0
\$\begingroup\$

In an opamp circuit for audio purposes like the following, I'd like to know some informations about the capacitor C1:

  • which type is the best in this application (polyester, Al electrolytic, tantalum, bi-polar electrolytic);
  • if polarized, what is the correct polarity in the circuit and why.

schematic

simulate this circuit – Schematic created using CircuitLab

The choice of electrolytic capacitor is giustified because it needs a large value if you want to keep a good lower cutoff frequency (in the example, about 70Hz) and low resistor values to reduce noise; but I'd like to know if there are troubles with the polarization (and maybe the distortion due to non-ideality)

Moreover, is it correct that C1 reduces op-amp offset effects at the output?

\$\endgroup\$
  • \$\begingroup\$ Is this homework question? It sure looks like it. Where are you stuck and what have you done so far? \$\endgroup\$ – Justme Apr 16 at 10:52
  • \$\begingroup\$ This is not an homework question. I built up the circuit on a breadboard with an electrolytic capacitor and it works perfectly, but I'd like to have a deeper understanding. \$\endgroup\$ – Alessio Caligiuri Apr 16 at 11:01
  • \$\begingroup\$ There‘s no problem with polarization if your input voltage is always above GND. The Elco needs positive voltage. An elco is a passive linear type of component, so there should be no distortion effects. \$\endgroup\$ – Stefan Wyss Apr 16 at 11:10
  • \$\begingroup\$ @StefanWyss: The opamp has a bipolar supply. The signal can go below ground. Electrolytic capacitors don't do well when reverse polarized - distortion is a possibility. \$\endgroup\$ – JRE Apr 16 at 11:49
  • \$\begingroup\$ @JRE What happens when reverse-polarized by the negative portion of a small AC signal (like an audio signal, in this circuit)? Does the el. cap. behave like a forward-biased diode? \$\endgroup\$ – Alessio Caligiuri Apr 16 at 11:57
0
\$\begingroup\$

The circuit acts like a high pass filter, as only higher frequencies cause a current flow through the capacitor, meaning that the gain is scaled based on the frequency, it seems to be about a 60z high pass filter

at lower frequencies the gain is small, at higher frequencies it tends towards a gain of 50,

The capacitor will need to survive potentially +-15V to cover all use cases, ideally for an audio signal it only sees mV on average

I would recommend a bipolar electrolytic,

It does not remove the input offset of the op amp, but does reduce it, at the trade off of much higher DC distortion, however for audio this is not too bad of an issue.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. I placed that cap just to remove very low frequencies below 70Hz. About the offset, is it correct to say that the DC gain of this circuit is 1 (because G=(1 + Z2/Z1), but for DC Z1 is infinite), so any voltage offset at the input is not amplified? \$\endgroup\$ – Alessio Caligiuri Apr 16 at 11:54
  • \$\begingroup\$ Assume DC steady state, and the input held at ground, negative feedback will cause the output to trend towards the input offset voltage, not amplified, but still present, If you remove the capacitor for a DC only example, its a voltage follower, that voltage being the input offset voltage. \$\endgroup\$ – Reroute Apr 16 at 11:58
  • \$\begingroup\$ ok, thank you. Instead, if C1 was not present (R1 was directly connected to GND), there is no DC-blocking, so the offset voltage is amplified by 100 as AC signals. Right? \$\endgroup\$ – Alessio Caligiuri Apr 16 at 12:18
  • \$\begingroup\$ as DC offset for the AC signals yes, which may be mitigated by an AC coupling cap on the output usually. \$\endgroup\$ – Reroute Apr 16 at 12:21
0
\$\begingroup\$

Finally, I found a very similar circuit here: https://sound-au.com/articles/reverb.htm I wrote to the author, who kindly replied me, addressing my doubts.

Using polarised electrolytic caps only works if the voltage across the capacitor (AC and/or DC) remains below 0.5V (but the desing guarantees 0.1V maximum). This has always been the case, and relies on the voltage being below that which causes any reverse current.

A non-polarised cap can be used, but they are larger and more expensive. The audible difference is zero.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.