I have been looking all over for derivations of the expression for the differential mode gain of a simple single op-amp differential amplifier. One thing that I have found very interesting is that every derivation uses the superposition principle to find the differential mode gain. Is this the only way to calculate it? And if so, why?

Here is one example derivation that I have looked at: http://www.ntu.edu.sg/home/aschvun/FAQ/DiffAmp.html

  • \$\begingroup\$ No, for sure superpostion is not the only mode, any correct circuit analysis would do. Superposition is only much simpler :) \$\endgroup\$ – carloc Sep 16 '16 at 17:04

You can certainly find it without using the superposition principle.

If this is your opamp as a differential amplifier:


simulate this circuit – Schematic created using CircuitLab

You can always go use the following equation for the output voltage:


Where \$A\$ is the open loop gain (for an ideal opamp \$A\rightarrow\infty\$).

Now you need to find \$V^-\$ and \$V^+\$ in order to plug them in the previous equation and find an equation for the output voltage.

Recall that for an ideal opamp, the input current is zero, therefore for \$V^+\$:

$$ V^+=\frac{R_4}{R_3+R_4}V_1$$

For \$V^-\$,

$$ V^-=\frac{R_2}{R_1+R_2}V_2+\frac{R_1}{R_1+R_2}V_o$$

You can now plug those into the \$V_o=A(V^+-V^-)\$ equation:

$$V_o=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2-\frac{R_1}{R_1+R_2}V_o\bigg) $$

$$V_o\bigg(1+A\frac{R_1}{R_1+R_2}\bigg)=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o\bigg(\frac{R_1+R_2+AR_1}{R_1+R_2}\bigg)=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=A\bigg(\frac{R_1+R_2}{R_1+R_2+AR_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=\frac{A}{A}\bigg(\frac{R_1+R_2}{\frac{R_1+R_2}{A}+R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=\bigg(\frac{R_1+R_2}{\frac{R_1+R_2}{A}+R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

Recall that for an ideal opamp \$A\rightarrow\infty\$, then \$\frac{R_1+R_2}{A}\rightarrow0\$

$$V_o\approx\bigg(\frac{R_1+R_2}{R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

And finally,

$$V_o\approx\bigg(\frac{R_1+R_2}{R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}\bigg)V_1-\bigg(\frac{R_2}{R_1}\bigg)V_2 $$

From there you can express it in terms of the differential input, \$V_d\$, and the common mode, \$V_{cm}\$, just as they do in the link you provided.

EDIT: Finding \$V^-\$

There are several ways you could find this. Let's try KCL:

  • Consider the node at \$V^-\$, there are two currents related to it.
  • Let's assume the direction of those currents is into the node \$V^-\$
  • Then, there is one current that flows out of \$V_2\$, through \$R_1\$ into \$V^-\$ (let's call this \$I_1\$) and another that flows out \$V_o\$, through \$R_2\$ into \$V^-\$ (let's call this \$I_2\$).

Now, you can say, for the node at \$V^-\$:

$$ I_1+I_2=0$$ $$ \frac{V_2-V^-}{R_1}+\frac{V_o-V^-}{R_2}=0$$

That is just KCL. You can solve this for \$V^-\$: $$ \frac{V_2}{R_1}-\frac{V^-}{R_1}+\frac{V_o}{R_2}-\frac{V^-}{R_2}=0$$

$$ \frac{V_2}{R_1}+\frac{V_o}{R_2}=V^-\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)$$ $$ \frac{V_2}{R_1}+\frac{V_o}{R_2}=V^-\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)$$

Which leads to:

$$ V^-=\frac{R_2}{R_1+R_2}V_2+\frac{R_1}{R_1+R_2}V_o$$

  • \$\begingroup\$ How did you get the equation for V-? Could you explain that part a bit more? \$\endgroup\$ – Takide Sep 16 '16 at 13:49
  • \$\begingroup\$ @Takide Check the answer again. I added further explanation. \$\endgroup\$ – Big6 Sep 16 '16 at 15:36

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