Microelectronic circuit design 4th edition jaeger solutions manual

Microelectronic Circuit Design 4th Edition Jaeger Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/microelectronic-circuit-design-4th-edition-jaeger-solutions-manual/

CHAPTER 10 10.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Speed Distance traveled Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage amplitude voltage phase

current amplitude current phase power power factor spectrum Fan speed Humidity Lawn mower speed Light intensity Oven temperature Refrigerator temperature Sewing machine speed Stereo volume Stove temperature Time TV picture brightness TV sound level Wind velocity

10.2 (a) 20 log (120)= 41.6 dB | 20 log -60 = 35.6 dB | 20 log (50000)= 94.0 dB

20 log -100000 = 100 dB | 20 log(0.90)= −0.915 dB (b) 20 log (600)= 55.6 dB | 20 log (3000)= 69.5 dB | 20 log -106 = 120 dB 20 log(200000)= 106 dB | 20 log(0.95)= −0.446 dB (c) 10 log (2x109 ) = 93.0 dB | 10 log (4x105 ) = 56.0 dB 10 log (6x108 ) = 87.8 dB | 10 log(1010 ) = 100 dB



10-1

©R. C. Jaeger & T. N. Blalock 03/24/2010

10.3 Using the final case :

Av =

Vo 1.71V = = 17.1 AvdB = 20log(17.1)= 24.8 dB Vi 0.1V

Ai =

I o 1.71V = Ii 32

Ap =

Po 45.2mW = = 9.22x105 Vi I i 49.0nW

Checking :





0.1V = 5.45x104 102k

(

)

AidB = 20log 5.45x104 = 94.7 dB

(

)

ApdB = 10log 9.22x105 = 59.6 dB

24.8 + 94.7 dB = 59.8 dB 2

10.4 Av = 20log(Av ) | Using our calulator solver or MATLAB yields Av = 29.353 10.5 (a) 4 vO

2 0 vS

-2 -4 0

0.5

1

1.5

2

2.5

3

3.5

4

x10 -3

(b) 500 Hz : 10 o | 1500 Hz : 0.3330 o | 2500 Hz : 0.2000 o (c) 500 Hz : 230 o | 1500 Hz : 130 o | 2500 Hz : 130 o (d) 500 Hz : 230 o | 1500 Hz : 330 o | 2500 Hz : 530 o (e) Yes 

10-2

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.6

Vs = 0.0025V | PO = 30W | Vo = 2PO RL = 2(30)(8) = 21.9V 21.9 = 8760 | 20 log (8760)= 78.9 dB .0025 0.0025V V 21.9V Ii = = 45.5nA | I o = o = = 2.74A 5k + 50k 8 8 2.74A Ai = = 6.02x107 | 20log 6.02x107 = 156 dB 45.5nA 30W Ap = = 5.27 x 1011 | 10log 5.27x1011 = 117 dB .0025V (45.5nA) Av =

(

)

(

)

2 78.9 + 156 Checking : = 117 dB 2 10.7



Vi = 0.02 V | PO = 20 mW | Vo = 2PO RL = 2(.02)(8) = 0.566 V 0.566 = 28.3 | 20 log (28.3)= 29.0 dB .02 0.02V V 0.566V Ii = = 385 nA | I o = o = = 70.8mA 2k + 50k 8 8 70.8mA Ai = = 1.84 x 105 | 20 log 1.84 x 105 = 105 dB 385nA 0.02W Ap = = 5.06 x 106 | 10 log 5.06 x 106 = 67.0 dB .02V (395nA) Av =

(

)

(

)

2 Checking :

29.0 + 105 = 67.0 dB 2



10-3

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.8 Rth v th

+ -

(a) v = v = 0.760 2 = 1.08 V th

vo =

oc

 0.760 − 0.740 RL v −v v th → Rth = RL th o = 1040  = 28.1  Rth + RL vo 0.740  

(b) v = v = 0.768 2 = 1.09 V th

vo =

oc

 0.768 − 0.721 RL v −v v th → Rth = RL th o = 430  = 28.0  Rth + RL vo 0.721  

(c) 1.09 V and 1.08 V → 9% error and 8% error (d ) G4 Laptop - 1V, 28  10.9



(a) V = 2R P = 2(8)5 = 8.94 V o

L O

Pi =

Vi2 12 1V 8.94V = = 25.0 W | I i = = 49.9 A | I o = = 1.12 A 2Ri 2(20k + 32) 20k + 32 8

Av =

Vo 8.94V 5W 1.12A = = 8.94 | Ap = = 2.00x105 | Ai = = 2.24x104 Vi 1V 25W 49.9A

(b) V = 8.94 V ; recommend  10 - V or 12 - V supplies o

10.10



2P R The 32- case represents a good trade off between voltage and current for battery powered applications.

V = 2PR

| I=

 R ()

V (V)

I (mA)

8

1.27

158

32

2.53

79.1

1000

14.1

14.1

10-4

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.11

i g11 = 1 : v1 = 104 i1 + 126 i1 (220k)→ g11 = 3.61 x 10−8 S v1 i =0 2

i 220k g12 = 1 : i1 = − (i2 + 125i1)→ g12 = −7.93 x 10−3 i2 v =0 220k + 10k 1

g21 =

v2 : v2 = 126i1 (220k) | i1 = g11v1 → g21 = 1.00 v1 i =0 2

g22 =

v2 v2 v v : i2 = + 2 + 125 2 → g22 = 79.3  i2 v =0 220k 10k 10k 1



10.12

Two Transfer function analyses [V(I2)/V1 and I(V1)/I2] yield : 1 g11 = g12 = −7.934x10−3 g21 = 0.9996 g22 = 79.34  27.73 M



10-5

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.13

i 1 g11 = 1 = = 0.333 mS v1 i =0 2.0k +1k

i 1k | g12 = 1 =− = −0.333 i2 v =0 2.0k +1k

2

g21 =

1

v2 1k 2.0k : v2 = v1 + v1 (−0.04)(60k)→ g21 = −1600 v1 i =0 2.0k +1k 2.0k +1k 2

g22 =

(

) | v = −i (1k 2.5k)

v2 : v2 = (i2 − 0.04va )60k + i2 1k 2.5k i2 v =0

a

2

1

g22 = 1.78 x 106 



0.0 4 v a

2.5 k 

i1 +

60 k 

i2 i2

-

va

v2

1k 

v1 = 0

+

-

10.14

i g11 = 1 : v1 = 103 i1 → g11 = 1.00 x 10−3 S v1 i =0 2

i g12 = 1 : i1 = −i2 → g12 = −1.00 i2 v =0 1

g21 =

v2 : v2 = v1 + 0.1v1 (30k) | v1 = 3001v1 → g21 = 3001 v1 i =0 2

g22 =

v2 v : i2 = 2 → g22 = 30 k i2 v =0 30k 1



10-6

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.15

Two Transfer function analyses [V(I2)/V1 and I(V1)/I2] yield : 1 g11 = g12 = −1.000 g21 = 3001 g22 = 30.00 k 1.000 k 10.16



(a) Vo = Vi

Rin RL A Rin + RI RL + Rout

54

| A = 1020 = 501.2

106 16 501.2) = 485.5 | AvdB = 20 log(485.5)= 53.7 dB 3 6 ( 0.5 + 16 10 + 10 485.5Vi Io 16 Ai = = = 3.041 x 107 | AidB = 20 log 3.041 x 107 = 150 dB Vi Ii 3 10 + 106 485.5Vi 485.5V i VI 16 = 485.5 3.041 x 107 = 1.478 x 1010 Ap = o o = V Vi I i Vi 3 i 6 10 + 10 ApdB = 10 log 1.478 x 1010 = 102 dB Av =

(

(

(

)

)

)

2 O

(b) 1 = 2 16 → V = 5.657 V | V = 485.5 = 11.65 mV ( ) V

5.657

O



i

I o2 I2 1 RL → o = 2 2 16  0.01165 2 106 I o2 1 I i2 PR out = R = 0.5 = 31.25 mW | PR in = Rin =  3 = 67.7 pW  2 16 2 10 + 106  2

(c) R out and R L see the same current. P =

P = 31.25mW + 67.7 pW = 31.3 mW



10-7

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.17

54 Rin RL 1000  20  16  (a) Vo = Vi R + R A R + R = Vi 1000+ 100010  16 + 16 = 125 Vi in I L out  

1=

Vo2 5.66 → Vo = 5.66V | Vi = = 45.3 mV 125 2(16)

(b) Since R out and R L have the same current and same value, the power dissipated in R out is 1 W.  45.3mV 2 1 The power lost in R in is  = 0.257W | PD = 0.257W + 1.00W = 1.00 W   2  2(1000) 10.18



io =

2(0.1W ) 2Po = = 91.3 mA | v = io (28 + 24) = 4.75 V RL 24

(91.3mA) (28) = 117 mW iv i2 R PD = o = 217mW | PL = o out = 2 2 2 2

10.19 

io =

2(0.1W ) 2Po = = 14.1 mA | v = io (28 + 1000) = 14.5 V RL 1000

(14.1mA) (28) = 2.78 mW iv i2 R PD = o = 103 mW | PL = o out = 2 2 2 2

10.20 

Vo = Vi

Rin RL A Rin + RI RL + Rout

To maximize the output voltage, Rin =  and Rout = 0 

Vo = AVi = −2000Vi | Vo = 0.01V (−2000)= −20 V | P =

Vo2 202 12.5W = = 12.5 W | AP = = 2RL 2(16) 0

10.21



74

Av = −10 20 = −5010 | − 5010 = AdB = 20log(5440)= 74.7 dB

30k 2k A → A = −5440 30k + 1k 2k + 0.1k

10.22





i1 = ii

RI RI + Rin

io = i1

Rout Rout + RL

Ai = 

Rout RI Rout + RL RI + Rin

Rin = 0

Rout = 

Aimax = 

10.23 10-8

©R. C. Jaeger & T. N. Blalock 04/3/2010

10 A(4000) 10  = 80.0 mW | A = 80mW =  I R =0  | R = | P = R = in

out

2

2

−6

2 o

4

L

2

P

0

10.24





Io = Ii

RI Rout  RI + Rin Rout + RL

| 150 =

200k 300k  →  = 182 200k + 10k 300k + 47k

10.25 RS should be RI.

Vo  Vo 2 RI + Rin Vo I o RL AP = = =  Vi Vi I i RL  Vi  Vi RI + Rin  2   Vo 2   V R + R o I in  = 10log  + 10log RI + Rin  APdB = 10log  RL   Vi   Vi   RL  V   RL   RL  APdB = 20log o  −10log  = AvdB −10log   Vi   RI + Rin   RI + Rin  Vo

 I o 2 RL Vo I o I o RL I o AP = = =  Vi I i I i (RI + Rin )I i  I i  RI + Rin  2   I o 2  RL  Io RL   AidB = 10log   = 10log  + 10log   I i  RI + Rin   Ii   RI + Rin  I   RL   RL  APdB = 20log o  + 10log  = AidB + 10log   Ii   RI + Rin   RI + Rin  A + AidB Note : APdB = vdB 2



10-9

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.26

Vo = Vi

Rin Rin RL A A Rin + RI Rin + Rout RL + Rout

5000 5000 500 −1200) −1200) = +5.45 x 105 ( ( 5000+ 1000 5000+ 500 500 + 500 5 AvdB = 20log 5.45 x 10 = 115 dB Av =

(

)

(

5

)

I o 5.45x10 Vi 1 = = +6.54 x 106 | AidB = 20log 6.54 x 106 = 136 dB Vi Ii 500 6000 5 5.45x10 Vi +6.54x106 I i AP = = +3.56 x 1012 | APdB = 10log 3.56x1012 = 126 dB Vi I i Ai =

(



)

(

)

10.27 In the dc steady state, the internal circuit voltages cannot exceed the power supply limits. (a) +15 V (b) -9 V 10.28

(a) For VB = 0.6V , VO = +8V | Av =

dvO 12 − 4 = = −40 dvI v =0.6V 0.5 − 0.7 I

Av = 32 dB AV = 180

o

| VM  0.100 V for linear operation

(b) vI (t )= (0.6 + 0.1sin1000t ) V

vO (t )= (8 − 4sin1000t ) V

10.29



(a) For V = 0.8V , V = +3V | A = dv

dvO

B

O

=

v

I v I =0.8V

4−2 = −10 0.7 − 0.9

Av = 20dB AV = 180o | VM  0.100 V for linear operation vI (t )= (0.8 + 0.1sin1000t ) V

vO (t )= (0.3 − 0.2sin1000t ) V

(b) For V = 0.2V , V = +14V | A = dv

dvO

B

O

v

=0

I v I =0.2V

vI (t )= (0.8 + 0.2sin1000t ) V The output signal will be distorted for any value of VM . vO (t )= 14V + Fourier Series for the distorted waveform.



10-10

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.30

(a) For VB = 0.5V , VO = +12V |

dvO is different for positive and negative values of dvI

VM sin1000t. Thus, the gain is different for positive and negative signal excursions and the output will always be a distorted sine wave. This is not a useful choice of bias point for the amplifier. dvO (b) For VB = 1.1V , VO = +2V and = 0. The gain is zero for this bias point. dvI Thus this is also not a useful choice of bias point for the amplifier.

10.31



(a) For VB = 0.6V , VO = +8V | Av =

dvO 12 − 4 = = −40 dvI v =0.6V 0.5 − 0.7 I

Av = 32 dB AV = 180

o

| VM  0.100 V for linear operation

(b) vI (t )= (0.6 + 0.1sin1000t ) V

vO (t )= (8 − 4sin1000t ) V

14



vO

12 10 8 6 4

Time 2 0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

There are only two spectral components: 8 V at dc and -4 V at 159 Hz

10-11

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.32

For sin 1000t  0, VO = 12V and the gain (VTC slope) is Av =

dvO 12 − 4 = = −40 dvI v =0.6V 0.5 − 0.7 I

vI = (0.5 + 0.1 sin 1000t ) V vO = (12 − 4 sin 1000t ) V For sin 1000t  0, VO = 12V and the gain (VTC slope) is Av =

dvO 14 −12 = = −10 dvI v =0.6V 0.3 − 0.5 I

vO = (12 − 4 sin 1000t ) V 14

vO



12

10

8 Time 6 0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

Using the MATLAB FFT capability with a fundamental frequency of 1000/2 Hz: t=linspace(0,2*pi/1000,1000); y=12-4*sin(1000*t).*(sin(1000*t)>=0)-sin(1000*t).*(sin(1000*t)<0); z=fftshift(fft(y))/1000; yields the following series:

vO (t) =11.05 − 2.50sin(1000t ) + 0.638cos(2000t) + 0.127cos(4000t ) + 0.0546cos(6000t)Note: It is worth plotting this function to see if it is correct.



The Fourier coefficients may also be calculated directly using MATLAB. For example, for the cosine terms: Define a function: function y=four(t) y=cos(fn*1000*t).*(12-4*sin(1000*t).*(sin(1000*t)>=0)sin(1000*t).*(sin(1000*t)<0)); global fn fn=0; quad('four',0,pi/500)*1000/pi

10-12

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.33 At the input signal frequency : vi = 0.25sin1200t V | vo = 4sin1200t V

Av =

4V 0.82 + 0.42 = 16 | Second and third harmonics are present. THD = 100 % = 22.4 % 0.25V 4

10.34



vI = 0.004sin2000t V | Av = THD = 100

5V = 1250 | Third and fifth harmonics are present.

0.004V

0.52 + 0.202 % = 10.8 % 5



10-13

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.35 The problem should refer to vO2(t) in Fig. 10.15. Also needs new numbers! (a,b) EDU>> t=linspace(0,0.002,2048); EDU>> vo=min(10,(5+15*sin(2000*pi*t))); EDU>> plot(t,vo) EDU>> s=fft(vo)/2048; EDU>> smag=sqrt(s.*conj(s)); EDU>> smag(1:12) ans = Columns 1 through 12 2.4588 0.0038 0.1863 0.0023

5.3105

0.0066

1.3341

0.0026

0.4427

0.0028

0.0883

0.0012

The record length for the fft is 0.002 second so the fft coefficients are at 500 Hz intervals. The dc and 1st through 5th harmonics are underlined above. Note that these are only half the actual magnitudes except for the dc component. The other half of the FFT outputs are folded about the Nyquist frequency. Also note that the phases are not listed above. Some represent sin terms and some are cosine terms.

THD = 100%

(b)

2 1.3342 + 0.44272 + 0.08832 + 0.18632 = 26.8% 2(5.3105)





f (t )= a0 +  (an cosn ot + bn sin n ot ) with n=1

o =

2 T

T



1 2 T 2 T a0 =  f (t )dt | an =  f (t )cosn ot dt | bn =  f (t )sin n ot dt T 0 T 0 T 0 EDU>> ao='min(10,(5+15*sin(2000*pi*x)))' ; EDU>> a1='cos(2000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> a2='cos(4000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> a3='cos(6000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> b1='sin(2000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> b2='sin(4000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> b3='sin(6000*pi*x).*min(10,(5+15*sin(2000*pi*x)))' ; EDU>> aoinline=inline(ao); EDU>> a1inline=inline(a1); EDU>> a2inline=inline(a2); EDU>> a3inline=inline(a3); EDU>> b1inline=inline(b1); EDU>> b2inline=inline(b2); EDU>> b3inline=inline(b3); EDU>> ao=1000*quad(aoinline,0,.001) ao = 2.4588 10-14

©R. C. Jaeger & T. N. Blalock 04/3/2010

EDU>> a1=2000*quad(a1inline,0,.001) a1 = 0.0053 EDU>> a2=2000*quad(a2inline,0,.001) a2 =2.6700 EDU>> a3=2000*quad(a3inline,0,.001) a3 =-5.9509e-004 EDU>> b1=2000*quad(b1inline,0,.001) b1 =10.6828 EDU>> b2=2000*quad(b2inline,0,.001) b2 =0.0033 EDU>> b3=2000*quad(b3inline,0,.001) b3 =0.8918 ao, b1, a2, b3 agree with the fft coefficients.

10.36 t=linspace(0,0.002,512); y=max(-1,min(1,1.5*sin(1400*pi*t))); plot(t,y) 1.5

1

0.5

0

-0.5

-1

-1.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

w=1400*pi*linspace(0,0.001428571,512); y=max(-1,min(1,1.5*sin(w))); s=fft(y)/512; ymag=sqrt(s.*conj(s)); y2=ymag.*ymag; sumy2=sum(y2)-2*y2(1)-2*y2(2); thd=100*sqrt(sumy2)/(2*y2(2))

10-15

2 x 10 -3

thd = 18.3929 %

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.37 EDU>> t=linspace(0,.002,2048); EDU>> v=10*tanh(2*0.75*sin(2000*pi*t)); EDU>> plot(t,v) 10 8 6 4 2 0 -2 -4 -6 -8

EDU>> s=fft(v)/2048; EDU>> smag=sqrt(s.*conj(s)); EDU>> smag([1:12]) ans = -10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 -3

x 10

Columns 1 through 12 0.0000 0.0034 0.0940 0.0019

5.0534

0.0055

0.0025

0.0007

0.6091

0.0033

0.0021

0.0014

The dc and 1st through 5th harmonics are underlined above. The only significant terms are the odd order distortion terms (in blue), since tanh is an odd function. Note that these are only half the actual magnitudes. The other of the FFT outputs are folded about the Nyquist frequency.

THD = 100%

0.60912 + 0.09402 = 12.2% 5.0534

Reconstruction of the signal from the three dominant components appears below. 

EDU>> vo=2*(5.0534*sin(2000*pi*t)+0.6091*sin(6000*pi*t)+.0940*sin(10000*pi*t)); EDU>> plot(t,vo) 10 8 6 4 2 0 -2 -4 -6 -8

10-16

-10 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 -3

x 10

©R. C. Jaeger & T. N. Blalock 04/3/2010

10-17

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.38 60 20

A = 10 = 1000 | Vo = Vi Ii =

1M 1k V 1000) | Av = o = 883 or 59.7 dB ( 1M + 5k 1k + 25 Vi

Vi 883Vi and I o = 1M + 5k 1k

(

| Ai =

I o 883 = 1M + 5k)= 8.87x105 or 119 dB ( I i 1000

)

AP = Av Ai = 883 8.87x105 = 7.83x108 or 88.9 dB | Vi =

Vo 5V = = 5.66 mV Av 883

10.39



30 20

A = 20 = 31.6 | Vo = Vi Vi =

5k 1k V 31.6) and Av = o = 7.91 or 18.0 dB ( 5k + 5k 1k + 1k Vi

Vo 7.5V = = 0.948 V Since R out has the same value as R L , the power Av 7.91

dissipated in R out is also 0.5W. The power dissipated in R id will be  Vs 2   Vid2  2  V2 V PI = = = s where Vs = o and Vo = 2(0.5W )(1000) = 31.6V 2Rid 2Rid 8Rid 7.91

(31.6V 7.91) = 0.4mW. The total power dissipated in the amplifier is P = 8(5000) 2

I

P = 500mW + 0.4mW = 500 mW. 10.40



Io =

2(100W ) 50

I o2 Rout = 2 A and  2W or Rout  1.0  2

10.41



0.99mV  1mV

Rid → Rid  4.95 M Rid + 50k

11.42



120 20

A = 10

= 106 | vid =

vo 15V = = 15 V A 106

15V 15V  10−6V requires A  15x106 or 144 dB | i+ = = 15 pA A 1M

 10-18

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.43 106

A = 10 20 = 2x105 | vid =

vo 10V = = 50.0 V A 2x105

10V  10−6V requires A  107 or 140 dB A 10.44



(a) A = − R = − 4.7k = −46.8 | 20log(46.8)= 33.4 dB | R = R = 4.7 k | R = 0  R2

220k

v

in

1

out

1

(b) A = − R = − 47k = −46.8 | 20log(46.8)= 33.4 dB | R = R = 47 k | R = 0  R2

2.2 M

v

in

1

out

1

10.45



(a) A = − R = − 12k = −10.0 | 20log(10.0)= 20.0 dB | R = R = 12 k | R = 0  R2

120k

v

in

1

out

1

(b) A = − R = − 140k = −2.36 | 20log(2.36)= 7.45 dB | R = R = 140 k | R = 0  R2

330k

v

in

1

out

1

(c) A = − R = − 4.3k = −55.8 | 20log(55.8)= 34.9 dB | R = R = 4.3 k | R = 0  R2

240k

v

in

1

out

1

10.46



Av = − Ii =

R2 8200 =− = −10.9 | Vo = −10.9(0.05V )= −0.547V | vo (t )= −0.547sin(4638t ) V R1 750

Vi 0.05V = = 66.7A | ii (t )= 66.7sin(4638t ) A R1 750

10.47



(a) A = − 22k = −5 v = A v = 0 (b) V = A V = -5(0.22V )= −1.10 V (b) v = −5(0.15V )sin2500t = −0.75sin2500t V (d ) v = (−1.10 + 0.75sin2500t ) V 110k

v

o

v i

O

v I

o

o

0.15V 

(e) I = 22k = 10.0 A | i =  22k sin2500t = 6.82sin2500t A 0.22V

I

I

iI = (10.0 - 6.82sin2500t ) A

(f ) i = -i (g ) v = 0 o

i

IO = −10.0 A

io = −6.82sin2500t A

iO = (-10.0 - 6.82sin2500t ) A

-

 10-19

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.48

(a) A = − 10k = −10 V = A V = -10(0.5V )= −5.00 V V = A V = −10V v = (−5.00 −10V sin5000t ) V (b) -12  −5.00 −10V sin5000t → V  0.700 V | − 5.00 −10V sin5000t  +12 → V  1.70 V = min(0.700,1.70)= 0.700 V (c) V = A V = -10(−0.25V )= +2.50 V V = A V = −10V v = (2.50 −10V sin2000t ) V 100k

v

O

O

v I

o

v i

i

i

i

i

i

i

max

i

O

v I

o

v i

i

O

i

-12  2.50 −10Vi sin2000t → Vi  1.45 V | 2.50 −10Vi sin2000t  +12 → Vi  0.480

Vi max = min(0.480,1.45)= 0.480 V 10.49



(a) A = − 7.5k = −20 V = A V = -20(−0.2V )= +4.00 V V = A V = −20V v = (4.00 − 20V sin2000t ) V (b) -10  4.00 − 20V sin2000t → V  0.700 V | 4.00 − 20V sin2000t  +10 → V  0.300 V = min(0.700,0.300)= 0.300 V (c) V = A V = -20(−0.6V )= +12.0 V V = A V = −20V v = +10.0 V 150k

v

O

O

v I

o

v i

i

i

i

i

i

i

max

i

O

v I

o

v i

i

O

The - 0.6 V dc input saturates the output against the postive power supply rail.

10.50



(a) A = − 8.2k = −19.5 (b) The equivalent input source now has an 1.5 k  resistor in the circuit between v and R , 160k

v

I

and the the gain from vI to the output now becomes Av = − 10.51



Avnom = −

1

R2 160k =− = −16.5 RI + R1 1.5k + 8.2k

220k(1.1) 220k(0.9) R2 220k =− = −46.8 | Avmax = − = −57.2 | Avmin = − = −38.3 R1 4.7k 4.7k(0.9) 4.7k(1.1)

Rinnom = R1 = 4.70 k | Rinmax = 4.7k(1.1)= 5.17 k | Rinmin = 4.7k(0.9)= 4.23 k



10-20

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.52

(a) A

nom v

=−

120k(1.05) 120k(0.95) R2 120k =− = −10.0 | Avmax = − = −11.1 | Avmin = − = −90.5 R1 12k 12k(0.95) 12k(1.05)

Rinnom = R1 = 12.0 k | Rinmax = 12k(1.05)= 12.6 k | Rinmin = 12k(0.95)= 11.4 k

(b) A

nom v

=−

330k(1.05) 330k(0.95) R2 330k =− = −2.36 | Avmax = − = −2.61 | Avmin = − = −2.13 R1 140k 140k(0.95) 140k(1.05)

Rinnom = R1 = 140 k | Rinmax = 140k(1.05)= 147 k | Rinmin = 140k(0.95)= 133 k

(c) A

nom v

=−

240k(1.05) 240k(0.95) R2 240k =− = −55.8 | Avmax = − = −61.7 | Avmin = − = −50.5 R1 4.3k 4.3k(0.95) 4.3k(1.05)

Rinnom = R1 = 4.30 k | Rinmax = 4.3k(1.05)= 4.52 k | Rinmin = 4.3k(0.95)= 4.09 k 10.53 26



R Rin = R1 = 30 k | Av = − 2 = −1020 = −20.0 → R2 = 20.0R1 = 600k R1 Using the closest values from Appendix A, R1 → 30.1 k and R2 → 604 k The values for the final design are Av = −

604 k = −20.1 and Rin = 30.1 k 30.1 k

10.54 40



R Rin = R1 = 1.5 k | Av = − 2 = −10 20 = −100 → R2 = 100R1 = 150k R1 The resistors exist as standard values.

10.55 12



Rin = R1 = 100 k | Av = −

R2 = −1020 = −3.98 → R2 = 3.98R1 = 398 k R1

R1 is a standard value. 402 k  represents the closest value to R2 .

The values for the final design are Av = −

402k = −4.02 and Rin = 100 k 100 k

−−− A closer but less cost effective solution is to use three resistors: Using the closest 1% values from Appendix A, R1 →100 k and R2 → 301k + 97.6k The values for the final design are Av = −

398.6k = −3.97 and Rin = 100 k 100k

−−− If 5% resistors were utilized, the choices would be R1 = 100 k and R2 = 390 k.



10-21

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.56 The op amp is ideal, so Rout = 0, and the non - inverting input is a virtual ground. Thus Rin = 10 k.

vI , and iI must flow out through the feedback network. The Thevenin 10k equivalent looking into the left side of the feedback network is 20k v vTH = vO = O and RTH = 100k + 20k 100k = 117 k 20k + 100k 6 vo vI v = − 6 → Av = O = −70.2 | Rin = 10 k | Rout = 0 10k 117 k vI Input current iI =

(



)

10.57 The op amp is ideal, so the non - inverting input is a virtual ground, and

v input current iTH must flow out through feedback resistor R2 . iTH = − O → vO = −iTH R R 10.58



vI R1

iTH =

v v R | RTH = R1 | vO = −iTH R2 = − I R2 → Av = O = − 2 R1 vI R1

10.59



v 5V R=− O →R= = 2 M iTH 2.5A 10.60



Av = 1+

R2 680k = 1+ = 83.9 | 20log(83.9)= 38.5 dB | Rin =  | Rout = 0  R1 8.2k

10.61



Av = 1+

R2 8200 = 1+ = 10.0 | VO = 10.0(0.04V )= 0.400V | vo (t )= 0.400sin9125t V R1 910

10.62



(a) A = 1+ R = 1+ 24k = 6.00 | 20log(6.00)= 15.6 dB | R =  | R = 0  R2

120k

v

in

out

1

(b) A = 1+ R = 1+ 15k = 21.0 | 20log(21.0)= 26.4 dB | R =  | R = 0  R2

300k

v

in

out

1

(c) A = 1+ R = 1+ 3.3k = 110 | 20log(110)= 40.8 dB | R =  | R = 0  R2

360k

v

in

out

1



10-22

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.63

(a) A = 1+ 22k = +16.0 v = 16.0v = 0 (b) V = A V = 16.0(0.33V )= +5.28 V (b) v = 16.0(0.18V )sin3250t = 2.88sin3250t V (d ) v = (5.28 − 2.88sin3250t ) V (e) i = 0 330k

v

o

s

O

v S

o

O

I

(f ) i = + R + R vO

O

1

2

IO =

5.28V 2.88 = 15.0 A I o = = 8.18A io = 8.18sin3250t A 352k 352k

iO = (15.0 − 8.18sin3250t ) A

(g ) v = v = (0.33− 0.18sin3250t ) V -

I

10.64



(a) A

nom v

= 1+

R2 47k = 1+ = 262 | 20log(262)= 48.4 dB R1 0.18k

Idea op amp : Rin = 10k +  =  | Rout = 0 

(b) A

max v

= 1+

47k(1.1)

0.18k(0.9)

= 320 | Avmin = 1+

Avmax − Avnom 320 − 262 = = 0.22 | 262 Avnom

47k(0.9)

0.18k(1.1)

= 215

Avmin − Avnom 215− 262 = = −0.18 262 Avnom

(c) Tolerances: + 22%, −18% (d ) 215 = 1.49 :1 320



(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% .

10-23

©R. C. Jaeger & T. N. Blalock 04/3/2010

11.65 32

Av = 1+

R2 = 1020 = 39.8 → R2 = 38.8R1 R1

There are many possible pairs of values in Appendix A. Two choices are R1 → 3.09 k and R2 →121 k or R1 → 6.19 k and R2 → 243 k 121 k = 40.2 3.09 k

The values for the final design are Av = 1+ 10.66 6



R Av = 1+ 2 = 1020 = 2.00 → R2 = R1. Any value could theoretically be used, but R1 we don' t want a heavy load on a real op amp, so we should choose the resistors to be at least in the 10 - k range or so. A third 100 - k resistor is added in shunt with the input as in Fig. P11.23(b). R1 →10.0 k, R2 →10.0 k, and R3 →100 k 10.0 k = 2.00 and Rin = 100 k 10.0 k

The values for the final design are Av = 1+ 10.67 33



Av = 1+

R2 = 1020 = 44.7 → R2 = 43.7R1 R1

There are many possible pairs of values in Appendix A. Two choices are R1 → 4.99 k and R2 → 215 k or R1 → 2 k and R2 → 86.6 k 86.6 k = 44.3 2.00 k

The values for the final design are Av = 1+ 10.68



(a) A = 1+ R = 1+ 20k = +6.00 | R = 47k  = 47 k R2

100k

v

in

1

(b) Voltage follower : A = +1 | R = 560k  = 560 k v

in

11.69



vO = −

R3 R 47k 47k v1 − 3 v2 = − v1 − v2 = −47v1 − 23.5v2 R1 R2 1k 2k









vO (t )= −47(0.01) sin3770t + −23.5(0.04) sin10000t

vO (t )= (−0.470sin3770t − 0.94sin10000t ) V and v- (t ) 0

 10-24

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.70

 R



 R



 R







1 − v  + () 1 − v  + (0)− v  = −0.3750sin4000t V (a) v = (0)− 2R v  + ()  4R   8R   16R  O

I

 R

I



 R

R

I



 R

I







1 − v  + (0)− v  + () 1 − v  + () 1 − v  = −0.6875sin4000t V (b) v = ()  2R   4R   8R   16R  O

I

I

I

R

I

(c) v = − Asin4000t V where the amplitude A is in the table below : O



0000 0001 0010

0 -0.0625 V -0.1250 V

0100 0101 0110

-0.250 V -0.3125 V -0.3750 V

1000 1001 1010

-0.5000 V -0.5625 V -0.6250 V

1100 1101 1110

-0.7500 V -0.8125 V -0.8750 V

0011

-0.1875V

0111

-0.4375 V

1011

-0.6875 V

1111

-0.9375 V

10.71

1 | The worst - case condition for the switches occurs   ' W Kn  (VGS − VTN − VDS )  L

Ron =

for the one with VDS  0 and VSB = 0. If VREF = 3V and Ron = 0.01(10k)= 100, VD = VREF = 3V and VS = 3V

(

10k = 2.97V | VDS = 0.03V 10k + 100

)

VTN = 1+ 0.5 2.97 + 0.6 − .6 = 1.56V W   455 1 =  =    L  5x10−5 (100) (5 − 2.97)−1.56 − 0.03  1 





W   50  1 When the grounded transistor is on, VDS = 0 |   = =   −5  L  5x10 (100)5 −1− 0  1  10.72



(a ) Using Eq. 10.63, A = − R = − R = −10 R2

10R

v

1

(b) R = i

v2

= 11R = 110 k

in2

2 v1 =0

v | Rin1 = 1 = R = 10 k i1 v =0 2

(c) v = −10(v − v )= −10(3 −1.5cos8300t )= (−30 + 15cos8300t ) V (d ) v = −10(3 −1.5cos8300t −1.5cos8300t )= (−30 + 30cos8300t ) V O

1

2

O



10-25

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.73

(a) V = 3.2 V | V = 3.1 V | V = − R (V − V )= − R (3.2 − 3.1)= −1.00 V R2

1

2

O

10R

1

2

1

10R = 2.82 V | V- =V += 2.82 V R + 10R (2.82 + 1)V V −V (b) IO = −10R O = 1M = 3.82 A | V1 − V− 3.2 − 2.82 V V − V 3.1− 2.82 V = = 3.80 A | I2 = 2 + = = 2.80 A (c) I1 = 100k 100 k 1R 100 k V2 3.1V Checking : I2 = = = 2.82 A which agrees within roundoff error .

R + 10R 110k Also, IO = I1 within roundoff error.

V+ = 3.1V

10.74



vO = −

R2 10R v1 − v2 )= − ( (2sin1000t − 2sin1000t − 2sin2000t )= 20sin2000t V R1 R

v− = v+ = (2sin1000t + 2sin2000t )

10R = (1.82sin1000t + 1.82sin2000t ) V R + 10R

(2 −1.82)V sin1000t − 1.82V sin2000t = 12.0sin1000t −121sin2000t A v −v i1 = 1 − = ( ) R 15k 15k v 2V 2V i2 = 2 = sin1000t + sin2000t = (12.1sin1000t + 12.1sin2000t ) A 11R 165k 165k 10.75



3x109 s2

3x109 s2 (a) Ai (s)= s2 + 51s + 50 s2 + 13000s + 3x107 = s + 1 s + 50 s + 3000 s + 10000 ( )( )( )( )

(

)(

)

Zeros : s = 0, s = 0 | Poles: s = -1, s = -50, s = -3000, s = -10000

(

)

105 s2 + 51s + 50

(b) A (s)= s + 1000s + 50000s + 20000s + 13000s + 3x10 v



5

4

3

2

7

| Zeros : s = -1, s = -50

Using MATLAB to find the poles: Av=[1 1000 50000 20000 13000 3e7]; roots(Av) ans = -947.24 -52.13 -9.170 4.27 + j6.93 4.27 - j6.93

10-26

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.76

Av =

2 x 107 s

(s + 20 )(s + 2 x 10 ) 4

=

1000s | A mid = +1000 = 60 dB   s (s + 20 )1+ 2 x 104 

20 2 x 104 = 10 Hz | f H = = 10 kHz | BW = 10kHz −10Hz = 9.99 kHz 2 2 Bandpass Amplifier fL =

10.77



104 s | High - pass Amplifier | A mid = +104 = 80 dB s + 200 200 fL = = 100 Hz | f H =  | BW =  2

Av =

10.78



2 x 106 = s + 200

104 → Low - pass Amplifier | A mid = +104 = 80 dB s 1+ 200 200 f L = 0 Hz | f H = = 100 Hz | BW = 100Hz − 0Hz = 100 Hz 2

Av =

10.79



Av (s)=

7

s

o

s2 + s

o

5

10 s 10 s = 102 2 = Amid 5 14 s + 10 s + 10 s + 105 s + 1014 2

Bandpass Amplifier | Amid = 100 = 40 dB | f o =

Q Q

+  2o

10  107 = 1.592 MHz | Q = o5 = 5 = 100 2 10 10 7

1.592MHz = 15.92 kHz | For a high Q circuit : 100 BW fL  fo − = 1.592MHz −15.92kHz = 1.584 MHz 2 BW fH  fo + = 1.592MHz + 15.92kHz = 1.600 MHz 2

BW =



10-27

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.80[-

s2 + 1012 Av (s)= −20 2 = Amid s + 104 s + 1012

s2 +  2o s +s 2

o Q

+

| Notch Filter 2 o

    s o s o   Q Q  where Note : Av (s)= −201− is a bandpass function. o  s2 + s  o +  2  2 2 s +s + o  o Q Q   Amid = −20 = 20 dB but Av = 0 at f o | f o =

 o 106  = = 159.2 kHz | Q = o4 = 100 2 2 10

159.2kHz = 1.592 kHz | For a high Q circuit : 100 BW 1.592kHz fL  fo − = 159.2Hz − = 158.4 kHz 2 2 BW 1.592kHz fH  fo − = 159.2Hz + = 160.0 kHz 2 2 These values agree with direct calculation of the - 3 dB points with Excel.

The width of the null = BW =

10.81



Av (s) =

4  2 x 1014 s2 (s + 20 )(s + 50 )(s + 2 x 105 )(s + 2 x 106 )

Av (s) =

103 s2

   s s 1+ (s + 20 )(s + 50 )1+ 5  6  2 x 10  2 x 10 

| Amid = 1000 = 60 dB

Zeros: s = 0, s = 0 | Poles: s = -20 , s = -50 , s = -2 x 105, s = -2 x 106 103    s s 1+ 1+ 5  6  2 x 10  2 x 10  Since the two high frequency poles are separated in frequency by a decade, For s >> 50 , Av (s) 

2 x 105 = 100 kHz | However, we are not that lucky at low frequencies.

2 103 s2 103 2L 103 For s << 2 x 105 , Av (s) | Av (j L ) = = 2  2  2 2 (s + 20 )(s + 50 )  L + (20 )  2L + (50 )     fH 



 

 

 L4 − (20 ) + (50 )  2L − (20 ) (50 ) = 0 →  L = 178 | f L = 2

2

2

2

178 = 28.3 Hz 2

BW = f H − f L = 100 kHz



10-28

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.82 Using MATLAB: n=[2e7*pi 0]; d=[1 (20*pi+2e4*pi) 40e4*pi^2]; bode(n,d) 70 60

Gain dB

50 40 30 20 10 0 10 0

10 1

10 2

10 3 10 4 Frequency (rad/sec)

10 5

10 1

10 2 10 3 10 4 Frequency (rad/sec)

10 5

10 6

100 50

Phase d eg

0 -50

-100 -150 -200 10 0

10.83 Using MATLAB:

10 6

n=[1e4 0]; d=[1 200*pi]; bode(n,d) Bode Diagrams

80

70 60

50 40 100 80 60 40 20 0 1 10

10 2

10 3

10 4

Frequency (rad/sec)

10-29

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.84 Using MATLAB:

n=[2e6*pi]; d=[1 200*pi]; bode(n,d) Bode Diagrams

80 75 70 65 60 55 0 -20 -40 -60 -80 -100 2 10

10 3

10 4

Frequency (rad/sec)

10.85 Using MATLAB:

n=[1e7 0]; d=[1 1e5 1e14];

bode(n,d)

Gain d B

100 50 0 -50 -100 10 6

10 7 Frequency (rad/sec)

10 8

10 6

10 7 Frequency (rad/sec)

10 8

Phase deg

100 50 0 -50

-100

10-30

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.86 Using MATLAB:

n=[-20 0 -2e13]; d=[1 1e4 1e12];

bode(n,d)

Gain d B

30 20 10 0 10 5

10 6 Frequency (rad/sec)

10 7

10 6 Frequency (rad/sec)

10 7

Phase d eg

200

150

100 10 5

10.87 Using MATLAB: n=[4e14*pi^2 0 0]; p1=[1 20*pi]; p2=[1 50*pi]; p3=[1 2e5*pi]; p4=[1 2e6*pi]; d=conv(conv(p1,p2),conv(p3,p4)); bode(n,d) 60

Gain d B

40 20 0 -20 -40 -60 10 -1

10 0

10 1

10 2

10 4 10 5 10 3 Frequency (rad/sec)

10 6

10 7

10 8

10 0

10 1

10 2

10 4 10 5 10 3 Frequency (rad/sec)

10 6

10 7

10 8

200

Phase deg

100 0

-100 -200 10 -1

10-31

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.88 Using MATLAB: n=[2e7*pi 0]; d=conv([1 20*pi],[1 2e4*pi]); w=2*pi*[5 500 50000]; a=freqs(n,d,w); am=abs(a); ap=angle(a)*180/pi; Magnitudes: 447.2135 998.5526 196.1161 Phases: 63.4063 -1.7166 -78.6786 (a) vO = 0.894sin 10t + 63.4o V

( ) (b) v = 2.00sin(1000t −1.72 )V (c) v = 0.392sin(10 t − 78.7 )V o

O

5

o

O

10.89



Av (s)=

104 s s + 200

Av (j ) =

| Av (j )=

104 f

104 j 104 jf = j + 200 jf + 100

| Av (j )= 90o − tan −1

f 2 + 1002

f 100

(a) 1 Hz : A (j ) = 100 | A (j )= 89.4 | v = 0.03 sin(2t + 89.4 ) V (b) 50 Hz : A (j ) = 4472 | A (j )= 63.4 | v = 1.34 sin(100t + 63.4 ) V (c) 5 kHz : A (j ) = 9998 | A (j )= 1.15 | v = 3.00 sin(10 t +1.15 )V o

v

o

v

O

o

v

v

v

v

o

O

o



4

o

O

10.90 Using MATLAB: n=[2e6*pi]; d=[1 200*pi]; w=2*pi*[2 2000 200000]; a=freqs(n,d,w); am=abs(a); ap=angle(a)*180/pi Magnitudes: 9998 4472 5.000 Phases: -1.1458 -63.4349 -89.9714 (a) vO = 100sin 4t + 88.9o mV

( ) (b) v = 44.7sin(4000t + 2.86 )mV (c) v = 50.0sin(4x10 t + 0.0286 )V o

O

5

o

O



10-32

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.91 Using MATLAB: n=[-1e7 0]; d=[1 1e5 1e14]; w=2*pi*[1.59e6 1e6 5e6]; a=freqs(n,d,w); am=abs(a) ap=angle(a)*180/pi Magnitudes: 98.1550 1.0381 0.3542 Phases: -168.9767 -90.5948 90.2029 (a) vO = 0.392sin 3.18x106 t −169o V

(

)

(b) v = 4.15sin(2x10 t − 90.6 )mV (c) v = 1.42sin(10 t + 90.2 )mV 6

o

O

7

o

O



10.92 Using MATLAB: n=[-20 0 -2e13]; d=[1 1e4 1e12]; w=2*pi*[1.59e5 5e4 2e5]; a=freqs(n,d,w); am=abs(a) ap=angle(a)*180/pi Magnitudes: 3.8242 19.9999 19.9953 Phases: 101.0233 179.8003 -178.7570 (a) vO = 0.956sin 3.18x105 t + 101o V

( ) (b) v = 5.00sin(10 t + 180)V (c) v = 5.00sin(4x10 t −179 )V 5

O

5

o

O



10-33

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.93 Using MATLAB: n=[4e14*pi^2 0 0]; p1=[1 20*pi]; p2=[1 50*pi]; p3=[1 2e5*pi]; p4=[1 2e6*pi]; d=conv(conv(p1,p2),conv(p3,p4)); w=2*pi*[5 500 50000]; a=freqs(n,d,w); am=abs(a) ap=angle(a)*180/pi Magnitudes: 87.7058 998.5400 893.3111 Phases: 142.1219 3.6930 -29.3873 (a) vO = 175sin 10t + 142o mV

( ) (b) v = 2.00sin(1000t + 3.69 )V (c) v = 1.79sin(10 t − 29.4 )V o

O

5

o

O

10.94



(a) A

mid

26 20

20 s

= +10 = +20 | Av = 1+

(

2 x 5 x 106

)

20 2x108  = = s s + 107  1+ 7 10 

2x108 

(b) A = − s + 10  v

7

10.95 

(a) A

mid

40 20

= +10 = +100 | Av (s)= 100

(b) Av (s)= −

s

  s (s + 400 )1+ 2 x 105 

=

2 x 107 s

(s + 400 )(s + 2 x 10 ) 5

2 x 107 s

(s + 400 )(s + 2 x 10 ) 5



10-34

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.96

 3 A  1 3  1  = o Av (s) = Ao   | Av ( j H ) = Ao 2 2 2  s + 1    H + 1 

| BW = f H

1  f H 2  dB  dB 2 = 1+   → f H = f1 2 3 −1 = 25kHz(0.5098) = 12.8 kHz | 3-20  = −60  dec  dec  f1  1 3



Using MATLAB: n=2000*(50000*pi)^3; d=[1 3*50000*pi 3*(50000*pi)^2 (50000*pi)^3]; bode(n,d) Bode Diagrams

70 60 50 40 30

-60 dB/dec

20 10 200 150 100 50 0 -50 -100 4 10

10 5

10 6

Frequency (rad/sec)

10-35

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.97 (a) To avoid distortion of the waveform, the phase shift must be proportional to frequency. 30o at 1500 Hz and 50o at 2500 Hz. o o o (b) vO = 3.16sin 1000t + 10 + 1.05sin 3000t + 30 + 0.632sin 5000t + 50 (c) Using MATLAB: t=linspace(0,.004); a=10^(10/20); vs=sin(1000*pi*t)+0.333*sin(3000*pi*t)+0.200*sin(5000*pi*t); vo=A*(sin(1000*pi*t+pi/18)+0.333*sin(3000*pi*t+3*pi/18)+0.2*sin(5000*pi*t+5*pi/18)); plot(t,A*vs,t,vo)

(



)

(

)

(

)

3

2

1

0

-1

-2

-3 0

0.5

1

1.5

2

2.5

3

3.5

4 x 10 -3

10.98

    Vo R2  1  R2 1.5k = | Amid = = = 0.600 | Amid-dB = −4.44 dB s  Vs R1 + R2  R1 + R2 1k + 1.5k 1+   H  1 1 fH = = = 26.5 kHz 2 R1 R2 C 2 1k 1.5k 0.01F

(

)

(

)

10.99



    Vo R2  1  R2 100k = | Amid = = = 0.909 | Amid-dB = −0.828 dB s  Vs R1 + R2  R1 + R2 10k + 100k 1+   H  1 1 fH = = = 1.75 kHz 2 R1 R2 C 2 10k 100k 0.01F

(

)

(

)

10.100



10-36

©R. C. Jaeger & T. N. Blalock 04/3/2010

(a) A

 10

C=

1

mid

(

−0.5 20

= 0.944 |

)

2 R1 R2 f H

=

(

R2  0.944 → R2  9440 | Choose R2 = 10 k 560 + R2 1

)

2 560 10k 20kHz

(b) R = 10 k | C = 15.0 nF

= 0.015 F

2

10.101



 R2  1 a A = − ( ) v  R  s 1 1+

| Av (0)= −

H

R2 10k 1 1 =− = −5 | f H = = = 15.9 kHz R1 2k 2R2C 2 (10k)(0.001F )

(b) A (0)= − R = − 2.7k = −20.7 | f = 2R C = 2 56k 100 pF = 28.4 kHz ( )( ) R2

v

56k

1

1

H

1

2

10.102 20



(a) Rin = R1 → R1 = 10 k | Av (0)= −1020 = −10 → R2 = 10R1 = 100 k 1 1 = = 79.6 pF 2R2 f H 2 (100k)(20kHz)

C=

(b) If the input resistance specification is most critical, R1 = 10 k | R2 = 10R1 = 100 k | C = 82 pF → f H = 19.4 kHz If the cutoff frequency specification is most critical, C = 82 pF → R1 = 9.76 k | R2 = 10R1 = 97.6 k | f H = 19.9 kHz



10.103 2 x 106 Av = = s + 200



10.104 Vo R2  s  R2 20k = = = 0.709 | Amid-dB = −2.98 dB   | Amid = Vs R1 + R2  s +  L  R1 + R2 8.2k + 20k

104 → Low - pass Amplifier | A mid = +104 = 80 dB s 1+ 200 200 f L = 0 Hz | f H = = 100 Hz | BW = 100Hz − 0Hz = 100 Hz 2 The circuit is a low - pass amplifier.

fL =



1 1 = = 564 Hz 2 (R1 + R2 )C 2 (8.2k + 20k)0.01F

10-37

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.105 Vo R2  s  R2 78k = = = 0.886 | Amid-dB = −1.05 dB   | Amid = Vs R1 + R2  s +  L  R1 + R2 10k + 78k

fL =

1 1 = = 181 Hz 2 (R1 + R2 )C 2 (10k + 78k)0.01F

10.106



(a) A

mid

C=

 10

−0.5 20

= 0.944 |

R2  0.944 → R2  5560 | Choose R 2 = 7.5 k 330 + R2

1 1 = = 1.02 nF 2 (R1 + R2 )f L 2 (330 + 7500)20kHz

(b) R = 7.5 k | C = 1.0 nF | Checking : f = 20.3 kHz 2

L

10.107



(a) A

=−

R2 20k 1 1 =− = −4.76 | f L = = = 18.9 kHz R1 4.2k 2R1C 2 (4.2k)(0.002F )

(b) A

=−

R2 56k 1 1 =− = −20.7 | f L = = = 105 kHz R1 2.7k 2R1C 2 (2.7k)(560 pF )

mid

mid

10.108



20 20

(a) R = R → R = 10 k | A (0)= −10 = −10 → R = 10R = 100 k in

C=

1

1

v

2

1

1 1 = = 0.0159 F 2R1 f L 2 (10k)(1kHz)

(b) If the input resistance specification is most critical, R1 = 10 k | R2 = 10R1 = 100 k | C = 0.015 F → f H = 1.06 kHz If the cutoff frequency specification is most critical, C = 0.015 F → R1 = 10.5 k | R2 = 10R1 = 105 k | f H = 1.01 kHz



10-38

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.109

Av (s)=

104 s s + 200

Av (j ) =

| Av (j )=

104 f f 2 + 1002

104 j 104 jf = j + 200 jf + 100

| Av (j )= 90o − tan −1

f 100

(a) 1 Hz : A (j ) = 100 | A (j )= 89.4 | v = 0.03 sin(2t + 89.4 ) V (b) 50 Hz : A (j ) = 4472 | A (j )= 63.4 | v = 1.34 sin(100t + 63.4 ) V (c) 5 kHz : A (j ) = 9998 | A (j )= 1.15 | v = 3.00 sin(10 t +1.15 )V o

v

o

v

O

o

v

v

v

v

o

O

o

4

o

O

The circuit is a high - pass amplifier.

10.110



t 1 t 1 + vO (t ) = −  vS ( ) d + vC (0 )= − 10k(0.005F )  0.1sin2000 d + 0 = 0.318cos2000t V RC 0 0

10.111



vO

vS

1 ms

2 ms

t

5V

t -5 V 0

1 ms

2 ms

10.112

Z (s) (a) A (s)= − Z s () 2

v

1

(b)

Z2 =

1 sC

Z1 (s)= R

Av (s)= −

1 , an inverting integrator. sRC

Vi − V+ Vo − V+ R1 V + = sCV+ and V+ = V− = Vo = o R KR R1 + KR1 1+ K

Combining these expressions yields : Av (s)=

Vo 1+ K =+ , a noninverting integrator.

Vi sRC

10.113



V V sCVi = − o | T(s) = o = −sRC R Vi



10-39

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.114 vO (t )= −RC

10.115 

Av (s)= −

dvS (t ) dt

Z2 (s)

= −(100k)(0.02F )

Z2 (s)= R2

Z1 (s)

Z1 (s)=

d 2cos3000t = +37.7sin3000t V dt

Av (s)= −

R1 sCR1 + 1

R2  s  1 1+  for  L = R1   L  R1C

10.116



(a) A = − R = − 20k = −6.00 | R = R = 20 k R2

120k

v

in

Rout = 0

|

1

1

Note that the 100 - k resistor does not affect the circuit because v- = 0.

(Practice by solving problem using a Thevinen equivalent of v i and the 20 k and 100 k resistors.)

(b) A = 1+ R = 1+ 15k = +9.00 | R = 91k  = 91 k R2

120k

v

Rout = 0

|

in

1

(c) A = − R = − 160k = 0 | R = R = 160 k R2

0

v

in

1

Rout = 0

|

1

(This is not a very useful circuit except possibly as an " electronic ground".) 

10.117 The inverting terminal of the op amp represents a virtual ground (0 V).

V − (−V ) 0 − (−10) (a) I = I = I = R = 10 = 1 A (b) Saturation requires V  V − V where V = V − V = V 2() 1 2I V −V = = =2.83V → V  2.83 V −

O

D

EE

S

DS

GS

TN

DS

DD

−

DD

and

D

GS

TN

Kn

0.25

DD

1 10 = 10W. So the resistor must dissipate 10 W. (c) P = I R = () 2

2

R

A 15 - W resistor would provide a reasonable safety margin.



10-40

©R. C. Jaeger & T. N. Blalock 04/3/2010

10.118 The inverting terminal of the op amp represents a virtual ground (0 V).   V− − (−VEE )  30  0 − (−15)  =    = 0.484 A (a) IO = IC =  F I E =  F  R 1+  F    31 30 

(b) VO  V− + VBE = 0 + VBE

I 0.484 VBE = VT ln C = 0.025V ln −13 = 0.730V IS 10

VO = 0.730 V

(c) Forward - active region operation requires VCE  VBE but VCE = VCC − V− = VCC Therefore VCC  0.730 V. (d ) PR = I 2 R = (0.484) 30 = 7.03W. So the resistor must dissipate 7.03 W. 2

A 10W resistor would provide a reasonable safety margin.

0.484 PD = ICVCE + I BVBE = 0.484(15)+ 0.730 = 7.27 W. 30 10.119



VS = VO

Z1 = VO Z1 + Z2

R1 R1 +

= VO

R2 SC R2 +

(SCR + 1)R (SCR + 1)R + R 2

2

1

1

2

| Av (s)=

(

)

VO  R2  SC R1 R2 + 1 = 1+  VS  R1  SCR2 + 1

1 SC

10.120



Av (s)= − Avnom = − f Hnom = f Hmin =

R2 1 R 1 | Av (0)= − 2 | f H = R1 sCR2 + 1 R1 2CR2

330k(1.1) 330k(0.9) 330k = −33 | Avmax = − = −40.3 | Avmin = − = −27.0 10k 10k(0.9) 10k(1.1) 1

1 max = 4.82 kHz | f = = 10.7 kHz H 2 10−10 3.3x105 2 10−10 (0.5)3.3x105 (0.9)

( )

( )

2 10

−10

( )

1

(1.2)3.3x105 (1.1)

= 3.65 kHz



10-41

©R. C. Jaeger & T. N. Blalock 04/3/2010