Consider a dual-supply op amp circuit used to process (but not amplify) line level audio; for the sake of simplicity, imagine a unity gain follower. With the input AC coupled and referenced to ground, only a small DC offset, caused by the op amp, will appear at the output and this seems negligible at the first glance.

In this scenario, is it necessary to also put a coupling capacitor at the output? Here are some of my thoughts so far.

Pros of having an output coupling capacitor:

  • If a later stage is DC coupled and has gain, the coupling capacitor would prevent the DC offset from being amplified.
  • If a fault develops within the device and the op amp starts outputting a rail voltage (or any substantial DC level for that matter), the output capacitor would prevent this condition from negatively affecting later stages.

Cons of having an output coupling capacitor:

  • Since the input impedance of a later stage is unknown, a fairly large capacitance is required to achieve the desired low-frequency response.
  • The capacitor would have to be terminated with a resistor to ground. To avoid loading the output, it would have to be a high value resistor and this would introduce additional noise at the output.

Is there anything I'm missing here? As much as I'd like to omit the cap and the accompanying resistor for obvious reasons, I really wouldn't like to design a product that causes trouble for its users and interacts badly with other devices.

  • \$\begingroup\$ Personally I see no reason for the output cap. It's just going to roll off the low end. Anything the device is connected to is going to have an input DC blocking capacitor anyway. \$\endgroup\$
    – Ian Bland
    Commented Sep 24, 2016 at 23:01
  • 1
    \$\begingroup\$ @IanBland I agree any downstream device should be AC coupled; however, I can't be certain it will. Also, if the next stage is equipped with a polarized coupling capacitor and my device develops a fault as described in the question (starts outputting a DC level), it's possible for that cap to be reverse biased. In that case my device would be causing damage to the other device. I realize this isn't very probable, but that's precisely why I'm asking: in hope of someone more experienced assuring me of how and why it's usually done. \$\endgroup\$
    – uzde
    Commented Sep 24, 2016 at 23:44

2 Answers 2


The capacitor would have to be terminated with a resistor to ground. To avoid loading the output, it would have to be a high value resistor and this would introduce additional noise at the output.

Not really, a 1 Mohm resistor might introduce 18 uV across the full audio bandwidth (see this calculator) at ambient temperatures of 20 degC but this would be into an open circuit. That 1 Mohm resistor would not "see" an open circuit - the input impedance of the following stage might be 10 kohm so that turns 18 uV into 1.8 uV but, even then the driving amplifier at frequencies in the audio band would have a substantially lower impedance of a few ohms.

A 10 uF series output cap and 10 kohm shunt produce a low frequency 3 dB point of 1.6 Hz so there isn't even a problem about choosing an electrolytic because a neat little SMD cap will do the job.

So ,if the input impedance of the unknown next stage were 1 kohm, the low frequency cut-off would rise to 16 Hz - is this really an issue?

So how big could the offset be for a unity gain op-amp - maybe 5 mV - if this were put across a 10 ohm resistor there would be a current flow of 0.5 mA and would not really trouble nearly all op-amps capable of working to 20 kHz.

Is there anything I'm missing here?

Putting things into perspective? Considering reality? Provoking me to answer your question (by raising it LOL)?

  • \$\begingroup\$ Good point about the thermal noise. As for considering reality, it's an "audiophile" project and a low-noise op amp is specified so I'm just being cautious. \$\endgroup\$
    – uzde
    Commented Sep 25, 2016 at 7:03
  • \$\begingroup\$ 1Mohm resistor generated noise will not only be shunted by next stage input impedance, but mainly by "current stage" output impedance (probably less than one ohm at low frequencies). So don't care "audiophily noise" \$\endgroup\$
    – carloc
    Commented Sep 25, 2016 at 7:40
  • \$\begingroup\$ @carloc Well yes, that's how I intepreted this answer, "the driving amplifier" meaning "current stage" (correct me if I'm wrong). And that's the kind of answer I was hoping for as it pointed out something obvious that I had overlooked. I'm going to wait a little before accepting it though, just in case someone else chimes in with something helpful. \$\endgroup\$
    – uzde
    Commented Sep 25, 2016 at 8:24
  • 1
    \$\begingroup\$ I mentioned "driving amplifier" ie the outputting amplifier. \$\endgroup\$
    – Andy aka
    Commented Sep 25, 2016 at 9:04

There is ONE and only one compelling reason for an output capacitor. If the signal is to be switched further down the signal path, any DC offset will generate a 'click' when selected/deselected. That's it !

The value of the resistor to ground after the output coupling capacitor has no effect on noise whatever. The impedance here is dominated by the capacitor itself at LF, often of the value 10 - 100uF.

Zc = inv(2.pi.f.C) and taking f = 20Hz and C = 100uF we have 1/2.pi.20.100.10^-6 = 78.6 ohms


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.