# 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. Commented Jun 5, 2019 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 Commented Jun 5, 2019 at 12:17
• Nice observation sir. I missed that. Commented Jun 5, 2019 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. Commented Jun 5, 2019 at 12:30
• I wrote that (1-x)R is short circuited so that is same sa 0R Commented Jun 5, 2019 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: $0 = \frac{(V_a-V_2)}{R} + \frac{V_{1-} - V_{1+}}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ .. $0 = \frac{(V_a-V_2)}{R} - \frac{V_1}{(1-X)R} + \frac{(V_{1+} - V_-)}{R}$ Commented Jun 5, 2019 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

• Thanks. I'll give it a try Commented Jun 5, 2019 at 16:13