Output Voltage Of A Differential Operational Amplifier
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Output Voltage Of A Differential Operational Amplifier

The differential op-amp can be presented in more than one way. Figure 12.1 illustrates one of the ways.
Find the output voltage Vo of figure 12,1 and indicate the mathematical operation performed by the op-amp circuit. Assume the op-amp is an ideal op-amp.

The strings: S7P5A51 (change - physical).

The math:
Pj Problem of Interest is of type change (physical - change).

Consider figure 12.1. If the op-amp is an ideal op-amp then V - V2 = Vd = 0 approximately, V = V2, AOL is infinitely large, and current into op-amp, iin = 0.
Figure 12.1 can be viewed as a combination of a noniverting op-amp and an inverting op-amp.
Consider the ideal noninverting op-amp (first op-amp in figure 12.1):
output voltage = Vo1 = ( 1 + R1/R2)V1
So, V = (1 + R1/R2)V1 - i1R1
But V = V2
So, (1 + R1/R2)V1 - i1R1 = V2--------(1)

Consider the inverting op-amp (second op-amp i figure 12.1):
Current i1, through R1 of the inverting op-amp = current through R2 of the inverting op-amp
So, i1 = (V - Vo)/R2
So, i1 = [(1 + R1/R2)V1 - i1R1 - Vo]/R2
So, i1 = V1/R2 - Vo/(R1 + R2)------- (2)
So, substituting the value of i1 of eq(2) into eq (1), we have:
(R2 + R1)/R2)V1 - [V1/R2 - Vo/(R1 + R2)]R1 = V2
So, V1[((R2 + R1)/R2) - (R1/R2)] - (R1Vo)/(R1 + R2) = V2
So, V1 + (R1Vo)/(R1 + R2) = V2
So, Vo = (1 + R2/R1)(V2 - V1).
So, op-amp is a subtractor.

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