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On this webpage I read an explanation of an instrumentation amplifier:

enter image description here

The negative feedback of the top op-amp causes the voltage at Va to be equal to the input voltage V1. Likewise, the voltage at Vb is equal to the value of V2.

And the input amplifiers are marked "Voltage follower" to illustrate this, but I wonder if this is correct. Part of Vb is added, so Va can't be equal to V1, can it?

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    \$\begingroup\$ If R1 didn't exist, the description would be correct. \$\endgroup\$ – endolith Aug 16 '11 at 13:37
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You're right, this is wrong. If the output of the buffer amplifier would be equal to \$V_1\$ then there wouldn't be any voltage difference over R2, because the opamp will keep the inverting input at \$V_1\$ as well. So there wouldn't be any current through either R2. Yet there's a voltage difference \$V_2\$ - \$V_1\$ between both inverting inputs, causing a current through R1. Since no current flows in the opamp's inputs this would be a violation of KCL (Kirchhoff's Current Law).
So, the buffer amplifiers are not voltage followers and \$V_a\$ is not equal to \$V_1\$. The equation further on the page also mentions the factor \$\dfrac{R4}{R3}\$ a bit too early. It should read

\$\dfrac{V_b - V_a}{V_2 - V_1} = 1 + \dfrac{2 R2}{R1}\$

This follows from the current flowing between \$V_a\$ and \$V_b\$:

\$I = \dfrac{V_b - V_a}{R2 + R1 + R2}\$

and since the same current flows through R1:

\$I = \dfrac{V_2 - V_1}{R1}\$

hence

\$\dfrac{V_b - V_a}{R1 + 2 R2} = \dfrac{V_2 - V_1}{R1}\$

The factor \$\dfrac{R4}{R3}\$ is the amplification by the differential amplifier stage.

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Wrong: Yes, the explanation is incorrect.
(The statement is correct in the special case where Rgain is absent and input stages have unity gain. See next paragraph).

More wrong: In addition, the use of the term "buffer" is very misleading. The term "buffer" usually implies that a stage has unity gain, whereas the input stages usually have gain. In the special case where Rgain is open circuit (ie not used) then the input stages are indeed unity gain buffers and R1a and R1b are superfluous (ie could be short circuited). In cases where Rgain is present the input stages have gain.

Referring to your diagram

enter image description here

Correct statements would be that

  • the voltage at the left hand end of R2 upper = V1 and
  • the voltage at the left hand end of R2 lower = V2.

All following circuit references will be to the diagram below.

enter image description here <-- Small version - higher res version below

ie Due to the opamp providing negative feedback to drive the difference between its inputs to zero, (V1 = Vx) and (V2 = Vy)

A much better way of drawing the circuit - The amplifier can be much more easily understood intuitively if it is redrawn as shown below (and in smaller size version immediately above). This is done in some instances but usually the version in the original query is used. This is such a good way of showing it that it is strange that it is not used more often. Consider the diagram below.

  • The top of Rgain (Vx) is at voltage V1

  • The bottom of Rgain (Vy) is at voltage V2

  • The voltage across Rgain = Vx-Vy = (V1-V2) = Vin

Now the magic.

  • It can be seen 'by inspection' that the same current flows from Vw to Vx, then from Vx to Vy and then from Vy to Vz (or A1_out through R1a, then through Rgain and then through R1b to A2_out).

  • SO the voltages across R1a, Rgain & R1b must be proportional to their resistances (as they all carry the same current). So, for example, if R1 = 7 x Rgain it must have seven times the voltage across it that Rgain has.

  • BUT the voltage across Rgain = Vin = V1-V2 from above.

Use R1 for R1a or R1b as result is symmetrical.

  • So the voltage across R1 = Vin x R1/Rgain

  • So gain of 1st stage = (R1 + Rgain + R1 ) / Rgain
    = (2xR1 + Rgain)/Rgain or = 2xR1/Rgain + 1
    Which is the classic instrumentation amp gain expression. The gain of the input stage may be altered simply by altering Rgain.

The output stage is a standard differential amplifier with stage gain = R3/R2

Overall gain = (2 x R1/Rgain + 1) x R3/R2

enter image description here

The above explanation is mine but the marvellous redrawing of the standard circuit comes from Wikipedia's Instrumentation amplifier page. Look there for their explanation of the same circuit and a lot more.

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  • \$\begingroup\$ IMO a buffer doesn't need to have unity gain. High \$Z_{IN}\$ and low \$Z_{OUT}\$ are sufficient for me. And then "buffer amplifier" doesn't have to be a contradictio in terminis either. But "voltage follower" is definitely wrong here. \$\endgroup\$ – stevenvh Aug 16 '11 at 13:36
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The two input op-amps are acting as buffer amplifiers with gain. While the circuit does increase the gain equally for the differential input stages, it also increases the gain for common-mode signals. As the single gain resistor, R1, is connected between the summing junctions of the two input buffers, the full differential input voltage will appear across R1 because the voltage at the summing junction of each amplifier is equal to the voltage applied to their positive non-inverting input. Therefore, the amplified input voltages at the outputs of op-amp1 and op-amp2 will appear differentially across the three resistor network of, R2, R1, and R2 across the two op-amp outputs.

The voltage across R1 equals Vin, therefore the current through R1 will equal (Vin/R1). Op-amp1 and op-amp2 will therefore operate with gain and amplify the input signal. However, if a common-mode voltage is applied to the amplifiers inputs, the voltages on each side of R1 will be equal, and no current will flow through this resistor. Since no current flows through R1 (nor, therefore, through both R2 resistors, amplifiers 1 and 2 will operate as unity-gain followers (buffers). Therefore, common mode signals will be passed through the input buffers at unity gain, but differential voltages will be amplified by the factor (1 + (2 R2/R1)) as indicated below by Russell McMahon. The overall gain can be further increased by the ratio of R4/R3 set up by the differential amplifier, op-amp3. Then the transfer function correctly given in my tutorial will be:

  • Vout = V2 - V1 (1 + 2R2/R1).(R4/R3)

Also the 3 Op-amp instrumentation schematic presented by Russel McMahon as part of his Wikipedia explanation also matches exactly my own schematic representation of a high impedance buffered instrumentation amplifier. on the www.electronics-tutorials.ws site.

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  • \$\begingroup\$ While it may be of interest that you're the webmaster for the discussed page, I removed your signature because sigs are against the rules. If you want to publish contact information please do so on your profile page. \$\endgroup\$ – stevenvh Aug 16 '11 at 14:41
  • \$\begingroup\$ Your first sentence already contradicts the quoted text from the web page, which says that \$V_a = V_1\$. \$\endgroup\$ – stevenvh Aug 16 '11 at 16:52
  • \$\begingroup\$ @Wayne Storr - It appears that this webpage may be the source of the original diagram and query. The text statements on the webpage re the voltages at Va & Vb are incorrect as they stand BUT they are effectively a typo as the first formula below the diagram makes it clear that the voltage across R1 = Vin, and that R2upper and R2lower do have voltage drops. All that is required to make the webpage correct is a minor correction to bring the typo'd text into line with the correct formulas. \$\endgroup\$ – Russell McMahon Aug 16 '11 at 23:31

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