# How does this op amp circuit with adjustable gain work?

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?

• 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\$$.

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: -

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