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A basic instrumentation amplifier looks something like this:

enter image description here

My understanding is that you can feed this circuit with a differential signal while not loading the signal too heavily because of the high input impedance of the op-amp buffers.

But why would I want to use this circuit if I can achieve the same result with a non inverting or inverting configuration with a single buffer.enter image description here Just like this one, having an inverting gain of 100 and having high impedance inputs, while consisting of fewer components. What functionality am I losing when sacrificing a differential configuration over this one.

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    \$\begingroup\$ See this: e2e.ti.com/blogs_/b/analogwire/posts/… \$\endgroup\$
    – Rich S
    Dec 20, 2021 at 23:17
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    \$\begingroup\$ By the way, the input stage (in green) is not a "voltage follower"; it is an amplifier. \$\endgroup\$ Dec 21, 2021 at 12:43
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    \$\begingroup\$ You're trying to re-invent the "two opamp instrumentation amplifier". You're close, but it's a bit more complicated: analog.com/media/en/training-seminars/tutorials/MT-062.pdf \$\endgroup\$
    – John Doty
    Dec 21, 2021 at 22:27
  • \$\begingroup\$ @JohnDoty Could you write an answer which expounds the principles developed in the tutorial? I think such an answer would be worthwhile. \$\endgroup\$ Dec 23, 2021 at 0:34

4 Answers 4

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To start with, you do not have the same |gain| from each input in your circuit, it is +101 from one and -100 from the other. That is a serious issue when it comes to rejecting DC CM voltage changes. I'll add a bit on AC CMRR once I fix your circuit.

Here is the circuit (2 added resistors to correct the gain) compared to the 3 op-amp

enter image description here

You can see that with 1 V common mode noise at (only) 1 kHz the output is about 120 mV so there is not much rejection. That is because of the lack of symmetry - the signal lags from the inverting input because of the extra op-amp.

At close to DC (here 1Hz) the difference is still there, but not nearly as serious.

enter image description here

At mains frequency you'll see in between those two numbers, about 8mV out with 1V 60Hz CM voltage or only about -20 dB rejection.

In reality, if you control or low-pass filter the noise, and don't need the best rejection, either circuit can work for you. For example, here is the simulation with 1 V (a lot) of 60 Hz noise and 20 mV of signal at 10 Hz. As you can see there's not a lot of visible difference.

enter image description here

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Your question is "philosophical" and therefore requires a "philosophical" answer:-) You are actually asking, "What is the idea behind the classic op-amp instrumentation amplifier?" The best way to show it is by reinventing the circuit step by step. Here is my "building scenario"...

Step 1: Single-ended amplifier

It is assumed the two basic circuits of op-amp amplifiers with negative feedback have already been "invented":

  1. Non-inverting amplifier

schematic

simulate this circuit – Schematic created using CircuitLab

  1. Inverting amplifier

schematic

simulate this circuit

How this happened is another (no less interesting) story, but now let's just mention the trick with which this was done - in the basic circuits of a follower (K = 1) and inverter (K = -1) an attenuation is introduced in the negative feedback. Compensating for it, they have become amplifiers.

Step 2: Imperfect differential amplifier

We need a device that amplifies the difference between two input voltages. So we start looking for a second input with the opposite effect in each of the two configurations. We have two options to do so:

  1. In the circuit of the non-inverting amplifier we can insert the second source between the input (left) resistor and ground. Thus we obtain another but inverting input.

schematic

simulate this circuit

  1. In the circuit of the inverting amplifier we can insert the second source between the non-inverting input and ground. Now we obtain another but non-inverting input.

schematic

simulate this circuit

The result is the same "differential" configuration.

Step 3: Ordinary differential amplifier

But there is a problem - the gain of the non-inverting input is one unit higher than the gain of the inverting input (this is also interesting to explain... but let's not be now). We start looking for a way to equalize the gain of the two inputs. We have two options:

  1. We can decrease the gain of the non-inverting input by one through a voltage divider. Thus we get the classic 4-resistor circuit of a simple differential amplifier (shown on the right side of the OP's circuit).

schematic

simulate this circuit

  1. We can increase the gain of the inverting input by one through an additional non-inverting amplifier and we get Spehro's circuit solution.

schematic

simulate this circuit

At the time, they chose the first solution for obvious reasons - not only because it is more beautiful and symmetrical:-) but simply because operational amplifiers were expensive. Let's then follow them...

Step 4: Improved differential amplifier

But the simple 4-resistor differential amplifier has a significant drawback - low input resistance (in addition, it is different for both inputs). The worst thing in this case is that the difference in the internal resistances of the input sources unbalances the circuit.

The first thought that comes to mind to solve this problem, of course, is to put op-amp voltage followers before the circuit inputs.

schematic

simulate this circuit

This solves the input resistance problem... but a new problem arises when we need more significant gain - the ability of the second stage to amplify is limited and we must make the first stage also amplify.

Step 5: Instrumentation amplifier prototype

The straightforward solution is to make the input followers amplify, ie, to turn them into non-inverting amplifiers by inserting voltage dividers between the op-amp outputs and inverting inputs.

schematic

simulate this circuit

However, a new problem arises - the common-mode gain. Simply put, we would like the input stage to amplify the input voltages when they change in the opposite direction and not to amplify them when they change in the same direction with the same rate. How do we do it?

Step 6: Instrumentation amplifier

The famous "long-tailed pair" can help us. For this purpose, we simply combine the two lower resistors Rg/2 of the voltage dividers into one Rg = Rg/2 + Rg/2 (R1 in the OP's circuit).

schematic

simulate this circuit

Now the circuit has a different behavior depending on how we change the input voltages:

  1. Differential mode. When the voltages change in the opposite direction, the so-called "virtual ground" (zero voltage point) appears in the middle point inside Rg (as though it is connected to ground). The gain is maximum since the input op-amp stages act as non-inverting amplifiers (the same as above). You can "see" this "magic" point if you "open" Rg and touch the resistive film by the voltmeter probe... or you can just imagine the voltage distribution along Rg resistive film. I call this picture "voltage diagram" (I have considered in detail this visualization technique in one of my Codidact papers). See, for example, the picture below that I created in 2008 for my Wikibooks story about Ohm's experiment:

Voltage diagram - differential mode

  1. Common mode. If the input voltages change in the same direction with the same rate, the voltages of all op-amp inputs, all points inside Rg (including the middle point) and both op-amp outputs change in the same way. As a result, the input op-amp stages act as voltage followers (K = 1). See another picture of mine extracted from the Wikibooks story mentioned above:

Voltage diagram - common mode

You can see "live" voltage diagrams of these modes in the movie from my personal Google Photos. The computerized laboratory experiment shown in the movie is described in the same Wikibooks story.


This was my story about the famous op-amp instrumentation amplifier. See also my answer to a similar question.

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    \$\begingroup\$ This is a very great writeup going through the different stages and why they are used. \$\endgroup\$
    – Linkyyy
    Apr 27 at 11:36
  • \$\begingroup\$ @Linkyyy, Thanks! Such comments are so rare here ... but we need them for inspiration ... \$\endgroup\$ Apr 28 at 15:18
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In order to avoid CM noise issues using high DM gain and low CM gain is necessary for floating e-field from the lines at say 50V/m to amplify mV signal . Here your CMRR is 0 dB (CM = DM gain=40 dB) instead of 100 dB down or more with an INA. Using 3 Op Amps only gets you <<40 dB with 1% resistors, which will amplify SMPS CM noise if floating.

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When working with differential signals, it's very important to ensure that everything that happens before the final summing stage will affect both inputs equally. The easiest way to accomplish this is often to pass both signals through circuitry that is as close to identical as practical. While a two-op-amp circuit could work as well as a 3-op-amp circuit if the first-stage op-amp on the inverting input behaved in ideal fashion, and non-ideal behavior could be compensated for reasonably well using enough precisely-tweaked resistors and capacitors, a much easier way to compensate for imperfections on the inverting input's first-stage op amp is to put the non-inverting input through an identical op amp. Using resistors and capacitors for compensation would require knowing the exact real-world characteristics of the first stage op amp, but when using the "extra" op amp in the non-inverting input all that would matter is that the real-world characteristics of the front-end op amps are close enough to ideal, and that they match each other.

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