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Please see the attached circuit.

I'm trying to show \$-n\lt A_v \lt n\$.

I worked half part successfully:

When \$V_{+}=0\$, the op amp output works to put \$0\text{ V}\$ at \$V_{-}\$. KCL gives:
\$\dfrac{v_{\text{in}}-0}{R} = \dfrac{0-v_{\text{out}}}{nR} \Rightarrow \dfrac{v_{\text{out}}}{v_{\text{in}}} = -n\$


However I'm not able to get the correct answer when \$V_{+} = v_{\text{in}}\$.

Here is my failed attempt:

When \$V_{+} = v_{\text{in}}\$ the op amp output works to put \$v_{\text{in}}\$ at \$V_{-}\$. KCL gives:
\$\dfrac{v_{\text{in}}-v_{\text{in}}}{R} = \dfrac{v_{\text{in}}-v_{\text{out}}}{nR} \Rightarrow \dfrac{v_{\text{out}}}{v_{\text{in}}} = \color{red}{+1}\$.

But the correct answer is \$\color{red}{+n}\$.

Any help?

enter image description here

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  • When the potentiometer is at max:

$$ v_+ = v_- =v_i \rightarrow \frac{v_-}{nR/(n-1)}=\frac{v_o-v_-}{nR}\rightarrow v_o=n\ v_i \rightarrow A_v=n $$

  • When the potentiometer is at min:

$$ v_+ = v_- = 0 \rightarrow \frac{v_i-v_-}{R}=\frac{v_--v_o}{nR}\rightarrow v_o=-n\ v_i \rightarrow A_v=-n $$

Conclusion: \$-n \leq Av \leq n\$.

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When VIN+ receives the full Vin, resistor R has the same voltage on both its legs and so it passes no current and is therefore surplus to requirements and can be removed; it plays no role in this situation.

The gain of the op-amp then becomes: -

$$1 + \dfrac{nR}{\frac{nR}{n-1}} = 1+n-1 = n$$

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