I am using an instrumentation amplifier for my project . The differential signal applied to amplifier through sensor is 0-2mV

i was wondering if i replace 2k2(R5) resistor with a 5k pot ...would i be able to achieve a gain of 1000 i.e i want the output of this circuit to be 0-2 Volts .

p.s. i cannot use the instrumentation amplifiers like ad620...due to some reasons... Thank you .enter image description here

  • \$\begingroup\$ This doesn't address your question, but if you really need the high CMRR afforded by an instrumentation amp, you should buy an off the shelf amp and not cobble one together from op amps. Resistor mismatch will kill your CMRR. \$\endgroup\$ Jul 27, 2016 at 10:59
  • \$\begingroup\$ Also, often you don't really want a gain of 1000 in one stage, unless you can guarantee that there is no offset voltage on the input, or you'll just saturate \$\endgroup\$ Jul 27, 2016 at 11:02

1 Answer 1


With the values in the diagram, gain is about ten. If you want gain of 1000, it would work to bump R10 and R11 up to 100k, and LOWER the value of R5 to 202 ohms. Yes, R5 can be replaced with a pot, but those aren't terribly accurate nor stable.

  • \$\begingroup\$ Whenever you start measuring resistance in megaohms I get nervous. And do you mean 220 ohm? 202 is pretty non-standard! \$\endgroup\$
    – pipe
    Jul 27, 2016 at 7:27
  • \$\begingroup\$ I suspect that a resistor as high as 1M would add a lot of thermal noise. I would attempt it by only changing the gain resistor, R5 (but make sure all resistors are 1% or lower). \$\endgroup\$
    – Mark
    Jul 27, 2016 at 7:53
  • \$\begingroup\$ For a 100:1 gain in the first stage, you want a 99:1 resistor ratio; starting with 10k ohms, that means 101 ohms (because 99 * 101 = 10000); the R5 value is twice 101 ohms. I put too much gain, though, in the final stage; 100k ohms for R10 and R11 is correct. \$\endgroup\$
    – Whit3rd
    Jul 27, 2016 at 8:22

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