Closed Looped Gain Of A NonInverting Operational Amplifier
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Closed Looped Gain Of A NonInverting Operational Amplifier

Figure 9.1 illustrates a closed looped noninverting op-amp:
(a) Find the exact expression for the voltage gain Av.
(b) Find the expression for the voltage gain, Av if the op-amp is ideal.
(c) Evaluate both expressions derived in (a) and (b) if R1 = 1 kΩ, R2 = 10 kΩ, Rd = 1 kΩ, and AOL = -104. What is the percent difference?

The strings: S7P5A51 (change - physical+).

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

(a) Consider figure 9.1. By KCL, if = iin + i1 -------(1)
So, (vo - v1)/R2 = vd/Rd + v1/R1 --------(2)
So, vo/R2 - vd/Rd = v1(1/R1 + 1/R2) -------(3)
From the open loop ideal op-amp:
open loop gain, AOL = vo/vd
Where vd = vs - v1.
So, vd = vo/AOL
And v1 = vs + vd
So, substituting these values for vd and v1 in eq (3), we have:
vo[1/r2 - (R1 + R2)/(R1R2)AOL - 1/AOLRd] =vs(R1 + R2)/(R1R2)-------(4)
Multiply eq (4) by R1R2/vs, we have:
vo/vs[R1 - (R1 + R2)/AOL - (R1R2)/(AOLRd)] = (R1 + R2)
So, exact gain Av = vo/vs = (R1 + R2)/[R1 - (R1 + R2)/AOL - (R1R2)/(AOLRd)]-------(5)

(b) If op-amp is ideal, then iin = 0
So, vs equals v1 approximaly, so vd = 0
So, eq (2) becomes:
(vo - vs)/R2 = vs/R1 --------(6)
So, ideal Av = vo/vs = 1 + R2/R1 -------(7)

(c) Substituting the given values into eq (5) and eq (7) give exact Av = 10.977 and ideal Av = 11.
So, percent difference = 0.21%. So the ideal op-amp model is an ok approximation.

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