Post on 21-Apr-2018
Chapter 12 Problem Solutions 12.1 (a)
5
5
5 5
15 10120
1 (5 10 )120 (120)(5 10 ) 5 10
0.008331
fAAAβ
ββ
β
=+
×=+ ×
+ × = ×=
(b) 3
3
3 3
5 101201 (5 10 )
120 (120)(5 10 ) 5 100.008133
ββ
β
×=+ ×
+ × = ×=
12.2 (a) 0.151f
AAA
ββ
= =+
T Aβ= (i) T = ∞ (ii) 4 380 dB 10 1.5 10A A T= ⇒ = ⇒ = × (iii) 15T =
(i) 1 6.667fAβ
= =
(ii) 6.662fA = (iii) 6.25fA = (b) (i) T = ∞ (ii) 32.5 10T = × (iii) 25T =
(i) 1 4.00fAβ
= =
(ii) 3.9984fA = (iii) 3.846fA = 12.3 (a)
1 1251f
AAAβ β
= ≅ =+
0.0080β = (b)
[ ][ ]
(125)(0.9975) 124.6875
124.68751 (0.008)
124.6875 1 (0.008) A124.6875 1 0.9975
49,875
fA
AA
AA
A
= =
=+
+ =
= −=
12.4
(a) 4
4
3
101001 (10 )
9.9 10β
β −
=+
= ×
(b)
4
100 ( 0.10) 0.001 0.10%10
f f
f
f
f
dA A dAA A A
dAA
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= − = − ⇒ = −⎜ ⎟⎝ ⎠
12.5
4
4
4
3
0.001 ( 0.10)5 10
500
5 105001 (5 10 )
1.98 10
f f
f
f
f
dA A dAA A A
A
A
ββ −
⎛ ⎞= ⎜ ⎟⎝ ⎠
− = −×
=
×=+ ×
= ×
12.6 (a) For Fig. P12.6(a)
1
1 1 1
200 101 200 1 10
o o
i o
v vv vβ β
= =+ +
1 1
200 10 401 200 1 10fA
β β⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
21 1 1 150 (1 200 )(1 10 ) 1 210 2000β β β β= + + = + +
21 12000 210 49 0β β+ − =
1
1
210 44100 4(2000)(49)2(2000)
0.1126
β
β
− ± +=
=
For Fig P12.6(b)
2
2
(200)(10)401 (200)(10)0.0245
ββ
=+
=
Fig. P12.6(a) (b)
1 180180 10 (8.4634)(4.704)
1 (180)(0.1126) 1 (10)(0.1126)39.81
39.81 40 0.475%40
f
f
f
f
A
A
AdAA
=
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠=
−= ⇒ −
Fig. P12.6(b) (180)(10) 39.91
1 (180)(10)(0.0245)39.91 40 0.225%
40
f
f
f
A
dAA
= =+
−= ⇒ −
(c) Fig. P12.6(b) is a better feedback circuit. 12.7 (a) ( 10)( 15)( 20) 3000O
O S
V V VV V V
ε ε
ε β= − − − = −= +
So 3000( )O O SV V Vβ= − + We find
30001 3000
Ovf
S
VAV β
−= =+
For 3000120 0.0081 3000vfA β
β−= − = ⇒ =+
(b) Now ( 9)( 13.5)( 18) 2187OV V Vε ε= − − − = − Then
2187 2187 118.241 2187 1 2187(0.008)vfA
β− −= = = −+ +
120 118.24% change 100 1.47% change120
−= × ⇒
12.8
5(10 )(4) (50) 8B Bf f kHz= ⇒ = 12.9 (a) 5
3 3(50) (10 )(4) 8dB dBf f kHz− −= ⇒ =
(b) 53 3(10) (10 )(4) 40dB dBf f kHz− −= ⇒ =
12.10
30(50)(20 10 ) 5A× = so 5
0 2 10A = × 12.11
Low freq. 0
01fAAA β
=+
5000100 0.00981 (5000)
ββ
= ⇒ =+
Freq. response
1 2
1 2
1 2
1 2 1 2
1 2 1 2
5000
1 1
(5000)(0.0098)1 11 1
5000
1 1 49
5000
1 49
5000
50
f
f fj jf fAA
Af fj jf f
f fj jf f
f f jf jfj jf f f f
f f jf jfj jf f f f
β
⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= =
+ +⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=⎛ ⎞⎛ ⎞+ + + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Also 0 100
1 1 1
ff
A B A B A B
AA
f f f f f fj j j j j jf f f f f f
= =⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
So
1 2 1 2
100 10011 1
50 50 50A B A B
f f f f f f jf jfj j j j j jf f f f f f f f
=⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
Then
1 2
1 1 1 150 50A Bf f f f
+ = +
and 1 2
1 150A Bf f f f
=
1 10f = and 2 2000f = 1 1 1 1 0.002 0.000010 0.002010
50(10) 50(2000)A Bf f+ = + = + =
and
6
1 1 1(50)(10)(2000) 10
B
A B A
ff f f
= ⇒ =
Then 6
1 0.00201010
B
B
ff
+ =
6 2 3
6 2 3
10 1 2.01 1010 2.01 10 1 0
B B
B B
f ff f
− −
− −
+ = +
− × + =
3 6 6
6
3 4
6
2.01 10 4.0401 10 4(10 )(1)2(10 )
2.01 10 2.0025 102(10 )
B
B
f
f
− − −
−
− −
−
× ± × −=
× ± ×=
+ sign 31.105 10 HzBf = × + sign 29.05 10 HzAf = ×
12.12 (a) Fig. P12.6(a)
1
1
1
1 1
3
1
200
110
200 1 (10)(0.1126)1 (0.1126)1
200 (4.704)1 22.52
940.73 940.73 123.5223.52 1
(23.52)40 (23.
1(23.52)
f
dB
fjf
A
fjf
fjf
f fj jf f
fjf
f
−
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
+⎢ ⎥⎜ ⎟⎜ ⎟ ⎡ ⎤⎢ ⎥⎝ ⎠= ⎢ ⎥⎢ ⎥ +⎛ ⎞ ⎣ ⎦+⎢ ⎥⎜ ⎟⎢ ⎥+⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤= ⎢ ⎥⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
= = ⋅+ +
= =+
52)(100) 2.352 kHz⇒
Fig P12.6(b)
1
1
1
3
1
(200)(10)
12000
(0.0245)(200)(10)1 1 491
2000 1 (50)(100) 5 KHz50 1
(50)
f
dB
fjfA
fjf fjf
ffj
f
−
+= =
+ + ++
= ⋅ = ⇒+
(b) Overall feedback ⇒ wider bandwidth. 12.13
0 1 2 1
00
0
1000(100) (1) (100)(10) (1)(1) 10001
i n
i n
v A A v A vSv v vN
= +
= + = + ⇒ = =
12.14 (a)
(b)
Circuit (b) – less distortion 12.15 (a) Low input R ⇒ Shunt input Low output R ⇒ Shunt output Or a Shunt-Shunt circuit (b) High input R ⇒ Series input High output R ⇒ Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a) 4 5(max) (1 ) 10(1 10 ) (max) 10i i iR R T R k= + = + ⇒ ≅ Ω
34
10(min) 101 1 10
ii
RR k
T−= = ≅ Ω
+ +
Or (min) 1iR = Ω
(b) 4 4(max) (1 ) 1(1 10 ) (max) 10o o oR R T R k= + = + ⇒ ≅ Ω
44
1(min) 101 1 10
oo
RR k
T−= = ≅ Ω
+ +
Or (min) 0.1oR = Ω 12.17
Overall Transconductance Amplifier, og
i
iAv
= Series output = current signal and Shunt input = current
signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.
12.18
1
1 2
||||
ix
i
R RV V
R R Rε⎛ ⎞
= − ⋅⎜ ⎟+⎝ ⎠
12.19
3
50 48 2mV
5 2.5 10 V/V0.002
0.048 0.0096 V/V5
5 100 V/V0.05
b
b
i f
ov
f
o
orf
i
V V V
VA
VVVVAV
ε
ε
β
= − = − =
= = = ×
= = =
= = =
12.20
2 2
1 1
1 20 19vfR RAR R
⎛ ⎞≈ + = ⇒ =⎜ ⎟⎝ ⎠
d S iv i R=
0
1 2
( )S d s ds
v v v v viR R− − −
= + (1)
0 0 0
0 2
( ) 0L d s dv A v v v vR R
− − −+ = (2)
00
0 2 0 2
( )1 1 L d S dA v v vv
R R R R⎛ ⎞ −
+ = +⎜ ⎟⎝ ⎠
0
0 20
0 2
( )
1 1
L d S dA v v vR R
v
R R
−+=
⎛ ⎞+⎜ ⎟
⎝ ⎠
From (1):
0
2 0 2
1 2
0 2
0
0 22
2 21 2 1 2
0 0
( )1
1 1
111 1 1 1
1 1
L d S d
S d S dS
L
S S d
d S i
A v v vR R Rv v v v
iR R
R R
AR RRi v v
R RR R R RR R
v i R
⎡ ⎤−⋅ +⎢ ⎥− − ⎣ ⎦= + −⎛ ⎞
+⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞ −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= + − − + +⎜ ⎟⎜ ⎟+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠=
02 2
1 2 0 0 2 1 2 0 2
2 2
0 0
02 2 2
0 1 1 0 0 0 1 1 0 0
0 2
1 1 1 1 1 11 11
1 1
1 1 1 1 1 11
Li
S S
LS i S
S
AR RRR R R R R R R R R
i vR RR R
AR R Ri R vR R R R R R R R R R
i R R
⎧ ⎫⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + − + + −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠⎢ ⎥+ =⎨ ⎬ ⎢ ⎥⎪ ⎪+ +⎢ ⎥⎪ ⎪ ⎢ ⎥⎣ ⎦⎩ ⎭⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪+ + + ⋅ + + = + ⋅ +⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭
+ 0 02 20
1 1 1 1
1 1 (1)i L SR RR RR A vR R R R
⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪+ + + + = + +⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭
Let 2 190 kR ,= Ω 1 10 kR = Ω
5
7
0.1 0.10.1 190 100 20 10 2010 10
(1.000219 10 ) (20.01)
S S
S S
i v
i v
⎧ ⎫⎡ ⎤ ⎡ ⎤+ + ⋅ + + = +⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎩ ⎭× =
55 10 k 500 MSif if
S
vR Ri
= ≅ × Ω ⇒ ≅ Ω
Output Resistance
0
0 2 1
1
1 2
0 1
0 0 0 1 2 2 1
1
||||
||||1 1 1
( || ) ||
|| 10 ||100 9.09
X L d XX
i
id X
i
L iX
X f i i
i
V A v VIR R R R
R Rv V
R R RA R RI
V R R R R R R R R R
R R
−= +
+−
= ⋅+
⋅= = + +
+ +
= =
5
0
4
50 0
1 1 10 9.09 10.1 0.1 9.09 190 190 9.09
10 4.566 10 0.005022.19 10 k 0.0219
f
f f
R
R R−
⎛ ⎞= + ⋅ +⎜ ⎟+ +⎝ ⎠
= + × += × Ω ⇒ = Ω
12.21 a.
0
1 2
( )S d S dv v v v vR R− − −
= and 0d
vvA
=
0
1 2 2 1 2
0 0
2 1 2
0 2
1 2 2 1
2
10
2
1
1 1
1 1
1 1 11 1
1
11 1
S Sd
S
S
v v v vR R R R R
v vR A R R
v RvR R R A R
RRv
v RA R
⎛ ⎞+ = + +⎜ ⎟
⎝ ⎠⎛ ⎞
= + +⎜ ⎟⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞
+⎜ ⎟⎝ ⎠=⎛ ⎞
+ +⎜ ⎟⎝ ⎠
which can be written as 0
2
1
1 / 1vf
S
v AAv RA
R
= =⎡ ⎤⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
b. 2
1
1
1 RR
β =+
c. 5
5
10201 (10 )β
=+
So
5
5
10 120 0.0499910
β β−
= ⇒ =
Then 2 2
1 1
1 11 1 19.0040.04999
R RR Rβ
= − = − ⇒ =
d. 49 10A → × 4
4
9 10 19.999561 (9 10 )(0.04999)fA ×= =
+ ×
434.444 10 2.222 10 % 0.005%
20f f
f f
A AA A
−−Δ Δ− ×= = − × ⇒ = −
12.22
[ ]
1000 90.9 A/A1 1 (1000)(0.01)
1 90.91 1 (0.01)(1000)
(1 ) 10 1 (0.01)(1000) 110 k
iif
i i
iif if
i i
of o i i of
AAA
RR R
AR R A R
β
ββ
= = =+ +
= = ⇒ = Ω+ +
= + = + ⇒ = Ω
12.23
50 47.5 2.5 A
5 2000 A/A0.0025
0.0475 0.0095 A/A5
5 100 A/A0.05
i fb
oi
fbi
o
oif
i
I I II
AIIII
AI
ε
ε
μ
β
= − = − =
= = =
= = =
= = =
12.24 a.
Assume that 1V is at virtual ground.
0 fb FV I R= − Now
00 0
3 3
fb Ffb
fb S
I RVI I IR R
I I Iε
= + = −
= −
and 0
0 ii
II A IAε= =
so 0
fb Si
II IA
= −
From above
03
00
3
03 3
1
1
11 1 1
Ffb
FS
i
F FS
i
RI IR
I RI IA R
R RI IR A R
⎛ ⎞+ =⎜ ⎟
⎝ ⎠⎛ ⎞⎛ ⎞
− + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎣ ⎦
or
30
3
3
1
11 1
11
F
ifS F
i
iif
i
F
RRI
AI R
A R
A AARR
⎛ ⎞+⎜ ⎟
⎝ ⎠= =⎡ ⎤⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
= =+⎛ ⎞
+⎜ ⎟⎝ ⎠
b.
3
1
1i
FRR
β =⎛ ⎞
+⎜ ⎟⎝ ⎠
c. 5
5
10251 (10 ) iβ
=+
so
5
5
10 125 0.0399910i iβ β
−= ⇒ =
so 3 3
1 11 1 24.00.03999
F F
i
R RR Rβ
= − = − ⇒ =
d. 5 5 410 (0.15)(10 ) 8.5 10iA = − = ×
so 4
4
8.5 10 24.99891 (8.5 10 )(0.03999)ifA ×= =
+ ×
so 3
5 31.10 10 4.41 10 4.41 10 %25
if
if
AA
−− −Δ ×= − = − × ⇒ − ×
12.24 b.
( )0 1 2m gs m LV g V g V Rπ= + (1)
0i gsV V V= + (2)
1 12
2 2
10 0FEm
E E
V V h Vg V Vr R r R
ππ π
π π
⎛ ⎞++ + = ⇒ + =⎜ ⎟
⎝ ⎠ (3)
11
1
0m gsD
V V Vg Vr R
π π
π
−= + =
or
1 1 11
D m gsD
V VV R g V
r Rπ π
π
⎡ ⎤= + −⎢ ⎥
⎣ ⎦ (4)
Then
11
2 1
1 0FE Dm gs
E D
V Vh RV g Vr R r R
π ππ
π π
⎛ ⎞ ⎡ ⎤++ + − =⎜ ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦ (3)
1 11
2 2 2
1 1FE D Dm gs
E E E
h R RV g Vr R r R Rππ π
⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪+ + =⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎝ ⎠⎝ ⎠⎩ ⎭
Let
11
2
1 Dm gs
eq E
RV g VR Rπ
⎛ ⎞⋅ = ⎜ ⎟
⎝ ⎠
so 11
2
Dm eq gs
E
RV g R VRπ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
Then
10 1 1 2
2
Dm gs m m eq gs L
E
RV g V g g R V RR
⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (1)
so
( )10 1 2 0
2
1 Dm L m eq i
E
RV g R g R V VR
⎡ ⎤⎛ ⎞= + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
so
11 2
20
11 2
2
1
1 1
Dm L m eq
Ev
i Dm L m eq
E
Rg R g RRVA
V Rg R g RR
⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦= =⎡ ⎤⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
c. Set 0iV =
1 2X
X m gs mL
VI g V g VRπ+ + =
gs XV V= − From part (b), we have
1 11 1
2 2
D Dm eq gs m eq X
E E
R RV g R V g R VR Rπ
⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Then
11 2
0 2
1 1X Dm m eq
X L E
I Rg g RV R R R
⎛ ⎞= = + ⎜ ⎟
⎝ ⎠
or
01 1
1 22
1 1L
m Dm m eq
E
R Rg Rg g R
R
=⎛ ⎞⎜ ⎟⎝ ⎠
12.25
1,S fb iI I I V I Rε ε= + =
00
3fb
VI IR
= + and 0 1 fb FV V I R= −
00 i
i
II A I IAε ε= ⇒ =
Now
13
1
3 3
1 ( )
1
fb i fb F
Ffb i
fb S e
I A I V I RR
R VI A IR R
I I I
ε
ε
= + −
⎡ ⎤+ = +⎢ ⎥
⎣ ⎦= −
1
3 3
1
3 3 3
1
( ) 1R
1 1
FS e i e
F FS e i
i
R VI I A IR
R R VI I AR R R
VIRε
⎡ ⎤− + = +⎢ ⎥
⎣ ⎦⎡ ⎤⎡ ⎤ ⎛ ⎞
+ = + + +⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦
=
13 3 3
31
3 3
1 11 1
1
1 11
F FS i
i
F
ifS F
ii
R RI V AR R R R
RRVR
I R AR R R
⎧ ⎫⎡ ⎤⎡ ⎤ ⎛ ⎞⎪ ⎪+ = ⋅ + + +⎨ ⎢ ⎥ ⎬⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
⎛ ⎞+⎜ ⎟
⎝ ⎠= =⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪⋅ + + +⎨ ⎬⎢ ⎥⎜ ⎟
⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
The 31/ R term in the denominator will be negligible. Using the results of Problem 12.15:
5
4
251 (25) 102
5 10 k 0.5
if
if if
R
R R−
=⎧ ⎫⎡ ⎤+⎨ ⎬⎣ ⎦⎩ ⎭
≅ × Ω ⇒ = Ω
Output Resistance (Let 0)LZ =
3
X XX i
F i
X
F i
V VI A IR R R
VIR R
ε
ε
= + ++
=+
so
0 3 3
11 1 , 24iX F
X f F i
AI RV R R R R R
+= = + =
+
Let 240 k ,FR = Ω 3 10 R k= Ω 51 1 10 1
10 240 2ofR+= ++
so 30 0 05
240 2 2.42 10 k or 2.42 1 10 1
F if f f
i
R RR R RA
−+ +≈ = ⇒ ≈ × Ω ≈ Ω+ +
12.26
[ ] [ ][ ] [ ]
5 0.2 A/V1 1 (4.8)(5)
1 (10) 1 (4.8)(5) 250 k
1 (10) 1 (4.8)(5) 250 k
gfz
if i z
of o z
AgA Ag
R R Ag
R R Ag
βββ
= = =+ +
= + = + = Ω
= + = + = Ω
12.27
0
0
40 38 2.0 mV
8mA 4 A/V2mV
38mV 4.75 V/A8mA8mA 0.2 A/V40mV
i fb
og
fbz
gfi
V V V
IA
VVII
A V
ε
ε
β
= − = − =
= = =
= = =
= = =
12.28
0(1 ) SFE
EFE E
V VhI Ih R
ε−+= ⋅ =
Also 0 ( )FE gI h A Vε= so 0
FE g
IV
h Aε =
Then 0
0
0
0
0 0
1
1 1
(1 ) 1
11 (1 ) 1 ( )
SFE
FE E FE g E
SFE
FE FE g E E
g FE E S
FE g E E
FE g E FE g
S E g FE E S FE g E
V Ih Ih R h A R
Vh Ih h A R R
A h R VIh A R R
h A R h AI IV R A h R V h A R
+⋅ = −
⎡ ⎤++ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤+ +
=⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤= ⋅ ⇒ ≈⎢ ⎥
+ + +⎢ ⎥⎣ ⎦
b. z EB R=
c. 5
5
5 10101 (5 10 ) zβ
×=+ ×
5
5
5 10 110 0.099998 k5 10z z ERβ β
× −= ⇒ = = Ω
×
d. If 55.5 10gA → × then 5
5
5.5 10 10.00001821 (5.5 10 )(0.099998)gfA ×= =
+ ×
541.82 10 1.82 10 %
10gf
gf
AA
−−Δ ×= ⇒ ×
12.29
(1 ) ,E FE gI h A Vε= + E SE
VI IR
ε= − and ,S iV I Rε = S S S iV V V V I Rε ε= − = −
Now ( ) 11 ( )FE g S i S S i SE
h A I R V I R IR
+ = ⋅ − −
( ) i1 1 SFE g i S
E E
VRh A R IR R
⎡ ⎤+ + + =⎢ ⎥
⎣ ⎦
( )1 1S iif E FE g i
S E
V RR R h A R
I R⎡ ⎤
= = + + +⎢ ⎥⎣ ⎦
From Problem 12.16: ( ) 51 5 10 mSFE g FE gh A h A+ ≈ = ×
0.1 kER ≈ Ω
so 5 20(0.1) (5 10 )(20) 10.1ifR ⎡ ⎤= × + +⎢ ⎥⎣ ⎦
or 610 kifR = Ω
gV A Vr
πε
π
=
0
( )XX m
V VI g VR
επ
− −= + (1)
( )( || )X g E iV I A V R Rε ε= − + (2)
or 1 ( || ) ( || )g i X E iV A R R I R Rε ε⎡ ⎤+ = −⎣ ⎦ Now:
0 0
XX m g
VVI g A r VR R
επ ε= + + (1)
0 0
( || )11 ( || )
X E i XX m g
g E i
I R R VI g A rR A R R Rπ
⎡ ⎤⎛ ⎞ −= + +⎢ ⎥⎜ ⎟ +⎢ ⎥⎝ ⎠ ⎣ ⎦
0
00
( || )111 ( || )
Xf
X
E im g
g E i
VRI
R RR g A rR A R Rπ
=
⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪= + + ⎢ ⎥⎨ ⎬⎜ ⎟ +⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
55 10 mSm g FE gg r A h Aπ = = ×
Let 100FEh = so 35 10 mSgA = × || 0.1 || 20 0.1kE iR R = ≈ Ω
Then
50 3
1 0.150 1 5 1050 1 (5 10 )(0.1)fR
⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞= + × +⎨ ⎬⎜ ⎟ ⎢ ⎥+ ×⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
or 0 5.04 MfR = Ω
12.30
5 0.2 V/μA1 1 (4.8)(5)
1 40 1 1 (4.8)(5)
1 40 1 1 (4.8)(5)
zzf
g z
iif if
g z
oof of
g z
AAA
RR RA
RR RA
β
β
β
= = =+ +
= = ⇒ = Ω+ +
= = ⇒ = Ω+ +
12.31
40 38 2 μA
8 V42 μA
38 μA4.75 8 V8 V0.240 μA
i fb
oz
fbg
o
ozf
i
I I I
VA
IIVV
AI
ε
ε
β
= − = − =
= = =
= = =
= = =
12.32 a.
Assuming 1V is at virtual ground
0( ) fb FV I R− = − and 00( ) z
z
VV A I IAε ε− = − ⇒ =
fb SI I Iε= −
So 00 ( )S F S F F
z
VV I I R I R R
Aε⎛ ⎞
= − = − ⎜ ⎟⎝ ⎠
0 1 FS F
z
RV I RA
⎡ ⎤+ =⎢ ⎥
⎣ ⎦
so 0
1
F z Fzf
S z FF
z
V R A RAI A RR
A
= = =+⎡ ⎤
+⎢ ⎥⎣ ⎦
or 111
z zzf
z gz
F
A AAA
AR
β= =
+⎛ ⎞+ ⎜ ⎟
⎝ ⎠
b. 1g
FRβ =
c. 6
46
5 105 101 (5 10 ) gβ
×× =+ ×
6
45
6
5 10 15 10 1.98 10
5 10g gβ β −
× −×= ⇒ = ×
×
1 50.5 kF Fg
R Rβ
= ⇒ = Ω
d. 6 6(0.9)(5 10 ) 4.5 10zA = × = ×
64
6 5
4.5 10 4.994 101 (4.5 10 )(1.98 10 )zfA −
×= = ×+ × ×
34
55.4939 1.11 10 0.111%5 10
zf
zf
AA
−Δ= − = − × ⇒ −
×
12.33
1 0 0,i z xV I R V A I V A Iε ε ε= − = − ⇒ =
fb SI I Iε= − and 0 1 fb FV V I R− = −
1
1 11
1
1
( )
1
1
z S F
z S F Fi i
z FS F
i i
Fif
S z F
i i
A I V I I R
V VA V I R RR R
A RI R VR R
V RRI A R
R R
ε ε− = − −
⎛ ⎞ ⎛ ⎞− = − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎡ ⎤
= + +⎢ ⎥⎣ ⎦
= =⎡ ⎤
+ +⎢ ⎥⎣ ⎦
Or, using the results from problem 12.18.
[ ]
3
6 3
3 3
3
50.5 105 10 50.5 101
10 10 10 10
50.5 10 99.79 1 500 5.05
if
if
R
R
×=⎡ ⎤× ×+ +⎢ ⎥× ×⎣ ⎦
×= ⇒ = Ω+ +
12.34 Assume 0.2 CQI mA=
Then (100)(0.026) 13 0.2
r kπ = = Ω
(1) 1 2 2 1 2
1 1 1i A A o i oAA
i i i
V V V V V VV VR R R R R R R R
⎛ ⎞− −= + ⇒ + = + +⎜ ⎟
⎝ ⎠
Now 1 1 1 (12)
10 10 10 1 10i o
A i o AV V V V V V⎛ ⎞+ = + + ⇒ + =⎜ ⎟
⎝ ⎠
or 1 ( )12A i oV V V= +
(2) 2
(1 )v o o AFE
o
A V V V Vh
R r Rε
π
⎛ ⎞− −+ =⎜ ⎟+⎝ ⎠
where i AV V Vε = − Then
0 2
( )(1 )v i A o o A
FEA V V V V V
hR r Rπ
⎛ ⎞− − −+ =⎜ ⎟+⎝ ⎠
we find
2 2
(1 ) (1 ) (1 )v i FE o FE o v A FE A
o o o
A V h V h V A V h VR r R r R R r Rπ π π
+ + +− − = −
+ + +
Then 3 3(5 10 )(101) (101) (5 10 )(101) 114 14 10 14 10
i o oA
V V VV
⎛ ⎞× ×− − = −⎜ ⎟⎝ ⎠
Rearranging terms, we find
10.97
1 1( ) ( 10.97 ) 0.997512 12
ovf
i
i i iif i
i i Ai A
i
A i o i i i
VAV
V V VR RI V VV V
R
V V V V V V
= =
⎛ ⎞= = = ⎜ ⎟−⎛ ⎞− ⎝ ⎠
⎜ ⎟⎝ ⎠
= + = + =
Then 1 (10 ) 4
1 0.9975if ifR k R M⎛ ⎞= Ω ⇒ = Ω⎜ ⎟−⎝ ⎠
To find the output resistance:
2 1
( )(1 )||
v FE xx
o i
A V h VIR r R R Rε
π
++ =
+ +
1
1 2
||||
ix
i
R RV V
R R Rε⎛ ⎞
= − ⋅⎜ ⎟+⎝ ⎠
Now 1 || 1 ||10 0.909iR R = =
Then 0.0833 xV Vε = −
Now
{ }
3
3
(5 10 )(0.0833) 1 1(101)1 13 10 0.909
3.012 10 0.0917
x x
x
I V
V
⎧ ⎫⎡ ⎤× +⎪ ⎪= +⎨ ⎬⎢ ⎥+ +⎪ ⎪⎣ ⎦⎩ ⎭
= × +
Or 43.32 10 0.332x
of ofx
V R k RI
−= = × Ω ⇒ = Ω
12.35 a. Neglecting base currents
2 2
1 0
0.5 mA, 12 (0.5)(22.6) 0.7 V0.5 mA 0
C C
C
I VI v
= = − == ⇒ =
Then 3 2 mACI =
b. 1 21
(100)(0.026) 5.2 k0.5
FE T
C
h Vr rIπ π
⋅= = = = Ω
1 2
3
3
0.5 19.23 mA / V0.026
(100)(0.026) 1.3 k2
2 76.92 mA / V0.026
m m
m
g g
r
g
π
= = =
= = Ω
= =
( )
1 21 1 1 2
1 1
1 2 1 1 21
11 2 2
1
0
1( ) 0 1
( )
m m
m
i S b
V Vg V g Vr r
V V g V Vr
VV R r V V
r
π ππ π
π π
π π π ππ
ππ π
π
+ + + =
⎛ ⎞+ + = ⇒ = −⎜ ⎟
⎝ ⎠
= + − +
or 1 2 21
1 Si b
RV V V V
rπ ππ
⎛ ⎞= + − +⎜ ⎟
⎝ ⎠
But 2 1V Vπ π= − so
1 21
2 Si b
RV V V
rππ
⎛ ⎞= + +⎜ ⎟
⎝ ⎠ (2)
02 02 01 2
3
0mC
V V Vg VR rπ
π
−+ + = (3)
3 0 0 23 3
3 2
3 02 0
bm
L
V V V Vg V
r R RV V V
ππ
π
π
−+ = +
= −
so
202 0 0
3 2 2
1 1 1( ) bFE
L
VhV V Vr R R Rπ
⎛ ⎞ ⎛ ⎞+− = + −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ (4)
2 0 2 2
2 1 1
0b bV V V VR R r
π
π
−+ + = (5)
Substitute numbers into (2), (3), (4) and (5):
2 212
5.2i bV V Vπ⎛ ⎞= − + +⎜ ⎟⎝ ⎠
2 2(2.192)i bV V Vπ= − + (2)
02 2 01 1 1(19.23) 0
22.6 1.3 1.3V V Vπ
⎛ ⎞ ⎛ ⎞+ + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
02 2 0(0.8135) (19.23) (0.7692) 0V V Vπ+ − = (3)
02 0 2101 101 1 1 11.3 1.3 4 50 50bV V V⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
02 0 2(77.69) (77.96) (0.02)bV V V= − (4)
2 0 21 1 1 1 0
50 10 50 5.2bV V Vπ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 0 2(0.120) (0.020) (0.1923) 0bV V Vπ− + = (5)
From (2): 2 2 (2.192).b iV V Vπ= + Substitute in (4) and (5) to obtain:
02 0 2(77.69) (77.96) [ (2.192)](0.02)iV V V Vπ= − + (4 )′
2 0 2[ (2.192)](0.120) (0.020) (0.1923) 0iV V V Vπ π+ − + = (5 )′ So we now have the following three equations:
02 2 0(0.8135) (19.23) (0.7692) 0V V Vπ+ − = (3)
02 (77.69)V
0 2(77.96) (0.02) (0.04384)iV V Vπ= − − (4 )′
2 0(0.120) (0.4553) (0.020) 0iV V Vπ+ − = (5 )′ From (3): 02 0 2(0.9455) (23.64).V V Vπ= − Substitute for 02V in (4′) to obtain:
2 0 2(77.69)[ (0.9455) (23.64)] (77.96) (0.02) (0.04384)o iV V V V Vπ π− = − − or
0 20 (4.504) (0.02) (1836.5)iV V Vπ= − + Next, solve (5 )′ for 2 :Vπ
2 0(0.120) (0.4553) (0.020) 0iV V Vπ+ − =
2 0 (0.04393) (0.2636)iV V Vπ = − Finally,
0 0
0
0 (4.504) (0.02) (1836.5)[ (0.04393) (0.2636)]0 (85.18) (484.12)
i i
i
V V V VV V
= − + −= −
So 0 484.12 5.68
85.18vf vfi
VA AV
= = ⇒ =
12.36 a. 1 2|| 400 || 75 63.2 kTHR R R= = = Ω
2
1 2
1
1
75 (10) 1.579 V75 400
1.579 0.7 0.007106 mA63.2 (121)(0.5)0.853 mA
TH CC
BQ
CQ
RV VR R
I
I
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠−= =
+=
1
2
2
3
10 (0.853)(8.8) 2.49 V2.49 0.7 0.497 mA
3.610 (0.497)(13) 3.54 V3.54 0.7 2.03 mA
1.4
C
C
C
C
V
I
V
I
= − =−≈ =
= − =−≈ =
Then
1
1
2
2
3
3
(120)(0.026) 3.66 k0.8530.853 32.81 mA / V0.026
(120)(0.026) 6.28 k0.4970.497 19.12 mA / V0.026
(120)(0.026) 1.54 k2.032.03 78.08 mA / V
0.026
m
m
m
r
g
r
g
r
g
π
π
π
= = Ω
= =
= = Ω
= =
= = Ω
= =
b.
1 1 1 1i iV V V V V Vπ ε ε π= + ⇒ = − (1)
1 1 1 01 1
1 1m
E F
V V V Vg Vr R R
π ε επ
π
−+ = + (2)
2 1 1 1 2( )( || )m CV g V R rπ π π= − (3)
3 0 32 2
2 3
0mC
V V Vg VR r
π ππ
π
++ + = (4)
3 0 0 13 3
3 3m
E F
V V V Vg Vr R R
π επ
π
−+ = + (5)
Substitute numbers in (2), (3), (4) and (5): 0
1 11 1 132.81 ( )
3.66 0.5 10 10iVV V Vπ π
⎛ ⎞ ⎛ ⎞+ = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 1 0(35.18) (2.10) (0.10)iV V Vπ = − (2)
2 1(32.81) (88 || 6.28)V Vπ π= − or 2 1(120.2)V Vπ π= − (3)
3 0 32(19.12) 0
13 13 1.54V V V
V π ππ + + + =
or 2 3 0(19.12) (0.7263) (0.07692) 0V V Vπ π+ + = (4)
13 0
1 1 178.081.54 1.4 10 10
iV VV V ππ
−⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 3 0 1(78.73) (0.8143) (0.10) (0.10)iV V V Vπ π= − + (5)
Now substituting 2 1(120.2)V Vπ π= − in (4):
1 3 0(19.12)[ (120.2)] (0.7263) (0.07692) 0V V Vπ π− + + =
or 1 3 0(2298.2) (0.7263) (0.07692) 0V V Vπ π− + + =
Then 3 1 0(3164.3) (0.1059)V V Vπ π= −
Substituting 3 1 0(3164.3) (0.1059)V V Vπ π= − in (5):
1 0 0 1(78.73)[ (3164.3) (0.1059)] (0.8143) (0.10) (0.10)iV V V V Vπ π− = − + or 5
1 0(2.49 10 ) (9.152) (0.10)iV V Vπ × − = − Then
5 71 0 (3.674 10 ) (4.014 10 )iV V Vπ
− −= × − × Now substituting 5
1 0 (3.674 10 )V Vπ−= ×
7(4.014 10 )iV −− × in (2): 5 7
0(35.18)[ (3.674) 10 ) (4.014 10 )]iV V− −× − ×
or 0
0
(2.10) (0.10)(0.1013) (2.10)
i
i
V VV V
= −=
So 0 20.7i
VV
=
c. iif
i
VRI
= and 1 1i RB bI I I= +
11
iRB
B
VIR
=
11
1b
VIr
π
π
=
Now 5 7
14
1
(20.7 )(3.674 10 ) (4.014 10 )
(7.60 10 )i i
i
V V V
V Vπ
π
− −
−
= × − ×
= ×
Then
4
4
(7.60 10 )63.2 3.66
10.01582 2.077 10
iif
i i
VRV V −
−
=×+
=+ ×
or 62.4 kifR = Ω
d. To determine 0 :fR Equation (1) is modified to 1 1 0eV Vπ + = ( 0)iV = Equation (5) is modified to:
3 0 1(78.73) (0.8143) (0.10)XV I V Vπ π+ = + (5) Now
1 0(35.18) (0.10)V Vπ = − (2)
2 1(120.2)V Vπ π= − (3)
2 3 0(19.12) (0.7263) (0.07692) 0V V Vπ π+ + = (4) Now
1 0 (0.002843)V Vπ = − so
2 0
2 0
( )(0.002843)(120.2)(0.3417)
V VV V
π
π
= − −=
Then 0 3 0(0.3417)(19.12) (0.7263) (0.07692) 0V V Vπ+ + + =
or 3 0 (9.101)V Vπ = − (4) So then
0 (9.101)(78.73) XV I− +
0 0(0.8143) (0.10)( )(0.002843)V V= + − or
0 (717.3)XI V= (5) or
00 00.00139 k 1.39 f f
X
VR RI
= = Ω ⇒ = Ω
12.37 (a)
(1) 1 11
i A A OAm
E F
V V V VVg Vr R Rππ
− −+ = +
(2) 1 11 2
0B Bm
C
V Vg VR rπ
π
+ + =
(3) 2 22 3
0C C om
C
V V Vg VR rπ
π
−+ + =
(4) 3 33
C O O Am
F
V V V Vg Vr Rππ
− −+ =
1 2 3i A B C OV V V V V V V Vπ π π= − = = −
(1) ( ) 11
1 A OAi A m
E F
V VVV V gr R Rπ
⎛ ⎞ −− + = +⎜ ⎟
⎝ ⎠
(2) 11 2
1 1 ( ) 0B m i AC
V g V VR rπ
⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
(3) 22 3 3
1 1 0OC m B
C
VV g V
R r rπ π
⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
(4) ( ) 33
1 O AC O m
F
V VV V g
r Rπ
⎛ ⎞ −− + =⎜ ⎟
⎝ ⎠
1 1(100)(0.026) 14.30.182 K 550 mA/V
14.3 0.026mr gπ = = = =
2 2(100)(0.026) 4.620.563 K 178 mA/V
4.62 0.026mr gπ = = = =
3 3(100)(0.026) 4.470.582 K 172 mA/V
4.47 0.026mr gπ = = = =
(1) ( ) 1 5500.182 0.05 1.2
A OAi A
V VVV V −⎛ ⎞− + = +⎜ ⎟⎝ ⎠
(2) 1 1 (550)( ) 00.3 0.563B i AV V V⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
(3) 1 1 178 00.65 0.582 0.582
OC B
VV V⎛ ⎞+ + − =⎜ ⎟⎝ ⎠
(4) ( ) 11720.582 1.2
O AC O
V VV V −⎛ ⎞− + =⎜ ⎟⎝ ⎠
(1) ( )(555.5) (20) ( )(0.8333)i A A A OV V V V V− = + − (2) (5.109) 550( ) 0B i AV V V+ − = (3) (3.257) 178 (1.718) 0C B OV V V+ − = (4) ( )(173.7) ( )(0.8333)C O O AV V V V− = − (1) (555.5) (0.8333) (576.3)i O AV V V+ = (2) (5.109) 550 (550) 0B i AV V V+ − = (3) (3.257) 178 (1.718) 0C B OV V V+ − = (4) (173.7) (0.8333) (174.5)C A OV V V+ = From (2) (107.7) (107.7)B A iV V V= − From (4) (1.0046) (0.004797)C O AV V V= − Substitute into (3)
[ ](3.257) (1.0046) (0.004797)O AV V− (178)[ (107.7) (107.7)] (1.718) 0A i oV V V+ − − = (3.272) (0.01562) (19170.6) (19170.6) (1.718) 0O A A i OV V V V V− + − − = (19170.6) (19170.6) (1.554)A i OV V V= −
(1.00) (0.00008106)A i OV V V= − Substitute into (1)
[ ](555.5) (0.8333) (576.3) (1.00) (0.00008106)(576.3) (0.0467)
i o i o
i o
V V V VV V
+ = −= −
(0.880) (20.8)
23.6
o i
ovf
i
V VV AV
=
= =
Ideal 1.2 0.05 25.0
0.05F E
vfE
R RAR+ += = =
(b) iif
i
VRI
= and 1
1 1
i Ai
V V VIr r
π
π π
−= =
We have (1.00) (0.00008106)(1.00) (23.6) (0.00008106)(0.99809)
A i o
i i
A i
V V VV V
V V
= −= −=
Then (1 0.99809)
(0.01051)0.182
ii i
VI V
−= =
95.1 K(0.01051)
iif if
i
VR RV
= ⇒ =
To find ,ofR set 0iV =
33 3
3
3
33
1( )( )
x AX m
F
C X
X AX C X m
F
V V VI g V
r RV V V
V VI V V gr R
ππ
π
π
π
−+ + =
= −−
+ − + =
For 0,iV = we have (1.0046) (0.004797)
(576.3) (0.8333)(0.001446)(1.0046) (0.001446)(0.004797)(1.0046)
C X A
A X
A X
C X X
C X
V V VV VV VV V VV V
= −=
== −=
(1 0.004797)1(1.0046 1.0) 1720.582 1.2
(0.7991) (0.8293)(0.03024)
33.1 K
XX X
X X X
X X
Xof
X
VI V
I V VI V
VRI
−⎛ ⎞+ − + =⎜ ⎟⎝ ⎠
+ ==
= =
12.38
1 1 8 KD ER R= = ( )GG GS D C LV V I I R= + +
10 10 ( 0.7)8 8
D DD C
V VI I− − += =
( )DGG TN D C L
n
IV V I I Rk
= + + +
( )
2
2
10 10 9.34.5 1 (1.8) (1.8)8(0.4) 8 8
836 10 8 18 1.8 16.74 1.83.2
36 4.472 10 42.74 3.6
3.6 6.74 4.472 10
12.96 48.528 45.4276 19.999(10 )12.96 28.528 154.6 0
D D D
D D D
D D
D D
D D D
D D
V V V
V V V
V V
V V
V V VV V
− − −⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − + + − + −
= − + −
− = −
− + = −
− − =
28.528 813.847 4(12.96)(154.6)2(12.96)
4.726 V
D
D
V
V
± +=
=
10 4.726 0.6593 mA8
9.3 4.7268
0.5718 mA
D
C
I
I
−= =
−=
=
0.6593 1 2.284 V0.4GSV = + =
4.5 2.284 2.216VOV = − = 4.726 2.216 2.51VDS D OV V V= − = − =
( )( )sat 1.28VDSV = Also
( ) (0.6593 0.5718)(1.8) 2.216 O D C LV I I R V= + = + = OK
2 (4.726 0.7) 2.216 3.21 ECQV V= + − = (b)
(1) i gs oV V V= +
(2) 11 2
Am gs
D
VV g VR r
π
π
+ =
(3) 22 2
0Bm
E
VV g VR r
ππ
π
+ + =
B A gs i oV V V V V Vπ = − = −
(2) 11 2
( )A B Am i o
D
V V Vg V VR rπ
−+ − =
(3) 22 2
( ) 0B B Am B A
E
V V V g V VR rπ
−+ + − =
(4) 1 2
1 2( ) ( )
Om gs m
L
Om i o m B A
L
Vg V g VR
Vg V V g V VR
π+ =
− + − =
1 1
22
T
2
2 2 (0.4)(0.6593) 1.027 mA/V0.5718 21.99 mA/V0.026
(100)(0.026) 4.547 K0.5718
m n D
Cm
g K IIgV
rπ
= = =
= = =
= =
(2) (1.027)( )8 4.547
A B Ai o
V V VV V −+ − =
(3) (21.99)( ) 08 4.547B B A
B AV V V V V−+ + − =
(2) 0.125 1.027 1.027 0.2199 0.2199A i o B AV V V V V+ − = − (3) 0.125 0.2199 0.2199 21.99 21.99 0B B A B AV V V V V+ − + − = (2) 0.3449 0.2199 1.027 1.027A B i oV V V V− = − + (3) 22.21 22.335A BV V=
1.0056A BV V= Then (0.3449)(1.0056) 0.2199 1.027 1.027B B i oV V V V− = − + 0.1269 1.027 1.027B i oV V V= − +
8.093 8.093B i oV V V= − + 8.138 8.138A i oV V V= − +
Then
(4) [ ](1.027)( ) (21.99) 8.093 8.093 8.138 8.1381.8
oi o i o i o
VV V V V V V− + − + + − =
1.027 1.027 177.965 177.965 178.955 178.955 0.5556i o i o i o oV V V V V V V− − + + − = 2.017 2.5726 0i oV V− =
0.784 AVo
i
VV
= =
Set 0iV =
2 1
2 1( ) ( )
XX m m gs
L
XX m B A m X
L
VI g V g VR
VI g V V g VR
π+ + =
+ − + − =
8.0938.138
B X
A X
V VV V
==
[ ](21.99) 8.093 8.138 1.027( ) (0.5556)2.572
X X X X X
X X
I V V V VI V
+ − + − ==
0.389 KXof
X
V RI
= =
12.39 a. 2 2 (0.5)(0.5) 1 /m n DQg K I mA V= = =
0 ( )m gs SV g V R=
0i gsV V V= + so 0gs iV V V= − Then
0 0( )m S iV g R V V= − 1(2) 0.667
1 1 (1)(2)m S
v vm S
g RA Ag R
= = ⇒ =+ +
To determine 0 fR
XX m gs
S
VI g VR
+ = and gs XV V= −
0
1 1Xm
X f S
I gV R R
= = +
so 0 01 1 2 0.667 k
1f S fm
R R Rg
= = ⇒ = Ω
b. For 20.8 /nK mA V=
2 (0.8)(0.5) 1.265 mA / Vmg = = (1.265)(2) 0.7167
1 (1.265)(2)vA = =+
0.7167 0.667 7.45% increase0.667
f
f
AA
Δ −= ⇒
0
0
0
0
1 || 2 0.7905 || 21.2650.5666 k
0.5666 0.667 15.05% decrease0.667
f
f
f
f
R
RR
R
= =
= Ω
Δ −= ⇒
12.40 dc analysis:
1
1
150 || 47 35.8 k47 (25) 5.96 V
47 150
TH
TH
R ,
V
= = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠
2
2
33 || 47 19.4 k33 (25) 10.3 V
33 47
TH
TH
R ,
V
= = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠
1
1
5.96 0.7 0.0187 mA35.8 (51)(4.8)(50)(0.0187) 0.935 mA
B
C
I
I
−= =+
= =
2
2
10.3 0.7 0.03705 mA19.4 (51)(4.7)(50)(0.03705) 1.85 mA
B
C
I
I
−= =+
= =
1
2
(50)(0.026) 1.39 k ;0.935
(50)(0.026) 0.703 k1.85
r
r
π
π
= = Ω
= = Ω
1
2
0.935 35.96 mA / V0.0261.85 71.15 mA / V0.026
m
m
g
g
= =
= =
1S eV V Vπ= + (1)
1 01 1
1 1
e em
F
V V V Vg Vr R R
ππ
π
−+ = + (2)
2 2 21 1
1 2 2
0mC B
V V Vg VR R r
π π ππ
π
+ + + = (3)
0 02 2
2
0em
C F
V V Vg VR Rπ
−+ + = (4)
Substitute numerical values in (2), (3) and (4): 1e SV V Vπ= − (1)
11 1 0
1 1 1(35.96) ( )1.39 0.1 4.7 4.7SV V V V Vπ
π π⎛ ⎞ ⎛ ⎞+ = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 1 0(46.89) (10.213) (0.2128)SV V Vπ = − (2)
1 21 1 1(35.96) 0
10 19.4 0.703V Vπ π
⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
or 1 2(35.96) (1.574) 0V Vπ π+ = (3)
2 0 11 1 1(71.15) ( ) 0
4.7 4.7 4.7SV V V Vπ π⎛ ⎞ ⎛ ⎞+ + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 2 0 1(71.15) (0.4255) (0.2128) (0.2128) 0SV V V Vπ π+ − + = (4)
From (3): 2 1(22.85)V Vπ π= − Then substitute in (4):
1 0 1(71.15) (22.85) (0.4255) (0.2128) (0.2128) 0SV V V Vπ π− + − + = or
1 0(1625.6) (0.4255) (0.2128) 0SV V Vπ− + − = From (2): 1 0(0.2178) (0.004538)SV V Vπ = − Then
0 0(1625.6)[ (0.2178) (0.004538)] (0.4255) (0.2128) 0S SV V V V− − + − = or 0(354.3) (7.802) 0SV V− + = Finally
0 45.4S
VV
⇒ =
12.41 For example, use a z-stage amplifier. Each stage shown in Fig. 12.29. 12.42
For 3 33
:2n
nk WM K
L′ ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠
Let 3
25WL
⎛ ⎞ =⎜ ⎟⎝ ⎠
Then 23
0.080 (25) 1 /2nK mA V⎛ ⎞= =⎜ ⎟
⎝ ⎠
Want 0ov = for 0,iv = so that 2
3 2 30.4 1 ( )D Q GS TNI I V V= = = ⋅ −
Then 30.4 2 2.63 1GSV V= + =
For 3 3 22.63 12GS G D DV V V I R= ⇒ = − Or 2.63 12 (0.1) 93.7 D DR R k= − ⇒ = Ω
3 3 32 2 (1)(0.4) 1.26 /m n Dg K I mA V= = =
( ) ( )1
1 2
10 0.0769120 10A o o o
RV V V VR R
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
(Small amount of feedback) (1) 1 2i gs gs AV V V V= − +
(2) 1 2 1 20m gs m gs gs gsg V g V V V+ = ⇒ = − Then
2 212 ( )2i gs A gs A iV V V V V V= − + ⇒ = −
2 0.0384 0.5gs o iV V V= − (3) 2 [0.03846 0.5 ]B m gs D m D o iV g V R g R V V= − = − −
(4) 3gs B oV V V= − and [ ]3 3 1 2|| ( )o m gs LV g V R R R= + So
[ ]3 1 2|| ( ) ( )o m L B oV g R R R V V= + − Then
( ) ( )3 1 2|| 0.03846 0.5o m L m D o i oV g R R R g R V V V= + − − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ Or
[ ][ ] [ ][ ]3 1 2 3 1 21 || ( ) (0.03846) 1 || ( ) 0.5o m L m D m L m D iV g R R R g R g R R R g R V⎡ ⎤+ + + = + ⋅⎣ ⎦ Now
1 2|| ( ) 4 ||130 3.88LR R R+ = = So
[ ]1 (1.26)(3.88) (93.7)(0.03846) 1 (1.26)(3.88)(0.5) (93.7)o m m iV g g V⎡ ⎤+ + =⎣ ⎦ Rearranging terms, we find
229 10 1.11 /5.89 17.6
o mm
i m
V g g mA VV g
= = ⇒ =+
We have
1 2
0.0802 1.11 2 (0.1) 772 2n
m Dk W W W Wg I
L L L L′⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
12.43 Assuming an ideal op-amp, then from Equation (12.58)
0 1
2
201 1000.2S
I RI R
= + = =
Then 1
2
99RR
=
For example, set 2 5 kR = Ω and 1 495 kR = Ω 12.44
(a) 1 1
1
2
2
100 (0.2) 0.198 1 101
10 (0.198)(40) 2.08 2.08 0.7 1.38
1100 (1.38) 1.37 101
FEC E
FE
C
E
C
hI I mAh
V V
I mA
I mA
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠= − =
−= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
For 1 :Q
1
1
(100)(0.026) 13.1 0.198
0.198 7.62 /0.026m
r k
g mA V
π = = Ω
= =
For 2:Q
2
2
(100)(0.026) 1.90 1.37
1.37 52.7 /0.026m
r k
g mA V
π = = Ω
= =
(b)
1 1 1
1
eS
S F
V V V VIR r R
π π π
π
−= + + (1)
2 21 1
1 2
0em
C
V V Vg VR r
π ππ
π
++ + = (2)
2 12 2
2
e em
E F
V V V Vg Vr R R
π ππ
π
−+ = + (3)
Substitute numerical values in (1), (2), and (3):
1
1
1 1 1 110 13.1 10 10
(0.2763) (0.10)
S e
S e
I V V
I V V
π
π
⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − (1)
1 2
1 2
1 1 1(7.62) 040 1.90 40
(7.62) (0.5513) (0.025) 0
e
e
V V V
V V V
π π
π π
⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ + = (2)
2 1
2 1
1 1 1 152.71.90 1 10 10
(53.23) (1.10) (0.10)
e
e
V V V
V V V
π π
π π
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − (3) From (3),
2 1(48.39) (0.0909)eV V Vπ π= + Substituting into (1),
[ ]1 2 1(0.2763) (0.10) (48.39) (0.0909)SI V V Vπ π π= − + or
1 2(0.2672) (4.839)SI V Vπ π= − (1′) and substituting into (2),
( ) ( ) ( )1 2 2 1(7.62) (0.5513) 0.025 48.39 0.0909 0V V V Vπ π π π+ + + =⎡ ⎤⎣ ⎦ or
1 2 1 2(7.622) (1.761) 0 (0.2310)V V V Vπ π π π+ = ⇒ = − (2′) Then substituting (2′) into (1′), we obtain
2 2(0.2672)( )(0.2310) (4.839)SI V Vπ π= − − or
2 (4.901)SI Vπ= − Now
22 2
2
2 22(52.7) (42.16)
2 0.5
CO m
C L
RI g V
R R
V V
π
π π
⎛ ⎞= − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠
Then
(42.16)4.901
SO
II −⎛ ⎞= − ⎜ ⎟⎝ ⎠
or
8.60Oif
s
IAI
= =
(c) 1i
S
VRIπ= and ||i S ifR R R=
We had 1 2 (0.2310)V Vπ π= − and 2 (4.901)SI Vπ= −
so 1
1(4.901) (21.22)0.2310S
VI Vππ
−⎛ ⎞= − =⎜ ⎟⎝ ⎠
Then 1 1 0.04713
21.22iS
VRIπ= = =
Finally 10
0.04713 47.4 10
ifif
if
RR
R= ⇒ = Ω
+
12.45 (a)
(1) 1 1 2
1 1
ei
FS B
V V VIRR R r
π π
π
−= +
(2) 1 21 1
1 2 2
0Cm
C B
V Vg V
R R rπ
ππ
+ + =
(3) 2 2 2 12 2
2 2
e em
E F
V V V Vg Vr R R
π ππ
π
−+ = +
(4) 22 2
2
( ) Co m
C L
RI g V
R Rπ⎛ ⎞
= − ⎜ ⎟+⎝ ⎠
Now
(1)′ 1 2
1 1
ei
FS B F
V VIRR R r R
π
π
= −
So 2 11 1
Fe i F
S B F
RV V I RR R r R π
π
⎛ ⎞= ⎜ ⎟ −⎜ ⎟⎝ ⎠
Now, from (2) 2 2 2
1 11 2 2
0em
c B
V V Vg V
R R rπ π
ππ
++ + =
(2)′ 2 21 1
2 1 21 2 2
1 0em
C BC B
V Vg V
r R RR R rπ
ππ π
⎛ ⎞+ + + =⎜ ⎟
⎝ ⎠
Also
(3)′ 12 2 2
2 2
1 1 1m e
F E F
Vg V V
r R R Rπ
ππ
⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
And
(4)′ 22
2 2
O C L
m C
I R RV
g Rπ⎛ ⎞⎛ ⎞+
= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Substitute (1)′ into (2)′ and (3)′
(2)″ 21 1 1
2 1 21 2 2 1 1
1 1 0Fm i F
C BC B S B F
V Rg V V I Rr R RR R r R R r R
ππ π
π π π
⎡ ⎤⎛ ⎞ ⎢ ⎥+ + + − − =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦
(3)″ 12 2 1
2 2 1 1
1 1 1 Fm i F
F E F S B F
V Rg V V I Rr R R R R R r R
ππ π
π π
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎢ ⎥+ + = + − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎣ ⎦
Solve for 1Vπ from (2)″ and substitute into (3)″. Also use Equation (4)′. (b) 1 1 2 20 80 16 KBR R R= = =
120 (10) 2 V
100THV ⎛ ⎞= =⎜ ⎟⎝ ⎠
1
1
2 0.7 0.0111 mA16 (101)(1)1.11 mA
BQ
CQ
I
I
−= =+
=
2
2
15 85 12.75K
15 (10) 1.5V100
TH
TH
R
V
= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
2
2
1.5 0.7 0.01265 mA12.75 (101)(0.5)1.265 mA
BQ
CQ
I
I
−= =+
=
1
2
1.11 42.69 mA/V0.0261.265 48.65 mA/V0.026
m
m
g
g
= =
= =
1
2
(100)(0.026) 2.34K1.11
(100)(0.026) 2.06K1.265
r
r
π
π
= =
= =
Now 1 2 2 12.75 1.729KC BR R = =
1 1 1 1 16 2.34 10 1.695KS B F B FR R r R R r Rπ π≅ = =
1 2 2 1.729 2.06 0.940 KC BR R rπ = = Now
(2)″ 21 1
1 2
1 1 1042.69 (10)2.06 0.940 1.729 1.695
46.587 1.064 5.784 0
i
i
VV V I
V V I
ππ π
π π
⎛ ⎞ ⎡ ⎤+ + + ⋅ −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦+ − =
(3)″ 12 1
2 1
101 1 1 10 (10)2.06 10 0.5 10 1.695
49.03 12.29 21
i
i
VV V I
V V I
ππ π
π π
⎛ ⎞ ⎛ ⎞ ⎡ ⎤+ = + ⋅ −⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦= −
From (2)″ 1 2(0.1242) (0.02284)iV I Vπ π= − Then (3)″ [ ]2 249.03 12.29 (0.1242) (0.02284) 21i iV I V Iπ π= − −
249.31 19.47 iV Iπ = −
From (4)′ 24 4 (0.0411)
48.65 4o
oIV Iπ
+⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Then [ ](49.31) (0.0411) 19.47o iI I− = −
9.61oi
i
I AI
= =
12.46 a.
1
13.5 || 38.3 9.98 k13.5 (10) 2.606 V
13.5 38.3(120)(2.606 0.7) 1.75 mA
9.98 (121)(1)
TH
TH
C
R
V
I
= = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠−= =
+
( )( )1
2
10 1.75 3 4.75 V4.75 0.7 0.50 mA
8.1
C
C
V
I
= − =−≈ =
1
1
(120)(0.026) 1.78 k1.751.75 67.31 mA / V0.026m
r
g
π = = Ω
= =
2
2
(120)(0.026) 6.24 k0.500.50 19.23 mA / V0.026m
r
g
π = = Ω
= =
b.
1 1 1 2
1||S e
S B F
V V V V VR R r R
π π π
π
− −= + (1)
2 2 21 1
1 2
0em
C
V V Vg VR r
π ππ
π
++ + = (2)
2 2 2 12 2
2 2m
E F
V V V Vg Vr R R
π ε ε ππ
π
−+ = + (3)
and
0 2 2 2( )m CV g V Rπ= − (4) Substitute numerical values in (1), (2), and (3)
1 21
1 2
1 10.6 0.6 9.98 ||1.78 1.2 1.2
(1.67) (4.011) (0.8333)
S e
S e
V V VV
V V V
ππ
π
⎡ ⎤= + + −⎢ ⎥⎣ ⎦= − (1)
21 2
1 1(67.31) 03 6.24 3
eVV Vπ π⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
or 1 2 2(67.31) (0.4936) (0.3333) 0eV V Vπ π+ + = (2)
2 2 21
1 19.236.24 8.1 1.2 1.2
e eV V VV ππ⎛ ⎞+ = + −⎜ ⎟⎝ ⎠
or 2 2 1(19.39) (0.9568) (0.8333)eV V Vπ π= − (3)
From (1) 2 1(4.813) (2.00)e SV V Vπ= −
Then 1 2 1(67.31) (0.4936) (0.3333)[ (4.813) (2.00)] 0SV V V Vπ π π+ + − =
or 1 2(68.91) (0.4936) (0.6666) 0SV V Vπ π+ − = (2′)
and [ ]2 1 1(19.39) (0.9568) (4.813) (2.00) (0.8333)SV V V Vπ π π= − −
or 2 1(19.39) (3.772) (1.914)SV V Vπ π= − (3′)
We find 1 2(0.009673) (0.007163)SV V Vπ π= −
Then 2 2(19.39) (3.772)[ (0.009673) (0.007163)] (1.914)S SV V V Vπ π= − −
2 (19.42) ( 1.878)SV Vπ = − or 2 (0.09670)SV Vπ = − so that
0 (19.23)(4)( )(0.09670)SV V= − − Then
0 7.44S
VV
=
12.47
Using the circuit from Problem 12.32, we have 1if
S
VRIπ= .
Where 1SS
S
V VIR
π−=
From Problem 12.32 1 2(0.009673) (0.007163)
(0.009673) (0.007163)( )(0.09670)(0.01037)
S
S S
S
V V VV VV
π π= −= − −=
So (0.01037) (0.6) 0.00629 k
(0.01037)S
ifS S
VRV V
⋅= = Ω
−
or 6.29 ifR = Ω
12.48
1
1
1.4 ||17.9 1.298 k1.4 (10) 0.7254 V
1.4 17.90.7254 0.7 0.0196 mA
1.298(50)(0.0196) 0.98 mA
TH
TH
B
C
R
V
I
I
= = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠−= =
= =
Neglecting dc base currents, 2
2
2
10 (0.98)(7) 3.14 V3.14 0.7 3.25 mA0.25 0.550 (3.25) 3.19 mA51
B
E
C
V
I
I
= − =−= =+
⎛ ⎞= =⎜ ⎟⎝ ⎠
1
1
(50)(0.026) 1.33 k0.980.98 37.7 mA / V
0.026m
r
g
π = = Ω
= =
2
2
(50)(0.026) 0.408 k3.193.19 123 mA / V0.026m
r
g
π = = Ω
= =
1 1 1
1 2|| ||SF
V V VIR R r R
π π
π 1
−= + (1)
2 2 21 1
2 1
0em
C
V V Vg Vr R
π ππ
π
++ + = (2)
2 2 12 2
2 1
em
E
V V Vg Vr R
ππ
π
−+ = (3)
2 1 1 11
1 2
e
E E F
V V V VVR R R
π π− −= + (4)
Enter numerical values in (1), (2), (3) and (4): 1 1 1
17.9 ||1.4 ||1.33 5SV V VI π π −
= +
or
1 1(1.722) (0.20)SI V Vπ= − (1)
2 2 21(37.7) 0
0.408 7eV V V
V π ππ
++ + =
or 1 2 2(37.7) (2.594) (0.1429) 0eV V Vπ π+ + = (2)
2 2 12(123)
0.408 0.25eV V V
Vππ
−+ =
or 2 2 1(125.5) (4) (4)eV V Vπ = − (3)
2 1 1 11
0.25 0.50 5eV V V VV π− −
= +
or 2 1 1(4) (6.20) (0.20)eV V Vπ= − (4)
From (4): 2 1 1(1.55) (0.05)eV V Vπ= −
Then substituting in (3): 2 1 1 1(125.5) (4)[ (1.55) (0.05)] (4)V V V Vπ π= − −
or 2 1 1(125.5) (2.20) (0.20)V V Vπ π= − (3′)
and substituting in (2): ( ) ( ) ( ) ( ) ( )1 2 1 137.7 2.594 0.1429 1.55 0.05 0V V V Vπ π π+ + − =⎡ ⎤⎣ ⎦
or 1 2 1(37.69) (2.594) (0.2215) 0V V Vπ π+ + =
Now 1 1 2(170.16) (11.71)V V Vπ π= − −
Then substituting in (1): 1 1 2(1.722) (0.20)[ (170.16) (11.71)]SI V V Vπ π π= − − −
or 1 2(35.75) (2.342)SI V Vπ π= +
and substituting in (3′): 2 1 2 1(125.5) (2.20)[ (170.16) (11.71)] (0.20)V V V Vπ π π π= − − −
or 2 1
1 2
(151.3) (374.55)(0.4040)
V VV V
π π
π π
= −= −
so that
Then 2 2
2
(35.75)[ (0.4040)] (2.342)(12.10)
S
S
I V VI V
π π
π
= − += −
20 2 2
2
2 2
( )
2.2(123) (64.43)2.2 2
Cm
C L
RI g V
R R
V V
π
π π
⎛ ⎞= − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠
or 2 0(0.01552)V Iπ = − Then
0 01 5.33(0.01552)(12.10)S S
I II I
= ⇒ =
12.49
For example, use the circuit shown in Figure P12.30 12.50
1 2 3
1 2
3
6.24 k , 3.12 k , 1.56 k19.23mA / V 38.46 mA / V,76.92mA / V
m m
m
r r rg , gg
π π π= Ω = Ω = Ω= ==
1 1S eV V Vπ= + (1)
1 1 1 31 1
1 1
e e em
E F
V V V Vg Vr R R
ππ
π
−+ = + (2)
2 1 1 1 2( || )m CV g V R rπ π π= − (3)
3 3 32 2
2 3
0em
C
V V Vg VR r
π ππ
π
++ + = (4)
3 3 3 13 3
3 2
e e em
E F
V V V Vg Vr R R
ππ
π
−+ = + (5)
Enter numerical values in (2)-(5): 1
1 1 31 1 1(19.23)
6.24 0.1 0.8 0.8e eV V V Vπ
π⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 1 1 3(19.39) (11.25) (1.25)e eV V Vπ = − (2)
2 1 1(19.23) (5 || 3.12) (36.94)V V Vπ π π= − = − (3)
2 3 31 1 1(38.46) 02 1.56 2eV V Vπ π
⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 2 3 3(38.46) (1.141) (0.5) 0eV V Vπ π+ + = (4)
3 3 31 1 1 176.92
1.56 0.1 0.8 0.8e eV V Vπ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
or 3 3 1(77.56) (11.25) (1.25)e eV V Vπ = − (5)
From (1) 1 1S eV V Vπ = − Then
1 1 3( )(19.39) (11.25) (1.25)S e e eV V V V− = − or 1 3(19.39) (30.64) (1.25)S e eV V V= − (2′)
2 1(36.94) (36.94)S eV V Vπ = − + (3′)
1 3 3(38.46)[ (36.94) (36.94)] (1.141) (0.5) 0S e eV V V Vπ− + + + = (4′) From (5): 3 3 1(6.894) (0.1111)e eV V Vπ= + Then
1 3 1(19.39) (30.64) (1.25)[ (6.894) (0.1111)]S e eV V V Vπ= − +
or 1 3(19.39) (30.50) (8.6175)S eV V Vπ= − (2″)
and 1 3 3 1(1420.7) (1420.7) (1.141) (0.5)[ (6.894) (0.1111)] 0S e eV V V V Vπ π− + + + + =
or 1 3(1420.7) (1420.76) (4.588) 0S eV V Vπ− + + = (4″)
From (2″): 1 3(0.6357) (0.2825)e SV V Vπ= +
Then substituting in (4″): 3 3
3
(1420.7) (1420.76)[ (0.6357) (0.2825)] (4.588) 0(517.5) (405.95) 0
S S
S
V V V VV V
π π
π
− + + + =− + =
Now 0 3 3 376.92mI g V Vπ π= = or 3 0 (0.0130)V Iπ =
Then 0(517.5) (0.0130)(405.95) 0SV I− + = or
0 98.06 mA / VS
IV
=
12.52
1 2
1 2
3
3
(100)(0.026) 5.2 k0.5
0.5 19.23 mA / V0.026
(100)(0.026) 1.3 k2
2 76.92 mA / V0.026
m m
m
r r
g g
r
g
π π
π
= = = Ω
= = =
= = Ω
= =
1 21 1 2 2
1 2
0m mV Vg V g Vr r
π ππ π
π π
+ + + = (1)
Since 1 2r rπ π= and 1 2 ,m mg g= then 1 2V Vπ π= −
1 2 3 2 32S e eV V V V V Vπ π π= − + = − + (2)
3 3 32 2
3 2
0em
C
V V Vg Vr R
π ππ
π
++ + = (3)
3 3 23 3
3 2
em
F
V V Vg Vr R r
π ππ
π π
+ = + (4)
30 3 3
3
( )Cm
C L
RI g V
R R π⎛ ⎞
= −⎜ ⎟+⎝ ⎠ (5)
From (2): 3 22e SV V Vπ= +
(19.23) 3 32 2
1 ( 2 ) 01.3 18.6 18.6 SV V
V V Vπ ππ π+ + + + =
or 2 3(19.23) (0.8230) (0.05376) 0SV V Vπ π+ + = (3′)
23 2
1 176.92 ( 2 )1.3 10 5.2S
VV V V ππ π⎛ ⎞ ⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 3 2(77.69) (0.3923) (0.1) SV V Vπ π= + (4′)
0 3 32 (76.92) (51.28)
2 1I V Vπ π
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠ (5′)
From (3′): 2 3(0.04255) (0.002780) SV V Vπ π= − −
Then 3 3
3
(77.69) (0.3923)[ (0.04255) (0.002780) ] (0.1)(77.71) (0.0989)
S S
S
V V V VV V
π π
π
= − − +=
or 3 (0.001273) SV Vπ =
so that 0 (51.28)(0.001273) SI V= −
or 0 (0.0653) mA/VS
IV
= −
12.52
3
2
4 5
5 10
1 110 10
Af fj j
×=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
1 14 5Phase tan 2 tan
10 10f fφ − −⎛ ⎞ ⎛ ⎞= = − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
By trial and error, when 51.095 10 , 180f Hz φ= × ≅ ° For 1T = at 51.095 10 ,f Hz= ×
( ) ( )( )( )
3 33
2 2
4 5
5 10 5 101 1 4.84 10
10.996 2.1991 1
10 10f f
β ββ −
× ×= ⇒ = ⇒ = ×
⎡ ⎤⎛ ⎞ ⎛ ⎞+ ⋅ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
12.53 Use the basic circuit shown in Figure 12.27.
For the ideal case
0 1
i E
IV R
=
we want 30 10 A/V 1 mA/V
i
IV
−= =
Set 1 kER = Ω Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain is slightly less than the ideal. ER will need to be slightly decreased to increase the gain. 12.54 dc analysis
(on) 05 0.7 0.0343
100 (51)(0.5)(50)(0.0343) 1.71 mA
E E EB B B CC
B
C
I R V I R V
I
I
+ + + =−= =
+= =
Then (50)(0.026) 0.760 k1.71
rπ = = Ω
1.71 65.77 mA/V0.026mg = =
a.
To determine :ifR
0 ( ) 0||S
B F
V V VIR r R
π π
π
− −+ + = (1)
0 0 ( )m
C F
V V Vg VR R
ππ
− −= + (2)
Now from (2):
0
0
1 1(65.77)10 1 10
(65.67) (1.10)
VV V
V V
ππ
π
⎛ ⎞− = +⎜ ⎟⎝ ⎠
=
or 0 (59.7)V Vπ=
and from (1): 0 0
100 0.760 10 10(0.8543) (0.1)(59.7) 0
(6.824)
S
S
S
V V VI
I V VI V
π π
π π
π
+ + + =
+ + == −
Now ( ) 147 if if
S
VR RI
π−= ⇒ = Ω
To determine 0 :fR
X XX m
C F B
V VI g VR R R r π
π
= + −+
(3)
( )( )
( )B
XB F
R rV V
R r Rπ
ππ
⎛ ⎞−= ⎜ ⎟⎜ ⎟+⎝ ⎠
(4)
Now (100 0.760)
( ) (0.07014)(100 0.760) 10 X XV V Vπ
⎛ ⎞−= = −⎜ ⎟⎜ ⎟+⎝ ⎠
so 1 1 (65.77)(0.07014)1 10.754X XI V ⎛ ⎞= + +⎜ ⎟⎝ ⎠
0 0 175 Xf f
X
VR RI
= ⇒ = Ω
b. From part (a), we find
6.824SI
Vπ = −
then
0 (59.7)6.824
SIV −⎛ ⎞= ⎜ ⎟⎝ ⎠
or
0 8.75 kS
VV
= − Ω
c. If capacitance is finite, a phase shift will be introduced. 12.55 dc analysis: GS DSV V=
2
2
2
( )
10 (0.20)(8)( 2)
10 1.6( 4 4)
DD GSD n GS TN
D
GS GS
GS GS GS
V VI K V V
RV V
V V V
−= = −
− = −
− = − +
2
2
1.6 5.4 3.6 0
5.4 (5.4) 4(1.6)(3.6)3.95 V
2(1.6)10 3.95 0.756 mA
82 2 (0.2)(0.756) 0.778 mA/V
GS GS
GS
D
m n D m
V V
V
I
g K I g
− − =
± += =
−= =
= = ⇒ =
a.
0
0
0
1 1
gs S gs
S F
Sgs
S F S F
V V V VR R
V VV
R R R R
− −+ =
⎛ ⎞+ = +⎜ ⎟
⎝ ⎠ (1)
00
0
0
1 1 1
gsm gs
D F
gs mD F F
V VV g VR R
V V gR R R
−+ + =
⎛ ⎞ ⎛ ⎞+ = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (2)
So from (1): 01 1
10 100 10 100S
gsV VV ⎛ ⎞+ = +⎜ ⎟
⎝ ⎠
or 0
0
(0.11) (0.10) (0.010)(0.909) (0.0909)
gs S
gs S
V V VV V V
= +
= +
Then from (2):
0
0
0
0
1 1 1 0.7788 100 100
(0.135) ( 0.768)( 0.768)[ (0.909) (0.0909)]
(0.2048) (0.6981)
gs
gs
S
S
V V
V VV V
V V
⎛ ⎞ ⎛ ⎞+ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
= − += −
so 0 3.41vS
VAV
= = −
b. We have 0(0.909) (0.0909)
(0.909) (0.0909) (3.41 )0.599
gs S
S S
S
V V VV V
V
= +
= + −=
Now 0 0 ( 3.41 )
0.599S S
zfS gsS S S
S
V V V RAV VI V V
R
−= = =
− −
or ( 3.41)(10) 85.0 V/ma
0.401zf zfA A−= ⇒ = −
c. 0.401 0.599 (10) 14.9 k0.401
gs
gs S Sif if
S S S
VV V VR RI R V
= = = ⋅ ⇒ = Ω
d.
1010 100
(0.0909)
X XX m gs
D S F
Sgs X X
S F
X
V VI g VR R R
RV V VR R
V
= + ++
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=
00
1 1(0.778)(0.0909)8 10 100
1 0.2048 4.88 k
X X
Xf
X f
I V
I RV R
⎡ ⎤= + +⎢ ⎥+⎣ ⎦
= = ⇒ = Ω
12.56
As 0 100, 1010
Fm
S S
V RgV R
− −→ ∞ = = = −
To be within 10% of ideal, 0 10(0.9) 9S
VV
= − = −
From Problem 12.41, we had 0(0.909) (0.0909)
(0.909) ( 9 )(0.0909)0.0909
gs S
S S
S
V V VV V
V
= +
= + −=
Also from Problem 12.41, we had 0 (0.135) (0.010 )gs mV V g= −
or ( 9 )(0.135) (0.0909) (0.010 )
1.215 0.000909 0.0909S S m
m
V V gg
− = −− = −
or 13.36 mA/Vmg =
12.57 dc analysis
24 ||150 20.7 k24 (12) 1.655 V
24 150
TH
TH
R
V
= = Ω
⎛ ⎞= =⎜ ⎟+⎝ ⎠
1.655 0.7 0.00556 mA20.7 (151)(1)BQI −= =
+
so 0.834 mACQI =
(150)(0.026) 4.68 k0.834
0.834 32.08 mA / V0.026m
r
g
π = = Ω
= =
0
||S
S B F
V V V V VR R r R
π π π
π
− −= + (1)
0 0 0mC F
V V Vg VR R
ππ
−+ + = (2)
From (1):
01 1 15 5 20.7 || 4.68S
F F
V VV
R Rπ⎡ ⎤
= + + −⎢ ⎥⎣ ⎦
or
01(0.20) 0.4620SF F
VV V
R Rπ⎛ ⎞
= + −⎜ ⎟⎝ ⎠
From (2):
01 1 132.08 0
6F F
V VR Rπ
⎛ ⎞ ⎛ ⎞− + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
so
010.1667
132.08
F
F
VR
V
R
π
⎛ ⎞− +⎜ ⎟
⎝ ⎠=⎛ ⎞
−⎜ ⎟⎝ ⎠
(2)
Then
00
10.16671(0.20) 0.4620
132.08
FS
F F
F
VR V
VR R
R
⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥= + −⎜ ⎟ ⎢ ⎥⎛ ⎞⎝ ⎠ −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
Neglect the FR in the denominator term. Now
0 005 (0.20)
5SS
V VV VV
= − ⇒ = − = −
( )00 0
2
2
0.1667 1(0.20)(0.20) (0.4620 1)
32.08
1.283 (0.4620 1)(0.1667 1) 32.081.206 32.71 1 0
FF F
F
F F F F
F F
V RV R R V
R
R R R RR R
− +⎡ ⎤− = + −⎢ ⎥
⎣ ⎦− = − + + −
− − =
232.71 (32.71) 4(1.206)(1)2(1.206)FR
± +=
so that 27.2 kFR = Ω
12.58 dc analysis
4 ||15 3.16 k4 12 2.526 V
4 15
TH B
TH
R R
V
= = Ω =
⎛ ⎞= =⎜ ⎟+⎝ ⎠
2.526 0.7 0.002513.16 (181)(4)0.452 mA
BQ
CQ
I
I
−= =+
=
(180)(0.026) 10.4 k0.452
0.452 17.4 mA/V0.026m
r
g
π = = Ω
= =
1 1 1 0
||i
S B F
V V V V VR R r R
π π π
π
− −= + (1)
21 0
|| ||mC B
Vg VR R r
ππ
π
+ = (2)
32 0
|| ||mC B
Vg VR R r
ππ
π
+ = (3)
0 0 0 13 0m
C L F
V V V Vg VR R R
ππ
−+ + + = (4)
Now || || 8 || 3.16 ||10.4 1.86 k|| 3.16 ||10.4 2.42 k
C B
B
R R rR r
π
π
= = Ω= = Ω
Now substituting in (2): 2
1(17.4) 01.86V
V ππ + = or 2 1(32.36)V Vπ π= −
and substituting in (3): 3
2
31
(17.4) 01.86
(17.4)[ (32.36) ] 01.86
VV
VV
ππ
ππ
+ =
− + =
or 3 1(1047.3)V Vπ π= Substitute numerical values in (1):
01
1 1 110 10 2.42
i
F F
V VV
R Rπ⎛ ⎞
= + + −⎜ ⎟⎝ ⎠
or
01
1(0.10) 0.513iF F
VV V
R Rπ⎛ ⎞
= + −⎜ ⎟⎝ ⎠
Substitute numerical values in (4):
11 0
41 0
0
14
1 1 1(17.4)(1047.3) 08 4
1 11.822 10 0.375 0
10.375
11.822 10
F F
F F
F
F
VV V
R R
V VR R
VR
V
R
ππ
π
π
⎛ ⎞+ + + − =⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞
× − + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞− +⎜ ⎟
⎝ ⎠=× −
so that
00
4
10.3751(0.10) 0.513
11.822 10
Fi
F F
F
VR V
VR R
R
⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥= + −⎜ ⎟ ⎢ ⎥⎝ ⎠ × −⎢ ⎥⎢ ⎥⎣ ⎦
We have 0 80i
VV
= − or 0
80iV
V = −
4
10.375(0.10) 1 10.513
180 1.822 10
F
F F
F
RR R
R
⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥− = + −⎜ ⎟ ⎢ ⎥⎝ ⎠ × −⎢ ⎥
⎢ ⎥⎣ ⎦
Neglect that 1/ FR term in the denominator.
4
2 4
(0.513 1)(0.375 1)(0.00125 ) 11.822 10
22.775 (0.513 1)(0.375 1) 1.822 10
F FF
F
F F F F
R RRR
R R R R
+ +− = − −
×
= + + + ×
We find 2 422.58 1.822 10 1 0F FR R− × − =
4 4 21.822 10 (1.822 10 ) 4(22.58)(1)2(22.58)FR
× ± × +=
or 0.807 MFR = Ω
12.59
a. 1( )S d d
S F
V V V VR R
− − − −=
or
11 1 0Sd
S F S F
V VVR R R R
⎛ ⎞+ + + =⎜ ⎟
⎝ ⎠ (1)
0 1 11
1 2
( )d
F
V V V VVR R R− − −
= +
or
01
1 1 2
1 1 1 d
F F
V VV
R R R R R⎛ ⎞
= + + +⎜ ⎟⎝ ⎠
(2)
and 0 0L dV A V= or 0
0d
L
VVA
=
Substitute numerical values in (1) and (2): 0 14
1 1 05 10 5 1010
SV V V⎛ ⎞⋅ + + + =⎜ ⎟⎝ ⎠
or 4
0 1(0.3 10 ) (0.20) (0.10) 0SV V V−× + + = (1)
0 01 4
1 1 1 150 50 10 10 1010V VV ⎛ ⎞ ⎛ ⎞= + + + ⋅⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
or 5
0 1(0.02 10 ) (0.22)V V−− = (2)
Then 5
1 00.02 10
0.22V V
−⎛ ⎞−= ⎜ ⎟⎝ ⎠
and 5
40 0
4 50
0.02 10(0.3 10 ) (0.20) (0.10) 00.22
0.3 10 0.4545 10 0.00909 (0.20) 0
S
S
V V V
V V
−−
− −
⎡ ⎤⎛ ⎞−× + + =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
⎡ ⎤× − × + + =⎣ ⎦
Then 0 03
0.20 21.949.115 10S S
V VV V−
−= ⇒ = −×
b. ( )
d d Sif
S d S d
S
V V RRV V V V
R
− − ⋅= =
− − +
Now 04
0
21.9410
Sd
L
V VVA
= = −
Then 4
4
(21.94 10 )(5)1 21.94 10ifR
−
−
×=− ×
or 21.099 10 k 10.99 if ifR R−= × Ω ⇒ = Ω
c. Because of the 0L dA V source,
0 0fR =
12.60 For example, use the circuit shown in Figure 12.41 12.61 Break the loop
tI Iε=
Now 0 0
1
0i tF S i
V VA I
R R R R+ + =
+
0Sr
S i F S i
R VI
R R R R R⎛ ⎞
= ⋅⎜ ⎟+ +⎝ ⎠
or 0 ( )S ir F S i
S
R RV I R R R
R⎛ ⎞+
= ⋅ +⎜ ⎟⎝ ⎠
Then
1
1 1 ( ) 0S ii t r F S i
F S i S
R RA I I R R R
R R R R R⎛ ⎞ ⎡ ⎤⎛ ⎞+
+ + × + =⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟+ ⎝ ⎠⎣ ⎦⎝ ⎠
1
1 1 ( )
ir
t S iF S i
F S i S
AIT TI R R R R R
R R R R R
= − ⇒ =⎡ ⎤ ⎛ ⎞++ +⎢ ⎥ ⎜ ⎟+ ⎝ ⎠⎣ ⎦
12.62
1 1 1 01 1
1 1m
E F
V V V Vg Vr R R
π ε επ
π
−+ = + (1)
1 1 1 1 1 21 2
0 ( )( )rm r m C
C
Vg V V g V R rR rπ π π
π
+ = ⇒ = − (2)
2 tV Vπ = so that
3 0 32
2 3
0m tC
V V Vg VR r
π π
π
++ + = (3)
3 0 0 13 3
3 3m
E F
V V V Vg Vr R R
π επ
π
−+ = + (4)
From (4):
10 3 3
3 3
1 1 1m
E F F
VV V g
R R r Rε
ππ
⎛ ⎞ ⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
But 1 1V Vε π= −
so
13 3
30
3
1
1 1
mF
E F
VV gr R
V
R R
ππ
π
⎛ ⎞+ −⎜ ⎟
⎝ ⎠=⎛ ⎞
+⎜ ⎟⎝ ⎠
Then
13 3
31 1
1 1
3
11 1 1
1 1
mF
mE F
FE F
VV gr R
V gr R R
RR R
ππ
ππ
π
⎛ ⎞− + +⎜ ⎟⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎝ ⎠+ − + =⎢ ⎥⎜ ⎟ ⎜ ⎟
⎛ ⎞⎝ ⎠⎝ ⎠⎣ ⎦ ⋅ +⎜ ⎟⎝ ⎠
(1′)
and
13 3
32 3
2 32
3
11 1 0
1 1
mF
m tC
CE F
VV gr R
g V VR r
RR R
ππ
ππ
π
⎛ ⎞+ −⎜ ⎟⎛ ⎞ ⎝ ⎠+ + + =⎜ ⎟
⎛ ⎞⎝ ⎠ ⋅ +⎜ ⎟⎝ ⎠
(3′)
From (3′), solve for 3Vπ and substitute into (1′). Then from (1′), solve for 1Vπ and substitute into (2). Then
.r
t
VTV
= −
12.63
1
0r er r
S F
V VV VR r Rπ
−+ + = (1)
2 21
1 2
0em t
C
V V Vg VR r
π π
π
++ + = (2)
22 2
2
e e rm
E F
V V V Vg Vr R R
ππ
π
−+ = + (3)
Using the parameters from Problem 12.29, we obtain 1 1 1 0
10 15.8 10 10e
rVV ⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
or (0.2633) (0.10)r eV V= (1)
21 1(7.62) 040 2.28 40
et
VV Vπ⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
or 2(7.62) (0.4636) (0.025) 0t eV V Vπ+ + = (2)
21 1 152.7
2.28 1 10 10r
eVV Vπ
⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 2 (53.14) (1.10) (0.10)e rV V Vπ = −
Then 2 (0.0207) (0.001882)e rV V Vπ = − (3)
Substituting in (2): (7.62) (0.4636)[ (0.0207) (0.001882)] (0.025) 0t e r eV V V V+ − + =
or (7.62) (0.03460) (0.0008725) 0t e rV V V+ − =
From (1) (2.633)e rV V= Then
(7.62) (2.633)(0.03460) (0.0008725) 0(7.62) (0.09023) 0
t r r
t r
V V VV V
+ − =+ =
or 84.45r
t
VV
= −
Now
84.45r
t
VT TV
= − ⇒ =
12.64
tV Vπ = −
0 0m
C F B
V Vg V
R R R rππ
= ++
(1)
and
0B
rB F
R rV V
R r Rπ
π
⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠
(2)
Now
01 1(65.77)1 10 100 0.760
V Vπ
⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠
or 0(65.77) (1.0930)V Vπ = and
0 00.754 (0.07011)
10 0.754rV V V⎛ ⎞= =⎜ ⎟+⎝ ⎠
so 0 (14.26) rV V= Then (65.77)( ) (14.26) (1.0930)t rV V− =
4.22r
t
VV
= − so that 4.22T =
12.64 Want 1 12f MHz= for a phase margin of 45°
( ) ( ) 40 80 0 10dBT dB T= ⇒ = Then
( ) ( )
6
0
1 112 10PD
TT f
f fj jf
=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠
Set 1f f= and 1T = So
4
26
10112 101 2
PD
T
f
= =⎛ ⎞×+ ⋅⎜ ⎟⎝ ⎠
which yields 6 412 10 10 1.70
2 PDPD
f kHzf× = ⇒ =
12.65
a. 1 12 4tan 2 tan
5 10 10f fφ − −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠
or 1 1 4180 180
1802 4180 tan 2 tan 1.05 10 Hz5 10 10
f f f− −⎛ ⎞ ⎛ ⎞− = − − ⇒ ≈ ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠
b. 5
180 2 24 4
2 4
(10 )( ) 11.05 10 1.05 101 1
5 10 10
T f β= =⎡ ⎤⎛ ⎞ ⎛ ⎞× ×+ +⎢ ⎥⎜ ⎟ ⎜ ⎟× ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
5(10 )1(21.02)(2.105)
β= or
44.42 10β −= × 12.65
( )
40 0
0
0
80dB 10
01f
A AAA
Aβ
= ⇒ =
=+
or ( )4
4
105 0.21 10
ββ
= ⇒ ≈+
Then ( ) 400 0.2 10T Aβ= = ×
Inserting a dominate pole
1 1 16 7tan tan tan
10 10PD
f f ff
φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
If we want a phase margin of 45°, then 1 1
6 7135 90 tan tan10 10
f f− −⎛ ⎞ ⎛ ⎞− ° ≈ − ° − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
By trial and error, 0.845MHzf ≈ Then
4
2 226
0.2 10 110.8450.845 10 0.845 11 1 101PD
T
f
×= = ×⎛ ⎞ ⎛ ⎞× ⎛ ⎞ ++ + ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠
( )( )6 40.845 10 0.2 10
1.309 1.0036PDf× ×≈
so 555HzPDf = 12.66
(a) 3
2
4 5
(10 )
1 110 10
vT Af fj j
ββ= =⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 14 52 tan tan
10 10f fφ − −= − −
Set 180φ = − ° By trial and error, 44.58 10 Hzf = × (b) Set 1T = at 44.58 10 Hzf = ×
3
22 24 4
4 5
3
(10 )14.58 10 4.58 101 1
10 10
(10 )1 0.02417(21.976)(1.10)
β
β β
=⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⎜ ⎟+ +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= ⇒ =
(c) 3
3
10 39.71 (10 )(0.02417)ovfA = =
+
(d) Wait 1T < at 44.58 10 Hz,f = × so system is stable for smaller values of .β 12.67
1 1 14 4 5tan tan tan
10 5 10 10f f fφ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
At 48.1 10 Hz,f = × 180.28φ = − ° Determine ( )T f at this frequency.
3
2 2 24 4 4
4 4 5
3
1 1 1(10 )8.1 10 8.1 10 8.1 101 1 1
10 5 10 10
(10 )(8.161)(1.904)(1.287)
T β
β
= × × ×⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
a. For 0.005β = ( ) 0.250 1 StableT f = < ⇒
b. For 0.05β = ( ) 2.50 1 UnstableT f = > ⇒
12.68 (b)Phase margin 80 100φ= °⇒ = − °
1 13 4100 2 tan tan
10 5 10f fφ − −⎛ ⎞ ⎛ ⎞= − = − −⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠
By trial and error, 31.16 10 f Hz= ×
Then 3
22 23 3
3 4
34
(5 10 )11.16 10 1.16 101 1
10 5 10
(5 10 ) 4.7 10(2.35)(1.00)
T β
β β −
×= =⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⎜ ⎟+ ⋅ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠⎝ ⎠
×= ⇒ = ×
12.69
c. For 0.005,β =
( ) 1(0 dB)T f = at 42.10 10 Hzf ≈ × Then
4 4 41 1 1
3 4 5
2.10 10 2.10 10 2.10 10tan tan tan10 10 10
87.27 64.54 11.86
φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − − −
or 163.7φ = −
System is stable. Phase margin 16.3= ° For 0.05,β =
( ) 1 (0 dB)T f = at 46.44 10 Hzf ≈ ×
Then 4 4 4
1 1 13 4 5
6.44 10 6.44 10 6.44 10tan tan tan10 10 10
89.11 81.17 32.78
φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − − −
or 203.1 System is unstable.φ = − °⇒
12.70
5
4 5 5
(10 )
1 1 15 10 10 5 10
T Af f fj j j
ββ= =⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟× ×⎝ ⎠⎝ ⎠⎝ ⎠
Phase Margin 60 120φ= °⇒ = − ° So
1 1 14 5 5120 tan tan tan
5 10 10 5 10f f f− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
By trial and error, at 510 ,f Hz= 120φ ≅ − ° Then
( )( )( )( )
5
2 2 25 5 5
4 5 5
55
(10 )110 10 101 1 1
5 10 10 5 10
101 3.22 10
2.236 1.414 1.02
T β
ββ −
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ ⋅ + ⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= ⇒ = ×
12.71
(a) 5
35
10100 9.99 101 (10 )
ββ
−= ⇒ = ×+
3 5
2 2
3 5
2 2
3 5
5
(9.99 10 )(10 )1
1 110 10
9991
1 110 10
3.08 10 Hz
vT Af f
f f
f
β−×= = =
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= ×
Phase 1 1
3 5
5 51 1
3 5
tan tan10 103.08 10 3.08 10tan tan
10 1089.81 72.01161.8
f fφ
φ
− −
− −
= − −
× ×= − −
= − −= −
Stable (b) Phase Margin 180 161.8 18.2= − = ° 12.72
(a) 1 1 14 6 7tan tan tan
5 10 10 5 10f f fφ − − −= − − −
× ×
For 180φ = − ° By trial and error, 6
180 7.25 10 Hzf = ×
(b) 5
2 2 26 6 6
4 6 7
4
(0.10)(10 )
7.25 10 7.25 10 7.25 101 1 15 10 10 5 10
0(145)(7.319)(1.01)9.33
T
T
=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1=
=
System is unstable 4
2 2 2
4 6 7
7
101
1 1 15 10 10 5 10
2.14 10 Hz
Tf f f
f
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= ×
7 7 71 1 1
4 6 7
2.14 10 2.14 10 2.14 10tan tan tan5 10 10 5 10
89.87 87.32 23.17200.4
φ
φ
− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠= − − −= − °
(c) For 67.25 10 Hzf = × 5
2 2 26 6 6
4 6 7
(0.0010)(10 )
7.25 10 7.25 10 7.25 101 1 15 10 10 5 10
100(145)(7.319)(1.01)0.0933
T
T
=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
=
System is stable.
2 2 2
4 6 7
6
1001
1 1 15 10 10 5 10
2.13 10 Hz
Tf f f
f
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= ×
6 6 6
1 1 14 6 7
2.13 10 2.13 10 2.13 10tan tan tan5 10 10 5 10
88.66 64.85 2.44155.9
φ
φ
− − −× × ×= − − −× ×
= − − −= − °
12.73
(a) 1 1180 1803 5
5180
180 tan 2 tan5 10 10
1.05 10 Hz
f f
f
φ − −= − = − −×
= ×
(b) 3
2 25 5
3 5
0180
(0.0045)(2 10 )
1.05 10 1.05 101 15 10 10
9(21.02)(2.1025)
0.204
T
T f
×=⎡ ⎤⎛ ⎞ ⎛ ⎞× ×+ +⎢ ⎥⎜ ⎟ ⎜ ⎟× ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
=
= =
System is stable
2 2
3 5
4
91
1 15 10 10
3.88 10 Hz
Tf f
f
= =⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
= ×
4 41 1
3 5
3.88 10 3.88 10tan 2 tan5 10 10
82.66 42.41125.1
φ
φ
− −× ×= − −×
= − −= − °
(c) 3
2 25 5
3 5
(0.15)(2 10 )
1.05 10 1.05 101 15 10 10
300(21.02)(2.1025)6.79
T
T
×=⎡ ⎤⎛ ⎞ ⎛ ⎞× ×+ +⎢ ⎥⎜ ⎟ ⎜ ⎟× ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
=
=
System is unstable
2 2
3 5
5
3001
1 15 10 10
2.33 10 Hz
Tf f
f
= =⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
= ×
5 51 1
3 5
2.33 10 2.33 10tan 2 tan5 10 10
88.77 133.54222.3
φ
φ
− −× ×= − −×
= − −= − °
12.74 Phase Margin 45 135φ= °⇒ = − °
1 1 1 13 4 5 6
135
tan tan tan tan10 10 10 10
f f f fφ
− − − −
= − °
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
At 410 Hz,f = 135.6φ = − °
3
2 24 4
3 4
2 24 4
5 6
3
11 1(10 )10 101 110 10
1 1
10 101 110 10
(10 )1(10.05)(1.414)(1.005)(1.00)
T
β
β
=
= × × ×⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
× ×⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
or 0.01428β =
12.75
3
6 6
1 1500011 300 10
1 1
1 12 10 25 10
PD
Tff jj
f
f fj j
= × ×⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠
× ×⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠
Phase Margin 45 135φ= °⇒ = − ° at 300 kHzf = 3 3
1 13
300 10 300 10135 tan tan 0 0300 10
90 45PDf
− −⎛ ⎞ ⎛ ⎞× ×− ° = − − − −⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠= − ° − °
Now
23
2 23
3
1@ 300 kHz50001
300 101 2 1 1
300 10 500012
300 10 2 84.8 Hz5000
PD
PD
PD PD
T f
T
f
f
f f
= =
= ≈⎛ ⎞×+ ⋅ ⋅ ⋅⎜ ⎟⎝ ⎠
⎛ ⎞× ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
×≈ ⇒ =
12.76 (a) At 610 Hz,f =
6 61 1
4 6
10 10tan 2 tan10 10
89.4 90 180
φ − −= − −
= − − ≈ − °
3
2 26 6
4 6
3
10
10 101 110 10
10 5(100)(2)
T =⎡ ⎤⎛ ⎞ ⎛ ⎞
+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
= =
1T > at 6180 10 Hz,f f= = System is unstable.
(b) 3
2
4 6
10
1 1 110 10PD
Tf f fj j j
f
=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠
Phase margin 45 135φ= °⇒ = −
1 1 14 6135 tan tan 2 tan
10 10PD
f f ff
φ − − −= − = − − −
410 Hzf ≅ 3
2 2 24 4 4
4 6
3
4
4
3
10110 10 101 1 1
10 10
10110 (1.414)(1.0)
(10 )(1.414)10
14.14 Hz
PD
PD
PD
PD
T
f
f
f
f
= =⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦
≅⎛ ⎞⎜ ⎟⎝ ⎠
=
=
12.77
(a) 1 1 1180 180 1804 4 5180 tan tan tan
10 5 10 10f f fφ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
4180 8.06 10 Hzf ≅ ×
(b) 2 2 24 4 4
4 4 5
500
8.06 10 8.06 10 8.06 101 1 110 5 10 10
500(8.122)(1.897)(1.284)25.3
T
T
=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
=
(c)
4 4 5
500
1 1 1 110 5 10 10PD
Tf f f fj j j j
f
=⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
Phase Margin 60 120φ= °⇒ = − °
1 1 1 14 4 5120 tan tan tan tan
10 5 10 10PD
f f f ff
− − − −− = − − − −×
Assume 1tan 90PD
ff
− ≅ °
Then 34.2 10 Hzf ≅ ×
2 2 2 23 3 3 3
4 4 5
23
3
50014.2 10 4.2 10 4.2 10 4.2 101 1 1 1
10 5 10 10
50014.2 101 (1.085)(1.004)(1.0)
4.2 10 500(1.0846)(1.0035)(1.0)
9.14 Hz
PD
PD
PD
PD
T
f
f
ff
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × ×+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
=⎛ ⎞×+ ⎜ ⎟⎝ ⎠
× ≅
= 12.78
(a) 4
4
1050 0.01991 (10 )
ββ
= ⇒ =+
4
5
(0.0199)(10 )
1 110PD
Tf fj j
f
=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
Phase margin 45 135φ= °⇒ = − °
1 15135 tan tan
10PD
f ff
− −− = − −
510 Hzf = 4
2 25 5
5
5 4
(0.0199)(10 )110 101 1
10
10 (0.0199)(10 )1.414
711 Hz
PD
PD
PD
T
f
ff
= =⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
=
=
(b) 4
4
1020 0.04991 (10 )
ββ
= ⇒ =+
4
5
4
2 2
5
5
(0.0499)(10 )
1 1711 10
(0.0499)(10 )1
1 1711 10
1.76 10 Hz
Tf fj j
Tf f
f
=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= =⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= ×
5 51 1
5
1.76 10 1.76 10tan tan711 10
89.77 60.40150.2
φ
φ
− −⎛ ⎞ ⎛ ⎞× ×= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − −= −
Phase Margin 180 150.2 29.8= − = °
12.79 (a) 5100 dB 10O OA A= ⇒ =
5
5
1020 0.049991 (10 )
ββ
= ⇒ =+
5
6 7
(0.04999)(10 )
1 1 110 10PD
Tf f fj j j
f
=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠
Phase Margin 45 135φ= °⇒ = − °
1 1 16 7135 tan tan tan
10 10PD
f f ff
− − −− = − − −
610 Hzf ≈ 5
2 2 26 6 6
6 7
5
26
6 5
(0.04999)(10 )110 10 101 1 1
10 10
(0.04999)(10 )1101 (1.414)(1.005)
10 (0.04999)(10 )(1.414)(1.005)2.84 Hz
PD
PD
PD
PD
T
f
f
ff
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
=⎛ ⎞
+ ⎜ ⎟⎝ ⎠
=
=
(b) 5
5
105 0.199991 (10 )
ββ
= ⇒ =+
5
2 2 2
6 7
6
(0.19999)(10 )1
1 1 1284 10 10
2.25 10 Hz
Tf f f
f
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= ×
6 6 61 1 1
6 7
2.25 10 2.25 10 2.25 10tan tan tan284 10 10
89.99 66.04 12.68168.7
φ
φ
− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − − −= −
Phase Margin 180 168.7 11.3= − = ° 12.80 a. 5(0) 100 dB (0) 10T T= ⇒ =
5
6 6
10( )1 1 1
10 5 10 10 10
T ff f fj j j
=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟× ×⎝ ⎠⎝ ⎠⎝ ⎠
1T = =
5
2 2 2
6 6
1 1 110
1 1 110 5 10 10 10f f f
= × × ×⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠
By trial and error 0.976 MHzf =
61 1 10.976 10 0.976 0.976tan tan tan
10 5 1090 11.05 5.574 106.6
φ − − −⎛ ⎞× ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= − ° − ° − ° = − °
Phase Margin 180 106.6 73.4= ° − ° = °
b. 11
PF
fC
′ ∝ so 1
10 7520Pf
=′
or 1 2.67 HzPf ′ =
Now 1T = =
5
2 2
6
1 110
1 12.67 5 10
f f= × ×
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠
2
6
1
110 10
f×
⎛ ⎞+ ⎜ ⎟×⎝ ⎠
By trial and error 52.66 10 Hzf ≈ ×
then 5
1 1 12.66 10 0.266 0.266tan tan tan2.67 5 10
90 3.045 1.524 94.57
φ − − −⎛ ⎞× ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= − ° − ° − ° = − °
Phase Margin 180 94.57 85.4= ° − ° = ° 12.81
(a) 31
2dBfπτ− = where ( )1 2o i iR R Cτ =
3 12 7(500 1000) 10 2 10 6.67 10 sτ− −= × × × ⇒ = × Then
3 37
1 239 2 (6.67 10 )dB dBf f kHzπ− −−= ⇒ =
×
(b) For
10 Hz,PDf = 1 1 0.0159 2 2 (10)PD
sf
τπ π
= = =
Then ( ) ( )1 2o i i MR R C Cτ = +
( ) ( )30.0159 500 1000 10 i MC C= × × + or ( ) 8 124.77 10 2 10 0.0477 i M M MC C x C C Fμ− −+ = × = + ⇒ = 12.82 Assuming a phase margin of 45 .°
1 16 6135 90 tan tan
2 10 25 10f f− −⎛ ⎞ ⎛ ⎞− ° ≈ − ° − −⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠
By trial and error, 1.74 MHzf ≈ Then
26
2 2
115000
1.74 101
1 1
1.74 1.741 12 25
PD
T
f
=
= ×⎛ ⎞×+ ⎜ ⎟⎝ ⎠
× ×⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
or 61.74 10 5000
(1.325)(1.0024)PDf× ≈
so 462 HzPDf = 12.83
5
5
1020 0.049991 (10 )
ββ
= ⇒ =+
Phase Margin 60 120φ= ° ⇒ = − °
1 1 16 7120 tan tan tan
10 5 10PD
f f ff
− − −− = − − −×
Assume 1tan 90PD
ff
− = °
1 56tan 30 5.77 10 Hz
10f f− ≅ ° ⇒ = ×
5
2 2 25 5 5
6 7
3
25
(0.04999)(10 )15.77 10 5.77 10 5.77 101 1 1
10 5 10
4.999 1015.77 101 (1.155)(1.0)
PD
PD
T
f
f
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠⎝ ⎠
×=⎛ ⎞×+ ⎜ ⎟⎝ ⎠
5 35.77 10 4.999 10(1.155)(1.0)PDf
× ×
133.3 HzPDf =