1 + (1.414 )(3)
Aυ f =
ΔAυ f
Aυ f
× 100% = +19.1%
3
= 0.5723 k Ω
1 + (1.414 )(3)
ΔRo f
× 100% = −40.5%
Ro f
Ro f =
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU12.8 Computer Analysis
______________________________________________________________________________________
TYU12.9 Computer Analysis
______________________________________________________________________________________
TYU12.10
(h
1 + (h
)
)R
(180)(10 2 ) = 0.9999444 mA/V
1 + (180)(10 2 )(1)
FE Ag
E
I o = Ag f ⋅ Vi = (0.9999444 )(1.5) = 1.4999166 mA
(ii) V∈ = Vi − I o R E = 1.5 − (1.4999166 )(1) ⇒ V∈ = 83.4 μ V
(144)(10 2 ) = 0.9999306 mA/V
(b) (i) Ag f =
1 + (144 )(10 2 )(1)
I o = Ag f ⋅ Vi = (0.9999306 )(1.5) = 1.4998959 mA
(a) (i) Ag f =
FE
ΔAg f
Ag f
Ag
=
× 100% =
0.9999306 − 0.9999444
× 100% = −0.00138%
0.9999444
ΔI o
1.4998959 − 1.4999166
× 100% =
× 100% = −0.00138%
Io
1.4999166
(ii) V∈ = 1.5 − (1.4998959)(1) ⇒V ∈= 104 μ V
______________________________________________________________________________________
TYU12.11
(a) RTH = R1 R2 = 51 5.5 = 4.965 k Ω
⎛ 5.5 ⎞
VTH = ⎜
⎟(10) = 0.9735 V
⎝ 5.5 + 51 ⎠
0.9735 − 0.7
I BQ =
= 0.002865 mA
4.965 + (181)(0.5)
I CQ = 0.5157 mA
Then
0.5157
= 19.83 mA/V ,
0.026
(i) Vo = − g mVπ RC
gm =
rπ =
(180)(0.026) = 9.075 k Ω
0.5157
⎛ R1 R2 rπ ⎞
⎟ ⋅ V = ⎡ 4.965 9.075 ⎤ ⋅ V = 0.243V
Vπ = ⎜
i
⎜ R R r + R ⎟ i ⎢⎢ (4.965 9.075) + 10 ⎥⎥ i
S ⎠
⎦
⎣
⎝ 1 2 π
V
Aυ = o = −(0.243)(19.83)(10 ) = −48.187
Vi
(ii)
(1)
(2)
V − Vπ
Vo
+ g mVπ + o
=0
RF
RC
Vπ − Vi
Vπ
V − Vo
+
+ π
=0
RS
RTH rπ
RF
⎛ 1
V
1
1 ⎞⎟ Vo
+
+
Vπ ⎜
=
+ i
⎜R
⎟
⎝ S RTH rπ R F ⎠ R F RS
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛1
V
1
1 ⎞ V
+ ⎟= o + i
Vπ ⎜ +
⎜ 10 4.965 9.075 60 ⎟ 60 10
⎝
⎠
Vπ = Vo (0.038913) + Vi (0.233487 )
Now
(b) (i) I BQ
I CQ
⎛ 1
⎛
1 ⎞
1 ⎞
⎟⎟ + Vπ ⎜⎜ g m −
⎟=0
+
(1) Vo ⎜⎜
R F ⎟⎠
⎝
⎝ RC R F ⎠
1 ⎞ ⎛
1 ⎞
⎛1
Vo ⎜ + ⎟ + ⎜19.83 − ⎟[Vo (0.038913) + Vi (0.233487)] = 0
60
10
60
⎠
⎠ ⎝
⎝
Vo (0.887662) + Vi (4.62616) = 0
Then
V
Aυ = o = −5.2116
Vi
0.9735 − 0.7
=
= 0.004178 mA
4.965 + (121)(0.5)
= 0.5013 mA
0.5013
= 19.28 mA/V , rπ = 6.224 k Ω
0.026
rπ = 4.965 6.224 = 2.7618 k Ω
gm =
RTH
⎛ 2.7618 ⎞
Vπ = ⎜
⎟ ⋅ Vi = 0.2164Vi
⎝ 2.7618 + 10 ⎠
Aυ = − g mVπ RC = −(19.28)(0.2164)(10) = −41.722
V
1
1 ⎞ V
⎛1
+ ⎟= o + i
(ii) Vπ ⎜ +
⎝ 10 2.7618 60 ⎠ 60 10
Vπ = Vo (0.034811) + Vi (0.208877 )
Then
Vo (0.116667 ) + (19.26334 )[Vo (0.034811) + Vi (0.208877 )] = 0
Vo (0.787242) + Vi (4.023669) = 0
Then
V
Aυ = o = −5.1111
Vi
41.722 − 48.187
× 100% = −13.4%
(c) (i)
48.187
5.1111 − 5.2116
× 100% = −1.93%
(ii)
5.2116
______________________________________________________________________________________
TYU12.12
(a) I CQ = 0.5157 mA, g m = 19.83 mA/V, rπ = 9.075 k Ω
(i)
Ro = RC = 10 k Ω
(ii) RS RTH rπ = 10 4.965 9.075 = 2.43 k Ω
Ix =
Vx
Vx
+ g mVπ +
R F + 2.43
RC
⎛ 2.43 ⎞
Vπ = ⎜
⎟ ⋅ V x = (0.03892)V x
⎝ 2.43 + 60 ⎠
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ix
1
1
1
=
=
+ (19.83)(0.03892 ) +
62.43
V x Ro f 10
or
Ro f = 1.126 k Ω
(b) I CQ = 0.5013 mA, g m = 19.28 mA/V, rπ = 6.224 k Ω
R S RTH rπ = 10 4.965 6.224 = 2.164 k Ω
(i) Ro = 10 k Ω
Ix
1
1
1
⎛ 2.164 ⎞
=
+ (19.28)⎜
=
⎟+
V x 10
⎝ 2.164 + 60 ⎠ 62.164 Ro f
(ii)
or
Ro f = 1.270 k Ω
______________________________________________________________________________________
TYU12.13
From Example 12.15, for hFE = 100 , T = 4.10 .
Now for hFE = 150 ,
I BQ
I CQ
RTH = 4.965 k Ω , VTH = 0.9735 V
0.9735 − 0.7
=
= 0.003399 mA
4.965 + (151)(0.5)
= 0.5098 mA, g m = 19.61 mA/V, rπ = 7.650 k Ω
Req = R S RTH rπ = 10 4.965 7.65 = 2.314 k Ω
We find
⎛
Req
T = (g m RC )⎜
⎜ Req + RF + RC
⎝
or
⎞
2.314
⎞
⎟ = (19.61)(10)⎛⎜
⎟
⎟
⎝ 2.314 + 82 + 10 ⎠
⎠
T = 4.811
4.811 − 4.10
× 100% = +17.3%
4.10
______________________________________________________________________________________
Percent change =
TYU12.14
Vt = −V∈ , Vo = − AυVt
⎛ R1 Ri ⎞
⎟ ⋅V
Vr = ⎜
⎜ R1 Ri + R2 ⎟ o
⎠
⎝
V
1
1
= 10 4 ⋅
= 1.85 × 10 3
T = − r = Aυ ⋅
R2
20
Vt
1+
1+
5 50
R1 Ri
( )
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU12.15
⎛ f ⎞
⎛ f ⎞
− 135 = − tan −1 ⎜ 1353 ⎟ − 2 tan −1 ⎜ 1355 ⎟
⎝ 10 ⎠
⎝ 10 ⎠
f135 ≅ 4.25 × 10 4 Hz
T ( f 135 ) = 1 =
1=
β (3000 )
2
2
⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤
1 + ⎜ 1353 ⎟ ⎢1 + ⎜ 1355 ⎟ ⎥
⎝ 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦
β (3000)
⎛ 4.25 × 10 4
1 + ⎜⎜
3
⎝ 10
⎞
⎟
⎟
⎠
2
⎡ ⎛ 4.25 × 10 4
⎢1 + ⎜
5
⎢ ⎜⎝ 10
⎣
⎞
⎟
⎟
⎠
2
⎤
⎥
⎥
⎦
=
β (3000)
(42.51)(1.1806)
Then
β = 0.0167
______________________________________________________________________________________
TYU12.16
Αι 0 β
⎛
f ⎞⎛
f ⎞
⎜ 1+ j ⋅ ⎟⎜ 1 + j ⋅ ⎟
f
f
1 ⎠⎝
2 ⎠
⎝
⎡ −1 ⎛ f ⎞
⎛
f ⎞⎤
= − ⎢ tan ⎜ ⎟ + tan −1 ⎜ ⎟ ⎥
⎝ f1 ⎠
⎝ f2 ⎠⎦
⎣
T = Ai β =
Phase
Phase Margin = 60° ⇒ Phase = −120°
⎡
⎛ f ⎞
⎛ f ⎞⎤
−120° = − ⎢ tan −1 ⎜ 4 ⎟ + tan −1 ⎜ 5 ⎟ ⎥
⎝ 10 ⎠
⎝ 10 ⎠ ⎦
⎣
At f ′ = 7.66 × 10 4 Hz,
Phase = − ⎡⎣ tan −1 ( 7.66 ) + tan −1 ( 0.766 ) ⎤⎦
= − [82.56 + 37.45]
= −120°
T ( f ′) = 1 =
(10 ) β
5
1+ ( 7.66 ) × 1 + ( 0.766 )
2
(10 ) β
2
5
1=
( 7.725 )(1.26 )
⇒ β = 9.73 × 10−5
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU12.17
Phase Margin = 60° ⇒ Phase = −120°
⎛ f′ ⎞
= −120° = −3 tan −1 ⎜ 5 ⎟
⎝ 10 ⎠
Phase
⎛ f′ ⎞
tan −1 ⎜ 5 ⎟ = 40° ⇒ f ′ = 0.839 × 105 Hz
⎝ 10 ⎠
β (100 )
T ( f ′) = 1 =
3
2
⎡
f′ ⎞ ⎤
⎛
⎢ 1+ ⎜ 5 ⎟ ⎥
⎢
⎝ 10 ⎠ ⎥⎦
⎣
β (100 )
=
⇒ β = 0.0222
3
⎡ 1 + ( 0.839 )2 ⎤
⎣⎢
⎦⎥
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 13
Exercise Solutions
EX13.1
5 − 0.6 − 0.6 − (− 5)
= 0.352 mA
25
⎛ 0.352 ⎞
⎟
I C10 (5) = (0.026 ) ln⎜⎜
⎟
⎝ I C10 ⎠
I REF =
By trial and error
I C10 ≅ 16 μ A
Then
I
I C1 = I C 2 = C10 = 8 μ A
2
______________________________________________________________________________________
EX13.2
5 − 0.6 − 0.6 − (−5)
= 0.22 mA
40
= I C13 B = 0.75I REF = (0.75)(0.22) = 0.165 mA
0.165 (0.165)(0.1) + 0.6
=
+
200
50
= 0.000825 + 0.01233
= 13.2 μ A
I REF =
I C17
I C16
I C16
______________________________________________________________________________________
EX13.3
I C13 A = (0.25)(0.5) = 0.125 mA
0.6
I R10 =
= 0.012 mA
50
I C19 ≅ I C13 A − I R10 = 0.125 − 0.012 = 0.113 mA
I C18
I C19
0.113
⇒ 0.565 μ A
β
200
= I R10 + I B19 = 12 + 0.565 = 12.565 μ A
I B19 =
=
⎛ 0.113 × 10 −3 ⎞
⎟⎟ = 0.60185 V
V BE19 = (0.026 ) ln⎜⎜
−14
⎠
⎝ 10
⎛ 12.565 × 10 −6 ⎞
⎟⎟ = 0.54474 V
V BE18 = (0.026) ln⎜⎜
10 −14
⎠
⎝
V BB = 0.60185 + 0.54474 = 1.1466 V
⎛ 1.1466 ⎞
⎜
2 ⎟ ⇒ 0.113 mA
I C14 = 3 ×10 −14 exp⎜
⎟⎟
0
.
026
⎜
⎝
⎠
______________________________________________________________________________________
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX13.4
100
⇒ 10.5 MΩ
0.0095
ro 6 =
Then, using results from Example 13.4
Ract1 = ro 6 [1 + g m 6 ( R2 || rπ 6 ) ] = 10.5 [1 + (0.365)(1|| 547) ]
= 14.3 MΩ
ro 4 =
VA
100
=
⇒ 10.5 MΩ
I C 4 0.0095
⎛ 9.5 ⎞
Ad = − ⎜
⎟ (10.5 ||14.3 || 4.07 ) = −889
⎝ 0.026 ⎠
______________________________________________________________________________________
EX13.5
100
= 556 K
0.18
Ri 3 = rπ 22 + (1 + β P )[ R19 || R20 ]
= 7.22 + (51)(556 ||111) ⇒ 4.73 MΩ
R19 = ro13 A =
100
= 185 K
0.54
100
= 185 K
Ro17 =
0.54
−(200)(201)(50)(185 || 4730 ||185)
Av 2 =
4070[50 + [9.63 + (201)(0.1)]]
−182358786.9
=
32450.1
Av 2 = −562
Ract 2 =
______________________________________________________________________________________
EX13.6
100
100
= 185 K ro13 B =
= 185 K
0.54
0.54
= 185 [1 + (20.8)(0.1|| 9.63) ]
ro17 =
RC17
RC17 = 566 K
7.22 + 566 ||185
= 2.88 K
Re 22 =
51
100
= 556 K
RC19 =
0.18
0.65 + 2.88 || 556
Re 20 =
= 0.0689
51
= 68.9 Ω
RO = 22 + 68.9 = 90.9 Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX13.7
Ci = C1 (1+ | A2 |) = 30(1 + 562) = 16890 pF
Ri 2 = 4.07 MΩ
Ro1 = Ract1 || ro 4 = 14.3 ||10.5 = 6.05 MΩ
Req = Ro1 || Ri 2 = 6.05 || 4.07 = 2.43 MΩ
Then
Then
f PD =
1
1
=
2π Req Ci 2π (2.43 × 106 )(16890 × 10 −12 )
= 3.88 Hz
______________________________________________________________________________________
EX13.8
⎞⎛ W ⎞
⎟⎜ ⎟ = ⎛⎜ 0.04 ⎞⎟(20 ) = 0.40 mA/V 2
⎟⎝ L ⎠ 5 ⎝ 2 ⎠
⎠
5 − VSG 5 − (− 5)
2
K p 5 (VSG 5 + VTP ) =
150
2
60 VSG 5 − VSG 5 + 0.25 = 10 − VSG 5
⎛ k p'
K p5 = ⎜
⎜ 2
⎝
(
)
− 59VSG 5 + 5 = 0 ⇒ VSG 5 = 0.8897 V
10 − 0.8897
I REF = I Q =
⇒ 60.74 μ A
150
I D 7 = I D 8 = 60.74 μ A
I D1 = I D 2 = I D 3 = I D 4 = 30.37 μ A
______________________________________________________________________________________
60V
2
SG 5
EX13.9
⎛ 0.04 ⎞
2
K p1 = K p 2 = ⎜
⎟(20) = 0.40 mA/V
2
⎠
⎝
1
1
=
⇒ 1.646 M Ω
ro 2 = ro 4 =
λI D 2 (0.02)(0.03037)
Ad = 2 K p1 I Q (ro 2 ro 4 ) = 2(0.40)(0.06074) (1646 1646)
or
Ad = 181.4
Now
⎞⎛ W ⎞
0. 1 ⎞
⎟⎜ ⎟ I D 7 = 2 ⎛⎜
⎟(20 )(0.06074 ) = 0.4929 mA/V
⎟⎝ L ⎠
⎝ 2 ⎠
7
⎠
1
1
= ro8 =
=
= 823.2 k Ω
λI D 7 (0.02)(0.06074)
⎛ k'
g m 7 = 2 ⎜⎜ n
⎝ 2
ro 7
Aυ 2 = g m 7 (ro 7 ro8 ) = (0.4929)(823.2 823.2)
or
Aυ 2 = 202.9
Then
Aυ = Ad Aυ 2 = (181.4)(202.9 ) = 36,806 ⇒ 91.3 dB
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX13.10
⎛ k ' ⎞⎛ W ⎞ ⎛ I Q1 ⎞
0.08 ⎞
⎛ 0.2 ⎞
⎟ = 2 ⎛⎜
(a) g m1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ ⎜⎜
⎟ = 0.60 mA/V
⎟(22.5)⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠⎝ L ⎠1 ⎝ 2 ⎠
1
ro 2 =
= 1000 k Ω
(0.01)(0.1)
1
ro 4 =
= 666.7 k Ω
(0.015)(0.1)
Ad 1 = g m1 (ro 2 ro 4 ) = (0.6)(1000 666.7 ) = 240
⎛ k 'p ⎞⎛ W ⎞
⎛ 0.04 ⎞
g m 5 = 2 ⎜ ⎟⎜ ⎟ I D 5 = 2 ⎜
⎟(80 )(0.2 ) = 1.131 mA/V
⎜ 2 ⎟⎝ L ⎠ 5
⎝ 2 ⎠
⎝ ⎠
1
ro 5 =
= 333.3 k Ω
(0.015)(0.2)
1
ro 9 =
= 500 k Ω
(0.01)(0.2)
A2 = − g m5 (ro5 ro9 ) = −(1.131)(333.3 500) = −226.2
Then
Aυ = Ad 1 A2 = (240 )(− 226.2 ) = −54,288
(b)
⎛k'
K n 6 = ⎜⎜ n
⎝ 2
⎞⎛ W ⎞
⎟⎜ ⎟ = ⎛⎜ 0.08 ⎞⎟(25) = 1.0 mA/V 2
⎟⎝ L ⎠
⎝ 2 ⎠
6
⎠
I D 6 = K n 6 (VGS 6 − VTN )
2
0.04 = (1)(VGS 6 − 0.7 ) ⇒ VGS 6 = 0.90 V = V SG 7
Then
VGS 8 = 2(0.9) = 1.8 V
2
⎛ 0.08 ⎞⎛ W ⎞
⎛W ⎞
2
I D8 = 0.2 = ⎜
⎟⎜ ⎟ (1.8 − 0.7 ) ⇒ ⎜ ⎟ = 4.13
⎝ 2 ⎠⎝ L ⎠ 8
⎝ L ⎠8
______________________________________________________________________________________
EX13.11
I D1 = I D 2 = 25 μ A
g m1 = g m8 = 2
k p′ ⎛ W
⎜
2 ⎝L
⎞
⎛ 40 ⎞
⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m1 = g m8 = 224 μ A / V
⎠
⎝ 2 ⎠
⎛ k ′ ⎞⎛ W ⎞
⎛ 80 ⎞
g m 6 = 2 ⎜ n ⎟ ⎜ ⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m 6 = 316 μ A / V
⎝ 2⎠
⎝ 2 ⎠⎝ L ⎠
1
1
ro1 = ro 6 = ro8 = ro10 =
=
= 2 MΩ
λ I D (0.02)(25)
ro 4 =
1
λ ID4
=
1
=1 MΩ
(0.02)(50)
Ro8 = g m8 (ro8 ro10 ) = (224)(2)(2) = 896 M Ω
Ro 6 = g m 6 (ro 6 ) ( ro 4 || ro1 ) = 316(2) (1|| 2 ) = 421 M Ω
Then
Ad = g m1 ( Ro 6 Ro8 ) = 224 ( 421 896 ) ⇒ Ad = 64,158
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX13.12
(a) K p (VSG − VTP
)
2
=
(0.15)(VSG − 0.8)2
VSG − V BE
R1
=
V SG − 0.6
10
We obtain
2
1.5VSG
− 3.4VSG + 1.56 = 0 ⇒ VSG = 1.628 V
Now
1.628 − 0.6
I1 = I 2 =
= 0.1028 mA
10
VC 7 = V + − VEB − VBE = 5 − 0.6 − 0.6 = 3.8 V
VC 6 = V − + VSG = −5 + 1.628 = −3.372 V
VCB 7 = V BC 6 = VC 7 − VC 6 = 3.8 − (− 3.37 ) = 7.17 V
(b)
Set VCB 7 = V BC 6 = 0 ⇒ VC 7 = VC 6
V + − 1.2 = V − + 1.628 ,
Let V − = −V +
2V + = 2.828 , So that V + = −V − = 1.414 V
______________________________________________________________________________________
EX13.13
Ad = 2 K P I Q 5 ( Ri 2 )
= 2(1)(0.2)(26)
Ad = 16.4
______________________________________________________________________________________
EX13.14
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
rπ 13 =
β VT
I C13
=
(200)(0.026)
0.20
= 26 kΩ
Ri 2 = rπ 13 + (1 + β ) RE13 = 26 + 201(1)
= 227 kΩ
1
1
r010 =
=
= 500 kΩ
λ I D10 (0.02)(0.1)
r012 =
g m12 =
rπ 12 =
VA
50
=
= 500 kΩ
I C12 0.1
I C12
0.1
=
= 3.85 mA / V
VT
0.026
β VT
I C12
=
(200)(0.026)
= 52 kΩ
0.1
Ract1 = r012 [1 + g m12 (rπ 12 || R5 )]
= 500[1 + (3.85)(52 || 0.5)] = 1453 kΩ
Ad = 2 K n I Q 5 ⋅ ( ro10 || Ract1 || R12 )
= 2(0.6)(0.2) ⋅ (500 ||1453 || 227)
= (0.490)(141) ⇒ Ad = 69.1
______________________________________________________________________________________
EX13.15
From Example 13.15, f PD = 265 Hz
Av = Adi ⋅ A2 = (16.4)(1923) = 31,537
fT = f PD ⋅ Av = (265)(31,537) = 8.36 MHz
______________________________________________________________________________________
Test Your Understanding Solutions
TYU13.1 Computer Analysis
______________________________________________________________________________________
TYU13.2 Computer Analysis
______________________________________________________________________________________
TYU13.3
9.5
μ A I B1 = I B 2 = 47.5 nA
200
______________________________________________________________________________________
I B1 = I B 2 =
TYU13.4
Vi N (max) = V + − VEB (on) = 15 − 0.6 = 14.4 V
Vi N (min) ≅ 4VBE (on) + V +
= 4(0.6) − 15 = −12.6 V
−12.6 ≤ Vi N (cm) ≤ 14.4 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU13.5
a.
V0 (max) ≅ V + − 2VBE (on) = 15 − 2(0.6)
V0 (max) = 13.8 V
V0 (min) = 3VBE (on) + V − = 3(0.6) − 15
V0 (min)
b.
≅ −13.2 V
−13.2 ≤ V0 ≤ 13.8 V
V0 (max) = 5 − 1.2 = 3.8 V
V0 (min) ≅ 3VBE + V − = 3(0.6) − 5 = −3.2 V
−3.2 ≤ V0 ≤ 3.8 V
______________________________________________________________________________________
TYU13.6
5 − V EB12 − V BE11 − (− 5)
40
⎞
⎛ I
V EB12 = V BE11 = (0.026 ) ln⎜ REF 15 ⎟
⎝ 5 × 10 ⎠
Then by trial and error, I REF ≅ 0.218 mA, and V BE11 = V EB12 ≅ 0.637 V
I REF =
⎛ 0.218 ⎞
⎟
I C10 (5) = (0.026) ln⎜⎜
⎟
⎝ I C10 ⎠
By trial and error, I C10 ≅ 14.2 μ A
V BE10 = V BE11 − I C10 R4 = 0.637 − (0.0142)(5) = 0.566 V
I C10
= 7.1 μ A
2
⎛ 7.1 × 10 −6 ⎞
⎟ = 0.548 V
V BE 6 = (0.026 ) ln⎜⎜
−15 ⎟
⎠
⎝ 5 × 10
______________________________________________________________________________________
IC6 =
TYU13.7
10 − 0.6 − 0.6 − (−10)
⇒ I REF = 0.47 mA
40
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
I REF =
⎛ 0.47 ⎞
I C10 (5) = (0.026) ln ⎜
⎟
⎝ I C10 ⎠
By trial and error:
⇒ I C10 ≅ 17.2 μ A
IC 6 ≅
I C13 B
I C10
⇒ I C 6 = 8.6 μ A
2
= (0.75) I REF ⇒ I C13 B = 0.353 mA
I C13 A = (0.25) I REF ⇒ I C13 A = 0.118 mA
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU13.8
R0 = R6 + RE14
RE14 =
rπ 14 + R0 d R013 A
1 + βn
The diode resistance can be found as
⎛V ⎞
I D = I S exp ⎜ D ⎟
⎝ VT ⎠
⎛ 1 ⎞
⎛V ⎞ I
1 ∂I D
=
= I S ⎜ ⎟ ⋅ exp ⎜ D ⎟ = D
rd ∂VD
V
⎝ T⎠
⎝ VT ⎠ VT
or
rd =
VT
V
0.026
= T =
⇒ 144 Ω
I D I C13 A
0.18
RE 22 =
rπ 22 + R017 R013 B
1+ βP
R013 B = r013 B = 92.6 kΩ
R017 = r017 ⎡⎣1 + g m17 ( R8 rπ 17 ) ⎤⎦ = 283 kΩ
From previous calculations
RE 22 = 1.51 kΩ
R0 d = 2rd + RE 22 = 2(0.144) + 1.51 = 1.80 kΩ
R013 A = r013 A = 278 kΩ
rπ 14 =
β nVT
I C14
=
(200)(0.026)
= 1.04 kΩ
5
1.04 + 1.8 || 278
⇒ 14.1 Ω
201
R0 = R6 + RE14 = 27 + 14.1 ⇒ R0 ≅ 41 Ω
RE14 =
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU13.9
⎛ 40 ⎞
2
2
I D1 = 20.2 = ⎜ ⎟(12.5)(V SG1 + VTP ) = 250(V SG1 − 0.5) ⇒ V SG1 = 0.7843 V
⎝ 2 ⎠
V SG 5 = 0.9022 V, ⇒ V SD 6 (sat ) = 0.9022 − 0.5 = 0.4022 V
Then
υ CM (+ ) = 5 − 0.7843 − 0.4022 = 3.81 V
⎛ 100 ⎞
2
I D 3 = 20.2 = ⎜
⎟(6.25)(VGS 3 − 0.5) ⇒ VGS 3 = 0.7542 V
⎝ 2 ⎠
(
)
V SD1 sat = 0.7843 − 0.5 = 0.2843 V
Now
υ CM (− ) = V − + VGS 3 + V SD1 (sat ) − V SG1 = −5 + 0.7542 + 0.2843 − 0.7843 = −4.75 V
Then
−4.75 ≤ υ CM ≤ 3.81 V
______________________________________________________________________________________
TYU13.10
VSG 8 = VSG 5 = 0.9022 V
VSD8 (sat ) = 0.9022 − 0.5 = 0.4022 V
υ O (max ) = 5 − 0.4022 = 4.6 V
⎛ k ' ⎞⎛ W ⎞
2
I D 7 = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 7 − VTN )
L
2
⎝ ⎠⎝ ⎠ 7
⎛ 100 ⎞
2
40.4 = ⎜
⎟(12.5)(VGS 7 − 0.5) ⇒ VGS 7 = 0.7542 V
⎝ 2 ⎠
V DS 7 (sat ) = 0.7542 − 0.5 = 0.2542 V
υ O (min ) = −5 + 0.2542 = −4.75 V
Then −4.75 ≤ υ O ≤ 4.6 V
______________________________________________________________________________________
TYU13.11
10 − V SG 5
100
2
25VSG 5 − 24VSG 5 − 3.75 = 0 ⇒ VSG 5 = 1.097 V
10 − 1.097
I set = I Q =
⇒ 89.03 μ A = I D 7 = I D 8
100
Then I D1 − I D 4 = 44.52 μ A
(a) 0.25(V SG 5 − 0.5) =
2
(b) K p1 = K p 2 = 0.25 mA/V 2
ro 2 = ro 4 =
1
= 1123 k Ω
(0.02 )(0.04452 )
Ad = 2 K p1 I Q (ro 2 ro 4 ) = 2(0.25)(0.08903)(1123 1123) = 118.5
⎛ 0 .1 ⎞
g m7 = 2 ⎜
⎟(12.5)(0.08903) = 0.4718 mA/V
⎝ 2 ⎠
1
ro8 = ro 7 =
= 561.6 k Ω
(0.02 )(0.08903)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Aυ 2 = (0.4718)(561.6 561.6) = 132.5
Then Aυ = (118.5)(132.5) = 15,701
______________________________________________________________________________________
TYU13.12
⎛ k'
(a) g m1 = 2 ⎜⎜ n
⎝ 2
⎞⎛ W ⎞
⎟⎜ ⎟ I DQ = 2
⎟⎝ L ⎠
1
⎠
1
ro 6 = ro8 =
= 166.7
(0.02)(0.3)
⎛ 0. 1 ⎞
⎟(40 )(0.1) = 0.8944 mA/V
⎜
⎝ 2 ⎠
kΩ
Ad = Bg m1 (ro 6 ro8 ) = 3(0.8944)(166.7 166.7 ) = 223.6
(b) Ro = ro 6 ro8 = 166.7 166.7 = 83.33 k Ω
f PD =
1
1
=
⇒ f PD = 955 kHz
2πRo (C L + C P ) 2π 83.33 × 10 3 2 × 10 −12
(
)
(
)(
)
GBW = (223.6) 955 × 10 ⇒ 213.5 MHz
______________________________________________________________________________________
3
TYU13.13
⎛ 0.1 ⎞
(a) g m1 = 2 ⎜
⎟(40 )(0.1) = 0.8944 mA/V
⎝ 2 ⎠
1
ro 6 = ro8 = ro10 = ro12 =
= 166.7 k Ω
(0.02 )(0.3)
⎛ k'
g m12 = 2 ⎜⎜ n
⎝ 2
⎞⎛ W ⎞
0. 1 ⎞
⎟⎜ ⎟ I D12 = 2 ⎛⎜
⎟(40 )(0.3) = 1.549 mA/V
⎟⎝ L ⎠
⎝ 2 ⎠
12
⎠
⎛ 0.04 ⎞
g m10 = 2 ⎜
⎟(40 )(0.3) = 0.9798 mA/V
⎝ 2 ⎠
Ro10 = g m10 (ro10 ro 6 ) = (0.9798)(166.7 )(166.7 ) = 27,228 k Ω
Ro12 = g m12 (ro12 ro8 ) = (1.549 )(166.7 )(166.7 ) = 43,045 k Ω
Ad = (3)(0.8944)(27228 43045) = 44,751
(b)
Ro = Ro10 Ro12 = 16,678 k Ω
f PD =
1
⇒ 4.77 kHz
2π 16,678 × 10 3 2 × 10 −12
(
(
)(
)
)
GBW = (44,751) 4.77 × 10 ⇒ 213.5 MHz
______________________________________________________________________________________
3
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU13.14
d
m1 ( o 6
o8 )
(a)
From Example 13.11,
A =g
R || R
g m1 = 316 μ A / V , Ro8 = 316 M Ω
Now
Ro 6 = g m 6 ( ro 6 )( ro 4 || ro1 )
ro1 = 1 M Ω, ro 4 = 0.5 M Ω
I
50
⇒ 1.923 mA / V
gm6 = C 6 =
0.026
VT
ro 6 =
VA6 80
=
= 1.6 M Ω
I C 6 50
Then
Ro 6 = (1.923)(1600)(0.5 || 1) = 1026 M Ω
Ad = (316)(1026 || 316) ⇒ Ad = 76,343
1
⇒ f PD = 329 Hz
2π (316 || 1026) × 106 × 2 × 10−12
f PD ⋅ Ad = (329)(76,343) ⇒ 25.1 MHz
f PD =
(b)
______________________________________________________________________________________
TYU13.15
For Q7 and R1
VSG = VBE 7 + I1 R1 = 0.6 + I1 (5)
For M 8 :
I 2 = K p (VSG + VTP ) 2
I 2 = 0.3(VSG − 1.4) 2
By trial and error:
VSG = 2.54 V
I1 = I 2 = 0.388 mA
______________________________________________________________________________________
TYU13.16
For J 6 biased in the saturation region
⇒ I C 3 = I DSS = 300 μ A
Q1 , Q2 , Q3 are matched
⇒ I C1 = I C 2 = I C 3 = 300 μ A
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 14
Exercise Solutions
EX14.1
−40
−40
=
= −39.9918
1
41
1+
(41) 1 +
AOL
2 × 10 5
−40
(b) ACL =
= −39.9672
41
1+
5 × 10 4
39.9672 − 39.9918
× 100% = −0.0615%
(c) Percent change =
39.9918
______________________________________________________________________________________
(a) ACL =
EX14.2
(a)
(
)
1
1
1 1 + 5 × 10 4
=
+ ⋅
= 0.025 + 625.0
(1)
Ri f
40 80
so that
R i f = 1.6 Ω
1⎞
⎛
4
⎜1 + 5 × 10 + ⎟
1
1
1 ⎝
1 5.00011 × 10 4
10 ⎠
= 0.025 +
⋅
=
+
⋅
(b)
1
1 ⎞
(1.1125)
Ri f
40 80 ⎛
80
1
+
+
⎟
⎜
10
80
⎠
⎝
so that Ri f = 1.78 Ω
(
)
______________________________________________________________________________________
EX14.3
⎛ 40 ⎞
40 (1 + 10 4 ) + 99 ⎜ 1 + ⎟
1 ⎠
⎝
Ri f =
99
1+
1
4 × 105 + 4.059 × 103
≅
100
Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ
______________________________________________________________________________________
EX14.4
1+
a.
b.
R2
= 100
R1
1
1 ⎡ 105 ⎤
=
⎢
⎥ = 10 ⇒ R0 f = 0.1 Ω
R0 f 100 ⎣100 ⎦
1
1 ⎡ 105 ⎤
2
= ⎢
⎥ = 10
R0 f 10 ⎣100 ⎦
R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX14.5
(a) f T = 2 × 10 5 (5) = (30) f 3−dB ⇒ f 3− dB = 33.3 kHz
(b) υ O = ACLO ⋅ υ I = (30 )(100 sin (2πf t )) μ V
or
υ O = 3 sin (2πf t ) mV
(c) (i) υ O , peak ≅ 3 mV
(
)
(ii) υ O =
(iii) υ O =
3
⎛ 50 ⎞
1+ ⎜
⎟
⎝ 33.3 ⎠
3
2
= 1.663 mV
= 0.493 mV
2
⎛ 200 ⎞
1+ ⎜
⎟
⎝ 33.3 ⎠
______________________________________________________________________________________
EX14.6
(a) υ O = (SR ) ⋅ t
(i) υ O = (1.25)(2 ) = 2.5 V
(ii) υ O = (1.25)(4) = 5 ⇒ υ O = 4 V
(iii) υ O = 4 V
(b) υ O = 4 = (1.25)(t ) ⇒ t = 3.2 μ s
______________________________________________________________________________________
EX14.7
(a)
f max =
0.63 × 10 6
SR
=
⇒ 401 kHz
2πV PO
2π (0.25)
(b)
f max =
0.63 × 10 6
⇒ 50.1 kHz
2π (2)
0.63 × 10 6
⇒ 12.5 kHz
2π (8)
______________________________________________________________________________________
f max =
(c)
EX14.8
⎛I
VOS = VT ln⎜⎜ S 2
⎝ I S1
⎛I
2
= ln⎜⎜ S 2
26
⎝ I S1
⎞
⎟⎟
⎠
⎞
⎟
⎟
⎠
Then
⎛ 2 ⎞
I S 2 = I S1 exp⎜ ⎟ = 2.16 × 10 −15 A
⎝ 26 ⎠
2.16 − 2
× 100% = 8%
Percent change =
2
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX14.9
We need
iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V
By Equation (14.60(a))
⎡
⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC1 = I S1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜1 + ⎟
⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎣
⎡
⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 3 ⎢exp ⎜ EB 3 ⎟ ⎥ ⎜1 +
⎟
50 ⎠
⎝ VT ⎠ ⎦ ⎝
⎣
By Equation (14.60(b))
⎡
⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟
⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎣
⎡
⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 +
⎟
50 ⎠
⎝ VT ⎠ ⎦ ⎝
⎣
I S 1 = I S 2, take the ratio:
⎛v −v ⎞ I
exp ⎜ BE1 BE 2 ⎟ = S 3
VT
⎝
⎠ IS 4
⎛I ⎞
= VT ln ⎜ S 3 ⎟
⎝ IS 4 ⎠
= 0.026 ⋅ ln (1.05 )
vBE1 − vBE 2 = V0 S
⇒ V0 S = 1.27 m V
______________________________________________________________________________________
EX14.10
VOS =
0.020 =
I Q ⎛ ΔK n ⎞
1
⋅
⋅⎜
⎟
2 2Kn ⎝ Kn ⎠
1
150 ⎛ ΔK n ⎞
⋅
⋅
2 2 ( 50 ) ⎜⎝ 50 ⎟⎠
⇒ ΔK n = 1.63μ A / V 2
⇒
ΔK n 1.63
=
⇒ 3.26%
50
Kn
______________________________________________________________________________________
EX14.11
⎛ R5 ⎞ +
⎜
⎟V = 5 m V
R +R
Want ⎝ 5 4 ⎠
R5 R4 so
R5 =
R5
⋅ V + = 0.005
R4
( 0.005)(100 )
= 0.05 kΩ
10
⇒ R5 = 50 Ω
______________________________________________________________________________________
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lOMoARcPSD|14951455
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX14.12
R1′ = 25 1 = 0.9615 kΩ
R2′ = 75 1 = 0.9868 kΩ
I = 100 μ A ⇒ i = iC 2 = 50 μ A
C1
For Q
From Equation (14.75)
⎛ 50 × 10−6 ⎞
⎟ + ( 0.050 )( 0.9615 )
−14
⎝ 10
⎠
( 0.026 ) ln ⎜
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 )
⎝ IS 4 ⎠
0.58065 + 0.048075
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934
⎝ IS 4 ⎠
⎛i ⎞
ln ⎜ C 2 ⎟ = 22.284
⎝ IS 4 ⎠
50 × 10−6
= 4.7625 × 109
IS 4
I S 4 ≅ 1.05 × 10−14 A
______________________________________________________________________________________
EX14.13
(a) (i) Yes
(ii) R3 = R1 R2 = 20 120 = 17.14 k Ω
(b) (i) Yes
⎛
R ⎞
(ii) υ O = 0 = I B1 R2 − I B 2 R3 ⎜⎜1 + 2 ⎟⎟
R1 ⎠
⎝
(0.75)(120) = (0.85)R3 ⎛⎜1 + 120 ⎞⎟ ⇒ R3 = 15.13 k Ω
20 ⎠
⎝
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU14.1
v1CM ( max ) = V + − VSD1 ( sat ) − VSG1
v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1
We have:
I REF = 100 μ A, kn′ = 80 μ A / V 2 , k ′p = 40 μ A / V 2 ,
⎛W ⎞
⎜ ⎟ = 25
⎝L⎠
For M 1 :
2
⎛ 40 ⎞
I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP )
⎝ 2 ⎠
50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V
2
So
VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V
Then
vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V
For M 4 :
2
⎛ 80 ⎞
I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN )
⎝ 2⎠
100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V
2
So
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184
So
V − − 0.184 ≤ vCM ≤ V + − 1.13 V
______________________________________________________________________________________
TYU14.2
vo ( max ) = V + − VSD 8 ( sat ) − VSG10
vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat )
Now
VSG 8 = VSG10 =
50
+ 0.5 = 0.816 V
( 40/ 2 )( 25 )
VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V
So o (
Also
v max ) = V + − 0.316 − 0.816 = V + − 1.13
VGS 6 =
VGS 4 =
50
(80/ 2 )( 25 )
+ 0.5 = 0.724 V
100
+ 0.5 = 0.816 V
(80/ 2 )( 25 )
VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
So
vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54
Then
V − + 0.54 ≤ vo ≤ V + − 1.13 V
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU14.3
R2
250
=−
= −10.0
R1
25
(a) −
ACL = −(1 − 0.001)(10.0 ) = −9.99
Now
−10
− 9.99 =
⇒ AOL = 10,989
11
1+
AOL
(b) ACL = −(1 − 0.0005)(10 ) = −9.995
We find
−10
− 9.995 =
⇒ AOL = 21,989
11
1+
AOL
______________________________________________________________________________________
TYU14.4
ACL =
a.
ACL ( ∞ )
⎡ A (∞) ⎤
1 + ⎢ CL
⎥
⎣ A0 L ⎦
ACL ( ∞ ) = 1 +
ACL =
b.
R2
495
= 1+
= 100
5
R1
100
⇒
100
1+ 5
10
ACL = 99.90
ACL ( ∞ ) = 100
dACL
100
= 10 × 5 = 0.01%
ACL
10
ACL = 99.90 − ( 0.0001)( 99.90 )
⇒ ACL = 99.89
______________________________________________________________________________________
TYU14.5
ACL
=
(a)
ACL (∞ )
1
ACL (∞ )
1+
AOL
1
⇒ ACL (∞ ) = 20.02
0.999 =
A (∞ )
1 + CL 4
2 × 10
1
(b) 0.9995 =
⇒ ACL (∞ ) = 10.005
A (∞ )
1 + CL 4
2 × 10
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU14.6
a.
ii ⎛ Rif ⎞
=⎜
⎟
I1 ⎝ Ri ⎠
iI 0.1
=
= 1× 10 −5
i1 104
iI
10
= 4 = 1× 10 −3
i1 10
b.
______________________________________________________________________________________
TYU14.7
Voltage follower R2 = 0, R1 = ∞
Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 )
≅ 5 ×106 kΩ ⇒ Rif = 5000 MΩ
______________________________________________________________________________________
TYU14.8
(a) (i) f T = 5 × 10 4 (15) = (25) f 3−dB
or f 3− dB = f max = 30 kHz
(
(ii) V PO =
(
)
SR
0.8 × 10 6
=
= 4.24 V
2πf max 2π 30 × 10 3
)
(
)
(b) (i) f T = 5 × 10 (10) = (25) f 3− dB ⇒ f 3− dB = 200 kHz
5
SR
0.8 × 10 6
=
= 0.637 V
2πf max 2π 200 × 10 3
______________________________________________________________________________________
(ii) V PO =
(
)
TYU14.9
v0 = I B1 R3 = (10−6 )( 200 × 103 )
a.
⇒ v0 = 0.20 V
R4 = R1 R2 R3 = 100 50 200
⇒ R = 28.6 kΩ
4
b.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 15
Exercise Solutions
EX15.1
1
1
1
⇒ RC =
=
= 6.366 × 10 −6
2πRC
2πf 3− dB 2π 25 × 10 3
Set C3 = 50 pF = 1.414C ⇒ C = 35.36 pF
C 4 = (0.707 )C = 25 pF
f 3− dB =
(
)
6.366 × 10 −6
6.366 × 10 −6
=
⇒ R = 180 k Ω
C
35.36 × 10 −12
______________________________________________________________________________________
Then R =
EX15.2
1
1
=
⇒ Req = 8.33 M Ω
f c C 100 × 10 3 1.2 × 10 −12
1
1
(b) C =
=
⇒ C = 0.4 pF
3
f c Req
50 × 10 50 × 10 6
______________________________________________________________________________________
(a) Req =
(
)(
(
)(
)
)
EX15.3
T =−
C1
30
=−
= −6
5
C2
Low-frequency gain:
(100 ×103 )( 5 ×10−12 ) ⇒ f = 6.63 kHz
f C
f 3dB = C 2 =
3 dB
2π CF
2π (12 × 10 −12 )
______________________________________________________________________________________
EX15.4
fo =
1
2π 3 RC
Set R = 10 k Ω
⇒ RC =
(
1
2π 3 22.5 × 10 3
)
= 4.084 × 10 −6
4.084 × 10 −6
⇒ C = 408 pF
10 × 10 3
R2 = 8 R = 80 k Ω
______________________________________________________________________________________
C=
EX15.5
f0 =
C=
1
1
⇒C =
2π RC
2π f 0 R
1
2π ( 800 ) (104 )
⇒ C ≅ 0.02 μ F
R2 = 2 R1 = 2 (10 ) ⇒ R2 = 20 kΩ
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX15.6
⎛ R1
VTH = ⎜⎜
⎝ R1 + R2
Set R1 = 10 k Ω
⎞
⎟⎟ ⋅ V H
⎠
⎛ 10 ⎞
⎟⎟(9) ⇒ R2 = 170 k Ω
Then 0.5 = ⎜⎜
⎝ 10 + R2 ⎠
______________________________________________________________________________________
EX15.7
⎛ R1 ⎞
VTH − VTL = ⎜
⎟ (VH − VL )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
0.10 = ⎜
⎟ (10 − [ −10])
⎝ R1 + R2 ⎠
R
R
20
= 200 ⇒ 2 = 199
1+ 2 =
R1 0.10
R1
⎛ R2 ⎞
VS = ⎜
⎟ VREF
⎝ R1 + R2 ⎠
⎛
R ⎞
1 ⎞
⎛
VREF = ⎜1 + 1 ⎟ VS = ⎜1 +
⎟ (1) ⇒ VREF = 1.005 V
⎝ 199 ⎠
⎝ R2 ⎠
I=
VH − VBE ( on ) − Vγ
R + 0.1
10 − 0.7 − 0.7
= 43 kΩ
R + 0.1 =
0.2
R = 42.9 kΩ
______________________________________________________________________________________
EX15.8
At t = 0 − let υ O = −5 V so υ X = −2.5 V: For t > 0
⎛ −t ⎞
⎟⎟
⎝τ X ⎠
= 5 V, output switches
υ X = 10 + (− 2.5 − 10 ) exp⎜⎜
When υ X
⎛ −t
5 = 10 − 12.5 exp⎜⎜ 1
⎝τX
⎛ −t
exp⎜⎜ 1
⎝τX
⎞
⎟⎟
⎠
⎞ 10 − 5
5
⎟⎟ =
=
⎠ 12.5 12.5
⎛ + t ⎞ 12.5
⎛ 12.5 ⎞
⇒ t1 = τ X ln⎜
exp⎜⎜ 1 ⎟⎟ =
⎟ ⇒ t1 = τ X (0.916 )
τ
5
⎝ 5 ⎠
⎝ X ⎠
During the next part of the cycle
⎛ −t ⎞
υ X = −5 + [5 − (− 5)] exp⎜⎜ ⎟⎟
⎝τ X ⎠
When υ X = −2.5 V, output switches
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ −t
− 2.5 = −5 + 10 exp⎜⎜ 2
⎝τX
⎞
⎟⎟
⎠
⎛ −t
exp⎜⎜ 2
⎝τX
⎞ 5 − 2.5 2.5
⎟⎟ =
=
10
10
⎠
⎛ +t
exp⎜⎜ 2
⎝ τX
⎞ 10
⎛ 10 ⎞
⎟⎟ =
⇒ t 2 = τ X ln⎜
⎟ ⇒ t 2 = τ X (1.39 )
⎝ 2.5 ⎠
⎠ 2.5
Period = t1 + t 2 = T = (0.916 + 1.39)τ X = 2.31τ X , Frequency = f =
τ X = (50 ×10 3 )(0.01×10 −6 ) = 5 ×10 −4 s ⇒ f = 866 Hz
Duty cycle =
1
2.31τ X
t1
0.916
× 100% =
× 100% ⇒ Duty cycle = 39.7%
t1 + t 2
0.916 + 1.39
______________________________________________________________________________________
EX15.9
a.
τ X = RX C X
⎛
⎞
⎞
⎛
1
⎟⎟ ⋅υ O = ⎜
υ Y = ⎜⎜
⎟(12 ) = 1.2 V
⎝ 10 + 90 ⎠
⎝ R1 + R 2 ⎠
⎛
R
R
10
⎞
1
⎟⎟ = 0.10
β = ⎜⎜
R
+
⎝ 1 R2 ⎠
⎛ 1 + Vγ V P
T = τ X ln⎜⎜
⎝ 1− β
⎞
⎛ 1 + (0.7 ) (12 ) ⎞
⎟ = τ X ln⎜
⎟
⎟
⎝ 1 − 0.10 ⎠
⎠
T = 50 ×10 −6 = τ X ln(1.18) = τ X (0.162)
RX =
50 × 10 −6
⇒ R X = 3.09 k Ω
0.1× 10 − 6 (0.162 )
(
)
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
b. Recovery time
⎛ −t ⎞
⎟⎟
⎝τ X ⎠
υ X = V P + (− 1.2 − V P ) exp⎜⎜
When υ X = Vγ t = t 2
⎛ −t
0.7 = 12 + (− 1.2 − 12) exp⎜⎜ 2
⎝τX
⎞
⎟⎟
⎠
⎛ −t
exp⎜⎜ 2
⎝τX
t2 = τ X
⎞ 12 − 0.7
⎟⎟ =
= 0.856
13.2
⎠
⎛ 1 ⎞
ln⎜
⎟ = τ X (0.155)
⎝ 0.856 ⎠
τ X = (3.09 ×10 3 )(0.1×10 −6 ) = 3.09 ×10 −4 s, ⇒ t 2 = 48.0 μ s
______________________________________________________________________________________
EX15.10
(a) T = 1.1RC = (1.1) 20 × 10 3 0.012 × 10 −6 ⇒ T = 0.264 ms
(
(b) RC =
120 × 10
T
=
1. 1
1. 1
)(
−6
)
= 1.09 × 10 − 4
1.09 × 10 −4
⇒ R = 10.9 k Ω
0.01 × 10 −6
______________________________________________________________________________________
If C = 0.01 μ F, then R =
EX15.11
1
1
=
⇒ f = 802 Hz
0.693 ( RA + 2 RB ) C ( 0.693) ⎡⎣ 20 + 2 ( 80 ) ⎤⎦ × 103 × ( 0.01× 10−6 )
R + RB
20 + 80
× 100% =
× 100% ⇒ Duty cycle = 55.6%
Duty cycle = A
RA + 2 RB
20 + 2 ( 80 )
f =
______________________________________________________________________________________
EX15.12
P=
1 VP2
⋅
2 RL
VP = 2 RL P = 2 ( 8 )(1) ⇒ VP = 4 V
IP =
a.
b.
VP 4
= ⇒ I P = 0.5 A
RL 8
VCE = 12 − 4 = 8 V
I C ≈ 0.5 A
P = I C ⋅ VCE = ( 0.5)( 8) ⇒ P = 4 W
So
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX15.13
(a) (i) VP = 2 RL PL = 2(20)(5) = 14.14 V
IP =
(ii) V S =
V P 14.14
=
= 0.707 A
RL
20
πR L PS
, We have PS = 5 W
VP
π (20)(5)
= 22.2 V
VS =
14.14
(b) (i) V P = 2(8)(10 ) = 12.65 V
12.65
= 1.58 A
8
π (8)(10)
= 19.9 V
(ii) VS =
12.65
______________________________________________________________________________________
IP =
EX15.14
regulation =
Line
Now
dV0
dV dV
= 0⋅ Z
dV + dVZ dV +
dV0 ⎛ 10 ⎞
= ⎜1 + ⎟ = 2
dVZ ⎝ 10 ⎠
dVZ ⎛ rZ ⎞
10
=⎜
= 0.00227
⎟=
dV + ⎝ rZ + R1 ⎠ 10 + 4400
regulation = ( 2 )( 0.00227 ) = 0.00454
So Line
0.454%
______________________________________________________________________________________
EX15.15
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
V1 V0 − V1
⎛1 1⎞ V
=
⇒ V1 ⎜ + ⎟ = 0
10
10
⎝ 10 10 ⎠ 10
V
⎛ 2⎞ V
V1 ⎜ ⎟ = 0 ⇒ V0 = 2V1 ⇒ V1 = 0
2
⎝ 10 ⎠ 10
V0 − V1 V0 V0 − A0 L (VZ − V1 )
+
+
=0
10
RL
R0
V0 V0 V0 A0 LVZ V1 A0 LV1
+
+
−
= −
10 RL R0
10
R0
R0
=
V0
A V
− 0L 0
2(10) 2 R0
V0
V 1000 ( 6.3) V0 (1000 ) V0
+ I0 + 0 −
=
−
10
0.5
0.5
20
2 ( 0.5 )
V0 [0.10 + 2.0 − 0.05 + 1000] + I 0 = 12, 600
V0 (1002.05) + I 0 = 12, 600
For I 0 = 1 mA ⇒ V0 = 12.5732
For I 0 = 100 mA ⇒ V0 = 12.4744
V ( NL ) − V0 ( FL )
Load reg = 0
× 100%
V0 ( NL )
12.5732 − 12.4744
× 100%
12.5732
Load reg = 0.786%
=
______________________________________________________________________________________
EX15.16
a.
IC 3 =
IC 3 =
VZ − 3VBE ( on )
R1 + R2 + R3
5.6 − 3 ( 0.6 )
3.9 + 3.4 + 0.576
⎛I ⎞
I C 4 R4 = VT ln ⎜ C 3 ⎟
⎝ IC 4 ⎠
=
3.8
⇒ I C 3 = 0.482 mA
7.88
⎛ 0.482 ⎞
I C 4 (0.1) = (0.026) ln ⎜
⎟
⎝ IC 4 ⎠
By trial and error
I C 4 = 0.213 mA
VB 7 = 2(0.6) + (0.482)(3.9) ⇒ VB 7 = 3.08 V
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
b.
⎛ R13 ⎞
⎜
⎟ V0 = VB 8 = VB 7
⎝ R13 + R12 ⎠
⎛ 2.23 ⎞
⎜
⎟ (5) = 3.08
⎝ 2.23 + R12 ⎠
( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08 ) R12
11.15 = 6.868 = 3.08 R12 ⇒ R12 = 1.39 kΩ
______________________________________________________________________________________
Test Your Understanding Solutions
TYU15.1
1
1
1
⇒ RC =
=
= 7.958 × 10 − 4
2πRC
2πf 3− dB 2π (200)
For example, let C = 0.01 μ F
(a)
f 3− dB =
7.958 × 10 −4
⇒ R = 79.58 k Ω
0.01 × 10 − 6
79.58
R1 =
= 22.44 k Ω
3.546
79.58
R2 =
= 57.17 k Ω
1.392
79.58
R3 =
= 393.2 k Ω
0.2024
1
= 0.124 ⇒ −18.1 dB
(b) (i) T =
6
⎛ 200 ⎞
1+ ⎜
⎟
⎝ 100 ⎠
1
= 0.959 ⇒ −0.365 dB
(ii) T =
6
⎛ 200 ⎞
1+ ⎜
⎟
⎝ 300 ⎠
______________________________________________________________________________________
Then R =
TYU15.2
1
1
=
= 5.305 × 10 − 6
2πf 3− dB 2π 30 × 10 3
For example, let R = 100 k Ω
(a) RC =
(
)
5.305 × 10 −6
⇒ C = 53.05 pF
100 × 10 3
C1 = 1.082C = 57.4 pF
C 2 = 0.9241C = 49.02 pF
C 3 = 2.613C = 138.6 pF
C 4 = 0.3825C = 20.29 pF
Then C =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
8
⎛ f ⎞
⇒ ⎜ ⎟ = 0.020304
8
⎝ 30 ⎠
⎛ f ⎞
1+ ⎜ ⎟
⎝ 30 ⎠
which yields f = 18.43 kHz
______________________________________________________________________________________
(b)
(0.99 ) =
1
TYU15.3
1-pole
2-pole
3-pole
4-pole
1
T =
T =
T =
T =
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
2
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
4
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
6
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
8
⇒ −3.87 dB
⇒ −4.88 dB
⇒ −6.0 dB
⇒ −7.24 dB
______________________________________________________________________________________
TYU15.4
f cC =
1
1
=
= 4 × 10 −8
Req 25 × 10 6
For example, let f c = 50 kHz,
4 × 10 −8
⇒ C = 0.8 pF
50 × 10 3
______________________________________________________________________________________
C=
TYU15.5
fo =
1
⇒C =
1
⇒ C = 217 pF
2π 6 RC
2π 6 15 × 10 3 20 × 10 3
R2 = 29 R = (29 )(15) = 435 k Ω
______________________________________________________________________________________
(
)(
)
TYU15.6
f0 =
1
⎛ CC ⎞
2π L ⋅ ⎜ 1 2 ⎟
⎝ C1 + C2 ⎠
=
1
⎡ (10−9 ) 2
2π (10 ) ⎢
−9
⎣ 2 × 10
−6
⎤
⋅⎥
⎦
⇒ f 0 = 7.12 MHz
C2
= gm R
C1
gm =
C2 1
1
⋅ =
⇒ g m = 0.25 mA / V
C1 R 4 × 103
We have
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜ ⎟⎜ ⎟ (VGS − VTh )
⎝ 2 ⎠⎝ L ⎠
k ′ ≅ 20 μ A / V 2 , VGS − VTh ≅ 1 V
So
0.25 × 10 −3
W
=
= 12.5
L ( 20 × 10−6 ) (1)
and a value of W / L = 12.5 is certainly reasonable.
______________________________________________________________________________________
TYU15.7
⎛R ⎞
VTH = −⎜⎜ 1 ⎟⎟ ⋅ V L
⎝ R2 ⎠
⎛R ⎞
0.2 = −⎜⎜ 1 ⎟⎟(− 12)
⎝ R2 ⎠
Set R2 = 200 k Ω
R1
0.2
=
⇒ R1 = 3.33 k Ω
200 12
______________________________________________________________________________________
Then
TYU15.8
a.
VS =
⎛ R2 ⎞
⎛ 10 ⎞
⎜
⎟ VREF = ⎜
⎟ ( 2)
+
R
R
⎝ 1 + 10 ⎠
⎝ 1
2 ⎠
VS = 1.82 V
⎛ R1 ⎞
⎛ 1 ⎞
VTH = VS + ⎜
⎟ VH = 1.82 + ⎜
⎟ (10 )
⎝ 1 + 10 ⎠
⎝ R1 + R2 ⎠
VTH = 2.73 V
VTL =
⎛ R1 ⎞
⎛ 1 ⎞
VS + ⎜
⎟ VL = 1.82 + ⎜
⎟ ( −10 )
R
R
+
⎝ 1 + 10 ⎠
2 ⎠
⎝ 1
VTL = 0.91 V
b.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU15.9
VTH − VTL =
0.5 =
R1
(VH − VL )
R2
⎛
R1
[9 − (− 9)] = 18⎜⎜ R1
R2
⎝ R2
⎞
⎟⎟
⎠
Set R1 = 10 k Ω ,
(18)(10) = 360 k Ω
Then R2 =
0.5
⎛
R ⎞
Now VS = ⎜⎜1 + 1 ⎟⎟ ⋅ VREF
R2 ⎠
⎝
10 ⎞
⎛
− 2 = ⎜1 +
⎟ ⋅ VREF ⇒ VREF = −1.946 V
⎝ 360 ⎠
______________________________________________________________________________________
TYU15.10
1
1
=
= 454.5 Hz
3
2.2 R X C X (2.2) 20 × 10 0.05 × 10 −6
50% duty cycle
1
1
=
⇒ R X = 7.576 k Ω
(b) R X =
3
(2.2) f C X (2.2) 1.2 ×10 0.05 ×10 −6
______________________________________________________________________________________
(a)
f =
(
)(
(
)
)(
)
TYU15.11
β=
R1
20
=
= 0.333
R1 + R 2 20 + 40
τ X = R X C X = (10 4 )(0.01×10 −6 ) = 1×10 −4 s
⎛ 1 + Vγ V P ⎞
⎛ 1 + (0.7 ) 8 ⎞
⎟ = 10 − 4 ln⎜
T = τ X ln⎜⎜
⎟ ⇒ T = 48.9 μ s
⎟
1
−
β
⎝ 1 − 0.333 ⎠
⎠
⎝
Recovery time
⎛ R1 ⎞
⎛ 20 ⎞
⎟⎟ ⋅υ O = ⎜
υ Y = ⎜⎜
⎟(8) = 2.667 V
+
R
R
⎝ 20 + 40 ⎠
2 ⎠
⎝ 1
(
⎛ −t
0.7 = 8 + (− 2.667 − 8) exp⎜⎜ 2
⎝τX
)
⎞
⎟⎟
⎠
⎛ −t
exp⎜⎜ 2
⎝τX
⎞ 8 − 0.7
⎟⎟ =
= 0.6844
⎠ 10.66
⎛ 1 ⎞
t 2 = τ X ln⎜
⎟ ⇒ t 2 = 37.9 μ s
⎝ 0.6844 ⎠
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU15.12
f =
1
(0.693)(R A + 2RB )C
⇒ R A + 2 RB =
1
=
1
(0.693) fC (0.693)(103 )(0.01× 10 −6 )
R A + 2 RB = 1.443 × 10 5
Duty cycle = 55% =
(R A + RB ) × 100%
(R A + 2 R B )
1.443 × 10 5 − RB
1.443 × 10 5
RB = 1.443 × 10 5 (1 − 0.55) ⇒ RB = 64.9 k Ω and R A = 14.43 k Ω
______________________________________________________________________________________
0.55 =
(
)
TYU15.13
⎛
R ⎞ ⎛ 40 ⎞
(a) (i) For A1 : Aυ1 = ⎜⎜1 + 2 ⎟⎟ = ⎜1 + ⎟ = 3
R1 ⎠ ⎝ 20 ⎠
⎝
R
60
= −3
For A2 : Aυ 2 = − 4 = −
R3
20
(ii) V L ( peak ) = 12 − (− 12) = 24 V
24
I L ( peak ) =
= 48 mA
0.5
⎛ 24 ⎞⎛ 0.048 ⎞
⎟⎟ = 0.576 W
⎟⎟⎜⎜
PL (avg ) = ⎜⎜
⎝ 2 ⎠⎝ 2 ⎠
12
=4 V
(iii) υ O1 ( peak ) = 12 ⇒ υ I ( peak ) =
3
R ⎞
⎛
(b) Need Aυ1 = 6 = ⎜1 + 2 ⎟ ⇒ R2 = 100 k Ω
20 ⎠
⎝
R4
⇒ R4 = 120 k Ω
20
______________________________________________________________________________________
Aυ 2 = −6 = −
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 16
Exercise Solutions
EX16.1
⎛ k ' ⎞⎛ W ⎞
(a) υ O = V DD − i D R D = VDD − ⎜⎜ n ⎟⎟⎜ ⎟ R D 2(υ I − VTN )υ O − υ O2
⎝ 2 ⎠⎝ L ⎠
⎛ 0.1 ⎞
2
0.1 = 3 − ⎜
⎟(4)RD 2(3 − 0.5)(0.1) − (0.1) ⇒ R D = 29.6 k Ω
⎝ 2 ⎠
3 − 0.1
= 0.0980 mA
(b) i D,max =
29.6
PD ,max = (0.098 )(3) = 0.294 mW
[
[
]
]
(c) From Equation (16.9),
⎛ 0.1 ⎞
2
⎜
⎟(4)(29.6)VO t + VO t − 3 = 0
⎝ 2 ⎠
5.92VO2t + VO t − 3 = 0 ⇒ VO t = 0.632 V
VO t = V I t − VTN ⇒ V I t = 1.132 V
______________________________________________________________________________________
EX16.2
(a) (i) υ O = V DD − VTNL = 3 − 0.4 = 2.6 V
(ii) From Equation (16.21),
⎛ 0.1 ⎞
⎛ 0.1 ⎞
2
2
⎟(2)(3 − υ O − 0.4)
⎜
⎟(16) 2(2.6 − 0.4)υ O − υ O = ⎜
⎝ 2 ⎠
⎝ 2 ⎠
which yields
9υ O2 − 40.4υ O + 6.76 = 0 ⇒ υ O = 0.174 V
[
]
⎛ 0.1 ⎞
2
(b) i D ,max = ⎜
⎟(2)(3 − 0.174 − 0.4) = 0.589 mA
2
⎝
⎠
PD , max = (0.589 )(3) = 1.766 mW
(c) V I t =
VO t
(
3 − 0.4 + 0.4 1 + 8
) = 1.08 V
1+ 8
= 1.08 − 0.4 = 0.68 V
______________________________________________________________________________________
EX16.3
(a) From Equation (16.27(b)),
6
2
2(3 − 0.4)υ O − υ O2 = [− (− 0.8)]
2
or 3υ O2 − 15.6υ O + 0.64 = 0 ⇒ υ O = 0.0414 V
[
]
⎛ 0.1 ⎞
2
(b) i D ,max = ⎜
⎟(2)[− (− 0.8)] = 0.064 mA
⎝ 2 ⎠
PD ,mas = (0.064 )(3) = 0.192 mW
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(
)
6
V I t − 0.4 = −(− 0.8) ⇒ V I t = 0.862 V
2
Driver: V I t = 0.862 V, VO t = 0.462 V
(c)
Load:
V I t = 0.862 V, VO t = 3 − 0.8 = 2.2 V
______________________________________________________________________________________
EX16.4
VOH = 2.5 − 0.5 + (0.3) 0.73 + VOH − 0.73
{
[
]}
VOH − 2.2563 = −(0.3) 0.73 + VOH
2
VOH
− 4.5126VOH + 5.09098 = (0.09)(0.73 + VOH )
2
VOH
− 4.60264VOH + 5.02528 = 0 ⇒ VOH = 1.781 V
______________________________________________________________________________________
EX16.5
(a) (i)
[
]
[
]
KD
2
2(υ I − VTND )υ O − υ O2 = (− VTNL )
KL
⎛5⎞
2
2
⎜ ⎟ 2(1.8 − 0.4)υ O − υ O = [− (− 0.6)]
⎝1⎠
5υ O2 − 14υ O + 0.36 = 0 ⇒ υ O = 26 mV
[
]
⎛K ⎞
2
(ii) 2⎜⎜ D ⎟⎟ 2(υ I − VTND )υ O − υ O2 = (− VTNL )
K
L
⎝
⎠
⎛5⎞
2
2⎜ ⎟ 2(1.8 − 0.4)υ O − υ O2 = [− (− 0.6)]
⎝1⎠
[
]
10υ O2 − 28υ O + 0.36 = 0 ⇒ υ O = 12.9 mV
⎛ 100 ⎞
2
2
(b) i D ,max = K L (− VTNL ) = ⎜
⎟(1)[− (− 0.6)] = 18 μ A
2
⎝
⎠
P = i D ,max ⋅ V DD = (18 )(1.8) = 32.4 μ W
______________________________________________________________________________________
EX16.6
⎛ K ⎞
2
(a) ⎜⎜ D ⎟⎟ 2(υ I − VTND )υ O − υ O2 = (− VTNL )
⎝ 3K L ⎠
12
2
2(2.5 − 0.4 )υ O − υ O2 = [− (− 0.6 )]
(3)(1)
[
[
]
]
4υ O2 − 16.8υ O + 0.36 = 0 ⇒ υ O = 21.5 mV
4
2
(b)
2(2.5 − 0.4 )υ O − υ O2 = [− (− 0.6 )]
(3)(1)
[
]
1.333υ O2 − 5.6υ O + 0.36 = 0 ⇒ υ O = 65.3 mV
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX16.7
VDD 2.1
=
= 1.05 V
2
2
= VIt − VTD = 1.05 − (−0.4) = 1.45 V
VIt =
VOPt
VONt = VIt − VTN = 1.05 − 0.4 = 0.65 V
(a)
VIt =
VOPt
2.1 + (−0.4) + 0.5(0.4)
1 + 0.5
= 1.16 + 0.4 = 1.56 V
= 1.16 V
VONt = 1.16 − 0.4 = 0.76 V
(b)
VIt =
VOPt
VONt
2.1 + (−0.4) + 2(0.4)
= 0.938 V
1+ 2
= 0.938 + 0.4 = 1.338 V
= 0.538 V
(c)
______________________________________________________________________________________
EX16.8
2
P = f ⋅ CL ⋅ VDD
( 0.10 ×10 ) = f ( 0.5 ×10 ) ( 3)
−6
−12
2
f = 2.22 × 104 Hz ⇒ f = 22.2 kHz
______________________________________________________________________________________
EX16.9
V DD + VTP +
(a) V I t =
1+
Kn
⋅ VTN
Kp
1.8 − 0.4 + (0.4 )
=
Kn
Kp
1+
200
80
200
80
⇒ V I t = 0.7874 V
VOP t = V I t − VTP = 0.7874 + 0.4 = 1.187 V
VON t = V I t − VTN = 0.7874 − 0.4 = 0.3874 V
(b)
Kn
200
=
= 2 .5
80
Kp
VI L = 0.4 +
(1.8 − 0.4 − 0.4) ⎡2
(2.5 − 1) ⎢⎣⎢
V I H = 0. 4 +
(1.8 − 0.4 − 0.4) ⎡ 2(2.5) − 1⎤ ⇒ V
⎥
⎢
IH
(2.5 − 1) ⎣⎢ 3(2.5) + 1 ⎦⎥
⎤
2.5
− 1⎥ ⇒ VI L = 0.6323 V
2.5 + 3 ⎦⎥
= 0.8767 V
1
{(1 + 2.5)(0.6323) + 1.8 − (2.5)(0.4) + 0.4} ⇒ VOHU = 1.7065 V
2
(0.8767 )(1 + 2.5) − 1.8 − (2.5)(0.4) + 0.4 ⇒ V = 0.1337 V
VOLU =
OLU
2(2.5)
(c) NM L = 0.6323 − 0.1337 = 0.4986 V
NM H = 1.7065 − 0.8767 = 0.8298 V
______________________________________________________________________________________
VOHU =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX16.10
3 PMOS in series and 3 NMOS in parallel.
Worst Case: Only one NMOS is ON in Pull-down mode ⇒ same as the CMOS inverter ⇒ Wn = W .
⇒ W = 3(2W ) = 6W .
p
All 3 PMOS are on during pull-up mode
______________________________________________________________________________________
EX16.11
NMOS: Worst Case, M NA , M NB on, Wn = 2(W ) or M NC , M ND or M NC , M NE on ⇒ Wn = 2(W ).
PMOS: M PA and M PC on or M PA and M PB on ⇒ WP = 2(2W ) = 4W
If M PD and M PE on, need WP = 2(4W ) = 8W
______________________________________________________________________________________
EX16.12
(a) υ O = φ − VTN = 2.5 − 0.4 = 2.1 V
(b) For υ DS = 0 , υ O = 1.8 V
(c) υ O = φ − VTN = 2.5 − 0.4 = 2.1 V
(d) υ O = φ − VTN = 1.5 − 0.4 = 1.1 V
______________________________________________________________________________________
EX16.13
(a) υ I′ = φ − VTN = 3.3 − 0.5 = 2.8 V
[
]
KD
2
2(υ I′ − VTN )υ O − υ O2 = [V DD − υ O − VTN ]
KL
[
]
[
]
KD
2
2
2(2.8 − 0.5)(0.1) − (0.1) = [3.3 − 0.1 − 0.5]
KL
which yields
KD
= 16.2
KL
(b) υ I′ = φ − VTN = 2.8 − 0.5 = 2.3 V
KD
2
2
2(2.3 − 0.5)(0.1) − (0.1) = [3.3 − 0.1 − 0.5]
KL
which yields
KD
= 20.8
KL
______________________________________________________________________________________
EX16.14
16 K ⇒ 16384 cells
Total Power = 125 mW = (2.5) IT ⇒ IT = 50 mA
I=
Then, for each cell,
I≅
50 mA
⇒ I = 3.05 μ A
16384
V
2.5
VDD
R = DD =
⇒ R = 0.82 M Ω
I
3.05
R or
Now,
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU16.1
PD ,max = i D ,max ⋅ V DD
0.50 = i D , max (1.8) ⇒ i D , max = 0.2778 mA
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
i D ,max = 0.2778 = ⎜
⎟⎜ ⎟ (1.8 − 0.12 − 0.4 ) ⇒ ⎜ ⎟ = 3.39
⎝ 2 ⎠⎝ L ⎠ L
⎝ L ⎠L
KD
K
2
2
2(1.4 − 0.4 )(0.12 ) − (0.12 ) = (1.8 − 0.12 − 0.4 ) ⇒ D = 7.26
KL
KL
[
]
(W L )D (W L )D ⎛ W ⎞
KD
= 7.26 =
=
⇒ ⎜ ⎟ = 24.6
KL
3.39
(W L )L
⎝ L ⎠D
______________________________________________________________________________________
TYU16.2
PD ,max = i D ,max ⋅ V DD
0.2 = i D ,max (1.8) ⇒ i D ,max = 0.111 mA
⎛ 0.1 ⎞⎛ W ⎞
⎛W ⎞
2
i D ,max = 0.111 = ⎜
⎟⎜ ⎟ [− (− 0.6 )] ⇒ ⎜ ⎟ = 6.17
⎝ 2 ⎠⎝ L ⎠ L
⎝ L ⎠L
(W L )D (W L )D
KD
[− (− 0.6)]2
= 1.654 =
=
=
2
(W L )L
6.17
K L 2(1.8 − 0.4)(0.08) − (0.08)
⎛W ⎞
so ⎜ ⎟ = 10.2
⎝ L ⎠D
______________________________________________________________________________________
TYU16.3
(a) I = (0.098)(100,000) mA, ⇒ I = 9.8 A
P = (0.294 )(100,000 ) mW, ⇒ P = 29.4 W
(b) I = (0.589)(100,000) mA, ⇒ I = 58.9 A
P = (1.766 )(100,000) mW, ⇒ P = 176.6 W
(c) I = (0.064)(100,000) mA, ⇒ I = 6.4 A
P = (0.194 )(100,000 ) mW, ⇒ P = 19.2 W
______________________________________________________________________________________
TYU16.4
(a) P = i D ⋅ V DD
50 = i D (2.5) ⇒ i D = 20 μ A
⎛ 100 ⎞⎛ W ⎞
⎛W ⎞
2
i D = 20 = ⎜
⎟⎜ ⎟ [− (− 0.6 )] ⇒ ⎜ ⎟ = 1.11
⎝ 2 ⎠⎝ L ⎠ L
⎝ L ⎠L
KD
K
2
2
2(2.5 − 0.4)(0.05) − (0.05) = [− (− 0.6 )] ⇒ D = 1.735
KL
KL
[
]
(W L )D (W L )D ⎛ W ⎞
KD
= 1.735 =
=
⇒ ⎜ ⎟ = 1.93
KL
1.11
(W L )L
⎝ L ⎠D
[
]
2
= [− (− 0.6)]
(b) 3(1.735) 2(2.5 − 0.4)VOL − VOL
2
2
5.205VOL
− 21.861VOL + 0.36 = 0 ⇒ VOL = 16.5 mV
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU16.5
1 (W L )D
⎛W ⎞
2
2
⋅
(a)
2(2.1 − 0.4)(0.08) − (0.08) = (2.5 − 0.08 − 0.4) ⇒ ⎜ ⎟ = 15.4
2 (W L )L
⎝ L ⎠D
⎛ 100 ⎞
2
(b) i D ,max = ⎜
⎟(0.5)(2.5 − 0.08 − 0.4) = 102 μ A
⎝ 2 ⎠
P = i D ,max ⋅ V DD = (102 )(2.5) = 255 μ W
[
]
______________________________________________________________________________________
TYU16.6
1 (W L )D
⎛W ⎞
2
2
2(2.5 − 0.4)(0.08) − (0.08) = [− (− 0.6 )] ⇒ ⎜ ⎟ = 1.09
(a)
2 (W L )L
⎝ L ⎠D
⎛ 100 ⎞
2
(b) i D ,max = ⎜
⎟(0.5)[− (− 0.6 )] = 9 μ A
2
⎠
⎝
P = i D , max ⋅ V DD = (9 )(2.5) = 22.5 μ W
[
]
______________________________________________________________________________________
TYU16.7
a.
K n = K p = 50 μ A / V 2
VIt = 2.5 V
iD (max) = K n (VIt − VTN )2 = 50(2.5 − 0.8) 2 ⇒ iD (max) = 145 μ A
b.
K n = K p = 200 μ A / V 2
VIt = 2.5 V
iD (max) = (200)(2.5 − 0.8) 2 ⇒ iD (max) = 578 μ A
______________________________________________________________________________________
TYU16.8
5 + (− 2) + 0.8
= 1.9 V
1+1
= V I t − VTP = 1.9 − (− 2 ) = 3.9 V
(a) VI t =
VOP t
VON t = V I t − VTN = 1.9 − 0.8 = 1.1 V
3
[5 + (− 2) − 0.8] = 1.625 V
8
5
VI H = 0.8 + [5 + (− 2 ) − 0.8] = 2.175 V
8
1
VOLU = [2(2.175) − 5 − 0.8 − (− 2)] = 0.275 V
2
1
VOHU = [2(1.625) + 5 − 0.8 − (− 2)] = 4.725 V
2
(c) NM L = 1.625 − 0.275 = 1.35 V
NM H = 4.725 − 2.175 = 2.55 V
______________________________________________________________________________________
(b) VI L = 0.8 +
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU16.9
______________________________________________________________________________________
TYU16.10
NMOS − 2 transistors in series
Wn = 2 (W ) = 2W
PMOS − 2 transistors in series
W p = 2 ( 2W ) = 4W
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU16.11
The NMOS part of the circuit is:
______________________________________________________________________________________
TYU16.12
The NMOS part of the circuit is:
______________________________________________________________________________________
TYU16.13
Insert Figure X-TYU16.13
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 16
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU16.14
For NMOS, φ − υ O ≥ 0.4 V or φ − υ I ≥ 0.4 V
At υ I = 2.1 V, 2.1 = 2.5 − 0.2t ⇒ t = 2 s
NMOS conducting for 2 ≤ t ≤ 12.5 s
For PMOS, υ I − 0 ≥ 0.4 V
At υ I = 0.4 V, 0.4 = 2.5 − 0.2t ⇒ t = 10.5 s
PMOS conducting for 0 ≤ t ≤ 10.5 s
______________________________________________________________________________________
TYU16.15
1 K ⇒ 32 × 32 array
(a)
Each row and column requires a 5-bit word ⇒ 6 transistors per row and column, ⇒ 32 × 6 + 32 × 6 = 384
transistors plus buffer transistors.
4 K ⇒ 64 × 64 array
(b)
Each row and column requires a 6-bit word ⇒ 7 transistors per row and column ⇒ 64 × 7 + 64 × 7 = 896
transistors plus buffer transistors.
16 K ⇒ 128 ×128 array
(c)
Each row and column requires a 7-bit word ⇒ 8 transistors per row and column
⇒ 128 × 8 + 128 × 8 = 2048 transistors plus buffer transistors.
______________________________________________________________________________________
TYU16.16
From Equation (16.82)
(W / L )nA
(W / L )n1
=
2 (VDDVTN ) − 3VTN2
(VDD − 2VTN )
2
=
2(2.5)(0.4) − 3(0.4) 2
( 2.5 − 2(0.4) )
2
= 0.526
From Equation (16.84)
(W / L ) p
(W / L )nB
=
2
⎡ 2(2.5)(0.4) − 3(0.4) 2 ⎤
kn′ 2 (VDDVTN ) − 3VTN
⋅
= (2.5) ⎢
⎥ = 0.862
2
k ′p
(2.5 − 0.4) 2
(VDD + VTP )
⎣
⎦
⎛W ⎞
⎛W ⎞
⎜ ⎟
⎜ ⎟
L
So ⎝ ⎠ of transmission gate device must be < 0.526 times the ⎝ L ⎠ of the NMOS transistors in the
⎛W ⎞
⎛W ⎞
⎜ ⎟
⎜ ⎟
inverter cell. The ⎝ L ⎠ of the PMOS transistors must be < 0.862 times the ⎝ L ⎠ of the transmission
⎛W ⎞
⎜ ⎟
gate devices. Then the ⎝ L ⎠ of the PMOS devices must be < 0.453 times
⎛W ⎞
⎜ ⎟
⎝ L ⎠ of NMOS devices in cell.
______________________________________________________________________________________
TYU16.17
Initial voltage across the storage capacitor = VDD − VTN = 3 − 0.5 = 2.5 V .
Now
−I = C
dV
I
V = − ⋅t + K
dt or
C
2.5
V=
= 1.25 V ,
K
=
V
t
=
1.5
ms
,
2.5
,
2
where
and C = 0.05 pF . Then
I (1.5 × 10−3 )
1.25 = 2.5 −
⇒
(0.05 × 10−12 )
I = 4.17 × 10−11 A ⇒ I = 41.7 pA
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 17
Exercise Solutions
EX17.1
(a) i E =
− 0.7 − (− 1.8)
= 0.11 ⇒ RE = 10 k Ω
RE
iE
= 0.055 mA
2
1.8 − 1.45
RC =
= 6.364 k Ω
0.055
(b) (i) υ1 = 0.5 V, υ E = 0.5 − 0.7 = −0.2 V
− 0.2 − (− 1.8)
= 0.16 mA
iE =
10
υ O1 = 1.8 − (0.16 )(6.364) = 0.782 V
iC1 = iC 2 =
υ O 2 = 1.8 V
(ii) υ1 = −0.5 V, i E = 0.11 mA
υ O1 = 1.8 V
υ O 2 = 1.8 − (0.11)(6.364) = 1.1 V
(c) (i) i E = 0.16 mA, P = (0.16 )[1.8 − (− 1.8)] = 0.576 mW
(ii) i E = 0.11 mA, P = (0.11)[1.8 − (− 1.8)] = 0.396 mW
______________________________________________________________________________________
EX17.2
P(iCXY + iCR + i3 + i4 )(5.2)
v X = vY = logic 1 ⇒ iCXY = 3.22 mA
a.
iCR = 0
−0.7 + 5.2
= 3 mA
i3 =
1.5
−1.4 + 5.2
= 2.53 mA
i4 =
1.5
P = (3.22 + 0 + 3 + 2.53)(5.2) ⇒ P = 45.5 mW
v X = vY = logic 0 ⇒ iCXY = 0
iCR = 2.92 mA
i3 = 2.53 mA
i4 = 3 mA
P = (0 + 2.92 + 2.53 + 3)(5.2) ⇒ P = 43.9 mW
b.
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
EX17.3
υ B 5 = −1 + 0.7 = −0.3 V
0.3
R1 =
= 0.6 k Ω
0.5
− 0.3 − 1.4 − (− 3.3)
R2 =
= 3.2 k Ω
0.5
− 1 − (− 3.3)
R5 =
= 4.6 k Ω
0.5
______________________________________________________________________________________
EX17.4
I MAX = 1 =
−0.7 − (−5.2)
⇒ R3 = R4 = 4.5 K
R3
−0.7 − 0.7 − (−5.2)
= 3.22 mA
1.18
i5 = i1 = 1.40 mA
iCXY =
i3 = 1.0 mA
−1.4 − (−5.2)
= 0.844 mA
4.5
P = (3.22 + 1.4 + 1.4 + 1.0 + 0.844)(5.2) = 40.8 mW
i4 =
______________________________________________________________________________________
EX17.5
⎛ v − 0.7 − (−5.2) ⎞
iL = ⎜ oR
⎟ (10)
(1.18)(51)
⎝
⎠
voR − (−5.2)
i3 =
1.5
voR + 5.2 ⎛ voR + 4.5 ⎞
⎡ 0 − (VoR + 0.7) ⎤
⎢
⎥ (51) = 1.5 + ⎜ (1.18)(51) ⎟ (10)
0.24
⎣
⎦
⎝
⎠
⎡⎛ 51 ⎞ 1
1 − (6) ⎤ (0.7)(51) 5.2
4.5(10)
−voR ⎢⎜
+
+
+
⎟+
⎥=
0.24
1.5 (1.18)(51)
⎣⎝ 0.24 ⎠ 1.5 (1.18)(51) ⎦
−voR [212.50 + 0.6667 + 0.166168] = 148.75 + 3.4666 + 0.747757
−voR [213.3328] = 152.9644
voR = −0.7170
______________________________________________________________________________________
EX17.6
(100)(0.026)
= 2.6 K
1
1
g m3 =
= 38.46 mA / V
0.026
Vn
Vn
=
Ib3 =
RC 2 + rπ 3 + (1 + β ) R3 0.24 + 2.6 + (101)(4.5)
rπ 3 =
Ib3 =
Vn
457.34
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
⎛ Vn ⎞
VO = − I b 3 ( RC 2 + rπ 3 ) = − ⎜
⎟ (0.24 + 2.6)
⎝ 457.34 ⎠
VO = −0.00621 Vn
⎛ Vn ⎞
VO′ = (1 + β ) I b3 R3 = (101) ⎜
⎟ (4.5)
⎝ 457.34 ⎠
VO′ = 0.9938 Vn
______________________________________________________________________________________
EX17.7
P = I Q ⋅ VCC ⇒ 0.2 = I Q (1.7) ⇒ I Q = 117.6 μ A
QR on ⇒ v0 = 1.7 − I Q RC = 1.7 − 0.4 ⇒ RC =
VR =
0.4
⇒ RC = 3.40 kΩ
0.1176
1.7 + 1.3
⇒ VR = 1.5 V
2
______________________________________________________________________________________
EX17.8
(a)
v X = vY = 5 V
v1 = VBE ( sat ) + 2VY = 0.8 + 2(0.7) = 2.2 V
5 − 2.2
i1 =
= 0.70 mA
4
V − VCE ( sat ) 5 − 0.1
iRC = CC
=
= 1.225 mA
4
RC
P = (i1 + iRC )VCC = (0.70 + 1.225)(5)
or P = 9.625 mW
v X = vY = 0 ⇒ v1 = 0.70 V
V − v 5 − 0.70
i1 = CC 1 =
= 1.075 mA
R1
4
(b) P = i1 ⋅ VCC = (1.075)(5) ⇒ P = 5.375 mW
______________________________________________________________________________________
EX17.9
(a) υ X = υY = 0.1 V
υ B1 = 0.1 + 0.8 = 0.9 V
5 − 0.9
i1 = i B1 =
= 0.342 mA
12
iC1 ≅ 0, i B 2 = iC 2 = 0 , i Bo = iCo = 0
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b) υ X = υ Y = 5 V
υ B1 = 0.8 + 0.8 + 0.7 = 2.3 V
5 − 2.3
i1 = i B1 =
= 0.225 mA
12
i B 2 = iC1 = (1 + 0.2)i B1 = 0.27 mA
υ C 2 = 0.8 + 0.1 = 0.9 V
5 − 0.9
i 2 = iC 2 =
= 1.025 mA
4
i E 2 = i B 2 + iC 2 = 0.27 + 1.025 = 1.295 mA
0.8
0.8
i Bo = i E 2 −
= 1.295 −
= 0.895 mA
RB
2
5 − 0.1
i3 = iCo =
= 0.8167 mA
6
____________________________________________________________________________________
EX17.10
(a) υ X = υ Y = 3.6 V
5 − (0.8 + 0.8 + 0.7 )
i1 =
= 0.225 mA
12
i B 2 = (1 + 0.2 )(0.225) = 0.27 mA
5 − (0.8 + 0.1)
iC 2 =
= 1.025 mA
4
0.8
i Bo = 0.27 + 1.025 −
= 0.895 mA
2
5 − (0.1 + 0.8)
= 0.3417 mA
i L′ =
12
i L (max ) = β i Bo = N i L′
(25)(0.895) = N (0.3417 ) ⇒ N = 65
(b) i L (max ) = 12 = N i L′ = N (0.3417 ) ⇒ N = 35
______________________________________________________________________________________
EX17.11
5 − 0.4
= 2.044 mA
2.25
2 + 2.044
iC′ =
= 3.791 mA
1
1+
15
′
i
3.791
= 0.253 mA
i B′ = C =
15
β
i D = i B − i B′ = 2 − 0.253 = 1.747 mA
(a) iC =
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b) iC = 2.044 + 10 = 12.044 mA
2 + 12.044
= 13.166 mA
1
1+
15
13.166
i B′ =
= 0.878 mA
15
i D = 2 − 0.878 = 1.122 mA
iC′ =
(c)
i D = 0 , i B′ = 2 mA, iC′ = (2)(15) = 30 mA
i L = 30 − 2.044 ≅ 28 mA
______________________________________________________________________________________
EX17.12
v1 = 0.4 + 0.3 = 0.7,
i1 =
5 − 0.7
= 0.1075 mA
40
(a)
All transistor currents are zero.
P = (0.1075)(5 − 0.4) ⇒ 495 μ W
(b)
v1 = 1.4 V,
i1 = iB 2 =
5 − 1.1
= 0.325 mA
12
= 0.09 + 0.325 = 0.415 mA
vC 2 = 0.7 + 0.4 = 1.1 V,
iB 0 ≈ iB 2 + iC 2
5 − 1.4
= 0.090 mA
40
i2 = iC 2 =
iC 0 ≈ 0
P = (i1 + i2 )(5) = (0.09 + 0.325)(5) = 2.08 mW
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
TYU17.1
0.75 − 0.7 − (− 1.8)
⇒ RE = 2.31 k Ω
RE
1.8 − 1.1
RC 2 =
= 0.875 k Ω
0.8
1.1 − 0.7 − (− 1.8)
= 0.951 mA
(b) i E =
2.3125
1.8 − 1.1
RC1 =
= 0.736 k Ω
0.95135
______________________________________________________________________________________
(a) i E = 0.8 =
TYU17.2
logic 1 = −0.7 V
Q1 and Q2 on when v X = vY = −0.7 V
−0.7 − 0.7 − (−5.2)
= 2.5 ⇒ RE = 1.52 kΩ
iE =
RE
vNOR = −1.5 ⇒ RC1 =
0 − (−1.5 + 0.7)
⇒ RC1 = 320 Ω
2.5
−1.5 − 0.7
⇒ VR = −1.1 V
2
−1.1 − 0.7 − (−5.2)
= 2.237 mA
QR on ⇒ iE =
1.52
0 − (−1.5 + 0.7)
⇒ RC 2 = 358 Ω
RC 2 =
2.237
−0.7 − (−5.2)
⇒ R3 = R4 = 1.8 kΩ
R3 = R4 =
2.5
VR =
______________________________________________________________________________________
TYU17.3
State
1
2
3
4
5
6
7
8
A
0
1
0
0
1
1
0
1
B
0
0
1
0
1
0
1
1
C
0
0
0
1
0
1
1
1
Q01
Q02
Q03
Q1
Q2
QR
v0
off
“on”
off
off
on
on
off
on
off
off
on
off
on
off
on
on
off
off
off
on
off
on
on
on
off
off
off
on
off
on
on
on
on
on
on
off
on
off
off
off
on
on
“off”
on
off
“off”
on
off
0
0
1
0
1
1
0
1
( A AND C ) OR ( B AND C )
14243
14243
true for
true for
states 6 and 8 states 3 and 5
144444244444
3
Output goes high for these 4 states
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU17.4
A
0
1
0
0
1
1
0
1
B
0
0
1
0
1
0
1
1
C
0
0
0
1
0
1
1
1
v0
0
1
1
1
0
0
0
1
⇒ ( A ⊕ B) ⊕ C
______________________________________________________________________________________
TYU17.5
5 − (0.1 + 0.7 )
= 0.28 mA
15
i2 = i R = i B = i RC = 0
υO = 5 V
(b) Same as part (a).
5 − (0.8 + 0.7 + 0.7 )
= 0.1867 mA
(c) i2 = i1 =
15
0.8
iR =
= 0.0533 mA
15
i B = 0.1867 − 0.0533 = 0.1334 mA
5 − 0.1
i RC =
= 0.8167 mA
6
υ O = 0.1 V
______________________________________________________________________________________
(a) i1 =
TYU17.6
(a) i B = 0.1134 mA, i RC = 0.8167 mA
5 − (0.1 + 0.7 )
= 0.28 mA
15
+ N i L′ = β i B
i L′ =
i RC
0.8167 + N (0.28) = (30)(0.1134 ) ⇒ N = 9
(b) iC ,max = β i B = (30 )(0.1134 ) = 3.4 < 12 mA
⇒ N =9
______________________________________________________________________________________
TYU17.7
From EX17.9, i Bo = 0.895 mA, i3 = 0.8167 mA
υ O = 0.1 V, so υ B′ 1 = 0.1 + 0.8 = 0.9 V
5 − 0.9
= 0.3417 mA
12
i Co , max = β i Bo = i 3 + N i L′
i L′ =
(25)(0.895) = 0.8167 + N (0.3417 )
N = 63.1 ⇒ N = 63
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
______________________________________________________________________________________
TYU17.8
Q1 in saturation
5 − 0.9
iB1 =
⇒ iB1 = 0.683 mA
6
iC1 = iB 2 = iC 2 = 0
iB 0 = iC 0 = 0
vB 4 = 0.1 + 0.7 = 0.8 V
0.1
iB 4 =
⇒ iB 4 = 1.19 μ A
(21)(4)
iC 4 = 23.8 μ A
iB 3 = iC 3 = 0
______________________________________________________________________________________
TYU17.9
(a)
v X = vY = 0.4, vB1 = 0.4 + 0.7 = 1.1 V
5 − 1.1
= 1.393 mA
2.8
P = i1 (5 − 0.4) = (1.393)(5 − 0.4) = 6.41 mW
i1 =
(b)
v X = vY = 3.6 V
5 − 2.1
= 1.036 mA
2.8
5 − 1.1
i =
= 5.132 mA
vC 2 = 0.7 + 0.4 = 1.1 V, 2
0.76
0.4
i =
= 0.1143 mA
vE 4 = 1.1 − 0.7 = 0.4 V, R 4 3.5
⎛ β ⎞
⎛ 25 ⎞
iR 3 = ⎜
⎟ iR 4 = ⎜ ⎟ (0.1143) = 0.1099 ≈ 0.11 mA
1
β
+
⎝ 26 ⎠
⎝
⎠
vB1 = 2.1,
i1 =
P = (i1 + i2 + iR 3 )(5) = (1.036 + 5.132 + 0.11)(5)
or
P = 31.4 mW
______________________________________________________________________________________
TYU17.10
(a) υ X = 0.4 V, υ E1 = 0.4 + 0.7 = 1.1 V
5 − 1.1
i R1 =
⇒ 97.5 μ A
40
Q2 cutoff, i R 2 = 0
(b) υ X = 3.6 V, υ E1 = 3(0.7 ) = 2.1 V
5 − 2.1
i R1 =
⇒ 72.5 μ A
40
υ C 2 = 2(0.7 ) + 0.4 = 1.8 V
5 − 1.8
iR 2 =
⇒ 64 μ A
50
______________________________________________________________________________________
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Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 17
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU17.11
(a) υ X = 0.4 V, υ E1 = 0.4 + 0.7 = 1.1 V
3.5 − 1.1
i R1 =
⇒ 60 μ A
40
i R2 = 0
(b) υ X = 2.1 V, υ E1 = 2.1 V
3 . 5 − 2. 1
⇒ 35 μ A
40
υ C 2 = 1.8 V
3.5 − 1.8
iR2 =
⇒ 34 μ A
50
______________________________________________________________________________________
i R1 =
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