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 A_{v}.
**(b)** Find the expression for the voltage gain, A_{v} if the op-amp is ideal.
**(c)** Evaluate both expressions derived in (a) and (b) if R_{1} = 1 kΩ, R_{2} = 10 kΩ, R_{d} = 1 kΩ, and A_{OL} = -10^{4}. What is the percent difference?

**The strings**:
S_{7}P_{5}A_{51} (change - physical+).
**The math**:

Pj Problem of Interest is of type *change* (change-physical).

**(a)** Consider figure 9.1. By KCL, i_{f} = i_{in} + i_{1} -------(1)

So, (v_{o} - v_{1})/R_{2} = v_{d}/R_{d} + v_{1}/R_{1} --------(2)

So, v_{o}/R_{2} - v_{d}/R_{d} = v_{1}(1/R_{1} + 1/R_{2}) -------(3)

From the open loop ideal op-amp:

open loop gain, A_{OL} = v_{o}/v_{d}

Where v_{d} = v_{s} - v_{1}.

So, v_{d} = v_{o}/A_{OL}

And v_{1} = v_{s} + v_{d}

So, substituting these values for v_{d} and v_{1} in eq (3), we have:

v_{o}[1/r_{2} - (R_{1} + R_{2})/(R_{1}R_{2})A_{OL} - 1/A_{OL}R_{d}] =v_{s}(R_{1} + R_{2})/(R_{1}R_{2})-------(4)

Multiply eq (4) by R_{1}R_{2}/v_{s}, we have:

v_{o}/v_{s}[R_{1} - (R_{1} + R_{2})/A_{OL} - (R_{1}R_{2})/(A_{OL}R_{d})] = (R_{1} + R_{2})

So, exact gain A_{v} = v_{o}/v_{s} = (R_{1} + R_{2})/[R_{1} - (R_{1} + R_{2})/A_{OL} - (R_{1}R_{2})/(A_{OL}R_{d})]-------(5)
**(b)** If op-amp is ideal, then i_{in} = 0

So, v_{s} equals v_{1} approximaly, so v_{d} = 0

So, eq (2) becomes:

(v_{o} - v_{s})/R_{2} = v_{s}/R_{1} --------(6)

So, ideal A_{v} = v_{o}/v_{s} = 1 + R_{2}/R_{1} -------(7)
**(c)** Substituting the given values into eq (5) and eq (7) give exact A_{v} = 10.977 and ideal A_{v} = 11.

So, percent difference = 0.21%. So the ideal op-amp model is an ok approximation.

Math

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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