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In the following problems, ideal op-amps are assumed. This op-amp circuit appears in Dorf's Modern Control Systems, 12th Edition (pg. 136):

negative feedback op-amp circuit

With \$v_{in}\$ on the left and \$v_{o}\$ on the right. Dorf's answer key gives

$$ \frac{v_{o}}{v_{in}} = 1 + \frac{R_{2}}{R_{1}}. $$

How? A different problem gives a similar answer:

also negative feedback op-amp circuit

Where the answer is derived as follows:

$$ v_{o} = A(v_{+} - v_{-}) $$ $$ v_{o} = A(v_{in} - v_{o}\frac{R_2}{R_2 + R_1}) $$ $$ v_{o} = \frac{Av_{in}}{1 + A\frac{R_2}{R_2 + R_1}} $$ $$ \frac{v_{o}}{v_{in}} = \frac{R_2+R_1}{R_2} $$ $$ \frac{v_{o}}{v_{in}} = 1+\frac{R_1}{R_2} $$

because \$A>>1\$.

I don't see how the first circuit yields an answer so similar to the second when its configuration is apparently different.

I'm not even sure how to go about solving circuit one; normally I'd use KCL on the \$v_o\$ node but there's no current through \$R_2\$.

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The transfer function of this cicuit

is simply Vo = Vin.

R1 is connected between the opamp output and ground. It loads the opamp, but doesn't have any effect on the output voltage. The - input is driven from the output via R2. Since this is a ideal opamp, which therefore has infinite input impedance, there is no current thru R2 and therefore no voltage across it.

I suspect this drawing is a mistake, with R1 intended to be between the - input and ground.

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The first circuit is a unity-gain buffer, Vout/Vin=1. The answer key is incorrect. You are correct that there's no current through R2, so feedback will keep the + (Vin) and - nodes equal.

By Ohm's law with zero current through R2 the output will be equal to the (-) node and Vin.

If R1 were connected to the other side of R2, the answer key would be correct, it would be the same circuit as your second example.

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