Op Amps: Resonant EQ

[@Duongbacang]

Thanks verymuch, 😊

[@cansaded1688]

ty sir for the video¡

[@AnalogDude_]

21:34 How bout about CD4052, CD4053, CD4066 or some fancy (A)DG analog switch?

[@ElectronicswithProfessorFiore]

Sure.

[@AnalogDude_]

@@ElectronicswithProfessorFiore right will try that some day, Thnx. Roland used transistors to build a filter. Sallen Key.

[@rudygomez6996]

i am confused. makes more sense to me if values of C1 and C2 were swapped. since smaller C1 would set higher frequency range and a higher C2 would set lower frequency range. am i missing something??

[@ElectronicswithProfessorFiore]

It sounds like you're describing it correctly. Read the notes starting just before 4:00. In any case, think of the curve in two mirror image halves. C1 affects the upper half and C2 affects the lower half. If you decrease C1, it reacts with the pot at a higher frequency (shorting it out, so it no longer has an effect). This extends the upper part of the curve to a higher frequency. In contrast, C2 blocks the signal, so increasing its value extends the range lower. Thus, doing both makes the whole curve more broad (lower Q).

[@rudygomez6996]

@@ElectronicswithProfessorFiore but shouldn’t C1 be smaller than C2? for example C1=560p and C2=5.6n?

[@ElectronicswithProfessorFiore]

@@rudygomez6996 No, they're reacting against different resistances.

[@rudygomez6996]

@@ElectronicswithProfessorFiore C2 blocks low frequencies and therefore controls when signal begins to come through. Once signal is able to flow through C2, the freq-impedance curve of C1 will have steeper downward slope than C2 and the higher freq end of its curve having more impedance than C2 because it is in parallel with high resistance of P1 and 10k resistors. Therefore capacative reactance of C1 will be higher than P1/10k’s resistance. Once it reaches high enough freq, C1 reactance eventually becomes low enough so as to short and set high freq back to 0db. Is my understanding correct?

[@ElectronicswithProfessorFiore]

@@rudygomez6996 Sort of, I mean part of it is correct. For example, C2 is not "in parallel with P1 and 10k resistors". That's a series-parallel network. What is true is that C1 is in parallel with P1. Thus, at high frequencies, C1 shorts out P1. Think of replacing P1 with a wire. At those high frequencies, C2 will also be a short, thus, you end up with each 10k in parallel with a 1 Meg, for about 10k each. The left pair is Ri, and the right pair is Rf, thus the gain is one. In other words, above a frequency dictated by C1 (and P1), the position of the pot doesn't matter, and gain is unity. As the frequency is lowered, P1 can no longer be thought of as shorted by C1, and thus part of P1 adds with the left 10k to make Ri while the remainder of P1 adds to the other 10k making Rf. That sets some value of gain. At this point C2 is still an effective short, but as the frequency continues to drop, Xc2 increases, blocking the signal from the pot, and eventually reducing the gain back to unity.