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I have a simple force measuring circuit which consists of a strain gauge SG1 and an amplifier with a Wheatstone bridge.

The variable resistor RV1 is set to 200 Ω. If the strain gauge is fully deflected, the circuit gives an output of 50 mV.

How would I go about calculating the imbalance voltage at the bridge?

I found the voltage that goes into the op-amp's pin3:

$$R_{FSR}=\dfrac{1}{\dfrac{1}{1320}+\dfrac{1}{1320}}=660\ \Omega$$

$$V=\dfrac{680}{680+660}\cdot9=4.57\ \mathrm{V}$$

Any help will be appreciated.

enter image description here

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    \$\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

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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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