2
\$\begingroup\$

Either I suck at searching or there is not a lot information about calculating values of bypass caps for opamps(I found a bunch on the topic in general). The question is, what it says in the title, say I have an audio range circuit with variable gain with an op amp run off of +/- 4.5V ( voltage divider on a 9V cell ) output current is probably under 50mA, now how would I go about calculating the values for the supply bypass caps?

\$\endgroup\$
1
  • \$\begingroup\$ Bigger is better, but too big costs too much money. \$\endgroup\$ Nov 21 '14 at 20:18
3
\$\begingroup\$

Batteries are a special case because their internal resistance increases as they discharge.

Probably something like 100uF/16V in parallel with 1uF/16V ceramic. Here's what the impedance vs. frequency of a 100uF capacitor with 5 ohm ESR in parallel with a 1uF ideal capacitor looks like from 100Hz to 5MHz:

enter image description here

So, it's pretty low over the entire audio range (so it will prevent unnecessary changes in the battery voltage if you draw reasonable current) and very low up to where the audio op-amp gain should be below 1. 1000uF would be even better but they're physically large, and 100uF/16V is quite small.

\$\endgroup\$
4
  • \$\begingroup\$ That's a great answer, any reason for the 1uF in parallel with the electrolytic? \$\endgroup\$
    – Limiter
    Nov 21 '14 at 20:32
  • \$\begingroup\$ Yes; electrolytics don't bypass high frequencies very well. \$\endgroup\$
    – gbarry
    Nov 21 '14 at 20:38
  • \$\begingroup\$ The electrolytic I assumed has a 5 ohm ESR - it gives you the low end of the curve, but will be relatively high resistance at higher frequencies where stability issues might pop up. You could use a single 100uF/16V ceramic instead but it would cost several dollars rather than pennies and would only actually be something like 50uF at 9V. \$\endgroup\$ Nov 21 '14 at 21:14
  • 1
    \$\begingroup\$ Thanks! As You can probably tell I am a beginner at electronics, but I am also highly enthusiastic. I must say I love the helpfulness and general positivity of the community too! \$\endgroup\$
    – Limiter
    Nov 21 '14 at 21:38

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.