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I am just stuck with a problem. Here using an instrumentation circuit (see image below). I want to know the exact value of resistors for a gain of 10.86. I have calculated and found some values (25K and 5.1K), but I am not sure whether they are right or not.

enter image description here


EDIT (Circuit change prompted by a comment: gain resistor is now split)

enter image description here

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  • \$\begingroup\$ I get 10.804. Is that close enough for your application? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 10 '15 at 4:46
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Is this some kind of homework or exercise or is it a real design problem? I tell you this because a gain specified with 4 significant digits means all your circuit has to meet quite significant precision specs. If this is a precision design specifying your gain as simply 10.86 doesn't make sense, unless you add an error spec.

Assuming you want 4 exact digits, your gain must have an error less than \$\pm 0.01\$ over \$10.86\$, i.e. your gain must be precise to \$\pm \dfrac {0.01}{10.86} = \pm 0.092 \% = \pm 920 ppm\$.

This means that, without further error analysis and neglecting other sources of errors, your gain setting resistors must have a max tolerance of 0.1% (probably less). Moreover (as pointed out explicitly by @cowboydan in a comment) the feedback resistor R12 and R13 must be closely matched to avoid additional error due to offset voltages. Did you consider this?

For this kind of precision circuitry it is not, in general, advisable in real design to roll your own instrumentation amplifier (unless you really know what you are doing), but it's better to use an already made instrumentation amplifier IC (where all parts that have to be matched are matched for the best). For example, off the top of my head, the INA103, the INA128 or the AD623.

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  • \$\begingroup\$ Exactly - plus you would have to make sure that the feedback resistors in each loop are closely matched or else you will have pronounced offset error. \$\endgroup\$ – cowboydan Aug 10 '15 at 9:53
  • \$\begingroup\$ @cowboydan Yep, good point. I didn't explicit mention it, but it's worth pointing out. I'll integrate my answer. \$\endgroup\$ – Lorenzo Donati supports Monica Aug 10 '15 at 9:58
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I'm pretty sure that your gain calculation is wrong. The way that I calculate something like this is to split the amplifier into two halves. Think of the gain-set resistor (Rg) as being two resistors in series with the mid-point grounded. Then do the gain calculation.

I get a gain of 10.80392 as opposed to the 10.86 that you show.

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  • \$\begingroup\$ I'm pretty sure they used the formula in red in the middle. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 10 '15 at 5:23
  • \$\begingroup\$ Thanks for your reply.As per your suggestion I have modified the circuit. Instead of 5.1K resistor,now I have two 2.55K resistors in series. I hope this is what you recommend. What benefit do we accomplish with 2 resistors in series instead of having one with combined resistor values? Also, could you please educate me how to pick correct values for Gain resistor(Rg). I have attached latest circuit. \$\endgroup\$ – Vipu03 Aug 10 '15 at 6:52
  • \$\begingroup\$ Vipu03, I suppose you know that this gain applies to DC only? \$\endgroup\$ – LvW Aug 10 '15 at 6:54
  • \$\begingroup\$ yes. I know that \$\endgroup\$ – Vipu03 Aug 10 '15 at 7:02

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