I'm trying to amplify the voltage of my load cell (Wheatstone bridge I believe), but my calculated values are not the same as my experimental values.

The load cell outputs a differential voltage of 0.1mV - 5mV (measured with a voltmeter), and I want to boost it to 0V - 5V (initially, then from 0.3V - 3.3V).

The OP Amps I'm using are MCP6273 "170 μA, 2 MHz Rail-to-Rail Op Amp".


  1. First stage is amplifying the voltage - I put a 2.5KG mass on the load cell, and it output a voltage of 2.5mV differential voltage as expected. However, when I measured the differential voltage between the outputs of the first stage op-amp's (all resistors removed, unity gain) I get a differential voltage of 7.8mV. Why?

  2. Output of second stage is complete wrong. R1 and Rgain are 1Kohm each. R2 is 470 ohm, R3 is 100k ohm. This should give me a gain of 638 (i.e. 63.8mv - 3.19V at the output). However, even with no load (i.e. 0mV) the output is 3.3V. At 2.5KG (i.e. 2.5mV) it hits the 5V rail.

    • I have checked the connections multiple times
    • I have tried simply buffering the input (i.e without R1/Rgain/R1)
    • All the op amps are identical

What am I doing wrong?

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    \$\begingroup\$ Regardless of what you've been taught, you will not be able to build an instrumentation amplifier from parts, the way you did above, that's anywhere near as good in terms of performance as a commercial instrumentation amp, even a low cost option. I recommend analog.com/static/imported-files/data_sheets/AD622.pdf at about $6 \$\endgroup\$ Commented Oct 10, 2014 at 12:59
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    \$\begingroup\$ Are you powering the opamps from a single supply? (Negative voltage rail grounded.) (That could be trouble given the offset voltage as discussed by Spehro.) .... I think it's better (for CMRR reasons) to keep most of the gain in the first stage, and not the differential section. But that's sort of a higher level issue \$\endgroup\$ Commented Oct 10, 2014 at 13:07
  • \$\begingroup\$ @GeorgeHerold Yes, the negative rail is 0V ground. Why is this an issue for offset voltage? \$\endgroup\$
    – tgun926
    Commented Oct 10, 2014 at 20:37
  • \$\begingroup\$ @tgun926, Well I'm not quite sure how load load cell is wired. But if it's biased near ground, then a large offset on the negative opamp (V1 above) could drive it out of range. (near zero load at least) \$\endgroup\$ Commented Oct 11, 2014 at 1:27
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    \$\begingroup\$ @ScottSeidman Thanks for your suggestion. I had a MCN6N11 instrumentation amp with me (didn't have any breakout boards with me, so had to build it on a pcb) which I just wired up - worked like a charm! Shows that those "small" offset voltages really can throw off your results. \$\endgroup\$
    – tgun926
    Commented Oct 11, 2014 at 3:44

2 Answers 2


The offset voltage of each of the op-amps you're using can be as much as +/-3mV at room temperature. So, the difference between two outputs could be as much as 6mV different from the inputs with unity gain. You're seeing 5.4mV which is large, but within specifications and therefore plausible.

Since you don't have much gain in the first stage (only 3) you also have to consider the offset voltage in the second stage. In any case, 638 times your measured differential input offset voltage of 5.4mV + 2.5mV signal is almost 5V.

You can either use better op-amps (such as autozero or 'zero drift' types) or null out the offset voltage by some means (trimpot or reduce the gain and do it digitally). You should also consider the drift of the op-amps you're using which is not guaranteed, but is fairly reasonable typically (+/-1.7uV/K).

  • 1
    \$\begingroup\$ calibration of instrumentation op-amp setups like this could be done by scoping the output and using a pot on the "trim" or "offset" pin if the op-amp package/device has it yes? I've always wondered what the point of those were, but I can clearly see why they would be useful now! Calibration would drift or degrade over time too, so I can also see why technicians might have to re-tune equipment regularly. \$\endgroup\$
    – KyranF
    Commented Oct 10, 2014 at 11:59
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    \$\begingroup\$ @KyranF Yes indeed. Back in the dark ages, I even designed a computerized oven to test a couple hundred op-amps at a time so we could use cheap op-amps for precision use. No need for that these days, just buy what you need. Before that (when I were a wee lad) we even used custom hand-trimmed wirewound resistors instead of trimpots to compensate for offset. What a pain that was, but very stable and little chance of what we call "screwdriver drift". \$\endgroup\$ Commented Oct 10, 2014 at 12:11
  • \$\begingroup\$ MUCH better to simply use a proper instrumentation amplifier rather than build one with op amps. \$\endgroup\$ Commented Oct 10, 2014 at 12:56
  • \$\begingroup\$ Where does the value 5.4mV in your answer come from? Regarding nulling the offsets, is possible with an op-amp such as mine (it doesn't have any offset null pins)? \$\endgroup\$
    – tgun926
    Commented Oct 10, 2014 at 20:40
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    \$\begingroup\$ Must be round- off error due to using a phone calculator ;-) \$\endgroup\$ Commented Oct 11, 2014 at 5:56

Give separate +VCC & -VEE to all OPAMPs. Your requirement is to get 0-5V for 0-5mV input. So gain of instrumentation should be 1000. Calculate the resistor values for 1000 gain of instrumentation amplifier. Use one inverting amplifier at output if getting negative instrumentation output. For 1000 gain, R2=1k, R3=8.2k, Rgain=1k, R1=60k.


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