Once in a while in literature I see DC offset correction loop.

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What are the advantages of such DC offset removal loop, compared to an RC high-pass filter?
The RC filter would look like this.

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I work on one-off research devices for which the cost of an op-amp isn't a problem. I don't have an objective to economize the cost of one op-amp. I'm asking this question for curiosity and self-studying.


3 Answers 3


I think the use of such servo is extremely situational. It does not prevent the noise of the large resistor at DC, it only keeps the out-of-band output impedance low. Also it has less DC precision (due to additional op-amp offset) and more noise (due to additional op-amp) than the simple RC high-pass.

Which is why I say 'situational'. Either DC is out-of-band, then you don't usually care about the output impedance.

Or it is in-band, in which case both circuits suck, and you must use some chopping/switched capacitor circuit to prevent DC offsets without adding noise.

One application of the first circuit, that I could imagine is the following: Say you have a very high gain non-inverting output opamp (on the right hand side of your schematic), to amplify the AC-coupled signal. And you pick an ultra-low voltage noise BJT op-amp with humongous input bias currents. In this case, the large DC offset incurred by the input bias current through the grounding resistor can cause quite a bit of DC offset, which - due to the large gain - could saturate that amplifier. Here, it is indeed important to keep the out-of-band (DC) source impedance low, so the servo could replace the passive RC highpass. For the servo amp, you need something that has low input bias current and low input offset voltage, a typical "precision op-amp".

  • \$\begingroup\$ if DC is of interest, I don't think the OP would've chosen any of these as possible implementations. In that case, chopping might be the only way. Also, I don't understand why someone above said 1Mohm noise at DC is important, since, apparently, AC coupling is ok for the OP. That 1Mohm noise is shorted out by the capacitor and output impedance of the preceding stage, i.e. it's mostly irrelevant. At DC, the noise will 1.26uV/sqrt(Hz) for this resistor, but an offset due to 1nA ibias will be 1mV. I think DC offset due to the 1Mohm might be of more importance. \$\endgroup\$
    – Designalog
    Commented Apr 21, 2023 at 8:57
  • 1
    \$\begingroup\$ @ErnestoG completely agree.. That is why I really don't see a realistic advantage in the "active highpass" circuit. \$\endgroup\$
    – tobalt
    Commented Apr 21, 2023 at 10:13

The impedance of the R-C integrator is independent of the impedances in the signal path. Generally, this means you can have the effect of a very low frequency highpass filter, as in 1 Hz or less, without an enormous capacitor.

For example, audio. Something like this in an audio power amplifier can eliminate an input coupling capacitor, which some golden-eared wonders proclaim to be not just good, but essential. Sometimes called a DC servo, a 1M - 1 uF integrator gets you the effect of a 10 uF to 100 uF input coupling capacitor, without having a dreaded electrolytic capacitor in the signal path.

I'm picking on audio, but the benefits are real. The same holds true for instrumentation circuits.


The 1st circuit can, in theory, arbitrarily suppress any low-frequency drift or DC offset of any amplitude since it uses negative feedback. The output of the U2A integrator will do whatever it can to keep its input at 0V.

The 2nd circuit, on the other hand, will only attenuate the low-frequency drift by a fixed amount depending on frequency. In theory, if the low-frequency drift has an arbitrarily large amplitude, the 2nd circuit will only attenuate it, not dynamically zero it as the DC feedback loop does.

Since negative feedback is the only the technique that, in theory, can reduce the input error signal to 0 (in this case, reducing the output offset of 1st amplifier by comparing it to 0V, thus your error signal is (0V - AD620 DC output voltage). Therefore, the 1st circuit is more robust towards any sort of low-frequency drift compared to the 2nd circuit.

In practice, of course, the 2nd circuit will be limited by the common-mode range of the U2A. It cannot suppress arbitrarily large DC offsets because it's limited by its supply voltage rails and other imperfections.

I don't know what kind of application you're designing this in, but I'd say not to discard the 2nd option. I see that amplifying a DC signal is not of interest to you since you are AC coupling and redefining the DC level. The 2nd option is also a valid way to attenuate offset.

@Kuba said that the output impedance of the first stage is increased due to the RC. This is true, in principle, but it can be minimized by choosing a large capacitor in series such that its equivalent impedance is very low at your frequencies of interests.

Furthermore, even though 1Mohm resistor's noise is large at DC, at your frequencies of interest (I assume they are much higher than DC) becomes less relevant because the capacitor will be, basically, a short and the output impedance of the AD620 is a lot smaller than 1Mohm (hopefully). You can check this answer I wrote to prove to yourself how making the shunt resistor large will actually make it contribute less to the input-referred noise.

A disadvantage of making large, though, is that it can generate an offset towards the output of the 2nd stage due to its input bias current (Ibias*1Mohm). I don't know what kind of op-amp U6A is, but to give you some numbers, an Ibias of 1nA will cause a 1mV drop at the input of U6A.

This brings me to my final point. Since DC is not of interest to you (only that it's predictable and very close to 0V), your 2nd stage must be AC coupled in the feedback network as well. Why? Because, as it is right now, that 1mV drop due to an eventual 1nA of Ibias current will be amplified by a factor of (1+R9/R10). Also, not only will you be amplifying that offset, but you'll also be amplifying the U6A's input voltage offset as well towards the output. As that appears to be undesired by your application, you need to add a capacitor in series with R10 to ground. This will cause the branch R10 + Zcapacitor to be ana open at DC (because the capacitor is an open at DC), thus making U6A's closed-loop gain equal to 1 at DC. In that way, you're not amplifying any spurious DC voltage in your way.

  • \$\begingroup\$ The first circuit is no better or worse in suppressing DC offsets than the passive second version. Infact, for the same R and C, it will have the idential frequency response for its offset suppression than the second circuit. If you omit the capacitor in the first circuit, true, it would eliminate the offset entirely, but so it would with AC "offsets". You need the capacitor to roll off the DC elimination. So again, the DC suppression is fundamentally no different between the two circuits. \$\endgroup\$
    – tobalt
    Commented Apr 21, 2023 at 13:18
  • \$\begingroup\$ @tobalt you're right, at the output, the offset will be the same for either approach. I'll correct my answer. However, I'd still keep the robust word in my answer, but argument it differently. The reason being that the DC servo will remove DC deviations within the amplifier itself. If, for some reason, there's a DC shift inserted within the amplifier, the 1st circuit will eliminate it within the amplifier, which can help keeping an accurate operating point for all the internal circuitry. \$\endgroup\$
    – Designalog
    Commented Apr 22, 2023 at 10:33
  • \$\begingroup\$ @tobalt On the other hand, the 2nd circuit will not care whether there's an internal DC deviation within the amplifier, it will just remove it towards the output but keep the original AC signal, which might be contaminated by the consequences of the DC shift (distortion is the first thing that comes to mind). \$\endgroup\$
    – Designalog
    Commented Apr 22, 2023 at 10:35

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