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For the following instrumentation amplifier, I am having some confusion understanding the gain process for the 1st stage. I understand that the 2 input amps are essentially 2 non-inverting amplifiers tied together.

However what causes the gain for V1 and V2 to be the difference between V1 and the common-mode voltage?

**For example if V1 was 2.49V and V2 was 2.51 V, with the common-mode voltage being 2.5 V, if the gain from the 1st stage was 100, output labelled A would be 100*(2.5-2.49)+2.5 and hence it would be 3.5V. At least I believe that is what it is.

Essentially what I do not understand is why the gain of the 1st stage isn't just V1 * 100, but what causes output A. Since output A only multiplies the gain by the difference of the common-mode voltage from V1 or V2. Then the outputs add this value to the common-mode voltage.

Basically what causes the 1st stage (the 2 buffers) to only multiply the difference of the common-mode voltage from V1 by the gain? Then the output (A and B) adds the common-mode signal and this gain calculation. Why isn't it just V1 * Gain of a non-inverting amplifier for output A (like a normal single non inverting amplifier?)

Please include any textbooks or links that might possibly explain this as well as I lack some fundamental understanding for amplifiers.


3 Answers 3


The first stage is essentially differential with ideal op. amps. If you add the same constant voltage to \$V_1\$ and \$V_2\$, \$I_d\$ doesn't change, hence \$V_A - V_B\$ also doesn't change.

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\$I_d=\dfrac{(V_1-V_2)}{R_{gain}} = \dfrac{(V_A-V_1)}{R_{1}} = \dfrac{(V_2-V_B)}{R_{1}}\$

The differential gain expression, is the following:

\$V_A = V_1 + I_d \times R_1\$

\$V_B = V_2 - I_d \times R_1\$

\$V_A - V_B = V_1 + I_d \times R_1 - V_2 + I_d \times R_1\$

\$V_A - V_B = V_1 - V_2 + I_d \times (R_1 + R_1)\$

Substituting \$I_d\$:

\$V_A - V_B = (V_1 - V_2) + \dfrac{(V_1 - V_2)}{R_{gain}} \times (R_1 + R_1)\$

\$\dfrac{(V_A - V_B)}{(V_1 - V_2)} = 1 + \dfrac{2 \times R_1}{R_{gain}}\$

  • \$\begingroup\$ It might be worth mentioning that it would be possible to set the gain of the amplifier while omitting the two r1 resistors and rgain, but it would require having two precisely-matched gain-setting resistors; any mismatch would greatly degrade common-mode rejection. This circuit topology allows gain to be set with only one resistor, which need not be precisely matched with anything. \$\endgroup\$
    – supercat
    Aug 8, 2019 at 21:57

The output of the first stage is differential. You know the voltage across Rg (V1 - V2), so you know the current. The same current must go through each R1, so you can calculate the voltage scross the three series resistors.

  • \$\begingroup\$ so essentially for output A is it half the voltage drop of RG plus R1? \$\endgroup\$
    – Student
    Aug 8, 2019 at 10:54
  • 1
    \$\begingroup\$ Carry the equations through, but solve for Va-Vb. It will be good for your soul. \$\endgroup\$ Aug 8, 2019 at 10:55
  • \$\begingroup\$ why at Rgain is it V1-V2, what is the relationship/reason as to why this occurs \$\endgroup\$
    – Student
    Aug 8, 2019 at 10:58
  • 2
    \$\begingroup\$ For op amps with negative feedback, the two input terminals will be equipotential. \$\endgroup\$ Aug 8, 2019 at 11:00

I have added a couple of measurement points for clarity

3 amp in amp with added locations

The input amplifiers operate as non-inverting devices. Provided the inputs are within the common mode range, then the voltage (by feedback) at point C will be the same as at V1.

Note that for an instrumentation amplifier, the common mode range is a bit more involved.

The voltage at point D will be the same as V2.

Therefore there is a voltage across Rg of V1 - V2 assuming V1 is the more positive input; i.e. the input differential voltage.

Therefore a current flows through Rg of \$ I_{Rg} = \frac {V1 - V2} {Rg}\$.

Assuming no current into the inverting inputs of the amplifiers then this same current flows in both the R1 objects, then the voltage between points A and B must be \$I_{Rg} (2_{R1} + R_g)\$

So the actual voltage at points A and B (expanding back out) is \$(V1-V2)(1+\frac {2_{R1}} {R_g})\$

As the gain is given by \$\frac {V_{AB}} {V_{CD}}\$ then by factoring we yield \$G = 1+ \frac {2_{R1}} {R_g}\$

  • \$\begingroup\$ Thanks but what did you mean when you state common mode range \$\endgroup\$
    – Student
    Aug 8, 2019 at 11:30
  • 1
    \$\begingroup\$ Link for common mode range added \$\endgroup\$ Aug 8, 2019 at 11:32

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