I'm designing a circuit where I'm using instrumentation amplifier to amplify my input voltage
The formula for the amplifier is $$\frac{V_{\text{out}}}{V_{\text{in}}} = \left(1 + 2\frac{R_1}{R_{\text{gain}}}\right)\frac{R_3}{R_2}$$
In the circuit, there are an overall 7 resistors but only one controls the gain. From what I can find online, to find the gain of the amplifier what they did is they set all 6 resistors equal to each other and hence only one variable is left.
What I don't understand is why did they all equal to each other? What determines the value of the resistors? Why didn't they randomise the resistors and if they did what effect would it have on the amplifier?
Best Answer
R3/R2 must equal R3'/R2' for common-mode rejection. The values are not as important as the ratios, but there are good reasons for making the values the same.
R1, like R3 has some effect on how much current the op-amps must drive, so you probably want them within a certain range depending on the bandwidth, maybe 10K-ish for an amplifier that works at low frequencies, 1/10 or less for high frequencies and 10x or more for lower power consumption.
R1 and R1' could be different values but it will affect where the outputs of the left two op-amps saturate (combination of differential input voltage and common-mode voltage), which may be before the output actually gets to the rails. If the situation was asymmetrical (desired common-mode voltage not balanced wrt the supplies) it might be useful to have them different, but it will probably degrade the AC common-mode rejection.
Having R1 different from R2 would change the gain of the output stage and might have some advantage in bandwidth in some situations, and will also affect the saturation.
Saturation is one of those things you have to consider carefully with an instrumentation amplifier (usually noise is at least 40% worse than a similar simple op-amp too)- here is the permissible CM voltage vs. output voltage for a commercial 3-op-amp instrumentation amplifier: