Not understanding how the gain works in the 1st stage of an instrumentation amplifier

Question

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.

Answer 1

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.

\$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}}\$

Answer 2

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.

Answer 3

I have added a couple of measurement points for clarity

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}\$