Wah-wah pedal center (peak?) frequency

Started by ECistheBest, March 07, 2013, 06:38:08 AM

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ECistheBest

Hi everyone, I've been reading a whole bunch about the wah-wah pedal, and stuck on a few points trying to understand the circuit.

From RG Keen's article on his website, "The Technology of Wah Pedals" here http://www.geofex.com/article_folders/wahpedl/wahped.htm
I understand a lot of things, but on this part:
QuoteThe secret is this. The inductor looks to the second transistor like its far side is grounded, through the 4.7uF capacitor. To the inductor, the capacitor kind of looks like it's grounded because its far side is connected to the emitter of Q2. Q2's emitter has a low output impedance and therefore looks like "ground" if you ignore the signal coming out of the emitter. At the junction of the inductor, capacitor, and 1.5K resistor, the voltage looks like the voltage that would happen across a parallel L/C circuit. However - the current through the capacitor is NOT determined by the voltage across the inductor/capacitor, it is also determined by the voltage driving its "ground" side, and that voltage is increased or decreased by the position of the wah pot. If the wah pot setting increases, the capacitor will let more signal current through because the voltage driving it at Q2's emitter is bigger, so the capacitor has to let in more signal current. If the wah pot setting decreases, the capacitor will let in less signal current. A "capacitor" may be thought of as a special instance of ohm's law by the amount of signal current it lets through. The change in the effective current through the capacitor makes the capacitor look bigger to the inductor and rest of the circuit than it really is! We have a variable capacitor!

he explained a whole lot but i don't understand the relationship between the high/low AC voltage at the feedback capacitor's "ground" side and the capacitor "looking" bigger.


so the AC output signal controlled by the wah pot, buffered by Q2, is fed back through the 10n capacitor to the inductor. how do i calculate the exact resonance frequency? RG mentions the capacitor becomes bigger or smaller "looking" by changing the AC voltage from Q2. if i can get this, can i figure out the resonance frequency just by 1/[2*pi*sqrt(L*C)] Hz?



the other question i have is, lets say we have a given sound, the wah pot at 50% rotation, and the feedback capacitor at 22n. is there a point on the wah pot's rotation that gives the same exact resonance frequency with a 10n capacitor instead of 22n?


Thanks for looking, any insight for this would be great help to me as i'm trying to add fun things to a wah.