I realize that there are a lot of questions relating to op amp offset here on electronics.se; however, none of them seemed to answer my exact question. If some of them are too similar, feel free to close.

I have a very simple non-inverting single-supply 3.3v-0v LTC6240 circut. This op amp doesn't have an offset null pin. I am running this op amp at a very high gain (>1000) with a 40khz input signal. The circuit works perfectly fine most of the time; however, I have an unpredictable ~500uV offset on the negative input, which (with some op-amps and some populated boards) works out to an output offset that the rest of my circuit can't compensate for.

I thought the best solution was this. I put a 10uf cap in series to ground as the answer describes and somehow, my gain at 40khz went away. I've replicated this in ltspice, even though I don't understand why it happens; the impedance of a 10uf cap at 40khz is only ~0.5 ohm. Here's a schematic of how I connected it:

enter image description here

As for putting a potentiometer in to correct the offset as described here, unfortunately I don't have a negative voltage available on the board, so all the pot can do is raise the offset.

  • \$\begingroup\$ You should probably add a circuit to show how you're connecting things. I have a feeling that the reason why the cap at 40khz crushed your gain is because you just high-pass filtered your output at 40k. It's also possible you loaded the op-amp output down too much by creating a virtual ground at 40khz. \$\endgroup\$
    – horta
    Apr 9, 2016 at 17:06
  • \$\begingroup\$ @horta One schematic coming right up. \$\endgroup\$
    – 0xDBFB7
    Apr 9, 2016 at 17:11
  • \$\begingroup\$ @horta there we go. \$\endgroup\$
    – 0xDBFB7
    Apr 9, 2016 at 17:42
  • \$\begingroup\$ Is it really 7mOhm? Because if you look at the basic non-inverting amp equation, that results in a gain of (0.007/500+1)*V+ ~= V+ or a gain of 1. \$\endgroup\$
    – horta
    Apr 9, 2016 at 17:50
  • \$\begingroup\$ m = milli = \$10^{-3}\$. M = mega = \$10^6\$. Capitisation matters. Also "F" for farad and "Hz" for hertz. General rule for SI units is capital letter if abbreviation is used but lower case if spelt out. e.g., V for volt, A for ampere, K for kelvin. \$\endgroup\$
    – Transistor
    Apr 9, 2016 at 17:54

1 Answer 1


You need to use meg instead of m for the resistor value. You have specified a 0.007\$\Omega\$ resistor, which will tend to reduce the gain to 1.

Your amplifier has a GBW product of 18MHz meaning it will only have an open-loop gain of about 500 at 40kHz, not the 14,000 you appear to be expecting. However it will have extremely high gain at DC so any offset will be amplified - so only a few uV will saturate the amplifier.

You also need to bias the amplifier so the inputs and outputs are somewhere in the operating range- right now it will sit at the negative rail- you want it to be about Vdd/2.

It's better to have a few AC-coupled stages with gain of ~100 or less rather than one stage which needs astronomical gain.

  • \$\begingroup\$ Aha! Biasing the input! That's perfect. \$\endgroup\$
    – 0xDBFB7
    Apr 9, 2016 at 18:57

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