Just playing with circuit theory and tried to derive the equation for the differential amplifier:

enter image description here

$$ V_{out} = \frac{R_f}{R_1}(V_2 - V_1) \\ R_f/R_1 = R_g/R_2 $$

I've seen the solutions based on superposition and based on virtual ground, but I wanted to derive using the ideal op amp equation:

$$ V_{out} = A(V_+ - V_-) \\ V_+ = \frac{R_g}{R_2+R_g}V_2 \\ V_- = \frac{R_1}{R_1+R_f}(V_{out} - V_1) \\ $$

So I went:

$$ \frac{V_{out}}{A} = \frac{R_g}{R_2+R_g}V_2 - \frac{R_1}{R_1+R_f}(V_{out} - V_1) \\ \frac{V_{out}}{A} + \frac{R_1}{R_1+R_f}V_{out}= \frac{R_g}{R_2+R_g}V_2 + \frac{R_1}{R_1+R_f}V_1 $$

Here, A goes to infinity:

$$ \frac{R_1}{R_1+R_f}V_{out}= \frac{R_g}{R_2+R_g}V_2 + \frac{R_1}{R_1+R_f}V_1 \\ V_{out}= \frac{R_1+R_f}{R_1}\frac{R_g}{R_2+R_g}V_2 + V_1 $$

and it's now obvious I've gone wrong somewhere because the equation doesn't look like the expected answer above. There should be a subtraction between V2 and V1, but I have addition instead.

I've tried this style of derivation on a number of op-amp configurations and it seems to work. This derivation of this differential amp should be possible using just the op-amp's equations. Where did I go wrong?

  • \$\begingroup\$ Correct me if I'm wrong, but I think your equation for V- might be incorrect. Using the voltage divider equation is one approach, but you need to be careful where "ground" is referenced in this case because Vout is not = 0V, so the middle of the divider is not referenced to 0V in this case. -- Try re-writing it starting with V- = V1 - (I * R1) where I is the current through R1 and RF. \$\endgroup\$
    – Matt Egan
    Dec 14, 2020 at 23:50
  • \$\begingroup\$ @MattEgan you hit submit before I did. If you want to answer, I will delete mine. \$\endgroup\$ Dec 14, 2020 at 23:57
  • \$\begingroup\$ Yeah, that's it. Sorry guys, I had forgotten the voltage divider is always w.r.t. ground. I thought it was just w.r.t. the voltage difference. \$\endgroup\$
    – Roxy
    Dec 15, 2020 at 0:03
  • \$\begingroup\$ No worries @MathKeepsMeBusy! I didn't have time to write a full answer so was just leaving a comment :) \$\endgroup\$
    – Matt Egan
    Dec 15, 2020 at 17:40

3 Answers 3


I think your equation for \$V_-\$ is incorrect. If \$V_{out}=V_1\$ then \$V_-\$ should equal \$V_{out}\$. However, the equation you have gives 0.

  • 1
    \$\begingroup\$ Yep, should've been \$V_- = \frac{R_1}{R_1+R_f}(V_{out} - V_1) + V_1\$. Thanks! \$\endgroup\$
    – Roxy
    Dec 15, 2020 at 0:18

The equations for voltages at the inverting and non-inverting terminals can be derived using nodal analysis and voltage divider rule respectively.



Suppose that there are \$n\$ voltage sources passing through \$n\$ resistors and meeting up in one point (\$x\$).


simulate this circuit – Schematic created using CircuitLab

The voltage in that point, \$V_x\$ can be expressed as:

$$ V_x = \dfrac{1}{\sum_\limits{i=1}^{n} \dfrac{1}{R_i}} \cdot \left[ \sum_\limits{i=1}^{n} \dfrac{V_i}{R_i} \right] $$

Therefore, the voltage on the inverting input pin is:

$$ V_- = \dfrac{R_1 R_f}{R_1 + R_f} \left[ \dfrac{V_1}{R_1} + \dfrac{V_o}{R_f} \right] = \dfrac{R_f V_1 + R_1 V_o}{R_f + R_1} $$

I am copy pasting \$V_+\$ from your message:

$$ V_+ = \frac{R_g}{R_2+R_g}V_2 $$

Put these two expression for \$V_-\$ and \$V_+\$ into \$V_o = A \left( V_+ - V_- \right)\$ to find:

$$ \dfrac{R_f + R_1}{A R_1} V_o + V_o = \dfrac{R_g}{R_1} \dfrac{R_f + R_1}{R_2 + R_g} V_2 - \dfrac{R_f}{R_1} V_1 $$


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