A math question about attenuator bypass caps

[Mark Hammer]

I have a problem I'm trying to solve and I think my math is misguided.  I need straightening out.

Say you have a fixed resistive voltage divider, and you were interested in using a simple network to bypass the "input resistor" to yield some treble advantage.  The classic case is the compensation cap on a fender guitar volume control, or the "bright" switch on a Fender amp.

I have always assumed that if one had a resistive divider and a cap in parallel with the input leg of the attenuator network, the treble advantage could be calculated by treating the cap and the ground-leg resistor as a highpass filter.  So, if I had a pair of 100k resistors as a divider, and bypassed the input leg with a .001uf cap, that would provide treble advantage above 1 / [2*pi*.1meg*.001uf] = 1591hz.

Where it gets fuzzy and confusing for me is when I try to figure out how much treble advantage is achieved.  Clearly the full-bandwidth signal is attenuated by half via the pair of 100k resistors in the divider.  So how much more treble is there?

Now, the next step in my mission.

I know that over time, Fender replaced the cap-only compensation in guitar volume pots with a cap and fixed resistor in series.  This was apparently to yield even smoother compensation as the volume pot is turned down.  When an RC bypass is used, instead of just a cap, how does one calculate the point at which the treble advantage begins and the amount of treble advantage derived?

[R.G.]

I have always assumed that if one had a resistive divider and a cap in parallel with the input leg of the attenuator network, the treble advantage could be calculated by treating the cap and the ground-leg resistor as a highpass filter.  So, if I had a pair of 100k resistors as a divider, and bypassed the input leg with a .001uf cap, that would provide treble advantage above 1 / [2*pi*.1meg*.001uf] = 1591hz.

That's only true if the lower leg of the divider is less than 1/10 of the impedance of whatever uses the voltage at the output of the divider. If there is an impedance sucking current out of the junction, it must be included in the calculations. Notice that if the driving source has a significant impedance, like the approximately 62K plate resistance of a 12AX7, you have to worry about that one too.

Doing dividers like this with a circuit before and after the divider is a classic first circuits course trick question. What the prof is really asking is "did you remember that the circuit before and after the divider have impedances that affect things too?"

Where it gets fuzzy and confusing for me is when I try to figure out how much treble advantage is achieved.  Clearly the full-bandwidth signal is attenuated by half via the pair of 100k resistors in the divider.  So how much more treble is there?

Let's do a simple case. Divider driven by a voltage source (i.e. zero driving impedance) and the output sensed by a depletion mode MOSFET device with Zin in the teraohms, so it has no effect. At some very high impedance, the cap is a much lower impedance than either the top or bottom resistor, and so the cap is an effective short circuit. The full signal voltage, minus an eyelash, appears at the sensing point. At some lower frequency, say where the cap and the top resistor impedances are equal, the signal is 2/3 of full input. At a frequency where the cap impedance is say 10x the top resistor, you can ignore the cap.

I know that over time, Fender replaced the cap-only compensation in guitar volume pots with a cap and fixed resistor in series.  This was apparently to yield even smoother compensation as the volume pot is turned down.  When an RC bypass is used, instead of just a cap, how does one calculate the point at which the treble advantage begins and the amount of treble advantage derived?

Same way. The resistor in series with the cap puts a limit on how much treble gets "shorted" around the top resistor.

The frequency where these things happen is always F = 1/(2*pi*R*C) for the R and the C which are involved. You just have to be sure you're involving the right number of Rs and Cs.

[Mark Hammer]

Thank you sir.