# Operational Amplifier Problem

I'm trying to find $$\frac{V2}{V1}$$ as a function of x. I found this circuit but noone gave good explanation. I think I understand the way this potentiometer works, but I just can't work out the equations to get result. Also I tried to simulate this circuit and here is result (for simulation I assumed x=1) : • Why is R3 connected to the op-amp output? Study the original circuit carefully. – Andy aka Jun 5 '19 at 12:13
• Because Potentiometer R could be viewed as 2 resistors xR and (1-x)R . Since 1 of them is short circuited, there is just resistor xR left. Since I assumed x=1 then xR=R and in simulation that resistance is named R3 – Aleksandar Simonović Jun 5 '19 at 12:17
• Nice observation sir. I missed that. – Aleksandar Simonović Jun 5 '19 at 12:20
• No, The pot can't be viewed as xR and (1 - x)R since the wiper is connected to one end. It can only be considered as xR and 0R. The dot at the bottom of R3 should not be there. – Transistor Jun 5 '19 at 12:30
• I wrote that (1-x)R is short circuited so that is same sa 0R – Aleksandar Simonović Jun 5 '19 at 12:47

This is easier than it looks. I won't solve it all the way, but lets call the voltage at the top of the pot $$\V_a\$$ and the voltage at the bottom of the pot, $$\V_b\$$. Lets call the positive input $$\V_{1+}\$$ and the negative, $$\V_{1+}\$$. The input terminals of the op amp are $$\V_-\$$ and $$\V_+\$$.

Using the rule that no current enters the op amp, it is straightforward to show that $$\V_a=2V_- - V_{1+}\$$. Using the rule the $$\V_- = V_+\$$, it is straightforward to show that $$\V_b = 2V_- - V_{1-}\$$. Now, we know the voltage at either side of the pot, and the value of the pot, so $$\i_{pot} = \frac{(V_a-V_b)}{(1-X)R} = \frac{V_{1-} - V_{1+}}{(1-X)R}\$$

Application of Kirchoff's current law to the node at the top of the pot should give you $$\V_2\$$

• Kirchhoff's current law to node: $0 = \frac{(V_a-V_2)}{R} + \frac{V_{1-} - V_{1+}}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ .. $0 = \frac{(V_a-V_2)}{R} - \frac{V_1}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ – Aleksandar Simonović Jun 5 '19 at 18:31

Well, I think it is a tricky circuit - and it is not a simple task to derive the gain formula. Therefore, I recommend to split the calculation in separate steps:

(1) Set the lower input (we call it vi2) to zero vi2=0 and find the inverting gain as a function of the upper voltage (vi1): Vout1=f(vi1). This is relatively simple by applying the star-to-triangle conversion method for the negative feedback loop.

(2) Set the upper input to zero (vi1=0) and find the non-inv. gain as a function of the lower voltage vi2: Vout2=f(vi2). However, in this case, we have to face a negative as well as a positive feedback loop. Hence, the total feedback factor Hf again must be calculated in two sepate steps: Hf=Hf1+Hf2 (sum of both feedback factors, one of which must have a negative sign).

(3) Vout=Vout1+Vout2 with V1=Vi1-Vi2