ProbSolv_Chapter05

ProbSolv_Chapter05 - CHAPTER 5 OPERATIONAL AMPLIFIER List...

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Unformatted text preview: CHAPTER 5 - OPERATIONAL AMPLIFIER List of topics for this chapter : Operational Amplifiers Ideal Operational Amplifier Inverting Amplifier Noninverting Amplifier Summing Amplifier Difference Amplifier Cascaded Operational Amplifier Circuits Operational Amplifier Circuits with PSpice Applications OPERATIONAL AMPLIFIERS Problem 5.1 Calculate Vout for RL = 1 Q, 100 Q, 1 k9, 10 k9, 100 kg , given the 2k!) circuit in Figure 5.1. Figure 5.1 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. The value of the load resistor, RL , is varied over a wide range of values. The dependent voltage source (a voltage-controlled voltage source) has a large gain (100k). Obviously, the goal of the problem is to determine the output voltage in terms of the input (or source) voltage. > Establish a set of ALTERNATIVE solutions and determine the one that promises the . greatest likelihood of success. 77 Vout can be determined using nodal analysis, mesh analysis, or circuit analysis. Because we want to find a voltage, rather than a current, we will use nodal analysis. . ATTEMPT a problem solution. Use nodal analysis to find V in terms of VS and RL. out At the left node or Vin : Vin — Vs + Vin _ 0 Vin _ Vout 0 1k 100k + 2k " At the right node or Vout : Vout - Vin + Vout + 100k Vin Vout — O 0 2k 50 + RL ‘ Simplifying the left node equation, (100)(Vin — VS) + Vin + (50)(Vin — Vow) = O 151Vin ~100Vs — 50Vout = 0 Simplifying the right node equation, (V —Vm>+(40)(v out out 2k +100kVin)+—Vout = 0 RL 2k 41+— Vout +(4M—1)Vin = 0 1{L I [41+~2~k—] - RL V. = -———V in 4M _1 out Substituting the simplified right node equation into the simplified left node equation, _ 2k —(151) 41+— ~—————RL—V 100V 50V —0 4M—1 out 5— out — 2k -(151) 41+? —(50)(4M~1) L (4M_ 1) V0“, _100V5 (100)(4M — 1) 2k -(151)[41+1—{—)— (50)(4M—1) L V S V .2 out Substitute each value for RL into this equation to find Vout in terms of VS . For RL =1 9, Vout = -1.99692283 VS 784 For RL = 100 9, V0“, = —1.99990789V5 For RL = 1 k9, V06 = -1.99993507vs For RL =10 kQ, V... = -1.99993779vs For RL = 100 kQ, V0“, = -1.99993806VS > EVALUATE the solution and check for accuracy. First, the answers appear reasonable with the gain of the entire circuit approaching 2 as RL increases in size. In addition, even for RL : 1 Q , V0ut = 2VS is a good approximation Clearly, using an ideal op amp is reasonable. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. For RL =1 S2, V0“, = -1.99692283 Vs For RL = 100 9, v0“, = -1.99990789 Vs For RL =1kQ, Vout = -1.99993507Vs For RL =10 kg, V0th = -1.99993779Vs For RL = 100 kQ, V... = -1.99993806 VS Problem 5.2 [5.1] The equivalent model of a certain op amp is shown in Figure 5.2. Determine: (a) the input resistance, (b) the output resistance, (0) the voltage gain in dB. 60 Q (a) Rm = 1.5 MQ (b) Rom = 609 (c) A=8><104 79 AdB = 2010g10 (8 x10“) AdB = 98.06 . Problem 5.3 Solve for Vout when RL = 1 Q, 1 kg, 100 kQ, 1 M9 , given the circuit in Figure 5.3. Figure 5.3 Use nodal analysis to find Vout in terms of VS and RL. At the left node or Vin : . — V. — Vin V5 in O _ 0 . +——_———_ 1k 100k At the right node or Vout :. Vom + 100k Vin Vow — o ‘ ——————— + = 0 I 50 RL , Simplifying the left nodal equation, (100)(Vin — VS) + Vin : 0 101Vin =100VS Vin = fl Vs 1 0 1 . . . . . 1 1 Simplifying the right node equation, - + —— Vout + 2k Vin = 0 50 RL [RL+50]V — 2kV SORL out — - in V _[-100kRL]V out — RL +50 in 80 Substituting the simplified left node equation into the simplified right node equation, . -1001<RL 100 -10MRL Vout = _ VS : VS RL + 50 101 (101)(RL + 50) Substitute each value for RL into this equation to find Vout in terms of VS . For RL = 1 Q , Vout = -1,941.3v706VS For RL =1 k9, Vout = -94,295.1438VS For RL : 1'00 k9, Vout = -98,960.4208 VS For RL =1 MQ, Vout = -99,004.9507 VS IDEAL OPERATIONAL AMPLIFIER An ideal op amp has infinite open-loop gain, infinite input resistance, and zero output resistance. Problem 5.4 Looking at the circuit in Figure 5.4, what effect does RL have on the Value . of Vout ? RF Figure 5.4 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. Essentially, we are to determine if the value of RL affects the output voltage in any way. Thus, the goal of the problem is to solve for V0 in terms of the other variables. 81 Treat the operational amplifier as ideal. Due to infinite input resistance, we know that the currents into both input terminals are zero. The voltage across the terminals is negligibly . small or Va = Vb. Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. Because this is a circuit problem, we can use nodal analysis, mesh analysis, or basic circuit analysis. Nodal analysis typically works best for op amp circuits. ATTEMPT a problem solution. Referring to the circuit below, there are three unknown nodes. Write a node equation at node a. The node voltage at node b is already known, Vb = 0 . . Writing a node equation at node c will only introduce an additional unknown. This gives two equations with four unknowns. Solving for V0 in terms of VS (and the resistances) implies that we need one equation with two unknowns. Hence, we need a constraint equation to reduce the number of unknowns. Va ’VS Va _VC Atnodea, T+T+OZO At node b, VI) = 0 The constraint equation comes from a characteristic of the ideal op amp. Va = Vb Thus, ' Va = Vb = 0 Substitute the constraint into the node equation for node a to solve for V0 . _Vs +-Vc 0 R1 RF .— 82 Clearly, Vo = Vc = R F VS 1 We have shown that the value of RL has no effect on the value of V0 , assuming that RL is finite and not equal to zero. EVALUATE the solution and check for accuracy. Consider the following circuit. S a Using Ohm's law, I = From a characteristic of the ideal op amp, Va = Vb But Vb =0 ————-) Va =0 3 1—3 o, — R1 Al I Va -VC __ -VC so, —- RF — RF But Vc = V0 3 I 'V° o, — RF Thus, V5 'V0 V -RF V R1 RF 0 1 S 83 Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. The value of R]: has no effect on the value of V9, assuming that RE is finite and not zero. Problem 5.5 ‘ [5.11] Find v0 and i0 in the circuit in Figure 5.5. ZkQ 8kQ 3V0 4k§2 Figure 5.5 0) (3)=2V Atnodeb, vb=[10+5 Atnodea, 23: 88 ° ——> 12=5Va—Vo But va=vb=2V So, 12 = (5)(2_) — vo vo=-2V . v—v Vo—O_-2—2 -2__05 05 ° 8k 4k ’ 8k +4k""n"m 84 INVERTING AMPLIFIER An inverting amplifier reverses the polarity of the input signal while amplifying it. Problem 5.6 [5.19] Using the circuit in Figure 5.6, calculate Vo if Vs = 0. . Atnodea, 9—va Va—v0 Va-Vb = + 4k 8k 4k _ 18=5Va—Vo—2Vb (1) Atnodeb, Va—Vb_Vb—Vo 4k _ 2k Va=3Vb-2V0 (2) But vb=vs=0 Hence, (2) becomes Va = -2 VO and(1) becomes -18 = -10V0 ——Vo = -llV0 and v0 = -1-.6364 V 85 Problem 5.7 Express V0 in terms of VS for the circuit shown in Figure 5.7. RF Figure 5.7 ' Using nodal analysis, Va _Vs Va—Vo 0 ———————+—= R1 RF where Va = Vb = 0 is the constraint equation. - VS - Vo .._.—— + : R1 7 RF - R V = F V 0 R1 S NONINVERTING AMPLIFIER Problem 5.8 How does the circuit in Figure 5 .8 differ from the circuit in Figure 5.4? Figure 5.8 86 Using nodal analysis, . Va _ 0 Va — Vout + —— = 0 R1 RF where Va = Vb = V5 is the constraint equation. Simplifying, Vs Vs _ Vout 0 .—.— + _.__——_. = R1 RF _1__V {Lg}, RF out_ R1 R]: S V ~ [Bf— +1jv out R] s The significant difference between the two circuits is that the voltage gain for this circuit is , positive. In addition it should be noted that in the circuit of Figure 5.4, the relationship between the output voltage, V2, and the input voltage, V5, is a simple ratio of RE and R1. For the circuit in Figure 5.8, however, the gain can never be less than one. Since there is rarely a case where the gain is less than one, this is not normally a problem. .OW SUMMING AMPLIFIER A summing amplifier combines several inputs and produces an output that is the weighted sum of the inputs. Problem 5.9 [5.33] A four—input summing amplifier has R1 = R2 = R 3 = R 4 = 12 k9. What value of feedback resistor is needed to make it an averaging amplifier? Rf Rf' Rf Rf R In orderfor V0 =‘R-—1V1+R—2V2 +~l{—3V3 +EV4 =fi(V1+V2 +V3 +V4) tbecome v -1(v +v +V +v)~——>1 Rf 0 =—— _=____ ° 4 1 2 3 4 412k 11:23:21: 4 87 Problem 5.10 Express Vout in terms of V1 and V2 for the circuit shown in Figure 5.9. . What have we done here? IOkQ 10k!) Using nodal analysis, a + a + 3 out : 0 10k 10k 10k where Va = Vb = 0 is the constraint equation. - VI — V2 - Vout 10k+10k+10k “0 . Vout = -V1 _ V2 Vout = _(Vl + V2) We have constructed an inverting, summing amplifier. DIFFERENCE AMPLIFIER A difference amplifier amplifies the difference between two inputs but rejects any signals. common to the two inputs. Problem 5.11 Using an operational amplifier, can we construct a circuit where Vout : V2 _ V1 ? ¥_e§, we want to construct What is called a difference amplifier. We can do this using the circuit shown in Figure 5.10. 88 Figure 5.10 Now, verify that this circuit will amplify the difference of the two inputs. Using nodal analysis, va—vl+_\_/_g_—__y_o_ui=0 and vb—v2+vb 0 R1 R2 R 3 R4 where V,i = Vb is the constraint equation. Simplifying, R R “—2— + IJVa _ _2_V1 : Vout and Vb : R4 V2 R1 R1 R3 + R 4 Using the constraint equation to combine the two equations yields Vout= 52.4.1 LV2_B.2.VI R1 R3+R4 R1 01' +R R Vout=_11:_2— : 1/ 2 V2 —&V1 1 +R3/R4 RI R When 5 :31, Vout :——2-(V2 —Vl) R2 R4 R1 Ile =R2 and R3 =R4, V =V2—V1 out This was the desired case. 89 Problem 5.12 [5.39] Design a difference amplifier to have a gain of 2 and a common . mode input resistance of 10 k!) at each input. The input resistances are R1:R3=10k£2 For a gain of 2, R2 {:2 ~——> R2=2R1=20k£2 l A property of difference amplifiers is Thus, ‘ . R4 = R2 = 20 k9 Now, verify the results, _£<1+R1/R2> R2 0— V— v Rl (1+R3/R4) 2 R1 1 20k 1+0.5 20k V0 4 )V, _ 0 V RE 1+0.5 17%“ V0 =2(V2 _V1) This is the desired result. Therefore, R1:R3=10k§2 R2=R4=20k§2 CASCADED OPERATIONAL AMPLIFIER CIRCUITS A cascade connection is a head-to—tail arrangement of two or more op amp circuits such that the output of one op amp circuit is the input to the next op amp circuit. W Problem 5.13 [5. 45] Refer to the circuit in Figure 5.11. Calculate i() if: (a) V5 =12 mV (b) v5 = 10cos(377t) mV 90 . 12kQ 12kg Figure 5.11 This is a cascading system of two inverting amplifiers. (fir—12) -6 Vo — 4 6 Vs — Vs V . _ s = _3 1°_2x103 3x10 VS . (a) When Vs =12 mV, io=36uA (b) When vS = 10cos(377t) mV i0 = 30cos(377t) uA ¢==——-—————————=———-—-———-—-~————-————-—-———————==o OPERATIONAL AMPLIFIER CIRCUITS WITH PSPICE W Problem 5.14 Solve Problem 5.1 using PSpice. PSpice does not perform symbolic simulations. So, let VS = 1 V . Add a VIEWPOINT to the circuit to indicate the output voltage. Set the load resistor to the desired value, save the schematic and simulate. With the repetition of setting the load resistor and simulating the circuit, the output voltage for each load resistor in Problem 5.1 can be verified. 91 R2 —1.99892283V R4 GAIN=1UDI< From the schematic, for RL =1 .Q, Vout = -1.99692283 V By changing the load resistor and simulating the circuit, it can be shown that for RL = 100 9, V0“, = -1.99990789 V for RL =1ko, Vom = -1.99993507 V for RL =10 kQ, Vout = -1.99993779 V for RL = 100 kg, Vow = - 1.99993806 V I This matches the answers obtained in Problem 5.1 when VS = l V . Problem 5.15 Solve Problem 5.7 using PSpice. Consider the following schematic. 92 Because PSpice does not perform symbolic simulations, let VS = 1 V . We also need to choose values for the resistors. For the circuit above, let VS=1V, R1=le, and RF=10kQ which produces, - RF - 10k : V :— :— Vo R1 5 1k (1) 10V This verifies the answer obtained in Problem 5.7. One must realize that Problem 5.7 was performed assuming an ideal op amp; PSpice does not simulate an ideal op amp. Thus, the output voltage may not be an integer value even though the calculations from Problem 5.7 would predict an integer value. Also, the output voltage cannot be greater than V+ or less than V—, where V+ and V— are the power supply voltages of the op amp. There are three modes in which real op amps can operate. The most desirable is to have them give the desired output. The second mode is when the op amp goes into saturation, reaching its maximum output voltage and remaining there. The third mode is that the op amp can act like an oscillator; its output voltage can be some type of periodic signal such as a sine wave. Problem 5.16 Solve Problem 5.8 using PSpice. Consider the following schematic. 93 For the circuit above, VS=1V, R,=1k§2, and RF=10kQ and V —[§5+le —[~1-%+1)1—11V 0— R1 5— 1k ()" This verifies the answer obtained in Problem 5.8 Also, see the comments made concerning ideal versus real op amps in Problem 5.15. APPLICATIONS Problem 5.17 Use an operational amplifier to change an ideal voltage source to an ideal current source. . Consider the following circuit. Clearly, In the circuit above the 0 am will maintain the current throu h the black box in the feedback path at i1. Thus, the op amp is working like an ideal current source. 94 ...
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