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I solved the difference receiver circuit problem in the picture with the following formula but I am not sure of its correctness because I do not have an answer key. I would like to discuss whether it is right or wrong, or are the formulae I use correct?

$$ \begin{array}\\ V_o &= \frac {RF} {R1} \cdot (V2 - V1)\\[.5em] V_o &= \frac 1 1 \cdot (3 - 5)\\[.5em] V_o &= -2\,V \end{array} $$ Is this solution right? If it is wrong, what is the path I should follow?

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    \$\begingroup\$ To check whether you have used the correct formula, compute the voltage at the opamp's input (R1 and RF form a very simple voltage divider, so this is easy), and compare it against the voltage at the other input. \$\endgroup\$
    – CL.
    Commented Jun 16 at 15:37
  • \$\begingroup\$ I agree. In this case it's better to give insigths to solve rather than complete response. \$\endgroup\$ Commented Jun 16 at 17:41
  • \$\begingroup\$ so RF/F1 * (V1 - V2) is correct as I understand it? \$\endgroup\$ Commented Jun 16 at 19:18
  • \$\begingroup\$ @softcalculate You know that the (-) input of the opamp should be the same as the voltage at the (+) input. If so, what current would be present in R1 and what direction is the arrow of that current? Whatever you calculate for that must also be the same current (and direction) found in RF. If +5 is on the left of R1 and +3 is on the right of R1, and if the same left-right sign arrangement must be true for RF, what's the magnitude difference across RF and what does that mean for VO? \$\endgroup\$ Commented Jun 17 at 5:38

2 Answers 2

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You can apply super position theorem (SPST),for solving these kind of circuits.

Please see the answer below.

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  • \$\begingroup\$ I am grateful to you for your detailed explanation, thank you very much. it is a really academic explanation \$\endgroup\$ Commented Jun 18 at 7:36
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Your equation is incorrect, because it assumes the output to be referenced to zero volts (ground). In fact, the output will be offset from 0V by an amount dependent on V2, and will not be the simple difference described by your equation.

Here's the circuit with some annotations to help:

schematic

simulate this circuit – Schematic created using CircuitLab

The first rule of op-amps is that when negative feedback is present, the non-inverting and inverting inputs of the op-amp are equal in potential:

$$ V_Q = V_B $$

Then you consider the potential divider formed by R1 and R2, which relates \$V_Q\$, \$V_A\$ and \$V_{OUT}\$. The divider is easier to visualise like this:

schematic

simulate this circuit

I won't derive that relationship here, I'll let you do that. I will say, however, that there is an equation containing only the three variables \$V_Q\$, \$V_A\$ and \$V_{OUT}\$, (and the constants R1 and R2 of course).

Clearly, once you have an equation of that relationship, you can substitute \$V_Q=V_B\$, to obtain the correct equation in terms of \$V_A\$, \$V_B\$ and \$V_{OUT}\$, which is what you are looking for.

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