So I have this simple force measure circuit which consists of a strain gauge SG1 and the amplifier with the Wheatstone bridge, the variable resistor RV1 is set to 200Ohms.
If the strain gauge is fully deflected the circuit gives an output of 50mV.
How would I go about to calculate the imbalance voltage at the bridge.
What I did so far is I found the voltage that goes into the op-amp's pin3: $$R_{FSR}=\dfrac{1}{\dfrac{1}{1320}+\dfrac{1}{1320}}=660Ohms$$ $$V=\dfrac{680}{680+660}*9=4.57V$$ Any help will be appreciated. enter image description here

  • 1
    \$\begingroup\$ I am having a hard time understanding what you are asking here and an even harder time figuring out why you have the op-amp wired that way. It should be like this bing.com/images/… \$\endgroup\$
    – Trevor_G
    Nov 30, 2017 at 1:23
  • \$\begingroup\$ there is nothing wrong with the circuit. \$\endgroup\$
    – super95
    Nov 30, 2017 at 1:29
  • \$\begingroup\$ Sure.. its fine till you hook any significant load on the output.. \$\endgroup\$
    – Trevor_G
    Nov 30, 2017 at 1:47
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    \$\begingroup\$ The Supply sensitivity is too high and you need a differential Amp on the output, when you should have it here. Change all 1k2 to 120 and add an LDO for 15mA into 120 Ohms or 1.8V then use a rail to rail Op Amp with >10K load using Trevor's link. You should get 500mV then full scale then and less CM error and gain error. \$\endgroup\$ Nov 30, 2017 at 1:53
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    \$\begingroup\$ Your gain is ~200 so the full scale input is 250uV making it very prone to noise and errors. So there are a few things wrong \$\endgroup\$ Nov 30, 2017 at 1:59

1 Answer 1


Why don't you make it like this? enter image description here

All your calculations are much easier this way and it's "kind of linear".


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