Electrical Engineering Asked by MetrryBoros on August 27, 2020

How can a differential op-amp be configured in a way that it doesn’t amplify but only takes the difference between the signals at the inverting and non inverting input?

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 ...

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...

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":

Answered by Circuit fantasist on August 27, 2020

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 $text{R}_text{f}=text{R}_1$.

Answered by Jan on August 27, 2020

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 $R_2 = R_g$

Answered by Andy aka on August 27, 2020

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.

Answered by Neil_UK on August 27, 2020

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