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I'm trying to build a Instrumentation amplifier that would do a 10uV -> 10mV amplification. The motivation is to measure uA currents on a small enough shunt resistor (1-10Ohm).

For a proof of concept, I've built a circuit like the one below (ref.) with a distinction that I'm powering it with two 9V batteries and am using LM2902N as the op amps and Rg is somewhat different. It has much less that the required 1000x gain.

enter image description here

However, when measuring the results for different input voltages, the gain seems to vary:

enter image description here

I was expecting it to be constant.

  • What is a reasonable place to look for the causes of the variations?
    • Cheap resistors?
    • Inadequate op amp?
    • Breadboard prototyping?
    • 2x9V batteries as power supply?
    • other?
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  • \$\begingroup\$ 2 possibilities for falling "gain" with increasing input : (1) you're clipping (output is hitting a supply rail, or both. Unlikely with gain=10, and 5mv in). (2) You're making DC measurements, and confusing DC offsets with input signal. As DC offsets are likely to be around 5mv with low grade opamps, that's my bet. What bandwidth do you need, and have you looked at chopper stabilised opamps to eliminate DC offset? @Andyaka has one he recommends for this... \$\endgroup\$ – Brian Drummond Oct 6 '16 at 10:15
  • \$\begingroup\$ Here's the opamp I was thinking of ... analog.com/en/products/amplifiers/operational-amplifiers/… described in this Q&A electronics.stackexchange.com/questions/141162/… \$\endgroup\$ – Brian Drummond Oct 6 '16 at 10:24
  • \$\begingroup\$ @BrianDrummond Can you please elaborate in an answer how a constant DC offset would influence the gain difference? Why wouldn't it manifest itself as a constant error. \$\endgroup\$ – TheMeaningfulEngineer Oct 6 '16 at 10:25
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    \$\begingroup\$ It's rather better to plot output voltage versus input than gain versus input voltage, it gives us more information about what might be going wrong! \$\endgroup\$ – Neil_UK Oct 6 '16 at 10:38
  • \$\begingroup\$ Out of curiosity, why not just use an instrumentation amplifier IC instead of building one out of op amps and discrete resistors? Since the LM2902N is produced by TI, here's a link to the instrumentation amplifiers offered by TI: ti.com/lsds/ti/amplifiers/instrumentation-amplifiers/… \$\endgroup\$ – Null Oct 6 '16 at 14:34
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There are 2 obvious possibilities for falling "gain" with increasing input level :

  1. you're clipping (output is hitting either supply rail, or both. Unlikely with gain=10, and 5mv in).
  2. You're making DC measurements, and confusing DC offsets with input signal. As DC offsets are likely to be around 5mv with low grade opamps, that's my bet.

What you expect:
Vout = Vin * gain
And you are calculating gain as Gain = Vout/Vin

What you get:
Vout = (Vin + Voffset) * Gain.

So if Voffset = 5mv and gain = 5,
Vin = 5mv : Vout = (5mv + 5mv) * 5 = 50mv.
And you compute gain as Vout/Vin = 10.

Vin = 25mv : Vout = (25mv + 5mv) * 5 = 150mv
And you compute gain as Vout/Vin = 6.

This is a serious issue : it looks like you are attempting to amplify microvolts of DC with a device with millivolts of input offset voltage. This offset voltage is an error signal that will swamp the signal of interest.

What bandwidth do you need, and have you looked at chopper stabilised opamps to eliminate DC offset?

One opamp that may be useful to you is the ADA4528 which is further described in this Q&A which claims an offset voltage of only 2.5 uV.

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  • \$\begingroup\$ What bandwidth do you need It's for a variable signal from a power supply which fluctuates depending on the requirement of the load (uC). Don't know how to convert that to the minimal required frequency. \$\endgroup\$ – TheMeaningfulEngineer Oct 6 '16 at 21:45
  • \$\begingroup\$ Then the bandwidth you need depends how well you decouple the supply; you can control that to suit the amplifier you choose. \$\endgroup\$ – Brian Drummond Oct 6 '16 at 22:10

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