Comments on: How to Derive the Instrumentation Amplifier Transfer Function

By: luke

luke — Sat, 19 Nov 2011 00:43:33 +0000

Thank you. I was looking at the same thing. This clarifies. Great article by the way.

By: Adrian S. Nastase

Adrian S. Nastase — Tue, 15 Nov 2011 05:13:27 +0000

The notations are just a convention. Your U3 being turned upside down, is the same as saying “let’s call the upper transistors R3 and R4 and the lower transistors R1 and R2, and let’s switch V11 and V12 labels between them”. The resistor ratio is the same, since R4/R3 = R2/R1. Because we switched V11 and V12, then, yes, Vout1 = R2/R1 (V12-V11).

By: Gilbert

Gilbert — Mon, 14 Nov 2011 08:13:06 +0000

Hi, if U3 is up side down, means R4 connects to ground and R2 connects to Vout and U3 has the opposite sign. Will all the equation be not changed?
for example, will the equation 2 become Vout1=R2/R1(V12-V11)?
Thank you

By: Adrian S. Nastase

Adrian S. Nastase — Sun, 10 Jul 2011 00:06:48 +0000

Yes, it will be zero. But nothing is a perfect zero in this Universe. You will still have a few millivolts at the amplifier output due to offset, or due to V1 and V2 not being perfectly equal.

By: Robert

Robert — Fri, 08 Jul 2011 10:04:43 +0000

Hi,
If input voltages V1 and V2 are the same, does it mean that output voltage equals zero volt?

By: Adrian S. Nastase

Adrian S. Nastase — Thu, 17 Feb 2011 05:57:17 +0000

Grant, the two equations are identical, if R1 = R3 and R2 = R4 as stated two paragraphs above.

Equation (2) in this article is Vout1 = R2/R1 *(V11-V12).
Equation (1) in How to Derive the Differential Amplifier Transfer Function is Vout = V1 * R2/(R1+R2) * (1+R4/R3) – V2 * R4/R3.

If R1 = R3 and R2 = R4 then
R2/(R1+R2) * (1+R4/R3) = R2/(R1+R2) * (1+R2/R1) = R2/R1, and
R4/R3 = R2/R1

The inputs of the differential amplifier, which is the instrumentation amplifier output stage, are V11 instead of V1 and V12 instead of V2. We also note Vout with Vout1. Therefore, from the differential amplifier transfer function, as applied to the instrumentation amplifier output stage we get

Vout1 = V11 * R2/(R1+R2) * (1+R4/R3) – V12 * R4/R3 = V11 * R2/R1 – V12 * R2/R1 = R2/R1 * (V11 – V12),

exactly as shown in equation (2).

By: Grant

Grant — Thu, 17 Feb 2011 04:43:20 +0000

I looked at the derivation for the transfer function of the differential amplifier, as linked, but the transfer function proven on that page looks nothing like equation 2. How did you derive equation 2 of this page from the differential amplifier’s transfer function?

By: Adrian S. Nastase

Adrian S. Nastase — Tue, 14 Dec 2010 18:48:52 +0000

Arun, here are the calculations:

First, factorize V1*(1+R5/RG),
Vout1 = (R2/R1)*(V1*(1+R5/RG)*(1+R6/(R5+RG)))

Then, introduce 1 in each fraction,
Vout1 = (R2/R1)*(V1*(RG+R5)/RG*(R5+RG+R6)/(R5+RG))

Simplify RG+R5
Vout1 = (R2/R1)*V1*(R5+RG+R6)/RG

And, because R5=R6,
Vout1 = (R2/R1)*V1*(RG+2R5)/RG

Distribute RG, and this is the final result:
Vout1 = V1*(R2/R1)*(1+2R5/RG)

Let me know if you need anything else.

By: arun

arun — Tue, 14 Dec 2010 09:46:18 +0000

you did not solve equation number 6.how did u obtain equation 7 after solving equation 6