I would think a basic graphic equalizer could function as such: a bank of filters, each with their own connection through a potentiometer to a summer (such as an op-amp summer). However, I've run across several designs and it appears that while indeed it seems the boost is summed in this manner, the cut is also negatively summed at the input (for instance, here).

For reference, consider the following from the link, above: enter image description here

Why would one do this? Why do we have to or would we want to "sum" the cut at all?

I get in theory, if I'm correct, the cut is feedback, inverted to the input op-amp (U1B in the link above), and so cancels some of that frequency.

Is this necessary, or a feature? Why is the boost summed to the output op-amp (U2A) instead of say the same opamp as the cut?

My apologies if I don't understand this design, it's one of the things I'd really like to make as a project.

  • \$\begingroup\$ It is unclear what you mean by the cut and boost, but maybe I am here out of context. \$\endgroup\$ – sherrellbc Jul 19 '14 at 19:39

The amount of boost or cut for any given frequency is based on the ratio of the levels of signal applied to the boost and cut busses. For flat response, both busses get the same amount of signal.

If you eliminated the cut bus, anything that wasn't boosted would be at "full cut", which is not very useful. You want to be able to vary the boost/cut smoothly and continuously.

| improve this answer | |
  • 1
    \$\begingroup\$ I sense the schematic the OP links is pretty much a standard, I had a look at it and I'm not sure I fully understood how it works. Is it correct to say that "cut" cuts because it is summed with a minus sign? And why both cut and boost are fed in the chain with such an high gain (1000)? I'd really appreciate some insights on how this circuits work. \$\endgroup\$ – Vladimir Cravero Jul 19 '14 at 21:09
  • \$\begingroup\$ I agree. The above answer tells me it's necessary - and thank you kindly for that - but why? The link provides no real explanation and I've come up a little dry on "standard" equalizer theory. \$\endgroup\$ – MJXS Jul 19 '14 at 21:29
  • 1
    \$\begingroup\$ @VladimirCravero: If you're referring to those 10-ohm resistors on the cut and boost bus inputs, those don't set the gain. The gain is set by the 10K resistors around U1B and U2A, along with the 10K resistors at the output of each filter section. The gain of U2B controls the maximum amount of boost or cut, as described in the linked article. \$\endgroup\$ – Dave Tweed Jul 19 '14 at 22:18

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.