I'm trying to find $$\frac{V2}{V1}$$ as a function of x. Op amp

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) :

enter image description here

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    \$\begingroup\$ Why is R3 connected to the op-amp output? Study the original circuit carefully. \$\endgroup\$ – Andy aka Jun 5 '19 at 12:13
  • \$\begingroup\$ 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 \$\endgroup\$ – Aleksandar Simonović Jun 5 '19 at 12:17
  • \$\begingroup\$ Nice observation sir. I missed that. \$\endgroup\$ – Aleksandar Simonović Jun 5 '19 at 12:20
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    \$\begingroup\$ 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. \$\endgroup\$ – Transistor Jun 5 '19 at 12:30
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    \$\begingroup\$ I wrote that (1-x)R is short circuited so that is same sa 0R \$\endgroup\$ – 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\$

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    \$\begingroup\$ 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} \$ \$\endgroup\$ – 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

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