The questions about op-amp resistors are dime a dozen, forgive me for adding another one. I am not asking how to calculate them, I know at least that much. My questions approach the problem from different perspective. Specifically - hobbyist's choices when it comes to buying specific resistors.

Case in point: I have 1.6V differential signal biased with 1.1V common-mode voltage. I want to remove the bias and amplify it rail-to-rail. Pretty common task in itself. The obvious choice is to use differential configuration, which gives me 1.05K:28K, 16K:430K etc. combinations.

Now, in ideal world I would get something like precision voltage divider from Vishay and have 1 component on PCB in addition to op-amp itself. In reality, however, only industrial designers can afford $20,000 for 1K of these. What hobbyists usually end up with is an awful mess of resistors in series or trimmers taking up over PCB space.

This got me thinking about alternatives. Technically, I can take only one line of signal source and feed it to non-inverting amplifier with bias (to remove common-node). This should be less sensitive to resistor values giving me more choices for less money. And will not require trimmers.

Q1: Does this make sense?

Q2: Would the precision loss due to using non-differential signal be at least slightly alleviated by lesser sensitivity to resistor matching?

Q3: Or maybe somebody can suggest affordable source of voltage dividers or 0.1-0.05% resistors not in bulk amounts? Note, that while cheap 0.1% resistors are not that hard to find, for some reason they often sold with E24 decade values, which kinda defeats the purpose.

Update: Here are the circuits for comparison (and original text changed a bit) schematics

  • \$\begingroup\$ You can get plenty of resistors in every size you could need from a supplier like Digikey or Mouser. Also consider whether you need that precision, and whether there are other ways (a high-pass filter followed by a plain old buffer?) to remove the DC bias. \$\endgroup\$
    – Hearth
    May 11, 2018 at 19:35
  • \$\begingroup\$ Your scenario and choice of circuit is unclear to me. Given that your whole question depends on this it seems your questions are unanswerable. \$\endgroup\$
    – Andy aka
    May 11, 2018 at 19:46
  • \$\begingroup\$ Digikey is exactly where I got that outrageous lowest price for resistor divider network. It is true that individual precision resistors are much cheaper there, still not as cheap as bunches of 100 at many Chinese sites. But the main question I wanted to explore is plausibility of using part of signal instead of differential. Note that in my case it is DC-coupled circuit. \$\endgroup\$
    – Maple
    May 11, 2018 at 19:48
  • \$\begingroup\$ @Andyaka Which part is unclear? I have SI8920B amplifier which outputs 1.1±0.81V on two differential outputs. I want to decouple and amplify it to ± 24V. As simple as that. \$\endgroup\$
    – Maple
    May 11, 2018 at 19:53
  • 1
    \$\begingroup\$ Amplify it rail-to-rail? What circuit are you proposing and what are you trying to achieve. These questions and your answers and correct justifications form the backbone of your questions i.e. add context to your questions. At the moment it just sounds like you have a functional idea what you want (but aren't telling) and that you expect anyone reading this to understand that (maybe via telepathy) so that we comprehend that your circuit idea (unshown and unclear) makes sense which leads to an appreciation of the questions. \$\endgroup\$
    – Andy aka
    May 11, 2018 at 20:34

2 Answers 2


You are approaching this the wrong way.

The correct way would be to define the tolerance for matching, the ratio for differential gain, and tolerance for common mode gain error which is a result of the first two parameters and diff gain ratio.

If you cannot find a ratio resistor array of 0.005% to give you the full swing that you desire you may need to stages. However the first stage should eliminate the common mode error and the second stage can give you the desired differential net gain.

Here is a sample of resistor arrays that should meet your requirements for being in stock low-cost and minimum quantity of one.


  • \$\begingroup\$ Yes, I was through that filter before. It becomes much less affordable if you check "through hole". But you are right, if I go with SMD then the selection suddenly becomes more appealing :) \$\endgroup\$
    – Maple
    May 11, 2018 at 20:05
  • \$\begingroup\$ I also realize that my approach is rather simplistic. Let me rephrase that: given the fixed pair of source and amplification op-amps in two DC-coupled configurations and low signal bandwidth, is there clear preference between differential and non-inverting configurations? From the point of building/tuning it by non-professional with rudimentary knowledge of electronics as myself. Or is this over-simplification again? \$\endgroup\$
    – Maple
    May 11, 2018 at 20:37
  • \$\begingroup\$ Yes there is a clear difference with CMRR from ground shift or error in Vref = Vcm cancellation \$\endgroup\$ May 11, 2018 at 23:25
  • \$\begingroup\$ compute your common mode gain error \$\endgroup\$ May 11, 2018 at 23:42
  • \$\begingroup\$ OK, it took me a while and I am not sure the math is correct, but here is what I have (for op-amp with 95 dB CMRR): common mode error 0.521 mV, error due to 0.1% tolerance 1.6 mV. For inverting configuration an error due to 0.1% tolerance is about 0.041 mV. Anywhere close to reality? \$\endgroup\$
    – Maple
    May 12, 2018 at 2:42

For kicks, here is a non-inverting circuit that is not differential, cancels your 1.1V rail, steps +/- 0.81V up to +/- 24V, uses E24 resistors, and assuming the resistors are accurate, incurs 0.4% error on the 1.1k and 30k resistors.

Since the schematic editor does not support tablets, here goes:

  • run your voltage input to the non inverting input of your op amp.
  • run a voltage divider from 24V, through a 22k, 1.1k, to ground. Attach the middle of the divider to the inverting input.
  • run a feedback resistor between the output and inverting input of 30k.

But this is not really what you should be doing for a variable bench supply.

  • 1
    \$\begingroup\$ Wow... I really messed up my calculations when I ended up with negative bias voltage. You are right, both inverting and non-inverting configurations would be able to produce desired output. Yay, 1 resistor saved! And maybe this is not what I should be doing but I am learning a lot :) \$\endgroup\$
    – Maple
    May 12, 2018 at 10:31

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