I am using an MCP6N11-100 to amplify the signal out of a 2 mV/V load cell with a excitation voltage of 10 V, so the maximum output (differential) would be 20 mV. The load cell is rated for 4000 kg.

I am using the instrumentation amplifier in the following configuration. Note that at the inputs I have the outputs of the load cell connected directly. Rf = 9.9 kΩ and Rg = 55 Ω giving a gain of ~181 (both resistor values are measured). VREF is grounded.

My problem is that the gain doesn't seem to be 181 until there is sufficient difference between the inputs of the instrumentation amplifier. Suppose the difference is just 0.5 mV. The op-amp outputs 200 mV giving a gain of almost 200! However, as the difference is increased to 2 mV the output now is 366 mV to 368 mV - matching the gain much more accurately (183 vs 200). Why is this discrepancy present at low levels of input? My guess is that it's the offset voltage. The datasheet suggests that the MCP6N11-100 has a maximum Vos of 0.35 mV. Could this be the cause?

Enter image description here

  • \$\begingroup\$ 0.5 to 200 is a gain of 400, not 200. \$\endgroup\$
    – Kaz
    Commented Jul 8, 2013 at 15:54
  • 1
    \$\begingroup\$ The guess that it might be an offset requires an important clue: what is the output voltage if the inputs are tied together. \$\endgroup\$
    – Kaz
    Commented Jul 8, 2013 at 16:29
  • \$\begingroup\$ When you did the experiment with 0.5 mV applied, how did you measure that input voltage to know it was exactly 0.5 mV? \$\endgroup\$
    – The Photon
    Commented Jul 8, 2013 at 16:47
  • \$\begingroup\$ Voltage driven bridges are progressively non-linear too. Current driven bridges have about half the non-linearity. I'm pretty sure Kaz has nailed the problem though. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2013 at 16:49
  • \$\begingroup\$ @Kaz This is from memory, but I think it was 2.2mV. I'll have to take a second look at work tomorrow. \$\endgroup\$
    – Saad
    Commented Jul 8, 2013 at 16:55

1 Answer 1


Why don't you get an ac signal source, attenuate it 500:1 with a 499 ohm resistor and a 1 ohm resistor and insert this into one grounded element of the bridge. DC offset introduced can be countered with another 1 ohm resistor in the bottom leg of the other grounded element.

enter image description here

Also shown is a method that doesn't break-into the bridge - it applies a high value resistor across one of the grounded bridge elements in order to simulate a shift in bridge output. This can be controlled with a fet fed with a squarewave as shown. It's even possible to use an opto-coupler in this type of application should the return line of the bridge be very sensitive.

Now you can inject (say) 1kHz at a known measurable amplitude and see what comes out from the instrumentation amplifier. If the AC gain is fine then you have to look at the dc issues that may be causing your problem BUT until you rid yourself of the doubts about ac gain you might keep going round in circles.

  • \$\begingroup\$ I don't have access to the bridge. The Load Cell is pretty much sealed. \$\endgroup\$
    – Saad
    Commented Jul 8, 2013 at 17:30
  • \$\begingroup\$ @Saad, in that case you could ac couple the same attenuated signal to one of the sense wires from ground. Use a capacitor that has much lower impedance than the bridge elements at the frequency you use. Even if not you can calculate what effect the bridge components have. 10uF and a bridge of 1kohm will attenuate the cap output by about 1% which is enough to prove things. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2013 at 17:38
  • \$\begingroup\$ @Saad - alternative added. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2013 at 17:50

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