Integrator as DC servo

Question

I have used integrators in the feedback loop of standard inverting and noninverting amplifiers to eliminate DC off-set at the output. The circuit I'm talking about is something like this

schematic

^{simulate this circuit – Schematic created using CircuitLab}

I understand the function of the integrator. It samples the DC offset at the output and it starts ramping the output of the integrator is feedback to the inverting input which gets added to the original offset but with opposite polarity. The integrator stops ramping (and stays there) when the offset at the output is zero.

Thats if we focus on DC, but what happens to AC? Imagine a 1KHz sine is inserted at the input. That means that the output of the integrator will be a cosine which is being fed back to the inverting input and added with the original signal.

Isn't this detrimental? I mean adding a cosine signal will affect the original AC signal (by a small amount.) Am I missing something?

Voltage Spike · Accepted Answer · 2022-06-12 21:54:37Z

I sense the offset of the wideband amplifier (WBA) at its inverting input, where Vos resides without any dependence on Vin. I used am integrator and lowpass-filter (LPF) only to limit the noise of the auto-zero-amplifier (AZA). I chose a 1ms time constant for minimum total output noise and 0.5ms for the LPF. The advantage of sensing Vos at the inverting input is, the amplifier can be dc coupled.

Nice configuration, I should check it out, I guess it doesn't correct for the current offset voltage produced at R2. PS: Are you the same Kugelstadt from "Opamps for Everyone"? — S.s.
– S.s., Commented Jun 13, 2022 at 21:24

Andy aka · Accepted Answer · 2018-02-09 08:37:20Z

Yes, it will affect the AC signal by a small amount and that amount halves for every doubling of the AC frequency so, you design an integrator that is intended to remove DC so as to not cause a significant effect on the amplifier’s AC processing gain.

Look at R5 and C3; they form a low pass filter at about 0.2 Hz so, at 0.4 Hz the integrating effect is halved and at 0.8 Hz the effect has quartered. Do you follow where I am going with this?

Yes, I see what you mean, so adding a dc servo is not completely transparent audio wise, granted that at audible frequencies the effect contribution will be insignificant. — S.s.
– S.s., Commented Feb 9, 2018 at 0:15
@S.s. I think you mean, "adding a dc servo is completely transparent audio wise, granted that at audible frequencies the effect contribution will be insignificant." If the contribution is insignificant, doesn't that mean the effect will also be insignificant? — WhatRoughBeast
– WhatRoughBeast, Commented Feb 17, 2019 at 4:05

Thomas Kugelstadt · Accepted Answer · 2022-06-16 19:42:52Z

Yes, it does correct for the voltage across R2 due to bias current. The integrator continues to integrate until the voltage at inverting input is 0V (or Vcc/2) in a single-supply circuit. And Yes, I am the same Thomas Kugelstadt.

score 0 · Accepted Answer · 2018-02-09 05:17:26Z

I though you understood this type of servo-loop, which is used on high performance audio amps to avoid the phase shift caused by capacitors. Your time constant appears to be .16 HZ, so any 'event' at or under a .16 HZ rate will be cancelled out by the loop.

Frequencies above .16 HZ are ignored, as the loop is stable. By the way, the TL072 series is much more quiet. It is designed for audio, while the TL082 is a more general purpose op-amp

NOTE: Just to be accurate, because you have a 2 pole design with C3/R5, and C2/R4, roll off is 6 dB per octave, or 70.7% of 70.7%, which is 50%, or 1/2 intensity per octave. It is not a sharp roll off but it is continuous. At 16 HZ their is nothing left of any DC component.

How did you arrive to the 1.6Hz time constant, Im getting 0.16Hz, typo? — S.s.
– S.s., Commented Feb 9, 2018 at 0:10
Yep. Fixed them. I even have a look-up chart for that stuff. Mea Culpa. — user105652
– user105652, Commented Feb 9, 2018 at 0:14
BTW im not using the TL082, its just that I selected whatever default part to draw the circuit in the simulator. — S.s.
– S.s., Commented Feb 9, 2018 at 0:18

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