When we have a two-stage instrumentation amplifier, such as the following. enter image description here

Why do we need the first stage of the two Operational Amplifiers? Couldn't we just input V1 and V2 into the differential amplifier?

  • 2
    \$\begingroup\$ Look at the input impedances \$\endgroup\$
    – PlasmaHH
    Commented Sep 9, 2018 at 20:19
  • \$\begingroup\$ Infinite for the first stage, and R2 for second stage, right? \$\endgroup\$
    – Bee
    Commented Sep 9, 2018 at 20:20
  • \$\begingroup\$ Input resistance is one point which is sometimes important but increasing CMRR electronics.stackexchange.com/questions/343096/… is another key benefit \$\endgroup\$
    – carloc
    Commented Sep 9, 2018 at 20:34
  • \$\begingroup\$ No, different input impedences for the two inputs of stage 2 \$\endgroup\$ Commented Sep 9, 2018 at 20:42
  • \$\begingroup\$ If you have any imbalance in capacitance on Vin+ versus on Vin-, the common-mode rejection is in peril. \$\endgroup\$ Commented Sep 9, 2018 at 22:23

4 Answers 4


The 3 op-amp design has three main advantages over a single op-amp differential amplifier.

  1. The input impedance is much higher, since the inputs drive directly into an op-amp input rather than into a resistive divider.
  2. The gain can be set by changing a single resistor, so the critical parts can be easily integrated on to one chip (maximizing symmetry) with a single external resistor for setting the gain.
  3. In high gain configurations the common mode rejection is much better because the gain of the (highly symmetric) first stage effectively multiplies the common mode rejection of the (less symmetric) second stage.

Note that in general it is better to use a specific instrumentation amplifier chip than to try to build it yourself out of separate parts. Having everything on one chip improves symmetry and hence common mode rejection.

  • \$\begingroup\$ Excellent point about using a single IC; even using 1% discrete resistors lowers the theoretical CMRR to no more than 34dB \$\endgroup\$ Commented Sep 10, 2018 at 13:00

One of the biggest benefits of the 3 op amp INA is the equal and high input impedance. The op amp's non-inverting pins' input impedance can be up in the \$T\Omega\$ range. I'll leave it as an exercise for you, but if you look at the difference amplifier circuit, the input impedance of the negative input varies with the positive input.

  • \$\begingroup\$ Why do we want a high input impedance? \$\endgroup\$
    – Bee
    Commented Sep 9, 2018 at 20:29
  • 7
    \$\begingroup\$ @Bee so that we can measure from sources that have a high output impedance \$\endgroup\$
    – BeB00
    Commented Sep 9, 2018 at 20:33

In addition to input impedance concerns, gain in two stages offers better frequency response.

  • 1
    \$\begingroup\$ While this is true in principle. most instrumentation amplifiers have all the gain in the first stage with the second stage having unity gain. \$\endgroup\$ Commented Sep 10, 2018 at 22:59

There is a single stage, high input impedance, differential amplifier.


simulate this circuit – Schematic created using CircuitLab

Without Rg, this has a gain of (f+1). Rg can be used to increase the gain.

However it has performance compromises over the 3 amplifier version.

a) It has less open loop gain, so very high gains are not as stable
b) The two signal paths have different phase shift, so common mode rejection only works to low frequencies. You can improve this a bit with a strategically placed capacitor though
c) At low gains, common mode range is limited by the power supply headroom
d) It's more complicated to draw correctly than the 3 amplifier version. It's only recently that I've created a mnemonic to get the resistors in the right places from memory.

But, if you have only 2 amplifiers left, then it does work. It does allow you control of the gain with a single variable resistor, just like the 3 amplifier version, but unfortunately, just the 3 amp version, this resistor is floating.


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