COMPARATOR

CHAPTER 17 Fourier Method of Waveform Analysis 420

5.14 COMPARATOR

The circuit of Fig. 5-30 compares the voltagev₁with a reference levelv_o. Since the open-loop gain is very large, the op amp outputv₂is either atþV_cc(ifv₁>v_o) or atV_cc(ifv₁<v_o). This is shown by v₂ ¼V_ccsgn½v₁v_owhere ‘‘sgn’’ stands for ‘‘sign of.’’ Forv_o¼0, we have

v₂¼V_ccsgn½v₁ ¼ þV_cc v₁>0 V_cc v₁<0

EXAMPLE 5.23 In Fig. 5-30, letV_cc¼5 V,v_o¼0, andv₁¼sin!t. Findv₂. For 0<t< =!,

v₁¼sin!t>0 v₂¼5 V For=! <t<2=!,

v₁¼sin!t<0 v₂¼ 5 V

The outputv₂is a square pulse which switches betweenþ5 V and5 V with period of 2=!. One cycle ofv₂is given by

v₂¼ 5 V 0<t< =!

5 V =! <t<2=!

EXAMPLE 5.24 The circuit of Fig. 5-31 is a parallel analog-to-digital converter. TheþV_ccandV_ccconnections are omitted for simplicity. Let V_cc¼5 V,v_o¼4 V, andv_i¼t(V) for 0<t<4 s. Find outputs v₃;v₂; andv₁. Interpret the answer.

The op amps have no feedback, and they function as comparators. The outputs with values atþ5 or5 V are given in Table 5-2.

The binary sequencesfv₃;v₂;v₁gin Table 5-2 uniquely specify the input voltage in discrete domain. How- ever, in their present form they are not the binary numbers representing input amplitudes. Yet, by using a coder we could transform the above sequences into the binary numbers corresponding to the values of analog inputs.

Fig. 5-30

Table 5-2

time, s input, V outputs, V

0<t<1 0<v_i<1 v₃¼ 5 v₂¼ 5 v₁¼ 5 1<t<2 1<v_i<2 v₃¼ 5 v₂¼ 5 v₁¼ þ5 2<t<3 2<v_i<3 v₃¼ 5 v₂¼ þ5 v₁¼ þ5 3<t<4 3<v_i<4 v₃¼ þ5 v₂¼ þ5 v₁¼ þ5

Solved Problems

5.1 In Fig. 5-3, letv_s¼20 V, R_s¼10,R_i¼990,k¼5, andR_o¼3. Find (a) the The´venin equivalent of the circuit seen byR_land (b) v₂and the power dissipated inR_lforR_l¼0:5, 1, 3, 5, 10, 100, and 1000.

(a) The open-circuit voltage and short-circuit current at A–B terminal are v_o:c:¼5v₁ and i_s:c:¼5v₁=3, respectively.

We findv₁by dividingv_sbetweenR_sandR_i. Thus, v₁¼ R_i

R_sþR_iv_s¼ 990

10þ990ð20Þ ¼19:8 V Fig. 5-31

Fig. 5-32

Therefore,

v_o:c:¼5ð19:8Þ ¼99 V v_Th¼v_o:c:¼99 V i_s:c:¼99=3¼33 A R_Th¼v_o:c:=i_s:c:¼3 The The´venin equivalent is shown in Fig. 5-32.

(b) With the loadR_lconnected, we have v₂¼ R_l

R_lþR_Th v_Th¼ 99R_l

R_lþ3 and p¼v²₂ R_l

Table 5-3 shows the voltage across the load and the power dissipated in it for the given seven values of R_l. The load voltage is at its maximum whenR_l¼ 1. However, power delivered toR_l¼ 1is zero.

Power delivered toR_lis maximum atR_l¼3, which is equal to the output resistance of the amplifier.

5.2 In the circuits of Figs. 5-4 and 5-5 letR₁¼1 kandR₂¼5 k. Find the gainsG^þ¼v₂=v_sin Fig. 5-4 and G¼v₂=v_s in Fig. 5-5 for k¼1, 2, 4, 6, 8, 10, 100, 1000, and1. Compare the results.

From (5) in Example 5.3, atR₁¼1 kandR₂¼5 kwe have G^þ¼v₂

v_s¼ 5k

6k ð31Þ

In Example 5.4 we found

G¼v₂

v_s¼ 5k

6þk ð32Þ

The gainsGandG^þare calculated for nine values ofkin Table 5-4. Askbecomes very large,G^þand Gapproach the limit gain of5, which is the negative of the ratioR₂=R₁and is independent ofk. The circuit of Fig. 5-5 (with negative feedback) is always stable and its gain monotonically approaches the limit gain. However, the circuit of Fig. 5-4 (with positive feedback) is unstable. The gainG^þ becomes very large askapproaches six. Atk¼6,G^þ¼ 1.

Table 5-3 R_l; v₂;V p;W

0.5 14.14 400.04

1 24.75 612.56

3 49.50 816.75

5 61.88 765.70

10 76.15 579.94

100 96.12 92.38

1000 98.70 9.74

Table 5-4

k G^þ G

1 1:0 0:71

2 2:5 1:25

4 10:0 2:00

6 1 2:50

8 20:0 2:86

10 12:5 3:12

100 5:32 4:72

1000 5:03 4:97

1 5:00 5:00

5.3 LetR₁ ¼1 k,R₂¼5 k, andR_i¼50 kin the circuit of Fig. 5-33. Findv₂=v_sfork¼1, 10, 100, 1000,1and compare the results with the values ofG in Table 5-4.

This problem is solved by application of KCL at nodeA(another approach which uses the The´venin equivalent is suggested in Problem 5.30). Thus,

v₁v_s

1 þv₁v₂ 5 þv₁

50¼0 ð33Þ

From the amplifier we obtain

v₂¼ kv₁ or v₁¼ v₂=k ð34Þ

Replacingv₁in (34) into (33) and rearranging terms, we obtain v₂

v_s¼ 50k

61þ10k¼ 5k

6:1þk ð35Þ

Values ofv₂=v_sin (35) are shown in Table 5-5 as functions ofk. The 50-kinput resistance of the amplifier reduces the overall gain very slightly, as seen by comparing Tables 5-4 and 5-5. The feedback has made the input resistance of the amplifier less effective in changing the overall gain.

5.4 Let againR₁¼1 kandR₂¼5 kin the circuit of Fig. 5-33.

ðaÞ Find v₂=v_s as a function of kandR_i:

ðbÞ Let R_i¼1 k. Findv₂=v₁ fork¼1;10;100;1000;1. Repeat forR_i¼ 1:

ðcÞ Discuss the effects of R_i andkon the overall gain. Show that, fork¼ 1andR_i6¼0;

the gain of the amplifier is independent of R_i and is equal to R₂=R₁: (a) Apply KCL to currents leaving nodeAto obtain

v₁v_s

1 þv₁v₂ 5 þv₁

R_i¼0

From the amplifier we getv₂¼ kv₁orv₁¼ v₂=k. Substituting forv₁ in the KCL equation and rearranging terms we get

Fig. 5-33

Table 5-5

k v₂=v_s 1 0:704 10 3:106 100 4:713

1000 4:97

1 5:00

v₂

v_s¼ 5 ck

1þck wherec¼ R_i

5þ6R_i ð36Þ

(b) ForR_i¼1 k,c¼1=11 which, substituted into (36), gives v₂

v_s¼ 5k

11þk ð37Þ

ForR_i¼ 1we getc¼1=6 and so

v₂ v_s¼ 5k

6þk ð38Þ

Table 5-6 gives values ofv₂=v_sin (37) and (38) versusk. Note that (38) is identical with (32).

(c) Comparing the two columns in Table 5-6 we see that the smaller R_i reduces the overall gain G. However, as the open-loop gain k increases, the effect of R_i is diminished. As k becomes very large,v₂=v₁approaches5 unlessR_i¼0.

5.5 Let againR₁¼1kandR₂¼5kin the circuit of Fig. 5-33. Replace the circuit to the left of nodeAincluding v_s, R₁, andR_i by its The´venin equivalent. Then use (5) to derive (36).

The The´venin equivalent is given by

v_Th¼ R_iv_s

R₁þR_i¼ R_iv_s 1þR_i R_Th¼ R₁R_i

R₁þR_i¼ R_i 1þR_i where the resistors are in k:

From (5),

v₂¼ ð1bÞ k 1þbkv_Th b¼ R_Th

R_ThþR₂¼ R_i

6R_iþ5 and 1b¼5ð1þR_iÞ 6R_iþ5 where

Therefore,

v₂¼5ð1þR_iÞ

6R_iþ5 k

1þR_ik=ð6R_iþ5Þ R_i

1þR_iv_s¼ 5R_ik 6R_iþ5þR_ikv_s which is identical with (36).

5.6 Find the output voltage of an op amp withA¼10⁵andV_cc¼10 V forv¼0 andv^þ¼sint(V).

Refer to Figs. 5-7 and 5-8.

Table 5-6

k

v₂=v_s

R_i¼1 k R_i¼ 1 1

10 100 1000 1

0:31 2:38 4:51 4:95 5:00

0:71 3:12 4:72 4:97 5:00

Because of high gain, saturation occurs quickly at

jv₂j ¼10⁵jv_dj ¼10 V or jv_dj ¼10⁴V We may ignore the linear interval and write

v₂¼ þ10 V v_d>0 10 V v_d<0

wherev_d¼v^þv¼sint(V). One cycle of the output is given by v₂¼ þ10 V 0<t<

10 V <t<2

For a more exactv₂, we use the transfer characteristic of the op amp in Fig. 5-7.

v₂¼

10 v_d<10⁴V 10⁵v_d 10⁴<v_d< 10⁴V þ10 v_d> 10⁴V 8<

:

Saturation begins atjv_dj ¼ jsintj ¼10⁴V. Since this is a very small range, we may replace sintbyt. The outputv₂is then given by

v₂¼ 10⁵t 10⁴<t<10⁴s

v₂¼ 10 10⁴<t< 10⁴s

v₂¼ 10⁵ðtÞ 10⁴<t< þ10⁴s v₂¼ 10 þ10⁴<t<210⁴s

To appreciate the insignificance of error in ignoring the linear range, note that during one period of 2s the interval of linear operation is only 410⁴s, which gives a ratio of 6410⁶.

5.7 Repeat Problem 5.6 forv^þ¼sin 2t(V) andv¼0:5 V.

The output voltage is

v₂¼ 10 V whenv^þ>v v₂¼ 10 V whenv^þ<v

Switching occurs when sin 2t¼1=2. This happens att¼1=12, 5/12, 13/12, and so on. Therefore, one cycle ofv₂is given by

v₂¼ 10 V 1=12<t<5=12 s v₂¼ 10 V 5=12<t<13=12 s Figure 5-34 shows the graphs ofv^þ,v, andv₂.

5.8 In the circuit of Fig. 5-35v_s¼sin 100t. Findv₁andv₂. At nodesBandA,v_B¼v_A¼0. Then,

v₁¼ 30

20þ30v_s¼0:6 sin 100tðVÞ v₂¼ 100

30 v₁¼ 100

30 ð0:6 sin 100tÞ ¼ 2 sin 100tðVÞ v₂¼ 100

20þ30v_s¼ 2 sin 100tðVÞ Alternatively,

5.9 Saturation levels for the op amps in Fig. 5-31 areþV_cc¼5 V andV_cc¼ 5 V. The reference voltage isv_o¼1 V. Find the sequence of outputs corresponding to values ofv_ifrom 0 to 1 V in steps of 0.25 V.

See Table 5-7 whereL¼ 5 V andH¼ þ5 V.

5.10 Find vin the circuit of Fig. 5-36.

Apply KCL at nodeA,

Fig. 5-34

Fig. 5-35

Table 5-7

v_i, V v₃ v₂ v₁ 0 to 0.25 L L L 0.25^þto 0:5 L L H 0.5^þto 0.75 L H H

0.75^þto 1 H H H

ðvv₁Þg₁þ ðvv₂Þg₂þ ðvv₃Þg₃¼0 v¼v₁g₁þv₂g₂þv₃g₃

g₁þg₂þg₃ ¼v₁R₂R₃þv₂R₁R₃þv₃R₂R₁ R₁R₂þR₂R₃þR₃R₁ Then

5.11 In the circuit of Fig. 5-37 findv_C(the voltage at nodeC),i₁,R_in(the input resistance seen by the 9-V source),v₂, andi₂.

At nodesBandA,v_B¼v_A¼0. Applying KCL at nodeC, we get

ðv_C9Þ=4þv_C=6þv_C=3¼0 from which v_C¼3 V i₁¼ ð9v_CÞ=4¼1:5 A and R_in¼v₁=i₁¼9=1:5¼6 Then

From the inverting amplifier circuit we have

v₂¼ ð5=3Þv_C¼ 5 V and i₂¼ 5=10¼ 0:5 A

5.12 Find v₂ in Problem 5.11 by replacing the circuit to the left of nodes A-B in Fig. 5-37 by its The´venin equivalent.

R_Th¼3þð6Þð4Þ

6þ4¼5:4 and v_Th¼ 6

4þ6ð9Þ ¼5:4 V Thenv₂¼ ð5=5:4Þð5:4Þ ¼ 5 V.

5.13 Findv_C,i₁,v₂, andR_in, the input resistance seen by the 21-V source in Fig. 5-38.

From the inverting amplifier we get

v₂¼ ð5=3Þv_C ð39Þ

Note thatv_B¼v_A¼0 and so KCL at nodeCresults in Fig. 5-36

Fig. 5-37

v_C21 3 þv_C

6 þv_C

3 þv_Cv₂

8 ¼0 ð40Þ

Substitutingv_C¼ ð3=5Þv₂from (39) into (40) we getv₂¼ 10 V. Then v_C¼6 V

i₁¼ ð21v_CÞ=3000¼0:005 A¼5 mA R_in¼21=i₁¼21=0:005¼4200¼4:2 k

5.14 In the circuit of Fig. 5-38 change the 21-V source by a factor of k. Show that v_C, i₁, v₂ in Problem 5.13 are changed by the same factor butR_in remains unchanged.

Letv_s¼21k(V) represent the new voltage source. From the inverting amplifier we have [see (39)]

v₂¼ ð5=3Þv_C Apply KCL at nodeCto obtain [see (40)]

v_Cv_s 3 þv_C

6 þv_C

3 þv_Cv₂

8 ¼0

Solving forv_Candv₂, we have

v_C¼ ð6=21Þv_s¼6kðVÞ and v₂¼ ð10=21Þv_s¼ 10kðVÞ i₁¼ ðv_sv_CÞ=3000¼ ð216Þk=3000¼0:005k A R_in¼v_s=i₁¼21k=0:005k¼4200

These results are expected since the circuit is linear.

5.15 Find v₂ and v_C in Problem 5.13 by replacing the circuit to the left of node C in Fig. 5-38 (including the 21-V battery and the 3-kand 6-kresistors) by its The´venin equivalent.

We first compute the The´venin equivalent:

R_Th¼ð6Þð3Þ

6þ3¼2 k and v_Th¼ 6

3þ6ð21Þ ¼14 V

Replace the circuit to the left of nodeCby the abovev_ThandR_Th and then apply KCL at C:

v_C14 2 þv_C

3 þv_Cv₂

8 ¼0 ð41Þ

For the inverting amplifier we havev₂¼ ð5=3Þv_Corv_C¼ 0:6v₂, which results, after substitution in (41), inv₂¼ 10 V andv_C¼6 V.

Fig. 5-38

5.16 (a) Find the The´venin equivalent of the circuit to the left of nodesA-Bin Fig. 5-39(a) and then findv₂ forR_l ¼1 k, 10 k, and1. (b) Repeat for Fig. 5-39(c) and compare with part (a).

(a) The The´venin equivalent of the circuit in Fig. 5-39(a) is shown in Fig. 5-39(b).

v_Th¼ 6

6þ3ð15Þ ¼10 V and R_Th¼ð3Þð6Þ 3þ6¼2 k By dividingv_Th betweenR_Th andR_lwe get

v₂¼ R_l R_lþ2ð10Þ ForR_l¼ 1 k, v₂¼ 3:33 V

ForR_l¼10 k, v₂¼ 8:33 V ForR_l¼ 1 v₂¼ 10 V

The outputv₂depends onR_l. The operation of the voltage divider is also affected byR_l. (b) The The´venin equivalent of the circuit in Fig. 5-39(c) is shown in Fig. 5-12(d). Here we have

v_Th¼10 V and R_Th¼0

andv₂¼v_Th¼10 V for all values ofR_l, that is, the outputv₂depends onR₁,R₂, andv_sonly and is independent ofR_l.

Fig. 5-39

5.17 Find v₂ as a function ofi₁ in the circuit of Fig. 5-40(a).

Currenti₁ goes through resistorRproducing a voltageRi₁ across it from right to left. Since the inverting terminalBis zero potential, the preceding voltage appears at the output asv₂¼ Ri₁[see Fig. 5- 40(b)]. Therefore, the op amp converts the current i₁ to a voltage v₂ with a gain ofjv₂=i₁j ¼R. The current sourcei₁delivers no power as the voltagev_ABacross it is zero.

5.18 A transducer generates a weak currenti₁which feeds a loadR_l and produces a voltagev₁across it. It is desired thatv₁ follow the signal with a constant gain of 10⁸regardless of the value ofR_l. Design a current-to-voltage converter to accomplish this task.

The transducer should feed R_l indirectly through an op amp. The following designs produce v₁¼10⁸i₁independently ofR_l.

Design 1: ChooseR¼100 Min Fig. 5-40. However, a resistor of such a large magnitude is expensive and not readily available.

Design 2: The conversion gain of 10⁸V=Ais also obtained in the circuit of Fig. 5-41. The first op amp with R¼10⁶ converts i₁ to v₁¼ 10⁶i₁. The second amplifier with a gain of 100 (e.g., R₁¼1 k and R₂¼100 k) amplifiesv₁tov₂¼ 100v₁¼10⁸i₁. The circuit requires two op amps and three resistors (1 M, 100 k, and 1 k) which are less expensive and more readily available.

Design 3: See Fig. 5-42 and Problem 5.19.

5.19 Determine the resistor values which would produce a current-to-voltage conversion gain of v₂=i₁ ¼10⁸V=A in the circuit of Fig. 5-42.

Fig. 5-40

Fig. 5-41

Apply KCL at nodeC. Note thatv_B¼v_A¼0. Thus, v_C

Rþv_C

R₁þv_Cv₂ R₂ ¼0 Substitutingv_C¼ Ri₁and solving forv₂we get

v₂¼ R_eqi₁ whereR_eq¼R 1þR₂ R₁þR₂

R

For a conversion gain ofv₂=i₁¼R_eq¼10⁸V=A¼100 M, we need to find resistor values to satisfy the following equation:

R 1þR₂ R₁þR₂

R

¼10⁸

One solution is to chooseR¼1 M,R₁¼1 k, andR₂¼99 k. The design of Fig. 5-42 uses a single op amp and three resistors which are not expensive and are readily available.

5.20 Findi₂ as a function ofv₁ in the circuit of Fig. 5-43.

We have

v_B¼v_A¼0 i₁¼v₁=R₁ i₂¼i₁¼v₁=R₁

The op amp converts the voltage source to a floating current source. The voltage-to-current conversion ratio isR₁and is independent ofR₂.

5.21 A practical current source (i_s in parallel with internal resistanceR_s) directly feeds a loadR_las in Fig. 5-44(a). (a) Find load currenti_l. (b) Place an op amp between the source and the load as in Fig. 5-44(b). Find i_l and compare with part (a).

Fig. 5-42

Fig. 5-43

(a) In the direct connection, Fig. 5-44(a),i_l¼i_sR_s=ðR_sþR_lÞ, which varies withR_l. (b) In Fig. 5-44(b), the op amp forcesv_B to zero causing the current inR_sto become zero. Therefore,i_l¼i_s which is now independent ofR_l. The op amp circuit converts the practical current source to an ideal current source.

See Figure 5-44(c).

5.22 Find v_o in the circuit of Fig. 5-45.

The first op amp is a unity gain inverter withv₃¼ v₂. The second op amp is a summing circuit with a gain ofR₂=R₁for both inputsv₁andv₃. The output is

v_o¼ R₂

R₁ðv₁þv₃Þ ¼R₂

R₁ðv₂v₁Þ Fig. 5-44

Fig. 5-45

The circuit is a difference amplifier.

5.23 Findv_o in the circuit of Fig. 5-46.

Apply KCL at nodeB. Note thatv_B¼v_A¼v₂. Thus, v₂v₁

R₁ þv₂v_o R₂ ¼0 Solving forv_o, we getv_o¼v₂þ ðR₂=R₁Þðv₂v₁Þ.

5.24 Findv_o in the circuit of Fig. 5-47.

The left part of the circuit has a gain ofð1þR₁=R₂Þ. Therefore,v₃¼ ð1þR₁=R₂Þv₁. Using results of Problem 5.23 and substituting forv₃results in

v_o¼v₂þR₂

R₁ðv₂v₃Þ ¼ 1þR₂ R₁

v₂R₂ R₁ 1þR₁

R₂

v₁¼ 1þR₂ R₁

ðv₂v₁Þ

5.25 In Fig. 5-48 choose resistors for a differential gain of 10⁶ so thatv_o¼10⁶ðv₂v₁Þ.

The two frontal op amps are voltage followers.

v_A¼v₁ and v_B¼v₂ From (16), Sec. 5.9, we have

v_o¼R₂

R₁ðv_Bv_AÞ ¼R₂

R₁ðv₂v₁Þ

To obtain the required differential gain ofR₂=R₁¼10⁶, chooseR₁¼100andR₂¼100 M.

Fig. 5-46

Fig. 5-47

The circuit of Fig. 5-48 can have the same gain as that of Fig. 5-45, but its input resistance is infinite.

However, it employs two small and large resistors which are rather out of ordinary range.

5.26 Resistors having high magnitude and accuracy are expensive. Show that in the circuit of Fig. 5- 49 we can choose resistors of ordinary range so thatv_o¼10⁶ðv₂v₁Þ.

The two frontal op amps convey the input voltagesv₁andv₂to the terminals ofR_G, creating an upward current i¼ ðv₂v₁Þ=R_G in the resistor. The current also goes through the two R₃ resistors, creating voltage dropsiR₃across them. Therefore,

v_A¼v₁R₃i¼v₁R₃

R_Gðv₂v₁Þ v_B¼v₂þR₃i¼v₂þR₃

R_Gðv₂v₁Þ v_Bv_A¼ 1þ2R₃

R_G

ðv₂v₁Þ

v_o¼R₂

R₁ðv_Bv_AÞ ¼R₂

R₁ 1þ2R₃ R_G

ðv₂v₁Þ and

For a differential gain of 10⁶we must have v_o v₂v₁¼R₂

R₁ 1þ2R₃ R_G

¼10⁶ ChooseR₁¼R_G¼1 k,R₂¼100 k, andR₃¼5 M.

Fig. 5-48

Fig. 5-49

The circuit of Fig. 5-49 has an infinite input resistance, employs resistors within ordinary range, and uses three op amps.

5.27 Show that in the circuit of Fig. 5-50i₁¼i₂, regardless of the circuits ofN₁ andN₂.

NodesAandBare at the same voltagev_A¼v_B. Since the op amp draws no current,i₁andi₂flow through the two resistors and KVL around the op amp loopABCgivesRi₁Ri₂¼0. Therefore,i₁¼i₂. 5.28 LetN₁ be the voltage sourcev₁ and N₂ be the resistorR₂ in the circuit of Fig. 5-50. Find the

input resistanceR_in¼v₁=i₁.

From the op amp we obtain v_A¼v_B and i₁¼i₂. From connections to N₁ and N₂ we obtain v₁¼v_B¼v₂¼v_Aandv₂¼ i₂R₂, respectively. The input resistance isv₁=i₁¼ i₂R₂=i₂¼ R₂which is the negative of the load. The op amp circuit is a negative impedance converter.

5.29 A voltage follower is constructed using an op amp with a finite open-loop gainAandR_in¼ 1 (see Fig. 5-51). Find the gain G¼v₂=v₁. Defining sensitivity s as the ratio of percentage change produced inGto the percentage change in A, finds.

From Fig. 5-51 we havev₂¼Av_d. Applying KVL around the amplifier, obtain v₁¼v_dþv₂¼v_dþAv_d¼v_dð1þAÞ ¼v₂ð1þAÞ=A

G¼v₂ v₁¼ A

1þA Fig. 5-50

Fig. 5-51

The rate of change ofGwith respect toAis dG

dA¼ 1

ð1þAÞ² from which dG¼ dA ð1þAÞ² The percentage change produced inGis 100ðdG=GÞ.

dG G ¼ dA

ð1þAÞ²1þA

A ¼ 1

1þAdA A and the sensitivity is

s¼dG=G dA=A¼ 1

1þA

The percentage change inGdepends onA. Samples ofdG=dAandsare shown in Table 5-8.

For high values ofA, the gainGis not sensitive to changes inA.

Supplementary Problems

5.30 Repeat Problem 5.3 by replacing the circuit to the left of nodeB(includingv_s,R₁, andR_i) by its The´venin equivalent (see Fig. 5-33) Solve the problem by applying the results of Example 5.4.

5.31 Find the The´venin equivalent of the circuit to the left of nodesA-Bin Fig. 5-52 withk¼10 for (a) R₂¼ 1 and (b) R₂¼50 k. Ans: ðaÞ v_Th¼ 100 V;R_Th¼100; (bÞ v_Th¼ 31:22 V;R_Th¼37:48

5.32 Repeat Problem 5.31 forR₂¼50 kandk¼100. Ans: v_Th¼ 47:16 V;R_Th¼5:66

5.33 Determine the relationship between R, R₁, and R₂ in Fig. 5-41 such that the circuit has a gain of v₂=i₁¼10⁶V/A. Ans: RR₂=R₁¼10⁶

Table 5-8

A G¼v₂=v₁ dG=dA s

10 0.909 0.008 0.091

11 0.917 0.007 0.083

100 0.990 0.0001 0.01

1000 0.999 0 0

Fig. 5-52

5.34 In the circuit of Fig. 5-13,V_cc¼10 V,R₁¼2 kandv₁¼1 V. Find the maximum value ofR₂before the op amp is saturated. Ans: R₂¼20 k

5.35 Let the summing circuit of Fig. 5-14 have two inputs withv₁¼1 and v₂¼sint (V). Let R₁¼3 k, R₂¼5 k, andR_f ¼8 k. Apply superposition to findv_o. Ans: v_o¼ ð⁸₃þ⁸₅sintÞ

5.36 In Fig. 5-17 letR₁¼4 kandR₂¼8 k. Apply superposition to findv_oin terms of the input voltages.

Ans: v_o¼v₁þv₂þv₃

5.37 Find the input resistance seen byv_f in Fig. 5-19. Ans: R_in¼2R₁

5.38 Use superposition to findv_oin Fig. 5-20 forR₁¼2,R₂¼7,R₃¼10,R₄¼5, all values in k.

Ans: v_o¼1:5v₂3:5v₁

5.39 In the circuit of Fig. 5-20 find (a) v₀forR₁¼1,R₂¼3,R₃¼2, andR₄¼2, all values in k; (b) the input resistanceR_{2 in}seen byv₂; (c) i₁as a function ofv₁andv₂and show thatv₁sees a variable load which depends onv₂. Ans: ðaÞ v_o¼2v₂3v₁; ðbÞ R_{2 in}¼4 k; ðcÞ i₁¼v₁v₂=2

5.40 Using a single op amp, design an amplifier with a gain ofv₂=v₁¼3=4, input resistance of 8 k, and zero output resistance. Ans: See Fig. 5-53.

5.41 Show that, givenR₁¼ 1andR₂¼0, the noninverting op amp circuit of Fig. 5-15 and (12) is reduced to a voltage follower.

5.42 In the circuit of Fig. 5-22 let R_s¼10 k. (a) FindR_f such thati_s¼0. (b) IsR_f independent ofR_s? Discuss. Ans: ðaÞ 40 k;ðbÞ yes

5.43 The input to the circuit of Fig. 5-23 withRC¼1 isv₁¼sin!t. Write KCL at nodeBand solve forv₂. Ans: v₂¼ ð1=!Þcos!tþC

5.44 Show that the outputv₂in Fig. 5.54 is the same as the output of the integrator in Fig. 5-23.

Fig. 5-53

Fig. 5-54

5.45 Findv₂in the leaky integrator of Fig. 5-24 withR₁¼R_f ¼1 k,C¼1mF, andv₁¼ 1 V t>0 0 t<0

.

Ans: v₂ðtÞ ¼ 1þe^1000tðVÞ t>0

0 t<0

(

5.46 Repeat Problem 5.45 forv₁¼ 1 V t<0 0 t>0

. Ans: v₂ðtÞ ¼ e^1000tðVÞ t>0 1 V t<0

5.47 In the differential equation 10²dv₂=dtþv₂¼v_s,v_sis the forcing function andv₂is the response. Design an op amp circuit to obtainv₂fromv_s. Ans: See Fig. 5-24, withR₁¼R_f;RC¼10², andv₁¼ v_s.

5.48 Design a circuit containing op amps to solve the following set of equations:

y⁰þx¼v_s1 2yþx⁰þ3x¼ v_s2 Ans. See Fig. 5-55, withR₁C¼R₄C¼1 s,R₂C¼¹₃s,R₃C¼¹₂s.

Fig. 5-55

101

Waveforms and Signals