Subtractor op-amp configuration to find the diference of 2 signals

First story

The best way to reveal the idea of this clever circuit solution, is to show how it was invented (this approach is suitable for explaining every unknown circuit).

• Probably, some tried to convert the humble non-inverting amplifier into a differential amplifier by connecting the second voltage to the unused inverting input. But it turned out the gain of the inverting input is one unit smaller than the gain of the non-inverting input. So they connected an amplifier between the second input source and the inverting input.

• Probably, others tried to convert the humble inverting amplifier into a differential amplifier by connecting the second voltage to the unused non-inverting input. But now it turned out the gain of the non-inverting input is one unit larger than the gain of the inverting input. So they connected an attenuator between the second input source and the non-inverting input.

Of course, the second solution turned out to be simpler ...

Second story

The idea of this clever circuit solution is to equalize the gains of its two inputs since the gain of the non-inverting input is one unit larger than the gain of the inverting input. We can do it in two ways:

• By connecting a non-inverting amplifier before the inverting input (it needs another op-amp and resistors).

• By connecting an attenuator before the non-inverting input (it needs only two resistors).

Of course, we choose the second solution...

Third story

But at some point they realized that the simple 4-resistor solution had a significant drawback - it was influenced by the internal resistance of the input sources. Then they remembered the more complex solution with an additional op-amp... and decided to use it. They called it "instrumentation amplifier":

It is not hard to show that the output voltage of your circuit is:

$$text{V}_text{out}=text{V}_2cdotfrac{text{R}_text{g}}{text{R}_2+text{R}_text{g}}cdotfrac{text{R}_1+text{R}_text{f}}{text{R}_1}-text{V}_1cdotfrac{text{R}_text{f}}{text{R}_1}tag1$$

So, when we make $text{R}:=text{R}_text{f}=text{R}_1$ we get:

$$text{V}_text{out}=text{V}_2cdotfrac{text{R}_text{g}}{text{R}_2+text{R}_text{g}}cdotfrac{text{R}+text{R}}{text{R}}-text{V}_1cdotfrac{text{R}}{text{R}}tag2$$

Simplifying gives:

$$text{V}_text{out}=text{V}_2cdotfrac{2text{R}_text{g}}{text{R}_2+text{R}_text{g}}-text{V}_1tag3$$

So, we need:

$$frac{2text{R}_text{g}}{text{R}_2+text{R}_text{g}}=1spaceLongleftrightarrowspace2text{R}_text{g}=text{R}_2+text{R}_text{g}spaceLongleftrightarrowspacetext{R}_text{g}=text{R}_2:=text{k}tag4$$

So, we get:

$$text{V}_text{out}=text{V}_2cdotfrac{2text{k}}{text{k}+text{k}}-text{V}_1tag5$$

Simplifying gives:

$$text{V}_text{out}=text{V}_2-text{V}_1tag6$$

So, we need $text{R}_text{g}=text{R}_2$ and $text{R}_text{f}=text{R}_1$.

Using the superposition theorem: -

The gain through the inverting "channel" is $dfrac{R_f}{R_1}$

So, make $R_f = R_1$ to get an amplification of minus one.

The gain through the non-inverting "channel" is $dfrac{R_g}{R_2+R_g}cdot (1+dfrac{R_f}{R_1})$

So, if $R_f = R_1$ (as per above) then the non-inverting gain is $dfrac{2cdot R_g}{R_2+R_g}$

Hence for a non-inverting unity gain, make $R_2 = R_g$

We set the gain of a differential amplifier to unity (well, +1 from non-inverting input, -1 from inverting input), so that it 'doesn't amplify', while taking the difference between its inputs. For a conventional diff-amp, this means all four resistors equal.