Variable Q Mod for a CryBaby Wah - Two Methods

[drummer4gc]

The "vocal" mod on a CryBaby wah, switching the 33k resistor in parallel with the inductor with a higher value, seems to be one of the most popular beginning mods out there. There is plenty of information about how to sub the resistor for a pot to give you a variable Q.

I noticed this evening Dunlop's 535Q wah achieves this same effect by inserting a 1k pot in series with the inductor. Curious, I tried both methods on my wah and played around for a few minutes. They do the same sort of thing, but there is a difference I can't quite put my finger on yet.

Does anyone else do it this way? Is there a preference? I'm headed back to read "The technology of wah pedals" for a twentieth time, but I'd be interested if anyone wanted to explain the more technical differences between these two methods.

Thanks!

[R.G.]

The differences will be in how the resistor interacts with the internal capacitances of the inductor, I think.

This is actually pretty simple. The circuitry of the wah obscures it, though. The filtering of the wah is a result of the resonance of a cap and an inductor. The inductor is obvious, but the cap is varied electronically by the gain of the circuit, and that makes things hard to see.

A simple RLC circuit has a resonance determined by the value of L and C, and a Q (or sharpness of resonance) determined by the amount of energy in the stored L-C resonance being eaten up by the resistor. The resistor eats energy from the L and C whether it's in parallel or series with any of the elements.

If only those three elements were involved, the position of the resistor would make no difference at all. But the inductor has its own internal resistance and capacitance. In commerce, inductors are often rated for a resonant frequency and minimum Q. This is a nod to the internal R and C, and tells you the frequency above which you're not likely to get simple L responses. Good designers either take the internal R into account or use external parts that make it insignificant in good designs.

What you're hearing makes sense to me. There will be slight differences in response depending on whether you're damping across the inductor or in series with it. That's by best guess. It's possible that a series resistor allows the inductor's internal resonance to show through, being isolated a bit from the bigger outside influence of the external cap.

This is just a first guess, and some actual lab work would be needed to make more sense of it. But I think it's mostly what's going on. And the bottom line is - use whichever you like best. 

[drummer4gc]

Thanks RG! Between the explanation and your article, I'm starting to feel more confident pretending that I understand it all

FYI, for anyone interested in experimenting, implementing series resistance here is super easy on the newer CryBaby boards: there is a zero-ohm jumper resistor connecting the inductor to the 1k5 feedback resistor - just remove the jumper and install a 1k pot/trimmer and experiment away.

[joegagan]

i have not looked at a 535Q schematic. are you sure there isn't something else in line with the 1k pot? anything above 120 ohm total inductor R sounds like crap to me ( also looks bad in LT spice using  all the ' regular' wah schematics).

the only way i can see this working is if there is something else going on besides just a simple 1K series'd  with the inductor.

[drummer4gc]

This is what I was looking at:



There is definitely a sort of "muffling" effect that it produces, which sounds a little different than lowering the resistance in parallel, but the effects seem to be pretty similar, although I've spent very little time playing with them.

[joegagan]

wow, ok, i guess they knew what they were doing. i have one here, i will plug it in and see how i like that control.

[jatalahd]

The way I see it, the heart of the CryBaby wah is a simple parallel RLC circuit. When the Miller effect transformations in the small-signal model are made, the essential part of the circuit can be simulated with this model:



Here the impedance of the parallel RLC section defines how much of the signal is driven to ground at a certain frequency (here Z models the total impedance of the wah circuit forming a current divider with the RLC section). At the resonance frequency the impedance of the RLC is at maximum, so then most of the signal having that frequency will travel to the output. When plotting the impedance curves as  function of frequency (and capacitor C), we get something like this:



This looks quite similar to the actual frequency response of the whole circuit. Anyway, it is well known that the parallel RLC has a resonance at w0 = sqrt(1/LC) and the Q = R*sqrt(C/L). This applies to the basic CryBaby assuming that the inductor is ideal without any internal resistance (usually approximation applies nicely).

The modified schematic adds a resistor in series with the inductor. Continuing with the simplified models, this can be simulated with the following circuit:




Surprisingly the addition of one single resistor complicates the mathematical analysis considerably. It is relatively easy to calculate that the resonance frequency in this case is w0 = sqrt(1/LC - (RL/L)²). Since I assume this is quite basic circuit in electrical engineering, one might find an equation for Q directly from some handbook. I tried to evaluate the Q with my own calculations and I got an equation:

Q = Rx*sqrt(C/L - (C*RL/L)²)/(1+QL⁻²),

where Rx = [R*RL(1+QL²)]/[R + RL(1+QL²)] is the "effective" parallel resistance of R and RL and QL = (w0*L/RL) is the quality factor of the inductor with RL  [where w0 = sqrt(LC)]. So it seems that RL appears as a parallel resistance to R, mapped with the Q-factor of the inductor itself. Therefore RL is used to modify the value of R somewhat indirectly through the parallel connection.  The effect on Q is not linear though, because of the terms with QL are there to make things interesting. When RL is very small, then the term (1 + QL²) will grow significantly (and 1 + QL⁻² will equal 1) and the effective value of Rx = R. Making RL larger will only reduce the Q-factor of the circuit.

Well anyway, hope this helps, and remember not to trust too much on my evaluation on the Q-factor! But based on this analysis, RL must be designed very carefully to have certain values to fully control the Q of the circuit.

[joegagan]

i have spent many hours modelling wahs in LT spice. there is a parameter within the inductor spec to identify R. in stock wah circs, you can see the freq range ' wake up'  ie more pronounced, higher peaks when the r dials into the optimal range and no other component changes are made. that range is 35 to 55 ohm.

i try to base a lot of my research on existing circs to focus on selling parts that will work well in the millions of existing wahs.

you can move the optimal inductor R target optimal range up or down if you make other circuit changes to compensate, yes.

the most important thing to note is that in many many tests, the freq changes/ shapes that i see on the LT spice freq plot consistently bear out  or track very very similar in what i hear when i build the actual wah to the exact circ in spice. likewise, when tweaking components to solve a problem etc, these changes also track incredibly well from  model to real life. a most useful tool.

[PRR]

> one single resistor complicates the mathematical analysis considerably.

Precision analysis gets messy.

> I assume this is quite basic circuit in electrical engineering, one might find an equation for Q directly from some handbook.

Probably need better (older!) handbooks.

I never dug the math. Here's my cave-man approach.

The L and C will resonate at(/near) some frequency F. At this frequency they both have the same reactance (opposite signs) which is calculated the usual ways. (I like Reactance Chart, but abacus or sliderule work also.)

500mH and 0.01uFd resonate at 2,250Hz at 7KOhms.

With perfect parts the actual impedance is not 7K but infinite (parallel) or zero (series).

The coil always has resistance and at audio frequencies this resistance is not negligible. Series resistance, of course, because wire "is" resistance. Parallel resistance also, because of eddy currents in the core and wire.

Capacitors are usually perfect-enough for normal purposes.

Joe likes about 45 Ohms series resistance. To a cave-man approximation, the Q is the resonant impedance of either reactance divided-by the series resistance. 7000/45 gives Q of 155.(*)

Now it is proposed to add shunt resistance. If we only had shunt resistance, the Q is the shunt resistance divided-by resonant impedance of either reactance. For 33K, we have 33000/7000 or 4.7.

For both series and shunt: In this case where one resistor dominates, we can just round-down the lower Q and call it 4.6. Or we can reflect the 45 Ohms series resistance by the Q, twice, to get an effective 1,081,125 Ohm shunt resistor. Then use parallel resistance math to get effective 32,023 Ohm shunt and Q=4.57.

(*) I think it unlikely we get a Q over 100 with common parts and a single transistor. I don't think the 45 Ohm series resistance is the "whole" story, though certainly an important factor.

I don't even think we "want" Q like 100. It is hard to hit a note; when swept, it is mostly between notes/harmonics and unexcited. Q of 4 or 5 makes a lot more sense to me.

[guitarkill]

If you look at the crybaby patent they state that the only reason that the Q resistor is in the circuit is because they found that the Q of the circuit was too high and they found it desirable to add this resistor to lower it. In the same document it is stated that the inductor is what makes it a resonant circuit. No other special qualities are listed about the inductor.

I do not believe there is a magic wah pedal. Even if one is made to be exactly like the "holy grail" that so and so used. You are hearing much more the guitarist and how he uses it than an inductor with so many ohms of series resistance and this many millihenries. The same wah pedal may sound incredible for one guitarist and not that great with the next one. What one person considers to be perfect is not necessarily going to be the next guy's idea of perfection - hence all these different wah pedals on the market and all the ones that have been made in the past (which were all mostly based on the Crybaby circuit except for maybe the Maestro Boomerang and Schaller Wau Wau Yoy Yoy).

[gambitpawn1]

I know this is an old thread, but throwing my two cents in to maybe help the next guy. The 1K pot added in serious to the inductor is used in a few pedals that I know of for the variable Q, obviously the Dunlog and also in the Berhenger Hellbabe, and the 1k is probably overkill cause there is really only a very small window in the pot that is a usable range probably right around that 33 to 55 ohm area. While doing research for moding my Morley power wah PWO, it has a 47 ohm resistor in place of where Dunlop has the 1K pot thinking I might try a 33 ohm resistor and a 25 ohm pot in series there.