I'm trying to get the difference of two DC signals using an op-amp difference amplifier circuit. The gain will be 1. The difference will be around 1-5v. The inputs could be as high as 50v though.


1. Vin+ = 50v, Vin- = 48v, Vout = 2v
2. Vin+ = 2v,  Vin- = 0v,  Vout = 2v

The problem is this would need the op-amp's inputs to be capable of going up to 25v. Is there a way round this? I have absolutely no idea how to solve this!

Edit: Using two potential dividers before the op-amp circuit that reduce the signals by 90%, followed by the difference amplifier with a gain of 10 would work but this would degrade the accuracy of the result by a factor of 10? Edit 2: And those dividers would have to be very stiff compared to those in the differential circuit


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Why does the op-amp have to measure those inputs directly? Why don'y you just connect equivalent voltage dividers on both inputs? \$\endgroup\$
    – krb686
    Commented Feb 11, 2014 at 12:24

3 Answers 3


Make the two input resistors significantly bigger (say 10xR) - the gain will now be lower but you can added an extra amplifier onto Vout to restore the gain you need. Be aware of common mode problems with this type of circuit - you'll probably need to use 0.1% (or better) resistors to get an accurate representation of your 1-5V signal.


unlike Andy's answear i wouldn't use input resistors on your input. they add noise to your signal. of course its depending on your application.

what i would use is something like an AD8479 which is exactly what i think you want. high input offset voltage is the keyword here

  • 2
    \$\begingroup\$ While the 1Mohm resistors in the AD8479 are likely precision trimmed to maintain a good CMRR, they are not magic, and will add as much thermal noise as any 1Mohm resistor. That said, the output of a diff amp like that will have less common-mode noise than what could be build from parts. Instead of "add noise", "reduce CMRR" is more precise. \$\endgroup\$ Commented Feb 11, 2014 at 13:10
  • 1
    \$\begingroup\$ @ScottSeidman There are other kinds of noise besides Johnson-Nyquist. If you assume an op-amp voltage noise \$e_n\$ (including 1/f noise/drift), the presence of dividers at the inputs increases that noise (when referred to the inputs of the dividers) by the ratio. So Dreistein is not wrong on that point (though the AD8479 is not magic, as you point out, the same thing happens with it, so a bit wrong). The resistor ratio does not actually reduce the CMRR (Common Mode Rejection Ratio), rather this is an application that requires a large CMRR, a bit different. \$\endgroup\$ Commented Feb 11, 2014 at 13:48
  • \$\begingroup\$ Of course the AD8479 is not magic and adds noise too but a serious design avoids any additional noise. it's strongly application dependent. if you want to amplifier the differential of a few volts the resistors in front are no trouble but in the range of a few millivolts it's slightly different \$\endgroup\$
    – Dreistein
    Commented Feb 12, 2014 at 9:08

The voltages at the input terminals must be within the common mode input range of the op-amp, and the output must of course not exceed maximum output range. Unfortunately, all the 741 data sheets I can find don't actually have the common mode input range specced. I'd avoid that op-amp anyway, and move to a rail to rail for what little I know about your application


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.