R. Allan Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 11
Exercises
E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) (b) An inverting amplifier has negative gain. Thus v o (t ) = Avv i (t ) = -50v i (t ) = -5.0 sin(2000t ) E11.2
A = v
RL Vo 75 = Aoc = 500 = 375 v Vi Ro + RL 25 + 75 RL Ri V 75 2000 Avs = o = Avoc = = 300 500 Vs Rs + Ri Ro + RL 500 + 2000 25 + 75 R I 2000 Ai = o = Av i = 375 = 10 4 RL Ii 75
G = Av Ai = 3.75 10 6
E11.3
Recall that to maximize the power delivered to a load from a source with fixed internal resistance, we make the load resistance equal to the internal (or Thvenin) resistance. Thus we make RL = Ro = 25 . Repeating the calculations of Exercise 11.2 with the new value of RL, we have
A = v
RL Vo 25 = Aoc = 500 = 250 v Ro + RL Vi 25 + 25 R I 2000 Ai = o = Av i = 250 = 2 10 4 RL Ii 25
G = Av Ai = 5 10 6
306
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E11.4
By inspection, Ri = Ri 1 = 1000 and Ro = Ro 3 = 30 .
Vo 3 Ri 3 Ri 2 = Aoc1 Aoc2 A v v Vi 1 Ro1 + Ri 2 Ro2 + Ri 3 voc3 V 2000 3000 20 Aoc = o3 = 10 30 = 5357 v 100+ 2000 200+ 3000 Vi 1 Aoc = v
E11.5
Switching the order of the amplifiers of Exercise 11.4 to 3-2-1, we have Ri = Ri 3 = 3000 and Ro = Ro 1 = 100
Vo1 Ri 2 Ri 1 Aoc2 A = Aoc3 v v Vi 3 Ro3 + Ri 2 Ro2 + Ri 1 voc1 V 2000 1000 20 Aoc = o1 = 30 10 = 4348 v 300+ 2000 200+ 1000 Vi 3 Aoc = v
E11.6
Ps = (15 V) (1.5 A) = 22.5 W Pd = Ps + Pi - Po = 22.5 + 0.5 - 2.5 = 20.5 W P = o 100% = 11.11% Ps
The input resistance and output resistance are the same for all of the amplifier models. Only the circuit configuration and the gain parameter are different. Thus we have Ri = 1 k and Ro = 20 and we need to find the open-circuit voltage gain. The current amplifier with an open-circuit load is:
E11.7
307
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Avoc =
E11.8
v ooc Aiscii Ro AiscRo 200 20 = = = =4 vi Ri ii Ri 1000
For a transconductance-amplifier model, we need to find the shortcircuit transconductance gain. The current-amplifier model with a shortcircuit load is:
Gmsc =
iosc Aiscii Aisc 100 = = = = 0. 2 S 500 vi Ri ii Ri
The impedances are the same for all of the amplifier models, so we have Ri = 500 and Ro = 50 .
E11.9
For a transresistance-amplifier model, we need to find the open-circuit transresistance gain. The transconductance-amplifier model with an open-circuit load is:
Rmoc =
v ooc Gmscv i Ro = = GmscRo Ri = 0.05 10 10 6 = 500 k ii v i / Ri
The impedances are the same for all of the amplifier models, so we have Ri = 1 M and Ro = 10 .
E11.10
The amplifier has Ri = 1 k and Ro = 1 k . (a) We have Rs < 10 which is much less than Ri , and we also have load, the amplifier is approximately an ideal voltage amplifier.
RL > 100 k which is much larger than Ro . Therefore for this source and
308
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(b) We have Rs > 100 k which is much greater than Ri , and we also have
RL < 10 which is much smaller than Ro . Therefore for this source and
load, the amplifier is approximately an ideal current amplifier. (c) We have Rs < 10 which is much less than Ri , and we also have
RL < 10 which is much smaller than Ro . Therefore for this source and
load, the amplifier is approximately an ideal transconductance amplifier. (d) We have Rs > 100 k which is much larger than Ri , and we also have
RL > 100 k which is much larger than Ro . Therefore for this source and
load, the amplifier is approximately an ideal transresistance amplifier. (e) Because we have Rs Ri , the amplifier does not approximate any type of ideal amplifier.
E11.11
We want the amplifier to respond to the short-circuit current of the source. Therefore, we need to have Ri << Rs . Because the amplifier should deliver a voltage to the load that is independent of the load resistance, the output resistance Ro should be very small compared to the smallest load resistance. These facts ( Rs very small and Ro very small) indicate that we need a nearly ideal transresistance amplifier.
E11.12
The gain magnitude should be constant for all components of the input signal, and the phase should by proportional to the frequency of each component. The input signal has components with frequencies of 500 Hz, 1000 Hz and 1500 Hz, respectively. The gain is 530 o at a frequency of 1000 Hz. Therefore the gain should be 515o at 500 Hz, and 545 o at 1500 Hz.
E11.13
We have
v in (t ) = Vm cos(t ) v o (t ) = 10v in (t - 0.01) = 10 m cos[ (t - 0.01)] = 10 m cos(t - 0.01 ) V V
The corresponding phasors are Vin = Vm 0 and Vo = 10 m - 0.01 . Thus V the complex gain is V 10 m - 0.01 V Av = o = = 10 - 0.01 Vin Vm 0
309
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E11.14 E11.15
B
0.35
tr
=
0.35 = 5.247 MHz 66.7 10 -9
Equation 11.13 states Percentage tilt 200fLT
Solving for fL and substituting values, we obtain percentage tilt 1 fL = = 15.92 Hz 200T 200 100 10 -6 as the upper limit for the lower half-power frequency.
E11.16
(a) v o (t ) = 100v i (t ) + v i2 (t )
= 100 cos(t ) + cos2 (t ) = 100 cos(t ) + 0.5 + 0.5 cos(2t )
The desired term has an amplitude of V1 = 100 and a second-harmonic distortion term with an amplitude of V2 = 0.5. There are no higher order distortion terms so we have D2 = V2 /V1 = 0.005 or 0.5%.
D = D22 + D32 + D42 ... = D2 = 0.5%
(b) v o (t ) = 100v i (t ) + v i2 (t )
= 500 cos(t ) + 25 cos2 (t ) = 500 cos(t ) + 12.5 + 12.5 cos(2t )
The desired term has an amplitude of V1 = 500 and a second-harmonic distortion term with an amplitude of V2 = 12.5. There are no higher order distortion terms so we have D2 = V2 /V1 = 0.025 or 2.5%.
D = D22 + D32 + D42 ... = D2 = 2.5%
E11.17
dB. Then we have CMRR = 20 log(Ad / Acm ) = 20 log(500,000) = 114.0 dB.
E11.18
With the input terminals tied together and a 1-V signal applied, the differential signal is zero and the common-mode signal is 1 V. The common-mode gain is Acm = Vo /Vicm = 0.1 / 1 = 0.1, which is equivalent to -20
(a)
Thus Ad = (A1 + A2 ) / 2 .
v id = v i 1 - v i 2 = 1 V v icm = (v i 1 + v i 2 ) / 2 = 0 V v o = A1v i 1 - A2v i 2 = (A1 + A2 ) / 2 = Ad v id + Acmv icm = Ad
310
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(b)
Thus Acm = A1 - A2 . (c)
v id = v i 1 - v i 2 = 0 V v icm = (v i 1 + v i 2 ) / 2 = 1 V v o = A1v i 1 - A2v i 2 = (A1 - A2 ) = Ad v id + Acmv icm = Acm
Ad = (A1 + A2 ) / 2 = (100 + 101) / 2 = 100.5 Acm = A1 - A2 = 100 - 101 = -1
A A + A2 CMRR = 20 log d = 20 log 1 2A - A A cm 2 1 A + A2 100 + 101 = 20 log CMRR = 20 log 1 2 100 - 101 = 40.0 dB. 2A - A 2 1
E11.19
Except for numerical values this Exercise is the same as Example 11.13 in the book. With equal resistances at the input terminals, the bias currents make no contribution to the output voltage. The extreme contributions to the output due to the offset voltage are
AdVVoff = AdVoff
Rin Rin + Rs 1 + Rs 2
-3
100 103 = 500 ( 10 10 ) = 2.5 V (100 + 50 + 50)103 The extreme contributions to the output voltage due to the offset current are I R (R + Rs 2 ) AdVIoff = Ad off in s 1 2 Rin + Rs 1 + Rs 2 = 500 100 10 -9 100 10 3 (50 + 50) 103 = 1.25 V 2 (100 + 50 + 50)103
Thus, the extreme output voltages due to all sources are 3.75 V.
E11.20
This Exercise is similar to Example 11.13 in the book with Rs 1 = 50 k and
Rs 2 = 0. With unequal resistances at the input terminals, the bias
VoBias = Ad I B Rs 1Rin Rs 1 + Rin
currents make a contribution to the output voltage given by
= 500 400 10
-9
50 10 3 100 10 3 = +6.667 V 50 10 3 + 100 10 3
311
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The extreme contributions to the output due to the offset voltage are
AdVVoff = AdVoff
Rin Rin + Rs 1 + Rs 2
100 103 = 3.333 V (100 + 50 + 0)10 3
= 500 ( 10 10 -3 )
The extreme contributions to the output voltage due to the offset current are
AdVIoff = Ad
I off Rin (Rs 1 + Rs 2 ) 2 Rin + Rs 1 + Rs 2
100 10 -9 100 10 3 (50 + 0) 10 3 = 0.8333 V 2 (100 + 50 + 0)10 3
= 500
Thus, the extreme output voltages due to all sources are a minimum of 2.5 V and a maximum of 10.83 V.
Problems
P11.1
An inverting amplifier has negative voltage gain. The output waveform is an inverted version of the input (usually with larger amplitude). A noninverting amplifier has positive voltage gain, and the output waveform is the same as the input except that it (usually) has larger amplitude. An amplifier can be characterized by its input impedance, output impedance and a gain parameter such as open-circuit voltage gain. Loading effects can occur either at the input or output of an amplifier. The output voltage decreases when a load is connected because the current drawn by the load causes a voltage drop across the output impedance of the amplifier. The voltage at the source terminals decreases when the amplifier is connected because the current drawn by the amplifier results in a voltage drop across the internal (Thvenin) resistance of the source.
P11.2
P11.3
312
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.4*
The equivalent circuit is:
RL Vo 4 = A oc = 100 = 50 v Vi 4+4 Ro + RL RL Ri V A = o = A oc vs v Ri + Rs Ro + RL Vs A = v
100 10 3 4 3 3 100 10 + 50 10 4 + 4 = 33.33 Ri 10 5 Ai = Av = 50 = 1.25 10 6 RL 4 = 100
G = Av Ai = 62.5 106
P11.5*
Ai = Ri =
G 5000 = = 100 Av 50 Ai 100 RL = 100 = 200 Av 50
P11.6
The equivalent circuit is:
vi (t ) =
Ri + Rth
Ri
vs =
10 6 [3 10 -3 cos(200t )] = 10 -3 cos(200t ) V 10 6 + 2 10 6
313
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
v o (t ) = Avocv i Pi = V
2 i - rms
Ro + RL
RL
= -10 4 [10 - 3 cos(200t )]
1000 = -5 cos(200t ) 1000 + 1000
Ri
=
(10 -3 / 2 )2 = 0.5 10 -12 W 6 10
Vo 2 rms (5 / 2 )2 - = = 12.5 10 -3 W RL 103 P G = o = 25 10 9 Pi Po =
P11.7
The equivalent circuit is:
1 is = 3000[2 10 -3 cos(200t )] = 6 cos(200t ) V 1 / Ri + 1 / Rs RL 1000 vo (t ) = A vi = -10[6 cos(200t )] = -30 cos(200t ) V voc Ro + RL 1000 + 1000
vi (t ) =
2 Vi - rms (6 / 2 )2 Pi = = = 1.5 10 -3 W Ri 12000
Vo 2 rms (30 / 2 )2 - = = 450 10 -3 W 3 RL 10 P G = o = 300 Pi Po =
P11.8*
Before the 2-k resistance is placed across the input terminals, the output voltage is given by
Vo = 2 = A ocI s Ri v
Ro + RL
RL
(1)
After the resistance is placed in parallel with the input terminals, the output voltage is given by
314
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vo = 1.5 = A ocI s v
RL 1 (2) 1 / 2000 + 1 / Ri Ro + RL
Dividing the respective sides of Equation 2 by those of Equation 1, we have 1 Vo 1.5 = = Vo 2 Ri / 2000 + 1 Solving we obtain Ri = 666.7 .
P11.9
The equivalent circuit using the amplifier is:
We have
A = vs
Po = ( o )2 RL = 45.9 mW V
RL Ri Vo 10 6 50 = A oc =1 6 = 0.303 v 5 Vs Ri + Rs Ro + RL 10 + 10 100 + 50 Vo = AvsVs = 0.303 5 = 1.52 V rms
The equivalent for the load connected directly to the source without the amplifier is:
In this case, we have:
Vo =Vs
Po = 125 10 -9 W
Thus, the output voltage and output power is much higher when the amplifier is used, even though the open-circuit voltage gain of the amplifier is unity, because the amplifier alleviates source loading.
RL + Rs
RL
=5
50 = 2.50 mV rms 50 + 10 5
315
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.10
Because Equation 11. 3 states
Ai = Av
Ri RL
we conclude that if the current and voltage gains are equal, then the input and load resistances are equal.
P11.11
The equivalent circuit is:
Vo = 100 i V
Av =
RL Vo 10 k = 90 = 100 = 100 Vi Ro + RL Ro + 10 k
Ro + RL
RL
Solving, we find Ro = 1.11 k
P11.12
This problem is very similar to Problem 11.4. RL 8 V A = o = A oc = 1000 = 800 v v Ro + RL Vi 2+8
A = vs
RL Ri Vo = A oc v Vs Ri + Rs Ro + RL
= 1000
20 103 8 3 3 20 10 + 10 10 2 + 8 = 533.3 R 20 10 3 Ai = Av i = 800 = 2 10 6 RL 8
G = Av Ai = 1.6 10 9
P11.13* With the switch closed, we have:
Vo = 100 mV = Av
Ro + RL
RL
Vs
(1)
With the switch open, we have:
316
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vo = 50 mV = Av
Rin + 10 Ro + RL
6
Rin
RL
Vs
(2)
Dividing the respective sides of Equation (2) by those of Equation (1), we obtain:
0. 5 =
Solving, we that that Rin = 1 M.
P11.14
Rin + 10 6
Rin
Because we have G = Av Ai , we can have G = 10 for Av = 0.1 provided that
Ai = 100. Then because Ai = Av Ri / RL , we have Ri / RL = 100.
P11.15
The equivalent circuit for the cascaded amplifiers is:
We can write:
Vi 2 = A oc1Vi 1 v
Ri 2 Ri 2 + Ro 1
Vo 2 = A oc2Vi 2 = A oc2A oc1Vi 1 v v v
Ri 2 Ri 2 + Ro 1
Thus, the open-circuit voltage gain is:
A oc = v
P11.16
Vo 2 Ri 2 = A oc2A oc1 v v Ri 2 + Ro 1 Vi 1
For the amplifiers in the order A-B, the equivalent circuit is:
Thus, we have:
317
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ri = RiA = 3 k Ro = RoB = 2 k A oc = A ocAA ocB v v v
RiB RiB + RoA
10 6 10 6 + 400 = 49.98 10 3 = 100(500 ) For the amplifiers cascaded in the order B-A, we have:
Ri = RiB = 1M Ro = RoA = 400 A oc = A ocAA ocB v v v
3000 = 100 500 3000 + 2000 = 30 103
P11.17* The equivalent circuit for the cascade is:
RiA RiA + RoB
We have:
Ri = Ri 1 = 2 k Ro = Ro 3 = 3 k A oc = A oc1A oc2A oc3 v v v v
Ri 3 Ri 2 Ri 2 + Ro 1 Ri 3 + Ro 2
= 100(200 )(300 ) = 3.6 10 6
P11.18
4000 6000 4000 + 1000 6000 + 2000
Reversing the order of the amplifiers of Problem P11.17, we have Ri = Ri 3 = 6 k Ro = Ro 1 = 1 k
A oc = A oc1A oc2A oc3 v v v v
Ri 2 Ri 1 Ri 2 + Ro 3 Ri 1 + Ro 2
318
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
= 100(200 )(300 ) = 1.714 10 6
4000 2000 4000 + 3000 2000 + 2000
P11.19* The voltage gain of an n-stage cascade is given by
Ri RL n 1 n -1 1 A =A v R + R R + R = 10 ( 2 ) ( 3 ) i o L o in which we have assumed that n 2. Evaluating for various values of n we have:
n voc
n -1
n
2 3 4 5
Av
16.67 83.33 416.7 2083
Thus five amplifiers must be cascaded to attain a voltage gain in excess of 1000.
P11.20
The power efficiency of an amplifier is the percentage of the power from the dc power supply that is converted to output signal power. We can write:
=
where Po is the output signal power and Ps is the power taken from the power supply. The remainder of the supplied power is converted to heat and is called dissipated power Pd . We have where Pi is the power supplied by the signal source. Normally, Pi is very
Po 100% Ps
Pd = Ps + Pi - Po
small compared to Ps or Po .
P11.21
Po = ( o )2 RL = 12.5 W V Ps = Vs I s = 30 W Pd = Ps + Pi - Po = 17.5 W
Pi = ( i )2 Ri = 10 -7 W V
319
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
=
Po 100% = 41.67% Ps
P11.22* The two 15-V sources deliver power:
P1 = (15 V ) (1 A) = 15 W P2 = (15 V ) (2 A) = 30 W
On the other hand, the 5-V source absorbs power:
P3 = (5 V ) (- 1 A) = -5 W
Thus, the new power supplied to the amplifier is: Ps = P1 + P2 + P3 = 40 W
P11.23
Pi = I i 2Ri = (10 -6 ) 2 10 5 = 0.1 W
Po = ( o )2 RL = 10 W V Ps = Vs I s = 18 W Pd = Ps + Pi - Po = 8 W P = o 100% = 55.56% Ps
P11.24
Pi = Pi 1 =
Vi 12- rms (2 10 -3 )2 = = 4 pW 10 6 Ri 1 V2 (12)2 Po = Po 2 = o 2 - rms = = 18 W RL 8 Psupply = 2 + 22 = 24 W Po = 4.5 1012 Pi Pdissipated = Psupply - Po = 24 - 18 = 6 W
=
G =
Psupply
Po
100% = 75%
P11.25
The voltage gain A oc is measured under open-circuit conditions. v The current gain Aisc is measured under short-circuit conditions. The transresistance gain Rmoc is measured under open-circuit conditions.
320
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The transconductance gain Gmsc is measured under short-circuit conditions. The amplifier models are:
P11.26* The equivalent circuit is:
We have:
I i =Vs (Rs + Ri ) = 454.5 A rms Vi = I i Ri = 9.090 mV rms Ro I o = Aisc I i = 0.909 A rms Ro + RL Vo = RL I o = 4.545 V rms Ai = I o I i = 2000 Av =Vo Vi = 500
Pi = ( i )2 Ri = 4.131 W V Ps = Vs I s = 12 2 = 24 W Pd = Ps + Pi - Po = 19.87 W P = o 100% = 17.2% Ps Po = ( o )2 RL = 4.131 W V
G = Ai Av = 10 6
321
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.27
We are given the parameters for the current amplifier model:
(a)
The open circuit voltage gain is: 500Ii Ro V A oc = ooc = = 50 v
Vi
Ri Ii
The voltage-amplifier model is:
(b)
The transresistance gain is: 500I i Ro V Rmoc = ooc = = 5000
Ii
Ii
The transresistance-amplifier model is:
(c)
The transconductance gain is: 500I i I Gmsc = osc = = 5S
Vi
Ri I i
The transconductance-amplifier model is:
322
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.28
The equivalent circuit is:
We can write:
I o = 10I i
Ai =
Io Ro = 8 = 10 Ii Ro + 50
Ro + RL
Ro
Solving, we find that Ro = 200
P11.29
Vo Rmoc (Vi / Ri ) Rmoc = = = 100 V/V Vi Vi Ri [R ( / R )] / Ro Rmoc V I Gmsc = o = moc i i = = 0. 1 S Vi Vi Ri Ro R I / Ro Rmoc I Aisc = o = moc i = = 10 A/A Ii Ii Ro A oc = v
P11.30
Vo GmscVi Ro = = GmscRo = 50 V/V Vi Vi G VR V Rmoc = o = msc i o = GmscRi Ro = 500 k Ii Vi / Ri G V I Aisc = o = msc i = GmscRi = 5000 A/A Ii Vi / Ri A oc = v
A oc = v Vo Aisc (Vi / Ri )Ro AiscRo = = = 30 V/V Vi Vi Ri A IR V = o = isc i o = AiscRo = 60 k Ii Ii A I A I = o = isc i = isc = 0.1 S Vi Ii Ri Ri
P11.31
Rmoc Gmsc
323
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.32* The equivalent circuit is:
10 8 Ii 500 10 100 V 200 + 50 = 2 10 7 Avoc = ooc = 10 6 Ii Vi Ri = RiA = 1 M
3
Ro = RoB = 500 k
Thus, the voltage-amplifier model is:
Then we can write Gmsc =
I osc (AvocVi ) Ro = = 40 S Vi Vi
Thus, the transconductance-amplifier model is:
P11.33
The equivalent circuit is:
Vooc = Vi Ri = RiB = 50 Ro = RoA = 200 A = voc
108
500 103 100Ii 10 6 + 500 103 = 6.667 10 7 50Ii
324
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Thus, the voltage-amplifier model is:
Then we can write Gmsc =
I osc (AvocVi ) Ro = = 333 10 3 S Vi Vi
Thus, the transconductance-amplifier model is:
P11.34
The circuit model for the amplifier is:
Vooc RmocIi Rmoc 200 k = = = = 20 Vi Ri Ii Ri 10 k R I / Ro Rmoc 200 k I Aisc = osc = moc i = = = 100 2 k Ii Ii Ro R I / Ro Rmoc I 200 k Gmsc = osc = moc i = = = 0.01 S Vi Ri I i Ri Ro (10 k)(2 k) A oc = v
P11.35* The circuit model for the amplifier is
Avoc =
Vooc GmscVi Ro = = GmscRo = (0.5 S)(200 ) = 100 Vi Vi
325
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Aisc =
Rmoc
I osc GmscVi = = GmscRi = (0.5 S)(1000 ) = 500 I i Vi / Ri G VR V = ooc = msc i o = GmscRo Ri = (0.5 S)(200 )(1000 ) = 100 k Vi / Ri Ii
P11.36
50 Vi Iosc / Gmsc Aisc = = = = 250 Ii Iosc / Aisc Gmsc 0.2 A V A V A Ro = voc i = voc i = voc = 500 Iosc GmscVi Gmsc G VR V Rmoc = ooc = msc i o = GmscRo Ri = (0.2 S)(500 )(250 ) = 25 k Ii Vi / Ri
Ri =
P11.37
Vi Iosc / Gmsc Aisc 50 = = = = 100 Ii Iosc / Aisc Gmsc 0.5 R I R 200 V Ro = ooc = moc i = moc = =4 GmscVi GmscRi Ii GmscRi (0.5)(100) R I R V 200 A oc = ooc = moc i = moc = = 2 V/V v Vi Ri Ii Ri (100) Ri =
The equivalent circuit is:
P11.38
ii =
Rs + Ri
vs
=
v o = Rmocii
Ro + RL
RL
2 10 -3 cos(200t ) = 0.6667 10 - 6 cos(200t ) A 1000 + 2000
= -4.444 cos(200t ) V
2
Pi = Ri I Po =
2 i - rms
Vo 2rms - RL P G = o = 22.22 10 6 Pi
0.6667 10 -6 = 444.4 pW = 2000 2 2 ( 4.444 / 2 ) = = 9.876 mW 1000
326
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.39
The input and output impedances of ideal amplifiers are given in Table 11.1 in the text. The equivalent circuit is:
P11.40
Rx =
Vin Vx 1 = =- = -10 I x - GmscVin Gmsc
Thus if Gmsc is positive, the circuit behaves as a negative resistance.
P11.41* The equivalent circuit is:
We can write:
Ii =
Vx Ri
(1)
I x = Ii +
have:
Using Equation (1) to substitute for I i in Equation (2) and solving, we
Vx - Rmoc I i Ro
(2)
Rx =
Vx 1 = = -2.23 I x 1 Ri + 1 Ro - Rmoc (Ri Ro )
327
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.42
We have Ri << Rs and Ro << RL . Thus, we have an approximately ideal transresistance amplifier. As in Example 11.7, we have: Rmoc = Avo Ri = 10
P11.43
We have Ri >> and Rs Ro >> RL . Thus, we have an approximately ideal transconductance amplifier. A 100 Gmsc = vo = 6 = 10 -4 S Ro 10
P11.44* To sense the open-circuit voltage of a sensor, we need an amplifier with very high input resistance (compared to the Thvenin resistance of the sensor). To avoid loading effects by the variable load resistance, we need an amplifier with very low output resistance (compared to the smallest load resistance). Thus, we need a nearly ideal voltage amplifier with a gain of 1000. P11.45* The input resistance is that of the ideal transresistance amplifier which is zero. The output resistance of the cascade is the output resistance of the ideal transconductance amplifier which is infinite. An amplifier having zero input resistance and infinite output resistance is an ideal current amplifier. Also, we have Aisc = RmocGmsc . P11.46
To sense the short-circuit current of a sensor, we need an amplifier with very low input resistance (compared to the Thvenin resistance of the sensor). To avoid loading effects by the potenitally variable load resistance, we need an amplifier with very low output resistance (compared to the smallest load resistance). Thus, we need a nearly ideal transresistance amplifier. To sense the short-circuit current of a sensor, we need an amplifier with very low input resistance (compared to the Thvenin resistance of the sensor). For the load current to be independent of the variable load resistance, we need an amplifier with very high output resistance (compared to the largest load resistance). Thus, we need a nearly ideal current amplifier. The input resistance is that of the voltage amplifier which is infinite. The output resistance of the cascade is the output resistance of the 328
P11.47
P11.48
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
transconductance amplifier which is infinite. An amplifier having infinite input resistance and infinite output resistance is an ideal transconductance amplifier. Also, we have Gmsc - cascade = A ocGmsc . v
P11.49* To sense the source voltage with minimal loading effects, we need Ri >> Rs . To force a current through the load independent of its
resistance, we need Ro >> RL . Thus, we need a nearly ideal transconductance amplifier. The equivalent circuit is:
We have I L =
Ri + Rs Ro + RL
Ri
Ro
GmscVs .
For the two given values of Rs , we require:
Ri + 1000 Ri + 2000 Solving, we have Ri = 98 k .
0.99
0.99
Ri
=
Ri
For the two given values of RL , we require:
Ro + 100
Ro
=
Ro + 300
Ro
Solving, we find:
Ro = 19.7 k
P11.50
We need Ri << 10 and Ro << 10 k . Because we want the chart calibration to be 1 mA/cm and the chart pen deflects 1 cm per volt applied, the required transresistance gain is
Rmoc =
1V = 1000 1 mA
Thus, a nearly ideal transresistance amplifier is needed. The equivalent circuit is
329
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
To achieve approximately 3% accuracy, we will allow 1% each for variations in the transresistance gain, in chart sensitivity, and in load resistance. From the equivalent circuit, we have
Vo =
Rs + Ri
Vs
Rmoc
Ro + RL
RL
Because are allowing a 1% change in Vo as RL varies from 10 k to an open circuit, we have: 10 k = 0.99 Ro + 10 k
Solving, we find Ro 100 Thus, we specify an amplifier having: Rmoc = 1000 1 %
Ri < 10 Ro < 100
P11.51
We need an amplifier with high input resistance, low output resistance, and a voltage gain of 10. Thus, a nearly ideal voltage amplifier is required. Let us allow 1% variation in the output voltage due to changes in Rs , in amplifier gain, and in load resistance. The equivalent circuit is:
We have:
330
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vo = Vs
We require: 0.99 0.99
Ri + Rs Ri
=
Ri
Avo
RL + Ro Ri
which yields Ri = 989 k
RL
Ri + 10
Ri + 10 4
10 6 10 4 = 4 which yields Ro = 102 10 6 + Ro 10 + Ro
Thus, we specify an amplifier having: Avo = 10 1% Ri 989 k Ro 102
P11.52
We need an amplifier with high input resistance, high output resistance, and a gain of Gm = (1 mA) (0.1 V ) = 10 -2 S. Thus, a nearly ideal transconductance amplifier is needed. The sensitivity of the recorder varies by 1% ; thus, we budget a total of 2% for variations in amplifier gain, source resistance, and load resistance. We will allow 0.667% for each of these. The equivalent circuit is:
I o = Vs
We require:
Ri + Rs
Ri
Gmsc
RL + Ro
Ro
which yields Ri = 1.49 M Ri + 10 Ri + 10 4 Ro Ro 0.9933 = which yields Ro = 14.9 k Ro + 0 Ro + 100 0.9933 = Thus, we specify an amplifier having: Gmsc = 10 2 3%
Ri
Ri
Ri 1.49 M Ro 14.9 k
331
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.53
P11.54
A wideband amplifier has constant gain over a wide range of frequency. (i.e., The ratio of the upper half-power frequency to the lower half-power frequency fH / fL is much greater than unity.) A narrowband amplifier has constant gain over a narrow range of frequency (i.e., fH / fL is nearly unity) and falls to zero outside that range.
P11.55* We are given v in (t ) = 0.1 cos(2000t ) + 0.2 cos(4000t + 30 o )
and
v o (t ) = 10 cos(2000t - 20 o ) + 15 cos(4000t + 20 o )
The phasors for the 1000-Hz components are Vin = 0.10 o and
V = 10 - 20 o . Thus the complex gain for the 1000-Hz component is o
Av =
V 10 - 20 o o = = 100 - 20 o 0.10 o Vin
Similarly, the complex gain for the 2000-Hz component is V 1520 o Av = o = = 75 - 10 o Vin 0.230 o
P11.56* The signal to be amplified is the short-circuit current of an electrochemical cell (or battery). This signal is dc and therefore a dccoupled amplifier is needed. (The dc gain of an ac-coupled amplifier is zero and the signal of interest would not be amplified. Thus an accoupled amplifier would not be appropriate.) P11.57
We need a midband voltage gain of (10 V)/(10 mV) = 1000 for the audio signal. Thus if a dc coupled amplifier were to be used the dc component of the output would be 2000 V. This is impractical, potentially dangerous, and would destroy most loudspeakers. Thus an ac-coupled amplifier is
332
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
needed. Appropriate values for the half-power frequencies are fL = 20 Hz or less and fH = 10 kHz or more.
P11.58* We are given that the gain of the amplifier as a function of frequency is: 1000 A (f ) = [1 + j (f / fB )]2
For f << fB , the gain magnitude is approximately 1000. This is the midband gain. As frequency increases, the gain magnitude decreases. At the half-power frequency fhp the gain magnitude is 1000 / 2 . Thus we can write 1000 1000 = A (fhp ) = 1 + (fhp / fB ) 2 2 Solving, we find fhp = fB
2 - 1 0.6436fB
P11.59
The equivalent circuit is:
(a)
A ocGmsc t vo (t ) = A ocvc (t ) = A oc Gmscvin (t )dt = v v v vin (t )dt C 0 C 0
1
t
(b)
A ocGmsc t V A G t v v Vm cos(2ft )dt = m2oc msc [sin(2ft )]0 C fC 0 V A G vo (t ) = m voc msc sin(2ft ) 2fC V A G - j m voc msc V 2fC = A ocGmsc = A ocGmsc - 90 v v A(f ) = o = 2fC V Vm j 2fC in vo (t ) =
333
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(c)
For the values given, we have A(f ) =
100
magnitude of the gain in dB is | A(f ) | = 40 - 20 log(f ) . The Bode Plots of dB magnitude and phase are:
f
- 90 . Then, the
P11.60
The equivalent circuit is:
(a) (b)
vo (t ) = Rmocic (t ) = RmocC
vo (t ) = A ocRmocC v
d [ cos(2ft )] = -2fAvocRmocCVm sin(2ft ) V dt m j 2fAvocRmocCVm V A(f ) = o = = 2fA ocRmocC90 v Vm V in
d (A ocvin ) dv v = A ocRmocC in v dt dt
(c)
For the values given, we have A(f ) = 0.1f90 . Then, the
magnitude of the gain in dB is | A(f ) | = -20 + 20 log(f ) . The Bode Plots dB of magnitude and phase are:
334
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.61
To avoid linear distortion, an amplifier must have constant gain magnitude and phase shift that is proportional to frequency over the frequency range of the input signal. The input signal is given as v i (t ) = 0.01 cos(2000 t ) + 0.02 cos(4000 t ) which has components with frequencies of 1000 Hz and 2000 Hz respectively. The gain of the amplifier, as a function of frequency, is given by 100 A= 1 + j (f 1000 ) Evaluating for f = 1000 and for f = 2000 , we have 100 A (1000 ) = = 70.71 - 45 o 1 + j (1000 1000 )
P11.62
A (2000 ) = 44.72 - 63.43o
Thus, the output is: v o (t ) = 0.7071 cos(2000t - 45 o ) + 0.8944 cos(4000t - 63.43o ) P11.63*
Thus, v in (t ) contains a 1000 Hz component and a 2000 Hz component. 335
v in (t ) = 0.01 cos(2000t ) + 0.02 cos(4000t )
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The gain magnitude must be the same for both components. The phase shift must be proportional to frequency. Therefore, the gain at 2000 Hz must be 100 - 90 o . The output signal is v o (t ) = 1 cos(2000t - 45o ) + 2 cos(4000t - 90 o ) The plots are:
P11.64
(a) The amplifier is linear because
[v inA (t ) + KvinA (t - td )] + [vinB (t ) + KvinB (t - td )] = [v inA (t ) + vinB (t )] + K [vinA (t - td ) + vinB (t - td )]
(b)
For vin (t ) = Vm cos(2ft ) , we have vo (t ) = vin (t ) + Kvin (t - td ) = Vm cos(2ft ) + KVm cos(2ft - 2ftd ) V V + KVm - 2ftd A(f ) = o = m = 1 + K exp( - j 2ftd ) Vm V in
(c) In MATLAB, a plot of the magnitude can be obtained with the commands: f=0:10:10000; A=1 + 0.5*exp(-i*2*pi*f*0.001); plot(f,abs(A)) The resulting magnitude plot is:
336
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Then a plot of the phase can be obtained with the command: plot(f, angle(A)) The resulting plot is
(d) This amplifer produces amplitude distortion because |A(f)| is not constant with f. The amplifier produces phase distortion because the phase is not proportional to frequency. P11.65 (a) The amplifier is linear because
337
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
dv (t ) dvinA (t ) ] + [vinB (t ) + K inB ] = dt dt d [vinA (t ) + vinB (t )] + K [v (t ) + vinB (t )] dt inA
[vinA (t ) + K (b) For vin (t ) = Vm cos(2ft ) , we have
dvin (t ) = dt Vm cos(2ft ) - KVm 2f sin(2ft ) V V + KVm 2f / 2 A(f ) = o = m = 1 + jK 2f Vm V in vo (t ) = vin (t ) + K
(c) In MATLAB, a plot of the magnitude can be obtained with the commands: f=0:10:10000; A=1 + i*f/1000; plot(f,abs(A)) The resulting magnitude plot is:
Then a plot of the phase can be obtained with the command: plot(f, angle(A)) The resulting plot is
338
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(d) This amplifer produces amplitude distortion because |A(f)| is not constant with frequency. The amplifier produces phase distortion because the phase is not proportional to frequency.
P11.66
The sketch is
The relationship between rise time and bandwidth is: 0.35 tr =
B
Percentage tilt, the lower half-power frequency, and the pulse duration are related by: percentage tilt 200fLT
339
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.67
(a)
Applying the voltage-division principle, we have 1 ( jC ) V 1 A= 2 = = V1 R + 1 ( jC ) 1 + jRC 1 Defining B = 1 / 2RC , we have A = 1 + j (f / B ) where B is the half-power bandwidth. The gain magnitude approaches zero as frequency becomes large.
(b)
The transient response is: v 2 (t ) = [1 - exp(- t RC )]u (t )
We have 0.1 = 1 - exp(- t10 RC ) and 0.9 = 1 - exp(- t90 RC ) Solving, we obtain: t10 = -RC ln(0.9) t90 = -RC ln(0.1) tr = t90 - t10 = RC ln(9) (c) Combining the results of parts (a) and (b), we obtain ln(9 ) 0.35 tr = 2B B This is the basis for the rule-of-thumb given in Equation 11.11. (a) Applying the voltage-division principle, we have V R 1 A= 2 = = V1 R + 1 ( jC ) 1 + 1 jRC
P11.68
Defining, fL = 1 2RC , we have 1 A= 1 - j (fL f ) (b)
fL = 1 2RC is the half-power frequency. The gain magnitude
approaches unity as frequency becomes large. At dc, the gain is zero. 340
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(c)
Solving for transient response, we obtain: v 2 (t ) = exp(- t RC ) for 0 < t <T
Thus, we have: P = 1 - exp(-T RC ) (1) P Percentage tilt = 100% = [1 - exp(-T RC )] 100% However,
P
T T T exp(-T RC ) = 1 - - - -L RC RC RC For T << RC T exp(-T RC ) = 1 - RC Using this to substitute in Equation (1), we have
2
3
Percentage tilt = (d)
This result is precise only for a first-order circuit for which RC >> T but it provides a useful rule-of-thumb for other high pass filters provided that the percentage tilt is small (i.e., less than 10%).
Combining the results of parts (b) and (c), we obtain Percentage tilt 2fLT 100%
T 100% RC
P11.69*
B fH = 15 kHz tr
0.35 Percentage tilt 2fLT 100% 18.8%
B
= 23.3 s 2 (15)(2 10 -3 ) 100%
P11.70* (a)
The upper half-power frequency is fH = 100 kHz , and we estimate 341
Because the amplifier is dc-coupled (i.e., fL = 0 ), the tilt is zero.
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
the rise time as 0.35 tr = = 3.5 s
fH
Because the frequency response shows no peaking, we do not expect overshoot or ringing. The gain magnitude is 100 in the passband, so we expect the output pulse to be 100 times larger in amplitude than the input. Thus, the output pulse is approximately like this:
(b)
We have fL = 20 Hz and fH = 50 kHz . Thus,
Percentage tilt 200fLT = 200 (20 )10 -3 = 12.6% 0.35 0.35 tr = = = 7 s fH 50 103
Because the frequency response shows no peaking, we do not expect overshoot or ringing. The gain magnitude is 200 in the passband, so we expect the output pulse to be 200 times larger in amplitude than the input. Thus, the output pulse is about like this:
(c)
The upper half-power frequency is fH 100 kHz , and we estimate the rise time as 0.35 tr = 3.5 s
Because the amplifier is dc-coupled (i.e., fL = 0 ), the tilt is zero.
fH
Because the frequency response displays peaking, we expect overshoot and ringing. Furthermore, the period of the ringing will be approximately 1 (100 kHz ) = 10 s . The gain magnitude is 100 in 342
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
the passband, so we expect the output pulse to be 100 times larger in amplitude than the input. Thus, the output pulse will be about like this:
P11.71
(a)
Part (a) of Figure P11.71 shows the input pulse. Thus there is no part (a) of the solution. Since tilt = 0 , we expect fL = 0 . 0.35 0.35 fH = = 3.5 MHz tr 0.1 10 -6 Because the output pulse is 100 times larger in amplitude than the input pulse, the midband gain of the amplifier is 100. Because the pulse shows little overshoot and ringing, we expect the amplifier gain to roll off smoothly versus frequency. Thus, we expect a gain versus frequency plot something like this:
(b)
(c)
2 - 1.6 = 20% P 2 percentage tilt 20 fL = = 31.8 kHz 200T 20010 -6 0.35 0.35 fH = = 7 MHz tr 0.05 10 -6 Percentage tilt = = Because the output pulse is 200 times larger in amplitude than the input pulse, the midband gain of the amplifier is 200. Because the pulse shows little overshoot and ringing, we expect the amplifier 343
P
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
gain to roll off smoothly versus frequency. Thus, we expect a gain versus frequency plot something like this:
(d)
Since tilt 0 , we expect fL = 0 . 0.35 0.35 fH = = 11.7 MHz tr 0.03 10 -6 Because the output pulse in 100 times larger in amplitude than the input pulse, the midband gain of the amplifier is 100. Because the pulse shows overshoot and ringing with a period of about 0.25 s, we expect the amplifier gain to display a gain peak at about f = 1 (0.25 s ) = 4 MHz . Thus, we expect a gain versus frequency plot something like this:
P11.72
When a sinewave input signal is passed through an amplifier having a nonlinear transfer characteristic, distortion consisting of harmonics is produced. Harmonics are components having frequencies that are integer multiples of the input frequency.
P11.73* We are given v in (t ) = 0.1 cos(2000t )
and
v o (t ) = 10 cos(2000t ) + 0.2 cos(4000t ) + 0.1 cos(6000t )
344
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
in which 10 cos(2000t ) is the desired term with an amplitude V1 = 10 V. The second term 0.2 cos(4000t ) is second harmonic distortion with an amplitude V2 = 0.2 V. Finally the term 0.1 cos(6000t ) is third harmonic distortion with amplitude V3 = 0.1 V. There is no fourth or higher order harmonic distortion. Then using Equations 11.17, 11.18, and 11.19 we have
V2 0.2 = = 0.02 10 V1 V 0. 1 D3 = 3 = = 0.01 V1 10 V 0 D4 = 4 = =0 V1 10 D2 = D = D22 + D32 + D44 + L = (0.02) 2 + (0.01) 2 = 0.02236 = 2.236%
P11.74
Substituting the input into the equation for the transfer characteristic, we obtain: v o (t ) = 20 cos(200t ) + 2.4 cos2 (200t ) + 3.2 cos3 (200t ) Applying the trigonometric identities suggested in the problem statement, we obtain
v o (t ) = 1.2 + 22.4 cos(200t ) + 1.2 cos(400t ) + 0.8 cos 3 (600t )
Thus the amplitude of the desired output term is V1 = 22.4, the amplitude of the second harmonic is V2 = 1.2 V, and the amplitude of the third harmonic is V3 = 0.8 V. The amplitudes of higher order terms are zero. Then using Equations 11.17, 11.18, and 11.19, we have
V2 1. 2 = = 0.05357 V1 22.4 V 0. 8 D3 = 3 = = 0.03571 V1 22.4 V 0 D4 = 4 = =0 V1 10 D2 =
345
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
D = D22 + D32 + D44 + L = (0.05357) 2 + (0.03571) 2
= 0.06438 = 6.438%
P11.75
A differential amplifier has two input terminals, one is known as the inverting input and the other is known as the noninverting input. If the input signals applied to the input terminals of a differential amplifier are denoted as v i 1 and v i 2 , the common-mode signal is v icm = (v i 1 + v i 2 ) 2 and
v o = Acmv icm + Ad v id where Ad is called the differential gain and Acm is
called the common-mode gain. Ideally, the common-mode gain is zero so the amplifier responds to only the differential signal.
the differential signal is v id = v i 1 - v i 2 . The output signal is
P11.76
The common-mode rejection ratio (in decibels) is defined as
CMRR = 20 log
Ad Acm
where Ad is the differential gain and Acm is the common-mode gain.
P11.77
v o = Ad v id = 10(v i 1 - v i 2 )
v icm = (v i 1 + v i 2 ) 2
346
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P11.78* With the input terminals tied together, the differential signal is zero and the common-mode signal is: v icm = (v i 1 + v i 2 ) 2 = 10 mV rms
The common-mode gain is: Acm = v o v icm = 20 10 = 2 The common-mode rejection ratio is:
CMRR = 20 log
P11.79
We want
Acm
Ad
= 20 log
500 = 47.96 dB 2
V 60 = 20 log od V ocm
AV = 20 log d id AcmVicm A (20 mV ) = 20 log d Acm (5 V ) A 20 mV = 20 log d + 20 log A 5V cm A = 20 log d - 48.0 A cm
Thus,
CMRR = 20 log
P11.80
Ad = 108 dB Acm
See Figure 11.40 in the text. The effect of these sources is to add a dc component to the output. See Figure 11.43 in the text.
P11.81
P11.82* The circuit is:
The bias currents are equal. However, because the resistances may not be equal, a differential input voltage is possible. In the extreme case, 347
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
one resistor could have a value of 1050 and the other could have a value of 950 . Then the differential input voltage is: v id = R1I bias - R2I bias = (R1 - R2 )I bias
= 100 100 nA = 10 V
Thus, the extreme values of the output voltage are: v o = Ad v id = 5 mV If the resistors are exactly equal, then the output voltage is zero.
P11.83* The equivalent circuit is:
The solution is similar to that for Example 11.13 in the text. The bias currents produce a common-mode input voltage of v icm = I B Rs 1 = I s Rs 2 = 10 mV . However, the common-mode gain is assumed to be zero so the common-mode input signal does not contribute to the output voltage. The differential input voltage due to the offset current is:
VIoff =
I off Rin (Rs 1 + Rs 2 ) = 1.667 mV 2 Rin + Rs 1 + Rs 2
The differential input voltage due to the offset voltage is:
VVoff = Voff
Rin = 1.667 mV Rin + Rs 1 + Rs 2
348
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
If these two voltages have the same polarity, the total differential input voltage is 1.667 + 1.667 = 3.333 mV in magnitude. Then the output voltage is: v o = Ad v id = 3.333 V Thus the output voltage can range from -3.333 to +3.333 V.
P11.84
The common-mode rejection ratio is:
CMRR = 20 log
Ad = 60 dB Acm
Using the fact that Ad = 1000 and solving, we find that Acm = 1 . As in Problem P11.83, the common-mode input voltage is 10 mV. Thus, the contribution to the output voltage is Acm v icm = 10 mV . The contributions due the offset current and offset voltage are the same as in Problem P11.83. Thus, the extreme output voltage is: v o = 0.01 + 3.333 = 3.343 V Thus, the contribution due to the common-mode gain is negligible.
349