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CHAPTER 12
12.1
1M
v
1000  1k
| A V  O  990
1M   5k 
1k  0. 5 
vS
vO  vS
iS 
or 59 . 9 dB
vS
990 v S
i
990
6
5
and i O 
| AI  O 
10  9. 9 x10 or 120 dB
1M   5k
1k
iS
1000


A P  A V A I  990 9. 9 x10 5  9. 8 x10 8 or 89 . 9 dB | v s 
12.2
vO  vS
vs 
5 k
1k 
31 . 6 
and A
5k   5k
1k  1k
v O 10 V

 1. 27 V Since R
A V 7. 91
dissipated in R
OUT
V

vO
5V

 5. 05 mV
A V 990
vO
 7. 91 or 18 . 0 dB
vS
has the same value as R
OUT
is also 0 . 5 W. The power dissipated in R
L,
ID
the power
will be
2
PI =
PI 
V 2ID
2R ID
 V S 


V 2S
 2 


2 R ID
8R ID
2
4
8 5000

where V
S

VO
7. 91
and V
O
 2 0. 5 W 1000   31 . 6 V
 0. 4 mW . The total power dissipated in the amplifier is
P = 500 mW + 0. 4 mW = 500 mW .
12.3
0. 99 mV  1mV
12.4
Io 
R ID
 R ID  4 . 95 M
R ID  50 k 
2 100 W
 2 A and
50 
I 2o R OUT
 5 W or R
2
OUT
 2. 5 
12.5
v ID 
12.6
vID 
12.7
Av  
v O 10 V

 0.1 mV |
A
10 5
10 V
6
 10 V requires A
A
 10
7
or 140 dB
vO 15 V
15 V
15V
6
6

 10 V requires A  15x10 | i  
 15 pA
6  15 V |
A
A
1M
10
R2
220 k

 46.8 | 20log 46.8   33. 4 dB | R IN  R 1  4.7 k | R OUT  0 
R1
4.7k
1
12.8
Av  
R2
8200 

 9. 01
R1
910
V O  9.010.05 V   0.451V
Is 
12.9
A v  1
12.10
A v  1
|
v o t  0.451 sin 4638t V
V s 0.05 V

= 54.9A | i s t  54.9 sin 4638t  A
R1
910
R2
680k
1
 83.9 | 20log 83. 9  38.5 dB | R IN   | R OUT  0 
R1
8. 2k
R2
8200
1
 10.0 | V O  10. 00.05V   0.500V | v o t   0.500 sin9125 t V
R1
910
12.11
vO  
R3
R
51k
51k
v1  3 v 2  
v1 
v 2  51v1  25.5v 2
R1
R2
1k
2k
v O t   510.01sin 3770t   25. 50. 04 sin 10000t 
v O t   0.510 sin 3770t  1.02 sin 10000t  V and v - t   0.
12.12
a A nom
1
v
R2
47k
1
 262 | 20log262  = 48. 4 dB
R1
0.18k
R IN  10k     | R OUT  0 
b  A max
1
v
47k1.1
47k0.9 
min
 320 | A v  1 
 215
0.18k0.9 
0.18k1.1
A max
 A nom
320  262
v
v

 0.22 |
nom
Av
262
c Tolerances: + 22%,  18% d 
A min
 A nom
215  262
v
v

 0.18
nom
Av
262
320
 1. 49:1
215
(e) function count=c;
c=0;
for i=1:500,
r1=180*(1+0.2*(rand-0.5));
r2=47000*(1+0.2*(rand-0.5));
a=1+r2/r1;
anom=1+47000/180;
if (a>=0.95*anom & a<=1.05*anom), c=c+1; end;
end
c
Executing this function twenty times yields 44% .
12.13
10R
 10
R
v
 11R  110 k | R1  1
 R  10 k .
i1 v  0
(a) Using Eq. 12.36: A V  
b  R 2 
2
v2
i2
v1 0
2
12.14
i TH 
v
v  vo
 
R TH
R
but v-  0 because it is a virtual ground (v -  v   0)
vo
| v o  i THR | V O  I THR | This circuit is known as a
R
current - to - voltage converter or transresistance amplifier.
i TH  
12.15
a A v  
R2
100k

 5.00 | R IN  R 1  20 k
R1
20k
b  A v  1 
c A v  
R2
100k
1
 6.00 | R IN  27k   27 k
R1
20k
R2
0

 0 | R IN  R 1  33 k This is not a very useful circuit.
R1
33k
12.16 The inverting terminal of the op-amp represents a virtual ground and
a I O  I DS 
V    V EE 
0  10 
1 A.
R
10
b  Saturation requires VDS  V GS  V TN where VDS  V DD  V   V DD and
V GS  V TN 
2I DS

Kn

21
 2.83V 
0.25
V DD  2.83 V.
2
c P R  I 2 R  1 10  10W. So the resistor must dissipate 10 W.
(A 15W resistor would provide a reasonable safety margin.)
12.17 The inverting terminal of the op-amp represents a virtual ground.
(a ) I O  I C   F I E 
 F V    V EE  30 0  15 

 0. 484 A.
1  F
R
31
30
(b) V O  V   V BE  0  V BE  V BE  V T ln
IC
IS
 0.025 V ln
0.484
 0.730V
10 13
(c) Forward - active region operation requires VCE  V BE but V CE  V CC  V   V CC
Therefore VCC  0.730 V.
(d) P R  I 2 R  0.484 2 30  7.03 W. So the resistor must dissipate 7.03 W.
(A 10W resistor would provide a reasonable safety margin.)
P D = I C V CE  I B V BE  0. 484 15  
0. 484
0.730  7.27 W.
30
12.18
VS  V VO  V
R1
VO

 sCV  and V +  V   V O

R
KR
R 1  KR 1 1  K
Combining these expressions yields: A V s  
Vo
1 K

, a non  inverting integrator.
Vs
sRC
3
12.19 (a) Applying ideal op-amp assumption 1, the voltage at the top end of R is v1 and the
voltage at the bottom end of R is v2. Applying op-amp assumption 2, the current io
must also equal the current in R, and
v  v2
iO  1
R
0V  0V
v
 0 | R OUT  X  
b  i X 
R
iX
vO1  A 0  v O1  vX 
vO2  A 0  vO 2   vO2  0
A
vX
| v1  vO1  v X 
1 A
1 A
v1  v2
vX
vX


| R OUT 
 1  A R
R
iX
1  A R
vO1  v X
iX
12.20
(a)
iO 
v5  vO
v  v
| v5  v   i R3  v  1
R 3  2v  v1 since R 1  R 3
R
R1
v 
v 2  vO
2
and v 5  v 2  v1  vO | i O 
iX
v2  v1  v O  v O
v  v1
 2
R
R
(b)
v X  v5
vX  vX
v


 0 | R OUT  X  
R
R
iX
v  v1
i TH  2
R
R TH  
An ideal current source!
12.21 (a) Note that voltages refer to the node numbers on the next page
Using voltage division since i +  0, v2  v 4  6
4.99k
4. 99k  5. 00k
4. 99k
 0. 5005v 4  2.997 V
4.99k  5.00k
4  v1
Since v id  0, v1  v2 and v 5  v1 
5.01k
5k
Solving for v 5 yields
v 5  1.992 V  1.002v 4
v2  v 4  6  v4 
io 
v5  v 4
 199A  2x10 7 v4
10k
v4 is unknown; let us assume 2x10 -7 v4  199 x10 6 A
which requires v 4  995V. So for v4  100 V, which should almost
always be true in transistor circuits, i O  199A.
4
For ZL = 10 k, v4 = 1.99 V, v2 = 3.99 V, v1 = 3.99 V, v5 = 3.99 V
Note that v5 - v4 = 2 V = (6V - 4V)
12.21 (b)
R OUT 
vX
iX
and i X 
vX  v5
10k
So we need to find iX, and hence v5, in terms of vX
5.00k
 0.5005 v X
4.99k  5.00k
v
v5  v1  i 5. 01k  v1  1 5.01k  2.002 v1  1. 002v x
5k
v X  v5
v X  1.002 v X
0. 002v X
iX 


and R OUT  5 M ! A negative output resistance!
10k
10k
10k
v1  v2  v X
12.22 Using ideal op-amp assumption 2,
V S  I SR 1  I L Z L , and using ideal op-amp
assumption 1, the voltage across R2 must equal the voltage across R1 which requires
R
I 2 R 2  I SR1 or I 2  I S 1 .
R2
 R 
V S  I SR 1  I S 1  1 Z L
 R 2 
and Z IN 
R OUT 
iX 
vX
iX
vX

iX
Note, for the case of finite gain A,
v
v  vO
R OUT  X where i X  X
iX
R2
vX
1 A
where
vX  vO vX  vX

0
R2
R2
R OUT 
vX  v O 
 R 
VS
 R 1  1  1 Z L
IS
 R 2 
and v O  v X
A
1 A
and R OUT  R 2 1  A 
 10k 
12.23 For ZL = 3.6 k, R IN  10k  3.6k1 
  49.6 k

1k 
12.24 Applying op-amp assumption 1 to the circuit on the next page, the voltage at the top
of R2 is vO2, and applying op-amp assumption 2,
vS
v
  O2
R1
R2
or
vO2  v S
R2
R1
Since the op-amp input currents are zero, and
R
v
R 
i  S , v O1  iR 2  iR 3    2  3 v S
R1
R 1 R1 
5
Alternatively, the voltage at the bottom of R2 is zero, so
 R 
 R  R 
R
R 
vO1  1  3 vO2  1  3  2 v S    2  3 vS
 R 2 
 R 2  R1 
 R1 R1 
12.25
 R
R 
1 1 
V O   V REF 

  3.2     1.2 V
4R 8R 
4 8 
R 
 R
1 1 
V O   V REF 

  3.2 
  1. 8 V
2R 16R 
2 16 
6
0000
0.000 V
0001
-0.200 V
0010
-0.400 V
0011
-0.600 V
0100
-0.800 V
0101
-1.00 V
0110
-1.20 V
0111
-1.40 V
1000
-1.60 V
1001
-1.80 V
1010
-2.00 V
1011
-2.20 V
1100
-2.40 V
1101
-2.60 V
1110
-2.80 V
1111
-3.00 V
12.26 Desire Ron ≤ 0.01(10 k) = 100 .
V S  3.2
10 4
 3.17 V
10 4  102
V TN  1  0.5
 0.6  3.17 

0.6  1.58 V
1
W

R on  K 'n
V GS  V TN  V DS

L

1
W

5x10 5
5  3.17  1.58  0.03 



L
W
 909
L
 100
12.27 Consider the error for each bit acting by itself:
 R 1  0   V REF
1   0  1  1 
  1 
V REF 
 V REF 
   V REF 0
 
2
2
2 
2
2R 1   1 

 R 1   0   V REF
   2 
V REF 
 V REF 0
 
4
4
4R 1   2 
 R 1   0   V REF
 0   3 
V REF 
 V REF

8
8
8R 1   3 
 R 1   0   V REF
   4 
V REF 
 V REF 0
 
16R
1

16
16
 4 

Adding these together yields
 0   1 
 0   2 
 0   3 
 0   4 
V REF
 V REF
 V REF
 V REF
 0.05V REF
2
4
8
16
15




 0  1  2  3  4  0.05
16
2
4
8 16
Giving each term the same weight:
15

 0  1% and  0  1. 07% | 1  1% and 1  2%  2  4%  3  8%  4  16%
16
2
12.28
An n-bit DAC requires (n+1) resistors. Ten bits requires 11 resistors.
210 R 210

or 1024:1
R
1
A wide range of resistor values is required but it could be done. For R = 1 k, 1024R = 1.024
M.
12.29 Taking successive Thévenin equivalent circuits at each ladder node yields:

12.30
V REF
 0.3V |
16

V REF
 0. 6V |
8

V REF
 1.2V |
4

V REF
 2. 4V
2
Combine amplifiers A & B; Combine amplifiers B & C
7
12.31
Av = 50, RIN = 24 k, ROUT = 0
12.32
12.33 (a) Driving the output of the circuit in Fig. 12.21 with a current source of value i X:
i X  iO  i2 | i2 
iO 
vX
R1  R 2
v X  Ai 2 R1
1  A
 vX
RO
RO
i X  vX
1  A
vX

RO
R1  R 2
; iO 
v X  Av id
RO
where  =
and R OUT 
; v id  i 2 R 1
R1
R1  R 2
vX
RO

R 1  R 2 
i X 1  A
(b) This approach places two potentially unequal value voltage sources in parallel which is not
permitted.
12.34 Assume i- << i2:
i    iX  
i1  i2 
8
vX

R ID 1  A 
0.1V
 48.0 pA

6
5 1 
10 1 10


48 
0.1V 1  47
vO

 100 A and i   i 2
R1  R 2
48k
12.35


2R 1
Setting v2  0, R IN1  R ID 1  A   500k1  4 x10 4
 785 M
2R 1  49R1 


RO
75
By symmetry , R IN2  R IN1  785 M | R OUT  R OUT3 

 3.75 m
1 A   4x104 
1 
2 


FGE  1 
12.36
A
1
4

 10
requires A  10,000 (80 dB)
1 A 1 A
Op-amp parameters: RID = 500 k, RO = 35 , A = 50,000
Amplifier Requirements: AV = 200, RIN ≥ 200 M, ROUT ≤ 0.2 .
12.37
We must immediately discard the inverting amplifier case. RIN = R1 requires R1 ≥ 200 M
which can be achieved, but then R2 must be 200 R1 ≥ 40 G which is out of the question (see
values in Appendix C). So, working with the non-inverting amplifier:
A V  200   
R OUT 
R1
1
50000

and A =
 250  1
R1  R 2 200
200
RO
35

 0.14 meets the specification
1  A 250
R IN  R ID 1  A   500k250   125M does not meet the requirements
.
So the specifications cannot be met using a single-stage amplifier built using the op-amp that
was given to us.
12.38 The non-inverting amplifier is the only one that can hope to achieve both the required
gain and input resistance (see Prob. 12.37):
1
10 4
A CL   200 and A  
 50

200
R IN  R ID 1  A   1M51  51 M - too small
R OUT 
RO
1  A 

100
 1.96 - too large
51
If the gain specification is met, the input and output resistance specifications will not be met.
 R  A 
12.39 The open circuit voltage is vth  v s  2 
. Checking the loop-gain:
 R1 1  A 
 R 
6.8k


110k 
A  5x10 4 
  2910  1 so vth  v S  2   v S 
  16.2v S
6.8k  110k 
6.8k 
 R1 


R th  R OUT 
RO
R
250
 O 
 85. 9 m
1  A A
2910
12.40 The open circuit voltage is vth 
A
v . Checking the loop-gain:
1  A s
9
 R 
0.39k
56k 



A  10 4 
  69.2  1 so v th  v S 1  2   v S 1 
  145 v S
0.39k  56k 
 0.39k 
 R 1 
or more exactly: v th  v S
A
10 4
RO
200
 vS
 143 v S | R th  R OUT 

 2. 85 
1  A
1  69.2
1  A
70. 2
12.41 Applying the definition of fractional gain error,

FGE 
FGE  1 
A
1  A
R 2  R 2 1     A
 

R1  R 1 1   1  A
1
R
 2
R1
A
1
A
2

2
1  A 1  A
1  A
For A  1, FGE 
1
A
2 
1
A
1
2x10 
5
1
1000
1    A
A
 1  1  2 


1

1

A
1

A




2 which must be  0.01
2  0. 01 | Taking the positive sign, 2  0.005 and   0.25%
12.42 Using the results from problem 12.41:
For A 
1
A
4x10 4
1
 800  1, so FGE 
50
A
2 
1
800
2 which must be  0.02
2  0.02 | Taking the positive sign, 2  0.01875
and   0.938%
12.43 One worst-case tolerance assignment is given below. The second is found by reversing
the resistor values.
9. 995
v  v
v  v
0.50025
 0.49975 v ic | i = ic
 ic

v ic
10.005  9. 995
9. 995k 9.995k 9.995k
0.50025
vO  v   i 10.005k  v   i 10. 005k  0.49975 vic 
v ic 10.005k
9. 995k
v   v ic
vO  0.001v ic and A cm 
vO
 0. 001 | The value of Adm  1 is not affected by the
vic
small tolerances. CMRR =
Adm
 1000 | CMRR dB  60 dB
A cm
12.44
10
5  5. 01
 5. 005V. The maximum equivalent input error is
2
V IC
5.005

 0.500 mV, but the sign is unknown. Therefore the meter reading
CMRR
10 4
may be anywhere in the range 9. 50 mV  V meter  10.5 mV.
V IC 
12.45
v1  v 2
 10 sin 120t V and v id  v1  v 2  0.50 sin 5000t V
2
99k
v  v
v  v
0.09258
v   v ic
 0.90742 vic | i = ic
 ic

v ic
10.1k  99k
9.9k
9.9k
9.9k
0.09258
vO  v   i 101k  v   i 101k   0.90742 v ic 
v ic 101k
9.9k
v
vO  0.037 v ic and A cm  O  0.037 | The value of Adm  10 is not
v ic
v ic 
affected by the small tolerances. CMRR =
A dm
 270 - a paltry 48.6 dB
A cm
v O  A dm v id  Acm v ic  0.370 sin 120t  5.00 sin 5000t V
12.46
2k


Setting v2  0, R IN1  R ID 1  A  2R IC  1M 1  7.5x10 4
 2500M  852 M

2k  24k 
By symmetry , R IN2  R IN1  852 M | R OUT  R OUT3 
RO
100

 2.67 m
1

A


 1  75000 


2 
12.47 See figure on next page for labels.
 R 
 R 
 R 
V O  V OS  I B1R1 1  2   I B2 R 3  2   V OS  I B1R 1 1  2   I B 2R 2
 R 3 
 R 3 
 R 3 
 10 6 
V O  0.001 10 7105 1 
 0.95x10 7106  .011.015 V
5 
10




Worst case VO = -0.026 mV, Ideal output = 0 V. Error = -26 mV
Yes R1 should be R2||R3 = 90.9 k.
12.48
v O  A V v ID  V OS | A V 
dv O 10  5  V

 7,500
dv ID
2  0 mV
When v O  0, v ID   V OS and so V OS   0.667 mV.
12.49
11
12.50
For I B2  0: Since v+ must = v - = V OS , the current through C is iC t  =
V OS
V
dt  V OS  OS t
R
RC
For V OS  0, i C t  = I B 2 since v-  v   0.
1 t
1 t
I
v O t    iC t dt   I B 2 dt  B2 t | Summing these two results yields
C 0
C 0
C
V
I
Eq . (12.100): v O t  V OS  OS t  B2 t | Note that vC 0   0 for both cases.
RC
C
v O t   V OS 
12.51
40dB  100  1 
R1 R 2

R1  R2
1
C
t
1
V OS
R
t
0 i C t  dt  V OS  C 0
R2
R2
or
 99 | For bias current compensation, R1 R 2  10k
R1
R1
R2
1M
 10k | R 2  10k1  99   1. 00 M and R1 
 10.1k
R2
99
1
R1
The nearest 5% values would be 1 M and 10 k.
12.52
v a  v b 4.99  5.01

 5V v dm  va  v b  4.99  5.01  0.02V
2
2
v  v2
0.02V
v2  v a  4.99V v 3  v b  5.01V i 32  3

 10 4 A
200
200
vcm 
v1  v2 10 4 4900   4.99  0.49  4.50 V
v4  v 3  10 4900   5.01  0.49  5.50V
4

v4


5.50 
i 2   i 32 
  10 4 
  3.75x10 4 A


10.01k  9. 99k 
20k 
9. 99k
v5  v6  v 4
 2. 747V
10.01k  9. 99k
v  v5
4.50  2.747
i 1  i 32  1
 10 4 
 75.5x10 4 A
9.99k
9.99k
v  v5
4.50  2.747
vO  v 5  1
10.01k  2.747 
10.01k  0.990 V
9.99k
9. 99k
The common-mode and differential mode inputs to the differential subtractor are
v  v2
vcms  1
 5.00V and vdms  v1  v 2  0.100V .
The subtractor outputs for the
2
common-mode and differential mode inputs are:
12
For the common - mode signal, v 5  v 6  5
9.99k
10.01k  9.99k
9.99k
5 5
v1  v 5
9. 99k
10.01k  9.99k 10.01k
vOcm  v 5 

10.01k  5


9.99k
10.01k  9. 99k
9.99k
.01
vOcm  . 0100V and A cm 
 .002
5
9. 99k
For the differential mode signal, v5  v6  0.5
10.01k  9. 99k
v  v5
0. 5  v5
vOdm  v 5  1
10. 01k  v 5 
10.01k
9. 99k
9.99k
vOdm  1.00V and A dm 
12.53
1 
a Ideal V O  0.005 V 

1. 00
A
 50 CMRR = dm  25, 000
. 02
A cm
100k 
  0. 460V
1.1k 
A
10 4
 0.005 V  0.001V 
 0.546 V
1  A
4 1.1
1  10
101.1
-0.460 -(-0.546)
c Error =
 0.187 or  18.7%
-0.460
b  V O  0.005 V  0.001V 
12.54
Inverting Amplifier: v O  A V v S  6.2v S as long as v O  10V as constrained
by the op- amp power supply voltages
(a ) V O  6.2 1 = -6.2V, feedback loop is working and V -  0
(b) V O  6. 23  = +18 V; V O saturates at VO  10V
The feedback loop is broken since the open- loop gain is now 0.
(The output voltage does not change when the input changes so =
A 0)
By superposition, V   3
6.2k
1k
 10
 1.19V
7.2k
10k
12.55
Non  inverting Amplifier: v O  A V v S  40v S as long as v O  15 V
(a ) V O  40 0.25V  = +10V, feedback loop is working and VID  0
(b) V O  400.5V  = 20V; V O saturates at VO  15 V
The feedback loop is broken since the open- loop gain is now 0.
(The output voltage does not change when the input changes so =
A 0)
V ID  V   V   0.5V 15
1k
 0.125V.
1k  39k
12.56
13
i O  i L  i 2 and
i O  1.5mA. The output voltage requirement gives i L 
which leaves 0.500mA as the maximum value of i2 . i 2 
The closed - loop gain of 40db (A V =100) requires
10V
gives
R1  R 2
10V
 1.00mA
10k
R 1  R 2   20k.
R2
 99.
R1
The closest ratio from the resistor tables appears to be
R2
 100 which is within 1%
R1
of the desired ratio. (This is close enough since we are using5% resistors.)
There are many many choices that meet both
R2
 100 and R1  R 2   20k.
R1
However, the choice, R1 = 200 and R 2  20k is not acceptable because its
minimum value does not meet the requirements
: 20. 2k1  0.05   19.2k.
The smallest acceptable pair is R 1 = 220 and R 2  22k.
12.57
15V
 3 mA so i2  1 mA
5k
v
15
i2  O  1 mA requires R 2 
 15k
R2
.001
To account for the resistor tolerance, 0.95R2  15k requires R2  15.8k . For AV = 46 dB
= 200, R2 = 200 R1, and one acceptable resistor pair would be R1 = 1 k and R2 = 20 k.
Many acceptable choices exist. An input resistance constraint might set a lower limit on R 1.
iO  iL  i2  4mA and iL 
12.58 The maximum base current will be 5 mA, and the maximum emitter current will be I E
= (F+1)IB = 51(5mA) = 255 mA. Since IE = 10V/R, R ≤ 10V/0.255A = 39.2 .
12.59 Referring to the figure used in the solution to problem 12.57:
10V
10 V
 2.5 mA and i L 
 2 mA so i 2  0. 5 mA
4k
5k
10V
R
R2 
 20k A V  46dB  2  200
0.5mA
R1
iO  iL  i2 
One possible choice would be R2 = 20 k and R1 - 100 . However, the op-amp would not be
able to supply enough output current if tolerances are take into acount. Better choices would
be R2 = 22 k and R1 - 110  or R2 = 200 k and R1 - 1 k which would give the amplifier a
much higher input resistance.
(b) V 
above.
14
v o max
200

10V
 50 mV
200
(c) RIN = R1 = 110  and 1 k for the two designs given
12.60
Using the expressions in Table12. 4:
5
10
24k
1
240k 11
First stage:  =
=
| A V1  
 10.0
5
24k  240k 11
24k
10
1
11
240k
100
R IN  24k  500k
 24. 0 k | R OUT 
 11. 0 m
1  10 5
105
1
11
105
10k
1
50k
6
Second stage:  =
=
| A V1  
 5.00
10k  50k 6
10k
10 5
1
6
240k
100
R IN  10k  500k
5  10.0 k | R OUT 
5  6.00 m
1  10
10
1
6
Overall amplifier:
10k
A V  10.0
5.00   50. 0 | R IN  24. 0 k | R OUT  6.00 m
10.0k 11.0m
For all practical purposes ,the numbers the same. R OUT  6.00 m is a good
approximation of 0.
12.61
Use the expressions in Table12.4. All three amplifier stages are the same.
5
10
2k
1
40k
21  20. 0
=
=
| A V1  
5
2k  40k 21
2k
10
1
21
40k
200
R IN  2k  250k
5  2. 00 k | R OUT 
5  42.0 m
1  10
10
1
21
2k
2k



For the overall amplifier: A V  20.0
20.0
20.0   8000

2k  42. 0m 
2k  42.0m 
R IN  2.00 k | R OUT  42.0 m | For all practical purposes ,the numbers the same.
R OUT  42.0 m is a good approximation of 0.
12.62
15
50 2  5000 < 50 3 | Three stages will be required to keep the gain of each stage  50.
However, the input and output resistance requirements could further constrain the
gains and must be checked as well. A
For R OUT 
85
= 10 20
 1.778 x 10
4
RO
100
1
:
 0.1  A  999    0. 0562   17.8.
1  A 1 + A

For R IN  R ID 1  A  2R IC: 1M1  A  2G  10M  A  9 
1
 1976

17.8 50 50   5000 so three stages is still sufficient.
12.63
VS  VO

12.64
A V s   
A nom

V
f nom

H
f min

H
12.65

 R  SC R1 R 2  1
R1
SCR 2  1R 1
V
 VO
| A V s   O  1  2 
R2
V S 
R1  SCR 2  1
SCR 2  1R 1  R 2
SC
R1 
1
R2 
SC
R2
1
R 1 sCR 2 1
A V 0   
R2
R1
fH 
1
2CR 2
330k
330k1.1
330k0. 9
 33 A max

 40.3 A max

 27.0
V
V
10k
10k0.9 
10k1.1

2 10

1
10
3.3x10

5
 4. 83kHz f max

H
1
2 10 10 1.2 3.3x10 5 1.1

1
2 10
10
0.53.3x10
5
0.9 
 10.7kHz
 3.65kHz
60db / decade requires 3 poles 3(-20db / decade). Using three identical
amplifiers: A V  3 1000  10 and f H1 
R 2C 

1
2 39.2x10 3

 4.06x10
6
f H3
1
23
 1.96 20kHz   39. 2kHz.
1
s. One possible choice would be
C = 200 pF, R 2  20 k and R1  1 k giving A V  10 and f H  39.8 kHz.
12.66
As  
T
RO
|  T  A o B | Z OUT 

s  B
1  A s 
Z OUT  R O
16
s  B
RO

s  B 1  A o 1  A o 
RO
s  B
 RO
T
s   B   T
1

s  B
s
s
1
B
RO
B

s
1  Ao  1  s
1
 B 1  A o 
T
1
12.67 Using MATLAB:
b=1/11; ro=100; wt=2*pi*1e6; wb=wt/1e5;
n=ro*[1 wb]; d=[1 b*wt];w=logspace(0,7);
r=freqs(n,d,w);
mag=abs(r); phase=angle(r)*180/pi;
subplot(212);semilogx(w,phase)
subplot(211);loglog(w,mag)
12.68
Z IN  R 1  R ID
s   B 
R2
R2
 R1  R ID
 R 1  R ID R 2
T
1 A s 
s  B  T
1
s  B
R IDR 2
Z IN  R 1 
s   B 
s  B   T
s  B 
R ID  R 2
 R1 
R IDR 2 s   B 
R ID s   B   T   R 2 s  B 
s  B  T
Z IN


s 
R2
s 
R ID R 2 B 1 
R ID

1 

1  A o 
  B 
 B 
 R1 
 R1 
R2
s
R ID  R 2
R ID B 1  A o   R 2 B  sR ID  R 2 
R ID 
1
R2
1

A

1

A

o
B
O R
ID 
1 A o 
Z IN

R 2 
 R 1  
R

ID

1  Ao 1 


s 
1

  B 
s
R ID  R 2
 B 1  A O  R  R 2
ID
1  A o 
12.69 Using MATLAB:
n1=1e6; d1=[1 2000];
n2=1e6; d2=[1 4000];
n3=1e12; d3=[1 6000 8e6];
w=logspace(2,5);
[m1,p1,w]=bode(n1,d1,w);
[m2,p2,w]=bode(n2,d2,w);
[m3,p3,w]=bode(n3,d3,w);
subplot(211)
loglog(w,m1,w,m2,w,m3)
subplot(212)
semilogx(w,p1,w,p2,w,p3)
12.70
17
AV  
Z 2 A
Z1 1  A
A V s   
A V s   
=
Z1
Z1  Z 2
A =
T
s + o
Z1  R1
Z2 
R2
sCR 2  1
R2
T
R1
 R

 R 
2
s R 2C  s 1  2  R 2 Co   T   o 1 2    T
 R 1

 R 1 
3.653x1013
s 2  3.142x10 7 s  1.916 x1012
Using MATLAB:
bode(-3.653e13,[1 3.142e7 1.916e12])
12.71
V s
V s 
1
a S  sCV O s  | A V s  O

which is the transfer function of an integrator
R
V S s
sRC
b  Generalizing Eq. 12.119: A V s  
=
R1
1
R1 
sC
A V s   
A V s   

sCR
T
| A s  
sCR  1
s  B
Z 2 A s
1
Z1
| Z2 
| Z1  R 1 |  =
Z1 1  As 
sC
Z1  Z 2
T
sRC
sRC
1
s   B sRC  1
| A V s  
T
sRC
sRC 1
sRC
1
s   B sRC  1
1
sRC T
T

sRC s  B sRC  1  sRC T
s  B sRC  1  sRC T
T
RC

s 2  s B   T

T
T
RC
RC


1   B





B


s   T s   RC
s   T s  A 1RC 

RC  RC




T
o
using dominant root factorization where it is assumed T   B and  T 
12.72
18
wrc=1/(1e4*470e-12); wt=2*pi*5e6; wb=2*pi*50;
n=wt*wrc; d=[1 wt+wb+wrc wb*wrc];
bode(n,d)
1
.
RC
12.73
20dB  A V  10  
C
R2
R1
R IN  R 1  20 k R 2  10R 1  200 k
1
1

 796 pF  820 pF using values in Appendix C
2f H R 2 2 1000 200k
12.74

2k
1
105

| A 
 4760  1
2k  40k 21
21
(a ) A V  
R2
40k
3x10 6 Hz

 20 | f H  f T 
 143kHz
R1
2k
21
3
(b) A V  20   8000 78dB  | f H3  0.51f H  72.9kHz
12.75 The table below follows the approach used in Table 12.8.
required gain = A V
85
 10 20
The only change is the
4
 1.778 x 10 .
Cascade of Identical Non-Inverting
Amplifiers
# of Stages
1
2
3
4
5
6
7
8
9
10
11
12
AV(0)
Gain per
Stage
1/
FH
Single Stage
2.00E+04
1.41E+02
2.71E+01
1.19E+01
7.25E+00
5.21E+00
4.12E+00
3.45E+00
3.01E+00
2.69E+00
2.46E+00
2.28E+00
5.00E+01
7.07E+03
3.68E+04
8.41E+04
1.38E+05
1.92E+05
2.43E+05
2.90E+05
3.33E+05
3.71E+05
4.06E+05
4.38E+05
FH
N Stages
RIN
ROUT
5.000E+01
4.551E+03
1.878E+04
3.658E+04
5.320E+04
6.717E+04
7.839E+04
8.724E+04
9.415E+04
9.951E+04
1.037E+05
1.068E+05
6.00E+09
7.08E+11
3.69E+12
8.41E+12
1.38E+13
1.92E+13
2.43E+13
2.90E+13
3.33E+13
3.71E+13
4.06E+13
4.38E+13
8.33E+00
7.06E-02
1.36E-02
5.95E-03
3.62E-03
2.60E-03
2.06E-03
1.72E-03
1.50E-03
1.35E-03
1.23E-03
1.14E-03
 x FT
We see from the spreadsheet that a cascade of seven identical stages is required to achieve
the bandwidth specification. Fortuitously, it also meets the input and output resistance
specs. For the non-inverting amplifier cascade:
AV  1 
R2
R
 4.12  2  3.12
R1
R1
A similar spreadsheet for the cascade of identical inverting amplifiers indicates that it is
impossible to meet the bandwidth requirement.
19
12.76
a From Problem 12.75, A V  1  R 2  4.12  R 2  3.12 | Exploring the 5% resistor
R1
R1
tables, we find R 2  62k and R 1  20k yields
R2
 3.10 as a reasonable pair.
R1
7
The nominal gain of the cascade is then AV = 4.10  = 1.948 x 10 4 .
A V = 86db  1dB  1.778 x 10 4  A V  2. 239 x 10 4 and the gain is well within this range.
Many amplifiers will probably fail due to tolerances with 5% resistors. A Monte Carlo
analysis would tell us. If we resort to 1% resistors to limit the tolerance spread,
R 2  30.9 k and R1  10. 0 k is one of many possible pairs.
b  For R 2  62k and R1  20k,  =
1
5x10 6
| f H1  f T 
 1.22 MHz
4.1
4.1
1
f H  1. 22MHz 2 7 1  394 kHz
12.77
Two stages - See problem 12.78
12.78 One possibility: Use a cascade of two non-inverting amplifiers, and shunt the input of
the first amplifier to define the input resistance.
60db  A V  1000 | A single - stage amplifier with a gain of 1000 would have a bandwidth
of only 5 kHz using this op - amp. Two stages should be sufficient if RIN and R OUT can also
be met. A design with f H2  f H1will be tried.
First stage: Non - inverting with bandwidth of 20 kHz
1 =
f H1 20kHz

 0. 004
fT
5MHz
A V1 
1
 250  A V2  4   2  0.25  f H2  1.25MHz
1
Since f H2  f H1 , f H  f H1  20kHz. A o  85dB  17800
R O2
100
Checking R OUT 

 0. 0225 which is ok.
1  A o 2 1  17800 0.25 
Choosing resistors from the Appendix, a possible set is
Amplifier 1: R 1  1. 2k, R 2  300k and shunt the input with R3 = 27k
Amplifier 2: R 1  3. 3k, R 2  10k
Checking gain: A V
12.79
20
17800

17800
1
251


 17800 
 17800   997.5  60.0dB
1


4.03 
function sg=Prob79(tol);
sg=0;
for j=1:10
ao=100000; ft=1e6*sqrt(2^(1/6)-1);
for i=1:500,
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g1=ao/(1+ao*beta); b1=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g2=ao/(1+ao*beta); b2=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g3=ao/(1+ao*beta); b3=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g4=ao/(1+ao*beta); b4=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g5=ao/(1+ao*beta); b5=beta*ft;
r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5));
beta=r1/(r1+r2);g6=ao/(1+ao*beta); b6=beta*ft;
gain(i)=g1*g2*g3*g4*g5*g6; bw(i)=(b1+b2+b3+b4+b5+b6)/6;
end;
sg=sg+sum(gain<1e5 | bw<5e4);
end;
end
(a) For 5000 test cases with tol = 0.05, 33.5% of the amplifiers failed to meet either the gain
or bandwith requirement.
(b) For 10000 test cases with tol = 0.015, 0.1% of the amplifiers failed to meet either the
gain or bandwith requirement.
12.80
N
1 
1
f
a    G    1 | f H1  f T | f H  T1
 
GN
GN
Z
Z ln 2
For 2  e
Setting
2Z  
Z
and G  e
Z ln G
d f H 
:
  
dz f T 

d f H 
Z 1
Z
   0  G  2 1
dz f T 
2



1
2 2Z
1
2N
1
f
1| H 
fT
1
GZ  2 Z 1
2



1
2 2Z
2N  1
1
GN
2

Z
G
1
2

1
Z
for Z =
1

ln 2  2Z  1 2 GZ ln G

G 2Z


1

Z
Z
Z
Z
ln 2  2  1 2 G ln G  0  2 ln 2  2 2  1 ln G




2 ln G
2 ln G


 Z ln 2  ln 

ln 2  2 ln G
 ln 2  2 ln G 


ln G
ln 

 ln G  ln 2 
Z
 N opt 
ln 2
b  N opt 
ln 2
5


ln10
ln 

5
ln10  ln 2 
1
N
ln 2

ln G

ln 

ln G  ln 2 
 22.7 which agrees with the spreadsheet in Table12.8
f Hopt  106.0 kHz
12.81
21

22k
1
5x104

A 
 7240  1
22k  130k 6.91
6.91
R
130k
(a ) A nom
1 2 1
 6.91
V
R1
22k
 nom 
max
AV
nom
fH
12.82
1
R2
130k1. 05
R
130k0.95 
min
1
 7.53 | A V  1  2  1 
 6.35
R1
22k0.95 
R1
22k1. 05
  nom f T 
10 6 Hz
10 6 Hz
10 6 Hz
max
min
 145 kHz | f H 
 157 kHz | f H 
 133 kHz
6.91
6.36
7.53
function [gain,bw]=Prob82a
ao=50000; ft=1e6;
for i=1:500,
r1=22000*(1+0.1*(rand-0.5));
r2=130000*(1+0.1*(rand-0.5));
beta=r1/(r1+r2);
gain(i)=ao/(1+ao*beta); bw(i)=beta*ft;
end;
end
[gain,bw]=prob82a;
mean(gain)
ans = 6.9140
std(gain)
ans = 0.2339
mean(bw)
ans = 1.4478e+05
std(bw)
ans = 4.8969e+03
Three sigma limits: 6.21 ≤ AV ≤ 7.62
130 kHz ≤ BW ≤ 159 kHz
function [gain,bw]=Prob82b
for i=1:500,
ao=100000*(1+1.0*(rand-0.5));
ft=2e6*(1+1.0*(rand-0.5));
r1=22000*(1+0.1*(rand-0.5));
r2=130000*(1+0.1*(rand-0.5));
beta=r1/(r1+r2);
gain(i)=ao/(1+ao*beta); bw(i)=beta*ft;
end;
end
[gain,bw]=prob82b;
mean(gain)
ans = 6.9201
std(gain)
ans = 0.2414
mean(bw)
ans = 2.8925e+05
std(bw)
ans = 8.5536e+04
Note that the bandwidth is essentially a uniform distribution.
3: 6.20 ≤ AV ≤ 7.64
98.9% of the values fall between: 146 kHz ≤ BW ≤ 439
kHz


12.83
SR  V O  15 V 2 2x10 4 Hz  1.89x10 6
12.84
f 
22
SR
10V
1

 159 kHz
2V o 10 6 s 20V
V
V
or 1.89
s
s
12.85
The negative transistion requires the largest slew rate
: SR 
V 20V
V

 10
t
2s
s
12.86
a For the circuit in Fig. 12.44: R ID  250k | R =1k  an arbitrary choice
B =

2 5x10
8x10
4
6
 =125 |
C=
1
1

 2.55F | R O  50 | A o  80,000
 BR 125 1000
b  Add a resistor from each input terminal to ground of value 2R IC  1G (See Prob. 12.87)
12.87 Two possibilities:
12.88
1: C =
1
1

 1.592 F  setting R1 arbitrarily to 100.
1R1 2 103 100 
 2: R 2 =
 
1
1

 1   Using the same value of C.
2 C 2 10 5 1.592 F 
 
23
12.89
12.90
*PROBLEM 12.89 - Six-Stage Amplifier
VS 1 0 AC 1
XA1 1 2 0 AMP
XA2 2 3 0 AMP
XA3 3 4 0 AMP
XA4 4 5 0 AMP
XA5 5 6 0 AMP
XA6 6 7 0 AMP
.SUBCKT AMP 1 2 7
RID 1 3 1E9
RO 6 2 50
E2 6 7 5 7 1E5
E1 4 7 1 3 1
R 4 5 1K
C 5 7 15.915UF
R2 2 3 130K
R1 3 7 22K
.ENDS
.TF V(7) VS
.AC DEC 40 1 1MEG
.PRINT AC V(1) V(2) V(3) V(4) V(5) V(6) V(7)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7)
.END
*PROBLEM 12.90 - Six Stage Amplifier
VS 1 0 AC 1
XA1 1 2 0 AMP
XA2 2 3 0 AMP
XA3 3 4 0 AMP
XA4 4 5 0 AMP
XA5 5 6 0 AMP
XA6 6 7 0 AMP
.SUBCKT AMP 1 2 8
RID 1 3 1E9
RO 7 2 50
E2 7 8 6 8 1E5
*Two dummy loops provide separate control of Gain & BW tolerances
G1 8 4 1 3 .001
R11 4 8 RG 1000
E1 5 8 4 8 1
RC 5 6 1000
C 6 8 CC 15.915UF
*
R2 2 3 RR 130K
R1 3 8 RR 22K
.ENDS
.MODEL RR RES (R=1 DEV=5%)
.MODEL RG RES (R=1 DEV=50%)
.MODEL CC CAP (C=1 DEV=50%)
.AC DEC 20 1E3 1E6
.PROBE V(7)
.PRINT AC V(7)
.MC 1000 AC V(7) MAX OUTPUT(EVERY 20)
*.MC 1000 AC V(7) MAX OUTPUT(RUNS 77 573 597 777)
.END
Maximum gain = 103 dB; Minimum gain = 98.5 dB
Maximum Bandwidth = 65 kHz; Minimum bandwidth = 38 kHz (These are approximate.)
24
12.91
A o  200,000 | R ID  1012  | R O unspecified | R =1 k
C


2 2.5x106
1

1
| o  T 
 25 | C 
 12.7 F
5
o R
Ao
2x10
25 1000 
12.92
A OL  15000
CMRR  80dB
PSRR  80dB V OS  13mV
I B  8nA I OS  2nA
12
Power supply voltages:  18 V R ID  lower bound not given (10  typical )
SR  lower bound not given (50 V / s typical )
GBW  lower bound not given (20 MHz typical )
12.93
a C 2  C  0.005F | C1  2C  0. 01F | R =
For Q =
1
2
: A V s 
2
 2o
2
s  2o s   o
1

2 O C
1
 1.13k
2 40000  0.005F
| A 0  1 | A j o  
1
2
 H   o! | f H  20kHz
C
b  For f O = 40kHz: C '  : C1'  0. 005F | C '2  0. 0025F | R  1.13k
2
12.94 (a)
(b)
r1=1130; r2=1130; c1=1e-8; c2=5e-9;
wo=1/sqrt(r1*r2*c1*c2)
n=wo*wo;
d=[1 2/(r1*c1) wo*wo];
bode(n,d)
*PROBLEM 12.94 - Low-pass Filter
VS 1 0 AC 1
R1 1 2 1.125K
R2 2 3 1.125K
C1 2 6 0.01UF
C2 3 0 0.005UF
ISEN 3 6 DC 0
EC 4 0 3 6 1
RC 4 5 1K
CC 5 0 15.915UF
E1 6 0 5 0 100000
.AC DEC 40 1 1MEG
.PRINT AC IM(VS) IP(VS) VDB(6) VP(6)
.PROBE I(VS) V(6)
.END
(c) The magnitude response is very similar to the ideal case. However, note the excess phase
shift as one approaches the fT of the amplifier. In this case, it is not causing a problem, but
for higher gain filters the situation would be different.
12.95
25
Using Eq . 12.134: V 1 s 
G1 V S s sC 2  G 2 
s C1C 2  sC 2 G1  G2   G1G 2
2
| I S  G1 V S  V1 
2


G1 sC 2  G2 
s C1C 2  sC 2 G2
I S  G1 V S 1  2
  G1V S 2
s C1C 2  sC 2 G1  G2   G1G 2
 s C1C2  sC 2 G1  G 2   G1G2 

2
2
s  s o  o
s2 C1C 2  sC2 G1  G2   G1G 2
VS
Q
Z S s  
 R1
 R1
IS
s 2C1C 2  sC 2 G2

1 
ss 

R 2 C1 

 2o 
26
1
|
R1R 2C1C 2
o
1  1
1 




Q
C1 R 1 R 2 
12.96
r1=2260; r2=2260; c1=2e-8; c2=1e-8;
wsq=1/(r1*r2*c1*c2);
n=r1*[1 2/(r1*c1) wsq];
d=[1 1/(r1*c1) 0];
bode(n,d)
12.97
G1V S  sC 1  G1  G2 V1  s KC1  G2 V 2
0  G 2 V1  sC 2  G 2 V 2
V O  KV 2 |
O 
VO
K
 2
VS
s R 1R 2 C1C 2  s R 1C1 1  K   C 2 R1  R 2  1
1
R1 R 2C1C 2

| Q

 3O
1 K

1





R 2 C 2 R 1 R 2 C1 
For R1  R 2  R and C1  C 2  C,  O 
1
RC
Q
 2O
3K
SQ
K 
K
3 K
12.98
o 
1
R o
R

| SR o  1
 1
1
R 1R 2 C1C 2
 o R1 o

S R o1  
3
1
 1 

R
 R 1  2  1
R 2 C1C 2  2 
o
 1  o 
1

  
2
 2 R1 
1
1

| By symmetry , S Co1  
2
2
12.99
a For C 1  C  C2 :  o 

SQo 
1
1
| Q=
C R 1R 2
2
R2
1
| o 
R1
2R1CQ
 Q  o 
Q o
Q 
1
o


  1 b  S Q  1

 
 o Q
o  2R 1CQ 2  o  Q 
12.100
As noted in Example 12.11, the maximally flat response corresponds to Q =
For Q =
1
2
.
 2o
1
1
: A V s  2
| A 0  1 | A j o  
 H   o |  f H  1kHz
2
2
s  2o s   o
2
R 1  R  R 2 and C1  2C 2  2C yields Q =
1
2
| o 
1
2R 2 C2

1
2RC
1
 1.125 x 10 4 | For C = 0.001 F, R =112.5 k. The
2 2 1000 
nearest 5% value is 110 k. The nearest1% value is 113 k. Using 1% values,
RC 
C1  0. 002 F | C 2  0.001 F | R1  R 2  113 k
12.101
27
Using the R 1  R 2  R and C1  C2  C case, A HP s  K
o 
1
RC
| Q=
1
3 K
s2

s2  s o  2o
Q
Q  1  K  2 | A HP s   2
|
s2
2
2
s  s o  o
We need to find the relationship between
 L and  o.
2
2
2
2
L 
2
L


2
2 2
o  L
  L o 
 2o  4o  4 4o
2
2
2
 4o  22o 2L  4L  2o 2L  2 4L |  4L  2o 2L   4o  0
2
 L   o
5 1
|  L  0.7862o | 2 20kHz   0.7862o
2
1
 6. 256x10 6 | For C = 270 pF, R = 23.17 k
o
The nearest1% resistor value is 23.2 k.
 o  1.599x105 | RC 
Final design: C1  C2  270 pF | R 1  R 2  23. 2 k
12.102
a Q = 1
2
R 2 1 200k 10
=

| fo 
R1
2
1k
2
1
3

5
2 10 2x10
2.2x10 
10 2
 51.2 kHz
fo
 7.23 kHz
Q
3.3k
b  K M 
 3.3 | R 1  3. 31k  3.3 k | R 2  3.3200k  660 k
1k
220pF
C1  C 2 
 66.7 pF
3.3
2f
220pF
 110 pF | R 1  1 k | R 2  200 k
b  K F  o  2 | C1  C2 
fo
2
BW 
28
12.103
(a ) BW 
f o 1000Hz
1

 200 Hz | C 1  C 2  C | Q =
Q
5
2
Choose R1  1 k  R 2  100 k | C =
o 
1
o R1R 2

R2
R
 5  2  100
R1
R1
1
  
2 10 3 10 4
 0.0159 F
1
: Checking the nearest standard values:
10R1C
C = 0.015 F  R 1  1.06k - not good; C = 0.01 F  R1  1.6 k | R 2  160 k
(b) C ' 
12.104
C
 0. 004 F | R 1  1.6 k | R 2  160 k
2. 25
r1=1000; r2=2e5; c=2.2e-10;
wo=1/sqrt(r1*r2*c*c); q=sqrt(r2/r1)/2;
n=[-2*q*wo 0]; d=[1 wo/q wo*wo];
w=logspace(4,7,300);
bode(n,d,w)
*PROBLEM 12.104 - Band-pass Filter
VS 1 0 AC 1
R1 1 2 1K
R2 3 6 200K
C1 2 6 220PF
C2 2 3 220PF
EC 4 0 0 3 1
RC 4 5 1K
CC 5 0 7.977UF
E1 6 0 5 0 50120
.AC DEC 500 1K 1MEG
.PRINT AC VDB(6) VP(6)
.PROBE V(6)
.END
SPICE yields: fo = 38.9 kHz, Q = 8, Center frequency gain = 38.8 dB. These values are off
due to the finite bandwidth of the op-amp and its excess phase shift at the center frequency
of the filter.
12.105
a BW 
o 1
rad
rad

| L  0. 833
| H  1.167
Q
3
s
s
1
BW '  BW 2 2  1  0. 215
C1  C 2  C: o 
rad
rad
1
| 'o  o  1
| Q' =
= 4.65
s
s
0.215
1
1
| Q=
C R1R 2
2
2
R2
1
|
 2Qo
R1
R1C
2




 2Qo

 6s 
b  A s  
  


s
s2  s o  1 
s2   1


Q
3


29
12.106
30
n=conv([-6 0],[-6,0]); d=conv([1 1/3 1],[1 1/3 1]); bode(n,d)
12.107
Using normalized frequency: 5 kHz   O  1 and 6 kHz  O  1.1
2
10s
12s
120s



As    2

 
s  0.2s  1s 2  0.24s  1.44  s 4  0.44s 3  2. 484s2  0.528s  1.44
At the new center frequency s = j O ,  0.443O  0.528 O  0  O  1. 095
and Aj O   1429  63.1 dB
The bandwidth points can be found using MATLAB:
w=linspace(.9,1.5,250);
[m,p,w]=bode([120 0 0],[1 .44 2.484 .528 1.44],w);
20*log10(max(m))
ans = 63.098
((20*log10(a))>60.098).*(w.');
From this last vector one can easily find: o = 1.095 -> 5.48 kHz, L = 0.970 -> 4.85 kHz,
H = 1.237-> 6.19 kHz, BW = 1.34 kHz, Q = 4.09
12.108
w1=2*pi*5000; q1=5; w2=2*pi*6000; q2=5;
n1=[-2*q1*w1 0]; d1=[1 w1/q1 w1*w1];
n2=[-2*q2*w2 0]; d2=[1 w2/q2 w2*w2];
n=conv(n1,n2); d=conv(d1,d2);
w=logspace(4,5,100);
bode(n,d,w)
12.109
Using A
V
 20 dB at the center frequency: R IN  R 1  10 k | 10 = KQ =
R 2  100 k | K =
12.110
R2
R1
10
1
1
 2 | R = KR 1  20 k | C =

 0. 0133 F.
Q
 oR 2 600Hz 20k
Q is set independently of C in the Tow-Thomas biquad.
Q
SC  0 .
12.111
12.112
12.113
12.114 In the first printing, the problem should state that there are 0.1pF stray capacitances
from each end of C1 to ground. CGS and CGD should be ignored.
31
a In this case, the sampling capacitor has an effective value of1.1pF.
AV  
1.1pF
1.0pF
 5.5 whereas the ideal gain is AV 
 5 | Gain error  0.5 or 10%
0.2pF
0.2pF
b  The stray insensitive integrator eliminates the effects of the stray capacitances
.
Therefore AV  5 and there is no gain error.
12.115 Note: It is very important to observe and discuss the effects of clock feedthrough in
the simulation results in both parts of this problem.
*PROBLEM 12.115(a) - SC Integrator
VCLK 2 0 DC 0 PULSE(0 5 0U 0.5US 0.5US 4US 10U)
VNCLK 4 0 DC 5 PULSE(0 5 5U 0.5US 0.5US 4US 10U)
VS 1 0 DC 1
M1 3 2 1 0 MOSN W=4U L=2U AS=16P AD=16P
C1 3 0 1PF
*Stray capacitance CS
CS 3 0 0.1PF
M2 3 4 5 0 MOSN W=4U L=2U AS=16P AD=16P
C2 5 6 0.2PF
E1 6 0 0 5 1E5
RID 5 0 1MEG
RL 6 0 10K
.OPTIONS VNTOL=1E-9 RELTOL=1E-6
.OP
.TRAN 0.025U 20U
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.PRINT TRAN V(3) V(6)
.END
Note that the output in part (a) is really only valid during Phase-1 times because of
clock feedthrough.
*PROBLEM 12.115(b) - Stray Insensitive SC Integrator
VCLK 2 0 DC 0 PULSE(0 5 0U 0.5US 0.5US 4US 10U)
VNCLK 4 0 DC 5 PULSE(0 5 5U 0.5US 0.5US 4US 10U)
VS 1 0 DC 1
M1 3 2 1 0 MOSN W=4U L=2U
M3 7 2 5 0 MOSN W=4U L=2U
C1 3 7 1PF
*Stray capacitances CS1 and CS2
CS1 3 0 0.1PF
CS2 7 0 0.1PF
M2 3 4 0 0 MOSN W=4U L=2U
M4 7 4 0 0 MOSN W=4U L=2U
C2 5 6 0.2PF
E1 6 0 0 5 1E5
RID 5 0 1MEG
RL 6 0 10K
.OPTIONS VNTOL=1E-9 RELTOL=1E-6
.OP
.TRAN 0.025U 30U
32
.MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99
+LAMBDA=.02 TOX=41.5N
+CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.PRINT TRAN V(3) V(6)
.END
Note that the output in part (b) is really only valid during Phase-2 times because of
clock feedthrough.
12.116
fo  fC
Q=
1pF 0.1pF 
C 3 C4
 10 5
 79.1 kHz
C1C 2
0.4pF 0.4pF 
C 3 C1C 2

C 4 C1  C2
0.4pF 0.4pF 
1pF
 1.58
0.1pF 0. 4pF  0. 4pF
12.117
12.118
V1
V o1
V
V
 I S exp
| V O1  V T ln 41
| V O2  V T ln 42
10k
VT
10 I S
10 I S

V
V 
V V
V O3   V O1  V O2   V T ln 41  ln 42   V T ln 18 22
10
I
10
I
10
IS

S
S 
4
4
V O  10 I D  10 I S exp
 V V 
VD
V V
4
 10 I S expln 18 22    14 2
VT
10 IS
 10 I S 
12.119
33
12.120
The waveform going into the low- pass filter is the same as that in Prob. 12.119
 8.2k 
except the amplitude will be VM  1V
  3.037 V.
 2.7k 
1 T
3.037 
 10k  2 2
The average value of the waveform isV  

 0.759 V
 10k 
T
12.121
The Fourier series converges very rapidly since only the even terms exist for n ≥ 2 and
the terms decrease as 1/n2.
Thus the RMS value will be dominated by the first term
(n = 1).
2
  

1

120
2
Require:
 0. 01 | 50   1    | o 

 2. 40 Hz
2
2
157
157
o 
  
1   
 o 
12.122
12.123
12.124
34
*PROBLEM 12.124 - RECTIFIER
VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0)
R1 1 2 10K
R2 4 5 10K
R3 5 6 10K
R4 2 4 10K
R5 1 5 20K
D1 3 2 DIODE
D2 4 3 DIODE
EOP1 3 0 0 2 1E5
EOP2 6 0 0 5 1E5
.MODEL DIODE D IS=1E-12A
.TRAN .01M 8M
.PRINT TRAN V(6)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
12.125
*PROBLEM 12.125 - RECTIFIER
VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0)
R1 0 2 10K
R2 4 5 10K
R3 5 6 20K
R4 2 4 10K
D1 3 2 DIODE
D2 4 3 DIODE
EOP1 3 0 1 2 1E5
EOP2 6 0 1 5 1E5
.MODEL DIODE D IS=1E-12A
.TRAN .01M 8M
.PRINT TRAN V(6)
.PROBE V(1) V(2) V(3) V(4) V(5) V(6)
.END
12.126
Simplify the circuit by taking a Thevenin equivalent of the5V source and two 10k
10k
 2.5V | R TH  10k 10k  5k
10k  10k
100k
5k
V O  5V  Using superposition: V +  2.5
5
 2.62V
100k  5k
100k  5k
100k
V O  0V: V +  2.5
 2.38V V N  2.62  2.38  0.24 V
100k  5k
resistors: V TH  5V
12.127
4.3k
 0.993 V
4.3k  39k
4.3k
V O  10V: V +  10
 0. 993 V
4.3k  39k
V N  0.993  0. 993  1.99 V
V O  10V: V +  10
12.128
4.3k
 0.487 V
4.3k  39k
4.3k
For V O  4.3  0.6  4. 9V: V +  4.9
 0. 487 V
4.3k  39k
V N  0.487  0. 487   0. 974 V
For V O  4.3  0.6  4. 9V: V +  4.9
35
12.129
Note: The design needs to use the circuit in Fig. 12.126.
For V O  0: V   V TH
R2
0.05
R4
1
 0. 975V | R TH  R 3 R 4 | V TH  5
R TH  R 2
2
R3  R 4
For V O  5: V   V TH
R2
R TH
0. 05
5
 1
 1. 025V
R TH  R 2
R TH  R 2
2
Subtracting: 5
R TH
R TH
R2
 0.05V 
 0.01 
 99
R TH  R 2
R TH  R 2
R TH
R2
R TH  R 2
0.975
R2
97.5

 V TH
 97.5  V TH 
 0.985 V
R TH
0. 01
R TH
99
R TH  R 2
V TH
0.985  5
R4
R
 3  4.077 | Choosing R 4  2k  R 3  8.154k
R 3  R4
R4
R TH  8.154k 2k  1.606k | R 2  991.606k  159k
Choosing standard values: R 2  160 k | R 3  8.2k | R 4  2 k
12.130
24k
3.4k
12
 6.74 V
3. 4k  24k
3. 4k  24k
24k
For v O = 0V: V +  6
 5. 26 V
3.4k  24k
 t 
v t   V F  V F  V I  exp

 RC 
For v O = +12V: V +  6


 T 
6.74
8
6.74  12  12  5.26  exp 1   T 1  6200 3. 3x10 ln
 50.7 s
 RC 
5.26


6.74
 T 
5.26 = 0  0  6.74 exp 2   T 2  6200 3.3x10 8 ln
 50.7 s
 RC 
5.26
1
f=
 9.86 kHz
50.7s + 50.7s
12.131 f = 0. The circuit does not oscillate. VO = 0 is a stable state.
12.132
36
a Let R1  R 2 |  =
R1
1
1+ 

| T = 2RC ln
 2RC ln 3  2.197RC
R1  R 2 2
1
During steady - state oscillation, the maximum output current from the op- amp is
I=
5  2.5
5
5
7.5

| Let R = R1  R 2 |

 1mA  R  10k
R1  R 2
R
2R
R
0. 001s
 4. 55x10 4 s | Selecting C = 0. 015F, R = 30. 3k  30 k 5% values
2.197
1
Final values: R = R 1  R 2  30 k | C = 0.015 F | f =
 1.01 kHz
2.19730 k 0. 015F 
RC 
b   =
1
R
1 2
R1
|  max 
1
1
= 0.525 |  min 
= 0. 475
30k 0.95 
30k 1.05 
1
1
30k 1.05 
30k 0.95 
1 0.525
 1. 213x10 3 s  f min  825 Hz
1 0.525
1 0. 475
T min  2 30k 0.95 0.015F 0.90  ln
 7.949x10 4 s  f min  1.26 kHz
1 0. 475
4.75
c For v O = +4.75V: V +  4.75 
 2.375 V
2
5.25
For v O = -5.25V: V +  5.25 
 2.625V
2
t 
v t   V F  V F  V I  exp


 RC 
T max  230k 1. 05 0. 015F 1.1 ln
7. 375
 T 
2.375  4.75  4.75  2. 625exp 1  T 1  RC ln
 1.133RC
 RC 
2. 375
 T 
7.625
2.625 = 5. 25  5.25  2.375 exp 2   T 2  RC ln
 1. 066RC
 RC 
2.625
T = 2.199RC = 2.19930k 0.015F  = 9.896 x10 -4 s  f = 1.01 kHz  Very little change
12.133
For a triangular waverform with peak amplitude VS and o  2000:
v t  

8V
 n2 S2
n1
sin
n
sin n ot
2
For the low - pass filter: A V s   
1
s
1
3000
| A V  jf  
1
2
 f 
1  

1500 
A V  j1000   0.832 | A V  j3000   0. 447 | A V  j5000   0. 287
1 5 2
 = 7.41 V
0.832
8

This series contains only odd harmonics: For n = 2, V 2000 = 0.
For a 5V fundamental:
8V S
2
0.832  5  V S =
 5  1
For f = 3 kHz , n = 3: V 3000  0. 447

 0.298 V
.832  3 2
 5  1
For f = 5 kHz, n = 5: V 5000  0.287 

 69.0 mV
.832  5 2
37
12.134
V on
0. 6
1
V CC
15k
10  841 s
T  RC ln
| =
 0. 357 | 51k0.033F  ln
1
15k  27k
1  0. 357
V
10 
1   CC
1  0. 357  
V EE
10 
T r  RC ln
 51k0.033F  ln
 416 s
V on
0. 6
1
1
V EE
10
1
12.135
V on
V
1   CC
V CC
V EE
| T r  RC ln
V
1
1  on
V EE
1
T  RC ln
V on
V
1   CC
V CC
T
V EE

ln
V
1
Tr
1  on
V EE
1
| ln
0.6
5
2
1 
10s
5
5 | ln 1.12   2 ln 1   | 1.12   1   
ln

ln






0.6
1 
5s
0.88
1   
1    0.88 
1
5
R
MATLAB gives  = 0.6998  1 = 2.33 | R 2  13 k | R 2  30. 3k  30 k
R2
1
RC =
38
10 5
 7.595s | C =150 pF | R = 50.6k  51 k
0.6
1
5
ln
1  0. 6998