Ch10s

Chapter 10
Problem Solutions

10.1
0 − 2Vγ − V −
a. I1 = I 2 =
R1 + R2
2Vγ + I 2 R2 = VBE + I C R3

2Vγ +
R2
R1 + R2
( −2Vγ − V − ) = VBE + IC R3
1⎧
⎪ − ⎛ R2 ⎞ ⎫
⎪
IC = ⎨2Vγ − ( 2Vγ + V ) ⎜ ⎟ − VBE ⎬
⎪
R3
⎩ ⎝ R1 + R2 ⎠ ⎪
⎭
b. Vγ = VBE and R1 = R2
1 ⎧ 1 ⎫
⎨2Vγ − ( 2Vγ + V ) − VBE ⎬
−
IC =
R3 ⎩ 2 ⎭
−V −
or I C =
2 R3
− ( −10 )
c. I C = 2 mA = ⇒ R3 = 2.5 kΩ
2 R3
−2 ( 0.7 ) − ( −10 )
I1 = I 2 = 2 mA = ⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ
R1 + R2

10.2
⎛I ⎞
(a) VBE1 = VT ln ⎜ C1 ⎟
⎝ IS ⎠
⎛ 10 × 10−6 ⎞
(i) I REF = I C1 = 10 μ A, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.5388 V
⎝ 10 ⎠
I O = 10 μ A
⎛ 100 × 10−6 ⎞
(ii) I REF = I C1 = 100 μ A, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.5987 V
⎝ 10 ⎠
I O = 100 μ A
⎛ 10−3 ⎞
(iii) I REF = I C1 = 1 mA, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.6585 V
⎝ 10 ⎠
I O = 1 mA
I REF
(b) IO =
2
1+
β
10 ⎛I ⎞
(i) IO = ⇒ I O = 9.615 μ A VBE1 = VBE 2 = VT ln ⎜ O ⎟
2 ⎝ IS ⎠
1+
50
⎛ 9.615 × 10−6 ⎞
= ( 0.026 ) ln ⎜ −14 ⎟
⎝ 10 ⎠
= 0.5378 V
100 ⎛ 96.15 × 10−6 ⎞
(ii) IO = ⇒ I O = 96.15 μ A VBE1 = ( 0.026 ) ln ⎜ ⎟
2 ⎝ 10
−14
⎠
1+
50
= 0.5977 V

1 ⎛ 0.9615 × 10−3 ⎞
(iii) IO = ⇒ I O = 0.9615 mA VBE1 = ( 0.026 ) ln ⎜ ⎟
2 ⎝ 10−14 ⎠
1+
50
= 0.6575 V

10.3
V + − VBE ( on ) − V − 3 − 0.7 − ( −3)
I REF = ⇒ 0.250 =
R1 R1
R1 = 21.2 K
I REF 0.250
I C1 = I C 2 = = ⇒ I C1 = I C 2 = 0.2419 mA
2 2
1+ 1+
β 60
I B1 = I B 2 = 4.03 μ A

10.4
V + − VBE ( on ) − V − 5 − 0.7 − ( −5 )
I REF = =
R1 18.3
I REF = 0.5082 mA
I REF 0.5082
I C1 = I C 2 = = ⇒ I C1 = I C 2 = 0.4958 mA
2 2
1+ 1+
β 80
I B1 = I B 2 = ( 6.198 μ A )

10.5
V + − VBE ( on ) − V − 15 − 0.7 − ( −15 )
(a) I REF = or R1 = ⇒ R1 = 58.6 k Ω
R1 0.5
V + − VBE ( on ) − V − 0 − 0.7 − ( −15 )
(b) R1 = = ⇒ R1 = 28.6 k Ω
I REF 0.5
Advantage: Requires smaller resistance.
(c) For part (a):
29.3
I O ( max ) = = 0.526 mA
( 58.6 )( 0.95 )
29.3
I O ( min ) = = 0.476 mA
( 58.6 )(1.05 )
ΔI O = 0.526 − 0.476 = 0.05 mA ⇒ ±5%
For part (b):
14.3
I O ( max ) = = 0.526 mA
( 28.6 )( 0.95)
14.3
I O ( min ) = = 0.476 mA
( 28.6 )(1.05 )
ΔI O = 0.05 mA ⇒ ±5%

10.6
⎛ 2⎞ ⎛ 2 ⎞
a. I REF = I 0 ⎜ 1 + ⎟ = 2 ⎜ 1 + ⎟ or I REF = 2.04 mA
⎝ β⎠ ⎝ 100 ⎠
15 − 0.7
R1 = ⇒ R1 = 7.01 kΩ
2.04

VA 80
b. r0 = = = 40 kΩ
I0 2
ΔI 0 1 ⎛ 1 ⎞
= ⇒ ΔI 0 = ⎜ ⎟ ( 9.3) = 0.2325 mA
ΔVCE r0 ⎝ 40 ⎠
ΔI 0 0.2325 ΔI
= ⇒ 0 = 11.6%
I0 2 I0

10.7
I 0 = nI C1
I C1 I0
I REF = I C1 + I B1 + I B 2 = I C1 + +
β β
⎛ 1 n⎞ ⎛ 1+ n ⎞
I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 + ⎟
⎝ β β⎠ ⎝ β ⎠
I0 ⎛ 1 + n ⎞ nI REF
= ⎜1 + ⎟ or I 0 =
n⎝ β ⎠ ⎛ 1+ n ⎞
⎜1 +
⎝ β ⎟⎠

10.8
I REF ⎛ 2 ⎞
IO = ⇒ I REF = ( 0.20 ) ⎜ 1 + ⎟ = 0.210 mA
2 ⎝ 40 ⎠
1+
β
5 − 0.7 4.3
R1 = = ⇒ R1 = 20.5 K
I REF 0.21

10.9
5 − 0.7
a. I REF = = 0.239 mA
18
0.239
I0 = ⇒ I 0 = 0.230 mA
2
1+
50
VA 50
b. r0 = = = 218 kΩ
I 0 0.230
1 ⎛ 1 ⎞
ΔI 0 = ⋅ ΔVEC = ⎜ ⎟ (1.3) = 0.00597 mA ⇒ I 0 = 0.236 mA
r0 ⎝ 217 ⎠
⎛ 1 ⎞
c. ΔI 0 = ⎜ ⎟ ( 3.3) = 0.01516 mA ⇒ I 0 = 0.245 mA
⎝ 217 ⎠

10.10
5 − 0.7 − ( −5 )
a. I REF = 1 = ⇒ R1 = 9.3 kΩ
R1
b. I 0 = 2 I REF ⇒ I 0 = 2 mA
5 − 0.7
c. For VEC 2 ( min ) = 0.7 ⇒ RC 2 = ⇒ RC 2 = 2.15 kΩ
2

10.11

I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA
⎛ 3⎞ ⎛ 3 ⎞
I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟
⎝ β⎠ ⎝ 60 ⎠
I REF = 0.2625 mA
2.5 − 0.7
R1 = ⇒ R1 = 6.86 K
0.2625

10.12

10 − 0.7
R1 = = 37.2 K
0.25

10.13
I 2 = 2 I1 and I3 = 3I1
(a) I 2 = 1.0 mA, I 3 = 1.5 mA
(b) I1 = 0.25 mA, I 3 = 0.75 mA
(c) I1 = 0.167 mA, I 2 = 0.333 mA

10.14
a.

IE3
I 0 = I C1 and I REF = I C1 + I B 3 = I C1 +
1+ β
VBE 2 I C1 VBE
I E 3 = I B1 + I B 2 + = +
R2 β R2
2 I C1 VBE
I REF = I C1 + +
β (1 + β ) (1 + β ) R2
VBE ⎛ 2 ⎞
I REF − = I 0 ⎜1 +
⎜ β (1 + β ) ⎟
⎟
(1 + β ) R2 ⎝ ⎠
VBE
I REF −
(1 + β ) R2
I0 =
⎛ 2 ⎞
⎜1 +
⎜ β (1 + β ) ⎟
⎟
⎝ ⎠
⎛ 2 ⎞ 0.7
b. I REF = ( 0.70 ) ⎜ 1 +
⎜ ( 80 )( 81) ⎟ + ( 81)(10 )
⎟
⎝ ⎠
I REF = 0.700216 + 0.000864
10 − 2 ( 0.7 )
I REF = 0.7011 mA = ⇒ R1 = 12.27 kΩ
R1

10.15
a.
I ES
I 0i = I CR and I REF = I CR + I BS = I CR +
1+ β
I ES = I BR + I B1 + I B 2 + ... + I BN = (1 + N ) I BR
(1 + N ) I CR
=
β
(1 + N ) I CR
Then I REF = I CR +
β (1 + β )
I REF
or I 0i =
⎛ (1 + N ) ⎞
⎜1 + β (1 + β ) ⎟
⎜ ⎟
⎝ ⎠
⎡ 6 ⎤
b. I REF = ( 0.5 ) ⎢1 + ⎥ = 0.5012 mA
⎢ ( 50 )( 51) ⎥
⎣ ⎦
5 − 2 ( 0.7 ) − ( −5 )
R1 = ⇒ R1 = 17.16 kΩ
0.5012

10.16

⎛ 2 ⎞ ⎡ 2 ⎤
⎜ β (1 + β ) ⎟ = ( 0.5 ) ⎢1 + ( 50 )( 51) ⎥ ⇒ I REF = 0.5004 mA
I REF = I 0 ⎜ 1 + ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦
5 − 2 ( 0.7 ) − ( −5 )
R1 = ⇒ R1 = 17.19 kΩ
0.5004

10.17

1
I 0 = I REF ⋅
⎛ 2 ⎞
⎜1 +
⎜ β (2 + β ) ⎟
⎟
⎝ ⎠
For I 0 = 0.8 mA
⎛ 2 ⎞
I REF = ( 0.8 ) ⎜ 1 +
⎜ 25 ( 27 ) ⎟ ⇒ I REF = 0.8024 mA
⎟
⎝ ⎠
18 − 2 ( 0.7 )
R1 = ⇒ R1 = 20.69 kΩ
0.8024

10.18

The analysis is exactly the same as in the text. We have
1
I 0 = I REF ⋅
⎛ 2 ⎞
⎜1 +
⎜ β (2 + β ) ⎟
⎟
⎝ ⎠

10.19
2
I 0 = 2 mA, I B 2 = = 0.0267 mA
75
1
I C1 = 1 mA, I B1 = = 0.0133 mA
75
I E 3 = I B1 + I B 2 = 0.0133 + 0.0267 = 0.04 mA
I E3 0.04
I B3 = = = 0.000526 mA
1+ β 76
I REF = I C1 + I B 3 ⇒ I REF = 1.000526 ≈ 1 mA
10 − 2 ( 0.7 ) 8.6
R1 = = ⇒ R1 = 8.6 kΩ
I REF 1

10.20
(a)
β ro3
Assuming RO ≈
2
VA V 100
rO 3 = = A = = 400 K
I O I REF 0.25
(100 )( 400 )
RO = ⇒ RO = 20 MΩ
2
(b)
ΔV ΔV 5
RO = ⇒ ΔI O = =
ΔI O 20 MΩ 20 MΩ
ΔI O = 0.25 μ A

10.21
V + − VBE1 − V − 5 − 0.7
I REF = =
R1 9.3
I REF = 0.4624 mA
VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞
IO = ln ⎜ ⎟= ln ⎜ ⎟
RE ⎝ I O ⎠ 1.5 ⎝ IO ⎠
⎛ 0.4624 ⎞
I O = 0.01733ln ⎜ ⎟
⎝ IO ⎠
By trial and error
I O 41.7 μ A
VBE 2 = 0.7 − I O RE
VBE 2 = 0.7 − ( 0.0417 )(1.5 )
VBE 2 = 0.6375 V

10.22
(a)

V + − VBE1 − V − 5 − 0.7 − ( −5 )
I REF = = ⇒ I REF = 93 μ A
R1 100
⎛I ⎞ ⎛ 93 × 10−3 mA ⎞
I O RE = VT ln ⎜ REF ⎟ ⇒ I O (10 ) = 0.026 ln ⎜ ⎟
⎝ IO ⎠ ⎝ IO ⎠
By trial and error, I O ≅ 6.8 μ A
Ro = ro 2 (1 + g m 2 RE )
′
Now
30
ro 2 = = 4.41 M Ω
6.8
0.0068
gm2 = = 0.2615 mA / V
0.026
(100 )( 0.026 )
rπ 2 = = 382.4 k Ω
0.0068
So
′
RE = rπ 2 || RE = 382 || 10 = 9.74 k Ω
Then
Ro = 4.41 ⎡1 + ( 0.262 )( 9.74 ) ⎤ ⇒ Ro = 15.6 M Ω
⎣ ⎦
(b) VBE1 − VBE 2 = I o RE = ( 0.0068 )(10 ) ⇒ VBE1 − VBE 2 = 0.068 V

10.23
I
ΔI 0 = ⋅ ΔVC
R0
R0 = r02 (1 + g m 2 RE )
′
V 80
r02 = A = = 11.76 MΩ
I 0 6.8
I 0 0.0068
gm2 = = = 0.2615 mA/V
VT 0.026
(80 )( 0.026 )
rπ 2 = = 306 kΩ
0.0068
′
RE = RE rπ 2 = 10 306 = 9.68 K
R0 = (11.76 ) ⎡1 + ( 0.2615 )( 9.68 ) ⎤ = 41.54 MΩ
⎣ ⎦
Now
⎛ 1 ⎞
ΔI 0 = ⎜ ⎟ ( 5 ) ⇒ ΔI 0 = 0.120 μ A
⎝ 41.54 ⎠

10.24
5 − 0.7 − ( −5 )
(a) I REF = = 0.50
R1
R1 = 18.6 K
⎛I ⎞
I O RE = VT ln ⎜ REF ⎟
⎝ IO ⎠
0.026 ⎛ 0.50 ⎞
RE = ln ⎜ ⎟
0.050 ⎝ 0.050 ⎠
RE = 1.20 K

(b) RO = rc 2 [1 + RE g m 2 ]
′
′
RE = RE rπ 2
( 75 )( 0.026 ) 0.050
rπ 2 = = 39 K gm2 = = 1.923 mA/V
0.050 0.026
VA 100
ro 2 = = ⇒ 2 MΩ ′
RE = 1.20 39 = 1.164 K
I O 0.05
RO = 2 ⎡1 + (1.164 )(1.923) ⎤ ⇒ RO = ( 6.477 ) MΩ
⎣ ⎦
ΔV 5
(c) ΔI O = = = 0.772 μ A
RO 6.477
ΔI O 0.772
× 100% = × 100 = 1.54%
IO 50

10.25
Let R1 = 5 k Ω, Then
12 − 0.7 − ( −12 )
I REF = ⇒ I REF = 4.66 mA
5
Now
⎛I ⎞ 0.026 ⎛ 4.66 ⎞
I O RE = VT ln ⎜ REF ⎟ ⇒ RE = ln ⎜ ⎟ ⇒ RE ≅ 1 k Ω
⎝ IO ⎠ 0.10 ⎝ 0.10 ⎠

10.26
⎛I ⎞
VBE = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A
−15

⎝ IS ⎠
⎛ 2 × 10−3 ⎞
At 2 mA, VBE = ( 0.026 ) ln ⎜ −15 ⎟
⎝ 2.03 × 10 ⎠
= 0.718 V
15 − 0.718
R1 = ⇒ R1 = 7.14 kΩ
2
V ⎛ I ⎞ 0.026 ⎛ 2 ⎞
RE = T ln ⎜ REF ⎟ = ⋅ ln ⎜ ⎟ ⇒ RE = 1.92 kΩ
I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠

10.27
a.
10 − 0.7
I REF ≈ = 0.465 mA
20
Let V − = 0
⎛I ⎞
VBE ≅ VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 ×10 A
−15

⎝ IS ⎠
Then
⎛ 0.465 × 10−3 ⎞
VBE ≅ ( 0.026 ) ln ⎜ −15 ⎟
= 0.680 V
⎝ 2.03 × 10 ⎠
Then

10 − 0.680
I REF ≅ ⇒ I REF = 0.466 mA
20
VT ⎛ I REF ⎞ 0.026 ⎛ 0.466 ⎞
b. RE = ln ⎜ ⎟= ⋅ ln ⎜ ⎟ ⇒ RE = 400Ω
I0 ⎝ I0 ⎠ 0.10 ⎝ 0.10 ⎠

10.28
10 − 0.7 − ( −10 )
I REF ≈ = 0.4825 mA
40
⎛I ⎞
VBE ≅ VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛ 10−3 ⎞
0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A
−15

⎝ IS ⎠
Now
⎛ 0.4825 × 10−3 ⎞
VBE = ( 0.026 ) ln ⎜ −15 ⎟
= 0.681 V
⎝ 2.03 × 10 ⎠
VBE1 = 0.681 V
So
10 − 0.681 − ( −10 )
I REF ≅ ⇒ I REF = 0.483 mA
40
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⎛ 0.483 ⎞
I 0 (12 ) = ( 0.026 ) ln ⎜ ⎟
⎝ I0 ⎠
By trial and error.
⇒ I 0 ≅ 8.7 μ A
VBE 2 = VBE1 − I 0 RE = 0.681 − ( 0.0087 )(12 ) ⇒ VBE 2 = 0.5766 V

10.29
VBE1 + I REF RE1 = VBE 2 + I 0 RE 2
VBE1 − VBE 2 = I 0 RE 2 − I REF RE1
For matched transistors
⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛I ⎞
VBE 2 = VT ln ⎜ 0 ⎟
⎝ IS ⎠
⎛I ⎞
Then VT ln ⎜ REF ⎟ = I 0 RE 2 − I REF RE1
⎝ I0 ⎠
Output resistance looking into the collector of Q2 is increased.

10.30
V + − VBE1 − V − 5 − 0.7 − ( −5 )
(a) I REF = = = 0.3174 mA
R1 + RE1 27.3 + 2
I O = I REF = 0.3174 mA
(b) Using the same relation as for the widlar current source.

RO = ro 2 ⎡1 + g m 2 ( RE rπ 2 ) ⎤
⎣ ⎦
VA 80 0.3174
ro 2 = = = 252 K gm2 = = 12.21 mA/V
I O 0.3174 0.026
(100 )( 0.026 )
rπ 2 = = 8.192 K RE || rπ 2 = 2 || 8.192 = 1.608 K
0.3174
RO = 252 ⎡1 + (12.21)(1.608 ) ⎤ ⇒ RO = 5.2 MΩ
⎣ ⎦
(c)
5 − 0.7 − ( −5 )
I O = I REF = = 0.3407 mA
27.3
V 80
RO = ro 2 = A = ⇒ RO = 235 K
I O 0.3407

10.31
Assume all transistors are matched.
a.
2VBE1 = VBE 3 + I 0 RE
⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟
⎝ IS ⎠
⎛I ⎞
VBE 3 = VT ln ⎜ 0 ⎟
⎝ IS ⎠
⎛I ⎞ ⎛I ⎞
2VT ln ⎜ REF ⎟ − VT ln ⎜ 0 ⎟ = I 0 RE
⎝ IS ⎠ ⎝ IS ⎠
⎡ ⎛I ⎞ 2
⎛I ⎞⎤
VT ⎢ ln ⎜ REF ⎟ − ln ⎜ 0 ⎟ ⎥ = I 0 RE
⎢ ⎝ IS ⎠
⎣ ⎝ IS ⎠⎥⎦
⎛ I 2 REF ⎞
VT ln ⎜ ⎟ = I 0 RE
⎝ I0 I S ⎠
b.
⎛ 0.7 ⎞
VBE = 0.7 V at 1 mA ⇒ 10−3 = I S exp ⎜ −15
⎟ or I S = 2.03 × 10 A
⎝ 0.026 ⎠
⎛ 0.1× 10−3 ⎞
VBE at 0.1 mA ⇒ VBE = ( 0.026 ) ln ⎜ −15 ⎟
= 0.640 V
⎝ 2.03 × 10 ⎠
0.640
Since I 0 = I REF , then VBE = I 0 RE ⇒ RE = or RE = 6.4 kΩ
0.1

10.32
(a)
5 − 0.7 − ( −5 )
I REF = = 0.80 mA
R1
R1 = 11.6 K
0.026 ⎛ 0.80 ⎞
RE 2 = ln ⎜ ⎟ ⇒ RE 2 = 1.44 K
0.050 ⎝ 0.050 ⎠
0.026 ⎛ 0.80 ⎞
RE 3 = ln ⎜ ⎟ ⇒ RE 2 = 4.80 K
0.020 ⎝ 0.020 ⎠
(b)
VBE 2 = 0.7 − ( 0.05 )(1.44 ) ⇒ VBE 2 = 0.628 V
VBE 3 = 0.7 − ( 0.02 )( 4.80 ) ⇒ VBE 3 = 0.604 V

10.33
(a)
VBE1 = VBE 2
V + − 2VBE1 − V −
I REF =
R1 + R2
Now
2VBE1 + I REF R2 = VBE 3 + I O RE
or
I O RE = 2VBE1 − VBE 3 + I REF R2
We have
⎛I ⎞ ⎛I ⎞
VBE1 = VT ln ⎜ REF ⎟ and VBE 3 = VT ln ⎜ O ⎟
⎝ IS ⎠ ⎝ IS ⎠
(b)
Let R1 = R2 and I O = I REF ⇒ VBE1 = VBE 3 ≡ VBE
Then
VBE = I O RE − I REF R2 = I O ( RE − R2 )
so
V + − V − − 2 I O ( RE − R2 )
I REF = I O =
2 R2
+
V −V −
⎛R ⎞
= − IO ⎜ E ⎟ + IO
2 R2 ⎝ R2 ⎠
Then
V + −V −
IO =
2 Rε
(c)
Want I O = 0.5 mA
5 − ( −5 )
So RE = ⇒ RE = 10 k Ω
2 ( 0.5 )
5 − 2 ( 0.7 ) − ( −5 )
2 R2 = = 17.2 k Ω
0.5
Then R1 = R2 = 8.6 k Ω

10.34
a.
20 − 0.7 − 0.7
I REF = = 1.55 mA
12
I 01 = 2 I REF = 3.1 mA
I 02 = I REF = 1.55 mA
I 03 = 3I REF = 4.65 mA
b.
VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V
VEC 2 = 10 − I 02 RC 2 = 10 − (1.55 )( 3) ⇒ VEC 2 = 5.35 V
VEC 3 = 10 − I 03 RC 3 = 10 − ( 4.65 )(1) ⇒ VEC 3 = 5.35 V

10.35
a. Ist approximation

20 − 1.4
I REF ≅ = 2.325 mA
8
⎛ 2.32 ⎞
Now VBE − 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = VEB = 0.722 V
⎝ 1 ⎠
Then 2nd approximation
20 − 2 ( 0.722 )
I REF ≅ = 2.32 mA
8
I 01 = 2 I REF = 4.64 mA
I 02 = I REF = 2.32 mA
I 03 = 3I REF = 6.96 mA
b.
At the edge of saturation,
VCE = VBE = 0.722 V
0 − 0.722 − ( −10 )
RC1 = ⇒ RC1 = 2.0 kΩ
4.64
10 − 0.722
RC 2 = ⇒ RC 2 = 4.0 kΩ
2.32
10 − 0.722
RC 3 = ⇒ RC 3 = 1.33 kΩ
6.96

10.36
10 − 0.7 − 0.7 − ( −10 )
I C1 = I C 2 = = 1.86 mA
10
I C 3 = I C 4 = 1.86 mA
⎛ 1.86 ⎞
I C 5 ( 0.5 ) = 0.026 ln ⎜ ⎟
⎝ IC 5 ⎠
By Trial and error.
⇒ I C 5 = 0.136 mA = I C 6 = I C 7
2 I C 3 ( 0.8 ) + VCE 3 = 10 ⇒ VCE 3 = 10 − 2 (1.86 )( 0.8 )
VCE 3 = 7.02 V
5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) − 10
VCE 5 = 5 + 10 − 0.7 − ( 0.136 )( 0.5 )
VCE 5 = 14.2 V
5 = VEC 7 + I C 7 ( 0.8 )
VEC 7 = 5 − ( 0.136 )( 0.8 )
VEC 7 = 4.89 V

10.37
10 − 0.7 − 0.7 − ( −10 )
I C1 = I C 2 = ⇒ I C1 = I C 2 = 1.86 mA
10
I C 4 = I C 5 = 1.86 mA
⎛I ⎞ ⎛ 1.86 ⎞
I C 3 RE1 = VT ln ⎜ C1 ⎟ ⇒ I C 3 ( 0.3) = 0.026 ln ⎜ ⎟
⎝ IC 3 ⎠ ⎝ IC 3 ⎠
By trial and error I C 3 = 0.195 mA
⎛I ⎞ ⎛ 1.86 ⎞
I C 6 RE 2 = VT ln ⎜ C 5 ⎟ ⇒ I C 6 ( 0.5 ) = 0.026 ln ⎜ ⎟
⎝ IC 6 ⎠ ⎝ IC 6 ⎠
By trial and error I C 6 = 0.136 mA

10.38
10 − 0.7
I REF = = 1 mA
6.3 + 3
VBE ( QR ) = 0.7 V as assumed
VRER = I REF ⋅ RER = (1)( 3) = 3 V
VRE1 3
VRE1 = 3 V ⇒ RE1 = = ⇒ RE1 = 3 kΩ
I 01 1
VRE 2 3
VRE 2 = 3 V ⇒ RE 2 = = ⇒ RE 2 = 1.5 kΩ
I 02 2
VRE 3 3
VRE 3 = 3 V ⇒ RE 3 = = ⇒ RE 3 = 0.75 kΩ
I 03 4
I 01 = 1 mA
I 02 = 2 mA
I 03 = 4 mA

10.38
VDS 2 ( sat ) = 2 V = VGS 2 − VTN 2 = VGS 2 − 1.5 ⇒ VGS 2 = 3.5 V
⎛1 ⎞⎛W ⎞
I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 2 )
2

⎝ 2 ⎠⎝ L ⎠2
⎛W ⎞ ⎛W ⎞
250 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 3.125
2

⎝ L ⎠2 ⎝ L ⎠2
⎛1 ⎞⎛W ⎞
I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 1 )
⎝2 ⎠ ⎝ L ⎠1
⎛W ⎞ ⎛W ⎞
100 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 1.25
2

⎝ L ⎠2 ⎝ L ⎠1
Now VGS 3 = 10 − VGS 2 = 10 − 3.5 = 6.5 V
⎛W ⎞ ⎛W ⎞
So 100 = ( 20 ) ⎜ ⎟ ( 6.5 − 1.5 ) ⇒ ⎜ ⎟ = 0.2
2

⎝ L ⎠3 ⎝ L ⎠3

10.39
2.5 − VGS ⎛ 0.08 ⎞
⎟ ( 6 )(VGS − 0.5 )
2
I REF = =⎜
15 ⎝ 2 ⎠
2.5 − VGS = 3.6 (VGS − VGS + 0.25 )
2

2
3.6VGS − 2.6VGS − 1.6 = 0
2.6 ± 6.76 + 23.04
VGS =
2 ( 3.6 )
VGS = 1.12 V (1.1193)
2.5 − 1.1193
I REF = ⇒ I REF = 92.0 μ A ( 92.05 )
15
I o = 92.0 μ A
VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5
VDS 2 ( sat ) = 0.619 V

10.39
a. From Equation (10.50),

⎛ 5 ⎞ ⎛ 5 ⎞
⎜ ⎟ ⎜ 1− ⎟
=⎜ 25 ⎟ 5 + ⎜ 25 ⎟ 0.5
VGS1 = VGS 2 ( ) ( )
⎜ 5 ⎟ ⎜ 5 ⎟
⎜ 1+ ⎟ ⎜ 1+ ⎟
⎝ 25 ⎠ ⎝ 25 ⎠
⎛ 0.447 ⎞ ⎛ 1 − 0.447 ⎞
=⎜ ⎟ (5) + ⎜ ⎟ ( 0.5 )
⎝ 1 + 0.447 ⎠ ⎝ 1 + 0.447 ⎠
VGS 1 = VGS 2 = 1.74 V
I REF ≅ K n1 (VGS 1 − VTN ) = (18 )( 25 )(1.74 − 0.5 ) ⇒ I REF = 0.692 mA
2 2

⎛1 ⎞⎛W ⎞
I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN ) (1 + λVDS 2 )
2
b.
⎝ 2 ⎠⎝ L ⎠2
I 0 = (18 )(15 )(1.74 − 0.5 ) ⎡1 + ( 0.02 )( 2 ) ⎤
2
⎣ ⎦
= ( 415 )(104 ) ⇒ I 0 = 0.432 mA
c. I 0 = ( 415 ) ⎡1 + ( 0.02 )( 4 ) ⎤ ⇒ I 0 = 0.448 mA
⎣ ⎦

10.40
(a)
⎛ 80 ⎞ ⎛ W ⎞
I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VGS − 0.5 )
2

⎝ 2 ⎠ ⎝ L ⎠1
2.0 − VGS
I REF = 0.050 =
R
Design such that VDS 2 ( sat ) = 0.25 = VGS − 0.5
VGS = 0.75 V
2 − 0.75
So 0.050 = ⇒ R = 25 K
R
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20
2

⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1
⎛W ⎞
⎜ ⎟
⎝ L ⎠1 I REF 20 50 ⎛W ⎞
= ⇒ = ⇒ ⎜ ⎟ = 40
⎛W ⎞ IO ⎛W ⎞ 100 ⎝ L ⎠ 2
⎜ ⎟ ⎜ ⎟
⎝ L ⎠2 ⎝ L ⎠2
1 1
(b) RO = = ⇒ RO = 667 K
λ I O ( 0.015 )( 0.1)
ΔV 1
(c) ΔI O = = ⇒ 1.5 μ A
RO 666
ΔI O ⎛ 1.5 ⎞
× 100% = ⎜ ⎟ × 100% ⇒ 1.5%
IO ⎝ 100 ⎠

10.41
⎛ 80 ⎞
I REF = 250 = ⎜ ⎟ ( 3)(VGS − 1)
2
(a)
⎝ 2⎠
VGS = 2.44 V
I O = 250 μ A at VDS 2 = VGS = 2.44 V
1 1
RO = = = 200 K
λ I O ( 0.02 )( 0.25 )

ΔV 3 − 2.44
(i) ΔI O = = ⇒ 2.8 μ A
RO 200
I O = 252.8 μ A
ΔV 4.5 − 2.44
(ii) ΔI O = = ⇒ 10.3 μ A
RO 200
I O = 260.3 μ A
ΔV 6 − 2.44
(iii) ΔI O = = ⇒ 17.8 μ A
RO 200
I O = 267.8 μ A
4.5
(b) IO = ( 250 ) = 375 μ A at VDS = 2.44 V
3
1 1
RO = = = 133.3 K
λ I O ( 0.02 )( 0.375 )
ΔV 3 − 2.44
(i) ΔI O = = ⇒ 4.20 μ A
RO 133.3
I O = 379.2 μ A
ΔV 4.5 − 2.44
(ii) ΔI O = = ⇒ 15.5 μ A
RO 133.3
I O = 390.5 μ A
ΔV 6 − 2.44
(iii) ΔI O = = ⇒ 26.7 μ A
RO 133.3
I O = 401.7 μ A

10.41
VSD 2 ( sat ) = 0.25 = VSG + VTP = VSG − 0.4 ⇒ VSG 2 = 0.65 V
k′ ⎛ W ⎞
I O = ⎜ ⎟ (VSG 2 + VTP )
p 2

2 ⎝ L ⎠2
40 ⎛ W ⎞ ⎛W ⎞
⎜ ⎟ ( 0.65 − 0.4 ) ⇒ ⎜ ⎟ = 20
2
25 =
2 ⎝ L ⎠2 ⎝ L ⎠2
(W /L )1 ⎛W ⎞
I REF = 75 μ A = ⋅ I O ⇒ ⎜ ⎟ = 60
(W /L )2 ⎝ L ⎠1
k′ ⎛W ⎞
⎜ ⎟ (VSG 3 + VTP )
2
I REF =
p

2 ⎝ L ⎠3
VSG 3 = 3 − 0.65 = 2.35 V
40 ⎛ W ⎞ ⎛W ⎞
⎜ ⎟ ( 2.35 − 0.4 ) ⇒ ⎜ ⎟ = 0.986
2
Then 75 =
2 ⎝ L ⎠3 ⎝ L ⎠3

10.42
(a)
I REF 0.5
VGS = VTN 1 + = 1+ =2V
K n1 0.5
2
⎛ I ⎞ ⎛ I REF ⎞
I O = K n 2 ⎜ REF ⎟ = Kn2 ⎜ ⎟
⎜ K ⎟
⎝ n1 ⎠ ⎝ K n1 ⎠

⎛ 0.5 ⎞
I 0 ( max ) = ( 0.5 )(1.05 ) ⎜ ⎟ ⇒ I 0 ( max ) = 0.525 mA
⎝ 0.5 ⎠
⎛ 0.5 ⎞
I 0 ( min ) = ( 0.5 )( 0.95 ) ⎜ ⎟ ⇒ I 0 ( min ) = 0.475 mA
⎝ 0.5 ⎠
So 0.475 ≤ I 0 ≤ 0.525 mA
(b)
2
⎡ I ⎤
I O = K n 2 ⎢ REF + VTN 1 − VTN 2 ⎥
⎢ K n1
⎣ ⎥
⎦
2
⎡ 0.5 ⎤
I 0 ( min ) = ( 0.5 ) ⎢ + 1 − 1.05⎥ ⇒ I 0 (min) = 0.451 mA
⎣ 0.5 ⎦
2
⎡ 0.5 ⎤
I 0 ( max ) = ( 0.5 ) ⎢ + 1 − 0.95⎥ ⇒ I 0 (max) = 0.551 mA
⎣ 0.5 ⎦
So 0.451 ≤ I 0 ≤ 0.551 mA

10.43

Vx − VA
(1) Ix = + g mVgs 2
ro
VA
(2) Ix = + g mVgs1
ro
Vgs1 = Vx , Vgs 2 = −VA
So
Vx ⎛1 ⎞
(1) Ix = − VA ⎜ + g m ⎟
ro ⎝ ro ⎠
VA
(2) Ix = + g mVx ⇒ VA = ro [ I x − g mVx ]
ro

Then
Vx ⎛1 ⎞
Ix = − ro ( I x − g mVx ) ⎜ + g m ⎟
ro ⎝ ro ⎠
Vx ⎡I g 2 ⎤
Ix = − ro ⎢ x + g m I x − m ⋅ Vx − g mVx ⎥
ro ⎣ ro ro ⎦
Vx 2
Ix = − I x − g m ro I x + g mVx + g m roVx
ro
⎡1 ⎤
I x [ 2 + g m ro ] = Vx ⎢ + g m + g m ro ⎥
2

⎣ ro ⎦
1
Since g m >>
ro
I x [ 2 + g m ro ] ≅ Vx ( g m )(1 + g m ro )
Vx 2 + g m ro
Then = Ro =
Ix g m (1 + g m ro )
1
Usually, g m ro >> 2, so that Ro ≅
gm

10.44
VDS 2 (sat) = 2 = VGS 2 − 0.8 ⇒ VGS 2 = 2.8 V
60 ⎛ W ⎞ ⎛W ⎞
⎜ ⎟ ( 2.8 − 0.8 ) ⇒ ⎜ ⎟ = 1.67
2
I O = 200 =
2 ⎝ L ⎠2 ⎝ L ⎠2
⎛W ⎞ ⎛W ⎞
I REF ⎜ L ⎟1
⎝ ⎠
⎜ ⎟
0.4 ⎝ L ⎠1 ⎛W ⎞
= ⇒ = ⇒ ⎜ ⎟ = 3.33
IO ⎛W ⎞ 0.2 (1.67 ) ⎝ L ⎠1
⎜ ⎟
⎝ L ⎠2
VGS 3 = 6 − 2.8 = 3.2 V
⎛ 60 ⎞ ⎛ W ⎞ ⎛W ⎞
I REF = 400 = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 0.8 ) ⇒ ⎜ ⎟ = 2.31
2

⎝ 2 ⎠ ⎝ L ⎠3 ⎝ L ⎠3

10.45
(a)
⎛ 60 ⎞ ⎛ 60 ⎞
I REF = ⎜ ⎟ ( 20 )(VGS 1 − 0.7 ) = ⎜ ⎟ ( 3)(VGS 3 − 0.7 )
2 2

⎝ 2 ⎠ ⎝ 2 ⎠
VGS1 + VGS 3 = 5
20
(VGS1 − 0.7 ) = 5 − VGS1 − 0.7
3
3.582VGS 1 = 6.107 ⇒ VGS1 = VGS 2 = 1.705 V
⎛ 60 ⎞
I O = ⎜ ⎟ (12 )(1.705 − 0.7 ) = 363.6 μ A at VDS 2 = 1.705 V
2

⎝ 2 ⎠
⎛ 60 ⎞
I REF = ⎜ ⎟ ( 20 )(1.705 − 0.7 ) = 606 μ A
2

⎝ 2 ⎠
1 1
(b) RO = = = 183.4 K
λ IO ( 0.015 )( 0.3636 )
ΔV 1.5 − 1.705
ΔI O = = ⇒ −1.12 μ A
RO 183.4
I O = 362.5 μ A

ΔV 3 − 1.705
(c) ΔI O = = ⇒ 7.06 μ A
RO 183.4
I O = 370.7 μ A

10.46
⎛ 50 ⎞ ⎛ 50 ⎞
I REF = ⎜ ⎟ (15 )(VSG1 − 0.5 ) = ⎜ ⎟ ( 3)(VSG 3 − 0.5 )
2 2

⎝ 2⎠ ⎝ 2⎠
VSG1 + VSG 3 = 10 ⇒ VSG 3 = 10 − VSG1
15
(VSG1 − 0.5) = 10 − VSG1 − 0.5
3
3.236VSG1 = 10.618 ⇒ VSG1 = 3.28 V
⎛ 50 ⎞
I REF = ⎜ ⎟ (15 )( 3.28 − 0.5 ) ⇒ I REF = 2.90 mA
2

⎝ 2⎠
I O = I REF = 2.90 mA
VSD 2 (sat) = VSG 2 + VTP = 3.28 − 0.5 ⇒ VSD 2 (sat) = 2.78 V

10.47
VSD 2 (sat) = 1.2 = VSG 2 − 0.35 ⇒ VSG 2 = 1.55 V
⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞
I O = 100 = ⎜ ⎟⎜ ⎟ (1.55 − 0.35 ) ⇒ ⎜ ⎟ = 2.78
2

⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2

I REF
=
( )
W
L1
⇒
200
=
( )
W
L1 ⎛W ⎞
⇒ ⎜ ⎟ = 5.56
IO ( )
W
L 2
100 2.78 ⎝ L ⎠1

VSG1 + VSG 3 = 4 ⇒ VSG 3 = 2.45 V
⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞
I REF = 200 = ⎜ ⎟⎜ ⎟ ( 2.45 − 0.35 ) ⇒ ⎜ ⎟ = 1.81
2

⎝ 2 ⎠⎝ L ⎠3 ⎝ L ⎠3

10.48
⎛ 80 ⎞ ⎛ 80 ⎞
I REF = ⎜ ⎟ ( 25 )(VSG1 − 1.2 ) = ⎜ ⎟ ( 4 )(VSG 3 − 1.2 )
2 2

⎝ 2⎠ ⎝ 2⎠
10 − VSG1
VSG1 + 2VSG 3 = 10 ⇒ VSG 3 =
2
25 10 − VSG1
Then (VSG1 − 1.2 ) = − 1.2
4 2
3VSG1 = 6.8 ⇒ VSG1 = 2.27 V
⎛ 80 ⎞
I REF = ⎜ ⎟ ( 25 )( 2.267 − 1.2 ) ⇒ I REF = I O = 1.14 mA
2

⎝ 2⎠
VSD 2 (sat) = VSG 2 + VTP = 2.27 − 1.2 ⇒ VSD 2 ( sat ) = 1.07 V

10.49
VSD 2 (sat) = 1.8 = VSG 2 − 1.4 ⇒ VSG 2 = 3.2 V
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
I O = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 1.4 ) = 100 ⇒ ⎜ ⎟ = 0.772
2

⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠2

I REF
=
( )
W
L1
IO ( )
W
L 2

200
=
( )
W
L1 ⎛W ⎞
⇒ ⎜ ⎟ = 1.54
100 0.772 ⎝ L ⎠1

Assume M3 and M4 are matched.
10 − 3.2
2VSG 3 + VSG1 = 10 ⇒ VSG 3 = = 3.4 V
2
⎛ 80 ⎞ ⎛ W ⎞
I REF = 200 = ⎜ ⎟ ⎜ ⎟ ( 3.4 − 1.4 )
2

⎝ 2 ⎠ ⎝ L ⎠3,4
⎛W ⎞
⎜ ⎟ = 1.25
⎝ L ⎠3,4

10.50
(a)
⎛ k′ ⎞⎛ W ⎞
I REF = ⎜ ⎟ ⎜ ⎟ (VSG1 + VTP )
p 2

⎝ 2 ⎠ ⎝ L ⎠1
⎛ k′ ⎞⎛ W ⎞
= ⎜ ⎟ ⎜ ⎟ (VSG 3 + VTP )
p 2

⎝ 2 ⎠ ⎝ L ⎠3
But VSG 3 = 3 − VSG1
So 25 (VSG1 − 0.4 ) = 5 ( 3 − VSG1 − 0.4 )
2 2

which yields VSG1 = 1.08 V
and VSG 3 = 1.92 V
I REF = 20 ( 25 )(1.08 − 0.4 ) ⇒ I REF = 231 μ A
2

IO (W / L )2 15
= = = 0.6
I REF (W / L )1 25
Then I O = ( 0.6 )( 231) = 139 μ A
(b)
VDS 2 ( sat ) = 1.08 − 0.4 = 0.68 V
VR = 3 − 0.68 = 2.32 = I O R
then
2.32
R= ⇒ R = 16.7 k Ω
0.139

10.51
VSD 2 (sat) = 0.35 = VSG 2 − 0.4 ⇒ VSG 2 = 0.75 V
⎛W ⎞ ⎛W ⎞
I O = 80 = ( 20 ) ⎜ ⎟ ( 0.75 − 0.4 ) ⇒ ⎜ ⎟ = 32.7
2

⎝ L ⎠2 ⎝ L ⎠2
W ( ) W ( )
I REF
=
L1
⇒
50
=
L1
⇒ W ( )
= 20.4
IO W ( )
L 2
80 32.7 L1

VSG 3 = 3 − 0.75 = 2.25
⎛W ⎞ ⎛W ⎞
I REF = 50 = ( 20 ) ⎜ ⎟ ( 2.25 − 0.4 ) ⇒ ⎜ ⎟ = 0.730
2

⎝ L ⎠3 ⎝ L ⎠3

10.52
I REF = K n (VGS − VTN )
2
a.
100 = 100 (VGS − 2 ) ⇒ VGS = 3 V
2

For VD 4 = −3 V, I 0 = 100 μ A

b. R0 = r04 + r02 (1 + g m r04 )
1 1
r02 = r04 = = = 500 kΩ
λ I 0 ( 0.02 )( 0.1)
g m = 2 K n (VGS − VTN ) = 2 ( 0.1)( 3 − 2 ) = 0.2 mA / V
R0 = 500 + 500 ⎡1 + ( 0.2 )( 500 ) ⎤
⎣ ⎦
R0 = 51 MΩ
1 6
ΔI 0 = ⋅ ΔVD 4 = ⇒ ΔI 0 = 0.118 μ A
R0 51

10.53

Vgs 4 = − I X r02
VS 6 = ( I X − g mVgs 4 ) r04 + I X r02
= ( I X + g m I X r02 ) r04 + I X r02
VS 6 = I X ⎡ r02 + (1 + g m r02 ) r04 ⎤ = −Vgs 6
⎣ ⎦
V − VS 6 VX ⎛ 1 ⎞
I X = g mVgs 6 + X = − VS 6 ⎜ g m + ⎟
r06 r06 ⎝ r06 ⎠
VX ⎛ 1 ⎞
IX = − IX⎜ g m + ⎟ ⎡ r02 + (1 + g m r02 ) r04 ⎤
r06 ⎝ r06 ⎠ ⎣ ⎦

⎧ ⎛
⎪ 1 ⎞ ⎫ V
⎪
I X ⎨1 + ⎜ g m + ⎟ ⎡ r02 + (1 + g m r02 ) r04 ⎤ ⎬ = X
⎣ ⎦
⎪ ⎝
⎩ r06 ⎠ ⎪ r06
⎭
VX
= R0 = r06 + (1 + g m r06 ) ⎡ r02 + (1 + g m r02 ) r04 ⎤
⎣ ⎦
IX
I 0 ≈ I REF = 0.2 mA = 0.2 (VGS − 1)
2

VGS = 2 V
g m = 2 K n (VGS − VTN ) = 2 ( 0.2 )( 2 − 1) = 0.4 mA / V
1 1
r02 = r04 = r06 = = = 250 kΩ
λ I0( 0.02 )( 0.2 )
R0 = 250 + ⎡1 + ( 0.4 )( 250 ) ⎤ × {250 + ⎡1 + ( 0.4 )( 250 ) ⎤ ( 250 )}
⎣ ⎦ ⎣ ⎦
R0 = 2575750 kΩ ⇒ R0 = 2.58 × 109 Ω

10.54
′
kn ⎛ W ⎞ ′
kn ⎛ W ⎞
⎜ ⎟ (VGS1 − VTN ) = ⎜ ⎟ (VGS 3 − VTN )
2 2

2 ⎝ L ⎠1 2 ⎝ L ⎠3
k′ ⎛ W ⎞
= ⎜ ⎟ (VGS 4 + VTP )
p 2

2 ⎝ L ⎠4
(1) 50 ( 20 )(VGS 1 − 0.5 ) = 50 ( 5 )(VGS 3 − 0.5 )
2 2

(2) 50 ( 20 )(VGS 1 − 0.5 ) = 20 (10 )(VGS 4 − 0.5 )
2 2

(3) VSG 4 + VGS 3 + VGS1 = 6
From (1) 4 (VGS 1 − 0.5 ) = (VGS 3 − 0.5 ) ⇒ VGS 3 = 2 (VGS 1 − 0.5 ) + 0.5
2 2

From (2) 5 (VGS1 − 0.5 ) = (VGS 4 − 0.5 ) ⇒ VSG 4 = 5 (VGS1 − 0.5 ) + 0.5
2 2

Then (3) becomes
5 (VGS1 − 0.5 ) + 0.5 + 2 (VGS 1 − 0.5 ) + 0.5 + VGS1 = 6
which yields VGS 1 = 1.36 V and VGS 3 = 2.22 V , VSG 4 = 2.42 V
k′ ⎛ W ⎞
Then I REF = n ⎜ ⎟ (VGS1 − VTN ) = 50 ( 20 )(1.36 − 0.5 ) or I REF = I O = 0.740 mA
2 2

2 ⎝ L ⎠1
VGS 1 = VGS 2 = 1.36 V
VDS 2 ( sat ) = VGS 2 − VTN = 1.36 − 0.5 ⇒ VDS 2 ( sat ) = 0.86 V

10.55
VDS 2 ( sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.5 ⇒ VGS 2 = 1 V
′
kn ⎛ W ⎞
⎜ ⎟ (VGS 2 − VTN )
2
I O = 50 μ A =
2 ⎝ L ⎠2
⎛W ⎞ ⎛W ⎞
= 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 4
2

⎝ L ⎠2 ⎝ L ⎠2
′
kn ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞
⎜ ⎟ (VGS 1 − VTN ) = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 12
2 2
VGS1 = VGS 2 = 1 V ⇒ I REF = 150 =
2 ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠1
VGS 3 + VSG 4 + VGS 1 = 6
2VGS 3 = 6 − 1 = 5 V ⇒ VGS 3 = 2.5 V
⎛W ⎞ ⎛W ⎞
I REF = 150 = 50 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ = 0.75
2

⎝ L ⎠3 ⎝ L ⎠3
k′ ⎛ W ⎞
⎜ ⎟ (VSG 4 + VTP )
2
I REF =
p

2 ⎝ L ⎠4
⎛W ⎞ ⎛W ⎞
150 = 20 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ 1.88
2

⎝ L ⎠4 ⎝ L ⎠4

10.56
a. As a first approximation
I REF = 80 = 80 (VGS 1 − 1) ⇒ VGS 1 = 2 V
2

Then VDS 1 ≅ 2 ( 2 ) = 4 V
The second approximation
80 = 80 (VGS1 − 1) ⎡1 + ( 0.02 )( 4 ) ⎤
2
⎣ ⎦
80
= (VGS 1 − 1) ⇒ VGS 1 = 1.962
2
Or
86.4

Then
I O = K n (VGS1 − VTN ) (1 + λnVGS 1 )
2

= 80 (1.962 − 1) ⎡1 + ( 0.02 )(1.962 ) ⎤
2
⎣ ⎦
Or I 0 = 76.94 μ A
b. From a PSpice analysis, I 0 = 77.09 μ A for VD 3 = −1 V and I 0 = 77.14 μ A for VD 3 = 3 V. The
change is ΔI 0 ≈ 0.05 μ A or 0.065%.

10.57
a. For a first approximation,
I REF = 80 = 80 (VGS 4 − 1) ⇒ VGS 4 = 2 V
2

As a second approximation
I REF = 80 = 80 (VGS 4 − 1) ⎡1 + ( 0.02 )( 2 ) ⎤
2
⎣ ⎦
Or VGS 4 = 1.98 V = VGS 1
I O = K n (VGS 2 − VTN ) (1 + λVGS 2 )
2

To a very good approximation I 0 = 80 μ A
b. From a PSpice analysis, I 0 = 80.00 μ A for VD 3 = −1 V and the output resistance is
R0 = 76.9 MΩ.
Then
For VD = +3 V
1 4
ΔI 0 = ⋅ VD 3 = = 0.052 μ A
R0 76.9
I 0 = 80.05 μ A

10.58
(a) VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0
k′ ⎛ W ⎞
I D = n ⎜ ⎟ (VGS 3 − VTN )
2

2⎝L⎠
⎛W ⎞ ⎛W ⎞
50 = 48 ⎜ ⎟ ( 0.2 ) ⇒ ⎜ ⎟ = 26
2

⎝L⎠ ⎝ L ⎠3
(b) VGS 5 − VTN = 2 (VGS 3 − VTN )
VGS 5 = 0.8 + 2 ( 0.2 ) ⇒ VGS 5 = 1.2 V
(c) VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) ⇒ VD1 ( min ) = 0.4 V

10.59
(a)
′
kn ⎛ W ⎞
⎜ ⎟ = 50 ( 5 ) = 250 μ A / V
2
K n1 =
2 ⎝ L ⎠1

1 ⎛ (W / L )1 ⎞
R= ⎜1 − ⎟
K n1 I D1 ⎜
⎝ (W / L )2 ⎟
⎠
1 ⎛ 5 ⎞
= ⎜1 −
⎜ ⎟ = ( 8.944 )( 0.6838 )
( 0.25 )( 0.05 ) ⎝ 50 ⎟
⎠
R = 6.12 k Ω
(b)

V + − V − = VSD 3 ( sat ) + VGS 1
VSD 3 ( sat ) = VSG 3 + VTP
I D1 = 50 = 20 ( 5 )(VSG 3 − 0.5 ) ⇒ VSG 3 = 1.207 V
2

Then VSD 3 ( sat ) = 1.21 − 0.5 = 0.707 V
Also I D1 = 50 = 50 ( 5 )(VGS1 − 0.5 ) ⇒ VGS 1 = 0.9472 V
2

Then (V + − V − ) = 0.71 + 0.947 = 1.66 V
min

(c)
⎛W ⎞ ⎛W ⎞
I O1 = 25 = 50 ⎜ ⎟ ( 0.947 − 0.5 ) ⇒ ⎜ ⎟ = 2.5
2

⎝ L ⎠5 ⎝ L ⎠5
⎛W ⎞ ⎛W ⎞
I O 2 = 75 = 20 ⎜ ⎟ (1.207 − 0.5 ) ⇒ ⎜ ⎟ = 7.5
2

⎝ L ⎠6 ⎝ L ⎠6

10.60
1
VGS 3 = ( 5 ) = 1.667 V
3
⎛1 ⎞⎛W ⎞
I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN )
2

⎝2 ⎠ ⎝ L ⎠3
⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞
100 = ( 20 ) ⎜ ⎟ (1.667 − 1) ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 11.25
2

⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ L ⎠5
⎛1 ⎞⎛W ⎞
I O1 = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN )
2

⎝2 ⎠ ⎝ L ⎠1
⎛W ⎞
⎜ ⎟
I REF ⎝ L ⎠3
Or =
I 01 ⎛W ⎞
⎜ ⎟
⎝ L ⎠1
⎛ W ⎞ ⎛ I 01 ⎞ ⎛ W ⎞ ⎛ 0.2 ⎞ ⎛W ⎞
⎜ ⎟ =⎜ ⎟⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 22.5
⎝ L ⎠1 ⎝ I REF ⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠ ⎝ L ⎠1
⎛ W ⎞ ⎛ I ⎞ ⎛ W ⎞ ⎛ 0.3 ⎞ ⎛W ⎞
And ⎜ ⎟ = ⎜ 02 ⎟ ⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 33.75
⎝ L ⎠ 2 ⎝ I REF ⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠ ⎝ L ⎠2

10.61
24 − VSGP − VGSN
I REF =
R
Also I REF = 40 (1)(VGSN − 1.2 )
2

I REF = 18 (1)(VSGP − 1.2 )
2

Then 40 (VGSN − 1.2 ) = 18 (VSGP − 1.2 )
6.325
which yields VSGP = (VGSN − 1.2 ) + 1.2
4.243

Then ⎡ 0.040 (VGSN − 1.2 ) ⎤ ⋅ R = 24 − VGSN − 1.49 (VGSN − 1.2 ) − 1.2
2
⎣ ⎦
which yields VGSN = 2.69 V
and VSGP = 3.43 V
24 − 3.42 − 2.69
Now I REF = ⇒ I REF = 89.4 μ A
200
89.4
I1 = = 17.9 μ A
5
I 2 = (1.25 )( 89.4 ) = 112 μ A
I 3 = ( 0.8 )( 89.4 ) = 71.5 μ A
I 4 = 4 ( 89.4 ) = 358 μ A

10.61
a. g m ( M 0 ) = 2 K n I REF
gm ( M 0 ) = 2 ( 0.25)( 0.2 ) ⇒ g m ( M 0 ) = 0.447 mA/V
1 1
r0 n = = ⇒ r0 n = 250 kΩ
λn I REF ( 0.02 )( 0.2 )
1 1
r0 p = = ⇒ r0 p = 167 kΩ
λ p I REF ( 0.03)( 0.2 )
b. Av = − g m ( r0 n || r0 p ) = − ( 0.447 )( 250 ||167 ) ⇒ Av = −44.8
c. RL = 205 ||167 = r0 n || r0 p or RL = 100 kΩ

10.62
We have VGSN = 2.69 V and VSGP = 3.43 V
10 − 2.69 − 3.43 3.88
So I REF = = ⇒ I REF = 19.4 μ A
R 200
Then I1 = ( 0.2 )(19.4 ) = 3.88 μ A
I 2 = (1.25 )(19.4 ) = 24.3 μ A
I 3 = ( 0.8 )(19.4 ) = 15.5 μ A
I 4 = 4 (19.4 ) = 77.6 μ A

10.63

I D2 =
(W L )2
⋅ I REF =
9
( 200 ) ⇒ I D 2 = 120 μ A
(W L )1
15

IO =
(W L )4 ⎛ 20 ⎞
⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 μ A
(W L )3
⎝ 9 ⎠

⎛ 40 ⎞
I O = 266.7 = ⎜ ⎟ ( 20 )(VSG 4 − 0.6 )
2

⎝ 2 ⎠
VSG 4 = 1.416 V
VSD 4 (sat) = 1.416 − 0.6 ⇒ VSD 4 ( sat ) = 0.816 V

10.64
⎛ 40 ⎞ ⎛ W ⎞
I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 )
2

⎝ 2 ⎠ ⎝ L ⎠1
1.75 − VSG1
I REF = = 50
R

VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V
1.75 − 0.95
R= ⇒ R = 16 K
0.05
⎛ 40 ⎞⎛ W ⎞ ⎛W ⎞
50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4
2

⎝ 2 ⎠⎝ L ⎠1 ⎝ L ⎠1

I O1 120
= =
W ( )
L 2
⇒
⎛W ⎞
= 49
I REF 50 ( 20.4 ) ⎜ L ⎟2
⎝ ⎠

I D 3 25
= =
W ( )
L 3 ⎛W ⎞
⇒ ⎜ ⎟ = 10.2
I REF 50 ( 20.4 ) ⎝ L ⎠3
VDS 5 (sat) = 0.35 = VGS 5 − 0.4 ⇒ VGS 5 = 0.75 V
⎛ 100 ⎞⎛ W ⎞ ⎛W ⎞
⎟⎜ ⎟ ( 0.75 − 0.4 ) = 150 ⇒ ⎜ ⎟ = 24.5
2
IO 2 = ⎜
⎝ 2 ⎠⎝ L ⎠5 ⎝ L ⎠5

I D4 I D3
= =
25
=
W
L 4 ( )⎛W ⎞
⇒ ⎜ ⎟ = 4.08
I O 2 I O 2 150 24.5 ⎝ L ⎠4

10.65
For vGS = 0, iD = I DSS (1 + λ vDS )
a. VD = −5 V, vDS = 5
iD = ( 2 ) ⎡1 + ( 0.05 )( 5 ) ⎤ ⇒ iD = 2.5 mA
⎣ ⎦
b. VD = 0, vDS = 10
iD = ( 2 ) ⎡1 + ( 0.05 )(10 ) ⎤ ⇒ iD = 3 mA
⎣ ⎦
c. VD = 5 V, vDS = 15 V
iD = ( 2 ) ⎡1 + ( 0.05 )(15 ) ⎤ ⇒ iD = 3.5 mA
⎣ ⎦

10.66
2
⎛ V ⎞
I 0 = I DSS ⎜ 1 − GS ⎟
⎝ VP ⎠
2
⎛ V ⎞
2 = 4 ⎜ 1 − GS ⎟
⎝ VP ⎠
VGS 2
= 1− = 0.293
VP 4
So VGS = ( 0.293)( −4 ) = −1.17 V
VS
Then I 0 = and VS = −VGS
R
−V ( −1.17 )
R = GS = − ⇒ R = 0.586 kΩ
I0 2
Finish solution:
See solution

10.66
Completion of solution
Need vDS ≥ vDS ( sat ) = vGS − VP
= −1.17 − ( −4 )
vDS ≥ 2.83 V
So VD ≥ vDS ( sat ) + VS = 2.83 + 1.17 ⇒ VD ≥ 4 V

10.67
⎛V ⎞
a. I REF = I S 1 exp ⎜ EB1 ⎟
⎝ VT ⎠
⎛I ⎞ ⎛ 1× 10−3 ⎞
or VEB1 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ −13 ⎟
⇒ VEB1 = 0.5568
⎝ I S1 ⎠ ⎝ 5 × 10 ⎠
5 − 0.5568
b. R1 = ⇒ R1 = 4.44 kΩ
1
c. From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V
⎛ 2.5 ⎞
−12 ⎛ VI ⎞ ⎡ 2.5 ⎤ −3
⎜ 1 + 80 ⎟
10 exp ⎜ ⎟ ⎢1 + ⎥ = 10 ⎜ 0.5568 ⎟
⎝ VT ⎠ ⎣ 120 ⎦ ⎜1+
⎜ ⎟
⎟
⎝ 80 ⎠
⎛V ⎞ ⎛ 1.03125 ⎞
1.0208333 × 10−12 exp ⎜ I ⎟ = (10−3 ) ⎜ ⎟
⎝ VT ⎠ ⎝ 1.00696 ⎠
Then VI = 0.026 ln (1.003222 × 109 )
So VI = 0.5389 V
− (1/ VT )
d. Av =
(1/ VAN ) + (1/ VAP )
1
−
−38.46
Av = 0.026 =
1 1 0.00833 + 0.0125
+
120 80
Av = −1846

10.68
⎛I ⎞ ⎛ 0.5 × 10−3 ⎞
a. VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ −12 ⎟ ⇒ VBE = 0.5208
⎝ I S1 ⎠ ⎝ 10 ⎠
5 − 0.5208
b. R1 = ⇒ R1 = 8.96 kΩ
0.5
c. Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each
other.
⎛ V ⎞⎛ V ⎞ ⎛ VBE ⎞ ⎛ VCE 2 ⎞
I CO = I SO exp ⎜ EBO ⎟⎜ 1 + ECO ⎟ = I C 2 = I S 2 exp ⎜ ⎟ ⎜1 + ⎟
⎝ VT ⎠⎝ VAP ⎠ ⎝ VT ⎠ ⎝ VAN ⎠
⎛ V ⎞ ⎛ 2.5 ⎞ ⎛ VBE ⎞ ⎛ 2.5 ⎞
5 × 10−13 exp ⎜ EBO ⎟ ⎜ 1 + −12
⎟ = 10 exp ⎜ ⎟ ⎜1 + ⎟
⎝ VT ⎠ ⎝ 80 ⎠ ⎝ VT ⎠ ⎝ 120 ⎠
⎛V ⎞ ⎛V ⎞
5.15625 × 10−13 exp ⎜ EBO ⎟ = 1.020833 × 10−12 exp ⎜ BE ⎟
⎝ VT ⎠ ⎝ VT ⎠
⎛V ⎞
exp ⎜ EBO ⎟
⎝ VT ⎠ = 1.9798 = exp ⎛ VEBO − VBE ⎞
⎜ ⎟
⎛V ⎞ ⎝ VT ⎠
exp ⎜ BE ⎟
⎝ VT ⎠
VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 )
VEBO = 0.5386 ⇒ VI = 5 − 0.5386 ⇒ VI = 4.461 V

− (1/ VT )
d. Av =
(1/ VAN ) + (1/ VAP )
1
−
0.026 = −38.46
Av =
1 1 0.00833 + 0.0125
+
120 80
Av = −1846

10.69
a. M1 and M2 matched.
For I REF = I 0 , we have VSD 2 = VSG = VSG 3 = VDS 0 = 2.5 V
For M1 and M3:
⎛1 ⎞⎛W ⎞
I REF = ⎜ μ p Cox ⎟ ⎜ ⎟ (VSG + VTP ) (1 + λPVSD )
2

⎝2 ⎠ ⎝ L ⎠1
⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞
100 = 10 ⎜ ⎟ ( 2.5 − 1) ⎡1 + ( 0.02 )( 2.5 ) ⎤ ⇒ ⎜ ⎟ = 4.23 = ⎜ ⎟ = ⎜ ⎟
2
⎣ ⎦
⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠3 ⎝ L ⎠ 2
For M0:
⎛1 ⎞⎛W ⎞
I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS − VTN ) (1 + λnVDS )
2

⎝2 ⎠ ⎝ L ⎠0
⎛W ⎞ ⎛W ⎞
100 = 20 ⎜ ⎟ ( 2 − 1) ⎡1 + ( 0.02 )( 2.5 ) ⎤ ⇒ ⎜ ⎟ = 4.76
2
⎣ ⎦
⎝ L ⎠0 ⎝ L ⎠0
b.
1 1
r0 n = r0 p = = = 500 kΩ
λ I0 ( 0.02 )( 0.1)
⎛1 ⎞⎛W ⎞
g m = 2 K n I O = 2 ⎜ μ n Cox ⎟ ⎜ ⎟ I O
⎝ 2 ⎠ ⎝ L ⎠o
=2 ( 0.02 )( 4.76 )( 0.1)
g m = 0.195 mA/V
Av = − g m ( r0 n || r0 p ) = − ( 0.195 )( 500 || 500 ) ⇒ Av = −48.8

10.70
I REF = K p1 (VSG + VTP )
2
a.
100 = 100 (VSG − 1) ⇒ VSG = 2 V
2

b. From Eq. 10.89
⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2
VO = ⎣ ⎦− n I TN

λn + λ p I REF ( λn + λ p )
⎡1 + ( 0.02 )(10 − 2 ) ⎤
⎦ − 100 (VI − 1)
2

5= ⎣
0.02 + 0.02 100 ( 0.02 + 0.02 )
(VI − 1)
2

5 = 29 −
0.04
(VI − 1)
2
= 0.96
VI = 1.98 V

c. Av = − g m ( r0 n || r0 p )
1 1
r0 n = r0 p = = = 500 kΩ
λ I REF ( 0.02 )( 0.1)
g m = 2 K n I REF = 2 ( 0.1)( 0.1) = 0.2 mA / V
Av = − ( 0.2 )( 500 || 500 ) ⇒ Av = −50

10.71
5 − 0.6 5 − 0.6
I REF = = = 0.22 mA
R1 20
From Eq. 10.96
⎛I ⎞ 0.22
−⎜ C ⎟ −
Av = ⎝ VT ⎠ = 0.026
⎛ IC 1 I C ⎞ 0.22 + 1 + 0.22
⎜ + + ⎟
⎝ VAN RL VAP ⎠ 140 RL 90
−8.4615 −8.4615
= =
1 1
0.0015714 + + 0.002444 0.004016 +
RL RL
(a) RL = ∞, Av = −2107
(b) RL = 250 K, Av = −1056
(c) RL = 100 K, Av = −604

10.72
5 − 0.6
I REF = = 0.1257 mA
35
Then
I CO = 2 I REF = 0.2514 mA
From Eq. 10.96
−0.2514
0.026 −9.6692
Av = =
0.2514 0.2514 1 1
+ + 0.002095 + 0.0031425 +
120 80 RL RL
−9.6692
Av =
1
0.0052375 +
RL
(a) RL = ∞ Av = −1846
(b) RL = 250 K, Av = −1047

10.73
(a) To a good approximation, output resistance is the same as the widlar current source.
R0 = r02 ⎡1 + g m 2 ( rπ 2 || RE ) ⎤
⎣ ⎦
(b) Av = − g m 0 ( r0 || RL || R0 )

10.74
Output resistance of Wilson source

β r03
R0 ≅
2
Then
Av = − g m ( r0 || R0 )
V 80
r03 = AP = = 400 kΩ
I REF 0.2
VAN 120
r0 = = = 600 kΩ
I REF 0.2
I REF 0.2
gm = = = 7.692 mA/V
VT 0.026
⎡ (80 )( 400 ) ⎤
Av = −7.69 ⎢600 ⎥ = −7.69 [ 600 ||16, 000] ⇒ Av = −4448
⎢
⎣ 2 ⎥
⎦

10.75
(a) I D 2 = I D 0 = I REF = 200 μ A
1 1
For M 2 ; ro 2 = = = 250 K
λP I D 2 ( 0.02 )( 0.2 )
⎛ 0.04 ⎞
gm2 = 2 K P I D2 = 2 ⎜ ⎟ ( 35 )( 0.2 )
⎝ 2 ⎠
g m 2 = 0.748 mA/V
1 1
For M 0 ; r∞ = = = 333 K
λn I Do ( 0.015 )( 0.2 )
⎛ 0.08 ⎞
g mo = 2 ⎜ ⎟ ( 20 )( 0.2 ) ⇒ g mo = 0.80 mA/V
⎝ 2 ⎠
(b) Av = − g mo ( ro 2 || roo ) = − ( 0.80 )( 250 || 333)
Av = −114.3
Want Av = −57.15 = −0.80 (142.8 || RL )
(c) 142.8 RL
142.8 || RL = 71.375 = ⇒ RL = 143 K
142.8 + RL

10.76
Assume M1, M2 matched
I REF = I D 2 = I Do = 200 μ A
1 1
ro 2 = = = 250 K
λ p I D 2 ( 0.02 )( 0.2 )
1 1
roo = = = 333 K
λn I D 0 ( 0.015 )( 0.2 )
Av = − g mo ( ro 2 roo )
−100 = − g mo ( 250 333) ⇒ g mo = 0.70 mA/V

⎛ 0.08 ⎞⎛ W ⎞
g mo = 2 ⎜ ⎟⎜ ⎟ ( 0.2 ) = 0.70
⎝ 2 ⎠⎝ L ⎠0
⎛W ⎞
⎜ ⎟ = 15.3
⎝ L ⎠0

⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′ ⎞ ⎛ W ⎞
Now ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟
p

⎝ 2 ⎠ ⎝ L ⎠0 ⎝ 2 ⎠ ⎝ L ⎠2
⎛ 80 ⎞ ⎛ 40 ⎞⎛ W ⎞
⎜ ⎟ (15.3) = ⎜ ⎟⎜ ⎟
⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ 2
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = 30.6
⎝ L ⎠ 2 ⎝ L ⎠1

10.77

Since Vsg 3 = 0, the circuit becomes

Vx − Vsg 2
′
I x = − g mVsg 2 + and Vsg 2 = I x ro 3
r02
Then
⎛ r ⎞ V
′
I x ⎜ 1 + g m ro3 + o3 ⎟ = x
⎝ ro 2 ⎠ ro 2
so that
Vx ⎛ r ⎞
′
= Ro = ro 2 ⎜1 + g m ro 3 + o 3 ⎟
Ix ⎝ ro 2 ⎠
or
Ro = ro 2 + ro 3 (1 + g m ro 2 )
′
vo
Av = = − g m1 ( ro1 Ro )
vi
Now
g m1 = 2 ( 0.050 )( 20 )( 0.10 ) = 0.632 mA / V
1 1
ro1 = = = 500 k Ω
λn I DQ ( 0.02 )( 0.10 )
′
g m = 2 K p I DQ = 2 ( 0.020 )(80 )( 0.1) = 0.80 mA / V
1 1
ro 2 = ro 3 = = = 500 k Ω
λ p I DQ ( 0.020 )( 0.1)
Then
Ro = 500 + 500 ⎡1 + ( 0.8 )( 500 ) ⎤ ⇒ 201 M Ω
⎣ ⎦
Av = − ( 0.632 ) ( 500 201000 ) ⇒ Av = −315

10.78
From Eq. 10.105
2
− gm
Av =
1 1
+
ro3 ro 4 ro1 ro 2
⎛ k′ ⎞⎛W ⎞
g m = 2 ⎜ n ⎟ ⎜ ⎟ I D1
⎝ 2 ⎠⎝ L ⎠
=2 ( 0.050 ) ( 20 ) ( 0.08 )
g m = 0.5657 mA / V
1 1
ro = = = 625 K
λ I D ( 0.02 )( 0.08 )
− ( 0.5657 )
2
−0.3200
Av = =
1
+
1 2 ( 0.00000256 )
( 625) ( 625)
2 2

Av = −62,500

10.79

Vπ 2 Vπ 2 V − ( −Vπ 2 )
(1) g m1Vi = + + g m 2Vπ 2 + O
rπ 2 ro1 ro 2
VO VO − ( −Vπ 2 )
(2) + + g m 2Vπ 2 = 0
RO 3 ro 2
⎛ 1 1 1 ⎞ V
(1) g m1Vi = Vπ 2 ⎜ + + gm2 + ⎟ + O
⎝ rπ 2 ro1 ro 2 ⎠ ro 2
⎛ 1 1 ⎞ ⎛ 1 ⎞
(2) VO ⎜ + ⎟ + Vπ 2 ⎜ + gm2 ⎟ = 0
⎝ RO 3 ro 2 ⎠ ⎝ ro 2 ⎠
1
g m >>
ro
⎛ 1 + β ⎞ VO
(1) g m1Vi = Vπ 2 ⎜ ⎟+
⎝ rπ 2 ⎠ ro 2
⎛ 1 1 ⎞
(2) VO ⎜ + ⎟ + Vπ 2 ⋅ g m 2 = 0
⎝ RO 3 ro 2 ⎠
VO ⎛ 1 1 ⎞
(3) Vπ 2 = − ⎜ + ⎟
gm2 ⎝ RO 3 ro 2 ⎠
Then
VO ⎛ 1 1 ⎞⎛ 1 + β ⎞ VO
(1) g m1Vi = − ⎜ + ⎟⎜ ⎟+
gm2⎝ RO 3 ro 2 ⎠⎝ rπ 2 ⎠ ro 2
⎛ 1 1 ⎞ ⎛ 1 + β ⎞ VO
= −VO ⎜ + ⎟⎜ ⎟+
⎝ RO 3 ro 2 ⎠ ⎝ β ⎠ ro 2
VO ⎛ 1 + β ⎞
≈− ⎜ ⎟
RO 3 ⎝ β ⎠
VO ⎛ β ⎞
= − g m1 RO 3 ⎜ ⎟
Vi ⎝1+ β ⎠
From Equation (10.20) RO 3 ≈ β rO 3
So

VO − g m1ro3 β 2 0.25
Av = = gm = = 9.615 mA/V
Vi 1+ β 0.026
80
ro3 = = 320 K
0.25
− ( 9.615 )( 320 )(120 )
2

Av = = −366,165
121

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