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v2 9=5 v1 9=5
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vo 6=1:2 v2 5 1:5vs 7:5vs v is i1 s A 0:166vs mA 6000 if 0 (b) Rf 40 k
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. From the inverting op amps we get vo 5v2 and v2 9=5 v1 so that vo 9v1 . Apply KCL to the currents leaving node B. v1 vs v1 v1 vo 0 19 1 5 40 Substitute vo 9v1 in (19) and solve for v1 to get v1 vs v2 9=5 v1 1:8vs vo 6=1:2 v2 5 1:8vs 9vs v v1 0 is s 1000 Apply KCL at node B. if i1 v1 v A s A 0:2vs mA 5000 5000
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The current i1 in the 5-k
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input resistor of the rst op amp is provided by the output of the second op amp through the 40-k
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feedback resistor. The current is drawn from vs is, therefore, zero. The input resistance of the circuit is in nite.
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5.11 INTEGRATOR AND DIFFERENTIATOR CIRCUITS Integrator By replacing the feedback resistor in the inverting ampli er of Fig. 5-13 with a capacitor, the basic integrator circuit shown in Fig. 5-23 will result.
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Fig. 5-23
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To obtain the input-output relationship apply KCL at the inverting node: v1 dv C 2 0 R dt and from which 1 RC t v1 dt
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dv2 1 v RC 1 dt (20)
v2
In other words, the output is equal to the integral of the input multiplied by a gain factor of 1=RC.
EXAMPLE 5.17 In Fig. 5-23 let R 1 k
, C 1 mF, and v1 sin 2000t. Assuming v2 0 0, nd v2 for t > 0.
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS t sin 2000t dt 0:5 cos 2000t 1
[CHAP. 5
v2
1 103 10 6
Leaky Integrator The circuit of Fig. 5-24 is called a leaky integrator, as the capacitor voltage is continuously discharged through the feedback resistor Rf . This will result in a reduction in gain jv2 =v1 j and a phase shift in v2 . For further discussion see Section 5.13.
Fig. 5-24
EXAMPLE 5.18 In Fig. 5-24, R1 Rf 1 k
, C 1 mF, and v1 sin 2000t. Find v2 . The inverting node is at zero voltage, and the sum of currents arriving at it is zero. Thus, v1 dv v C 2 2 0 R1 dt Rf 10 3 or v1 10 3 dv2 v2 0 dt 21
dv2 v2 sin 2000t dt
The solution for v2 in (21) is a sinusoidal with the same frequency as that of v1 but di erent amplitude and phase angle, i.e., v2 A cos 2000t B To nd A and B, we substitute v2 and dv2 =dt in (22) into (21). First dv=dt 2000A sin 2000t B . Thus, 22
But Therefore, A
10 3 dv2 =dt v2 2A sin 2000t B A cos 2000t B sin 2000t p 2A sin 2000t B A cos 2000t B A 5 sin 2000t B 26:578 sin 2000t p 5=5 0:447, B 26:578 and v2 0:447 cos 2000t 26:578 23
Integrator-Summer Ampli er A single op amp in an inverting con guration with multiple input lines and a feedback capacitor as shown in Fig. 5-25 can produce the sum of integrals of several functions with desired gains.
EXAMPLE 5.19 Find the output vo in the integrator-summer ampli er of Fig. 5-25, where the circuit has three inputs. Apply KCL at the inverting input of the op amp to get
CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
Fig. 5-25 v1 v v dv 2 3 C o 0 R1 R2 R3 dt  t  v1 v v vo 2 3 dt R2 C R3 C 1 R1 C
24
Initial Condition of Integration The desired initial condition, vo , of the integration can be provided by a reset switch as shown in Fig. 5-26. By momentarily connecting the switch and then disconnecting it at t to , an initial value of vo is established across the capacitor and appears at the output v2 . For t > to , the weighted integral of input is added to the output. 1 t v2 v dt vo 25 RC to 1
Fig. 5-26
Di erentiator By putting an inductor in place of the resistor in the feedback path of an inverting ampli er, the derivative of the input signal is produced at the output. Figure 5-27 shows the resulting di erentiator circuit. To obtain the input-output relationship, apply KCL to currents arriving at the inverting node: t v2 dt 0
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