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I've seen a number of different configurations for instrumentation amplifiers, including 2 opamp versions. This

enter image description here

is also one. But it's just a differential amplifier preceded by input buffers. When do you call it an instrumentation amplifier, in other words, what's so special about it that it deserves a separate name?

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    \$\begingroup\$ Nice question. Appreciate zebonaut answer. But, after all it looks like a good excuse for vendor's marketing to push few more opamps on designers with absolutely no real need in majority of apps. \$\endgroup\$ – user924 Jul 15 '11 at 22:20
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"An instrumentation amplifier is a precision differential voltage gain device [...]." One of the important words here is "gain". An OpAmp has infinite gain (in theory) and only gets a defined gain by adding circuitry around it. Usually, when using one OpAmp only, at least one of the inputs loses its extremely high input impedance because external resistors are necessary.

If you need two (differential) inputs with both a very high input impedance and a defined gain, you can use the two-OpAmp-InAmp you are talking about or the three-OpAmp-InAmp-configuration your picture shows. There are also readymade IC InAmps by such companies as Linear Technology or Analog Devices.

The three-OpAmp-InAmp circuit in in the picture of your question shows that two OpAmps are used as buffers, where they still have a high impedance at their otherwise unconnected non-inverting input pins ("+"). By feeding their outputs into another OpAmp, the upper non-inverting input ("+") becomes an inverting input ("-") because it is connected to the 3rd OpAmp's inverting ("-") input. The lower non-inverting input ("+") remains non-inverting due to its connection with the 3rd OpAmp.

Common three-OpAmp-InAmps use a slightly different configuration compared to your picture to set the gain with one resistor only (the external gain resistor in the case of completely integrated InAmps). Please refer to the links I've provided for more details.

With the three-OpAmp-InAmp, you get both a very high input impedance at two differential inputs (while you would get only one input with such a high input impedance with a regular OpAmp buffer) and you get a very good rejection of common-mode signals (that is achievable with one OpAmp, too, but at the cost of lowering the input impedance with the resistors you have to use to turn the OpAmp into a difference amplifier).

The two-OpAmp-InAmp circuit needs less parts, but at the cost of a not-so-good common mode rejection ratio (CMRR).

Here is a link to a very good book about InAmps by Analog's Charles Kitchin and Lew Counts where you can find a more in-depth look onto all these issues.

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  • \$\begingroup\$ How does the setup in Federico's schematic increase CMRR over a simple differential amplifier? \$\endgroup\$ – stevenvh Jun 26 '11 at 12:17
  • \$\begingroup\$ @ stevenvh: It doesn't. I have just tried to clarify my wording to include this piece of information. What the setup does is adding high-Z property to the inputs of the high-CMRR difference amplifier. \$\endgroup\$ – zebonaut Jun 26 '11 at 12:19
  • \$\begingroup\$ @stevenvh: Common mode noise rejection requires that the resistors be matched extremely well. If one uses a monolithic instrumentation amplifier with built-in resistors, the resistance matching may be better than would be readily achievable using discrete amplifiers and resistors. Not to say that devices using discrete parts can't be made very precise, but monolithic parts can inexpensively provide levels of precision that would require a lot of work (and possibly individual-unit calibration) with discrete parts. \$\endgroup\$ – supercat Jun 27 '11 at 15:06
  • \$\begingroup\$ nice find. Also I was suspecting and still suspect that its a bit of a marketing name or trick to sell more opamps \$\endgroup\$ – user924 Jul 15 '11 at 22:18
  • \$\begingroup\$ +1 The book A DESIGNER'S GUIDE TO INSTRUMENTATION AMPLIFIERS (3RD EDITION) by Charles Kitchin and Lew Counts is a very good read. IMHO well worth looking at the first few chapters, even for a 'digital guy'. Thank you. \$\endgroup\$ – gbulmer Sep 6 '14 at 16:29
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I agree with what Zebonaut said, but here are my criteria for being a "instrumentation amplifier" more concisely:

  • The gain must be finite and known. Sometimes the gain is fixed, like 10x or 100x. Other devices allow a selection of preset gains or you provide a resistor or something that sets the gain.

  • The inputs are differential. The common mode rejection is usually very good, generally much better than you could do with a opamp and discrete parts.

  • The inputs are high impedance. You can make a differential amplifier from a single opamp, but then the inputs are no longer high impedance, and one of them is partially driven with the signal on the other.

  • This is all in one itegrated package if it's sold as a "intrumentation amplifier" chip.

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  • \$\begingroup\$ is the last criterion required? This would make my schematic (and many others I encountered) not InAmps. \$\endgroup\$ – Federico Russo Jun 26 '11 at 12:42
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    \$\begingroup\$ @Fedrico Yes that was a bit misleading. You can make a instrumentation amp yourself from other parts. I have edited my post. \$\endgroup\$ – Olin Lathrop Jun 26 '11 at 13:48
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    \$\begingroup\$ Instrumentation amplifiers in chip behave different and may have bipolair/JFET transistor input stages to accomplish high gain. The JFET/bipolair differs whether voltage or current noise is more important. The gain in this part is the highest. The endstage (differential amplifier) would be similar. Thing to note is that monolithic solutions are much more precise: very high gain (up to 10000x can be bought), lower noise, higher CMRR and smaller to lay out on a board. \$\endgroup\$ – Hans Jun 26 '11 at 14:06
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An ideal opamp has a infinite input impedance and infinite amplification. Through feedback you can set the amplification to a realistic level, but this is at the expense of the high input impedance. An instrumentation amplifier (inamp) is a difference amplifier which solves this.
There are several instrumentation amplifier configurations, this one is probably the simplest to understand:

enter image description here

It's a regular differential amplifier with an opamp (the one on the right), with two voltage followers to buffer the inputs, so that they are high impedance. This inamp has an amplification of \$\times\$100 (100k\$\Omega\$/1k\$\Omega\$), and you have to change two resistors to get a different amplification.
Other inamp configurations let you set the amplification with a single resistor.

enter image description here

Here \$R_{GAIN}\$ sets the amplification:

\$V_{OUT} = \left(1 + \dfrac{2 R}{R_{GAIN}}\right) \times (V_2 - V_1) \$

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An instrumentation amplifier is a device which yields an output proportional to the voltage difference between two high-impedance inputs. An op amp is a device which attempts to yield an output which will make the difference between two (typically high-impedance) inputs zero. It is possible to use an op amp and some resistors to build a circuit which will produce an output which is proportional to the difference between two NON-HIGH-IMPEDANCE inputs. It is also possible to use an op amp to produce a low-impedance output which is equal to a high-impedance input. An instrumentation amplifier is conceptually similar to a pair of op amps which convert high-impedance input signals to low impedance outputs, and feeding into a third op amp which then produces an output proportional to the difference between those buffered signals. In practice, things are a little more complicated since the difference-measuring circuits require some precisely matched resistors, and there are some tricks to make them less sensitive to component variations, but conceptually an instrumentation amp is a pair of high-impedance buffers which then drive a difference-measurement circuit.

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