Re: LPF question

Posted by induction on Oct 08, 2021; 1:17am
URL: http://guitar-fx-layouts.238.s1.nabble.com/LPF-question-tp50267p50274.html

You've got it mostly right. The part of the circuit between c2 and c6 in your schematic is actually 1:1 with the relevant circuits in the links if you appropriately add the impedances in series and in parallel.

I'll use uppercase letters for the formulae in the links, and lowercase letters for components in your schematic:
For the corner frequency calculation, we have fc = 1/(2*pi*sqrt(R1*R2*C1*C2))
Comparing to your schematic, we have R1 = r4+rv1, R2 = r5+rv2, C1=c3, C2=c4.

rv1 and rv2 both range from 0 to 50k, so fc ranges from 298.6 Hz when rv1=rv2=50k, to 4822.9 Hz when rv1=rv2=0. These are pretty low frequencies for a low-pass filter, so I'm guessing this circuit is for bass guitar or a preamp for upright bass or something similar.

The gain is given by Av = 1+R3/R4. Without c5, the gain would be frequency-independent, but the impedance of a capacitor is given by Zc = 1/(2*pi*f*C), which is frequency-dependent. Another name for frequency dependent gain is EQ, so we really should account for c5 in the frequency calculations.

This is really only a slight complication though because c5 is tiny. The impedance of a cap and resistor in parallel is |Z| = 1/sqrt(R^-2+(2*pi*f*C)^2), which is gives R3=10k at 20 Hz and 9.9k at 20 kHz. This gives a gain of 1.45 at all frequencies that humans can hear, which means we can ignore it for the purpose of calculating the corner frequency of the filter. (At f=160 kHz, the gain is about 1.3, and as the frequency approaches infinity, the gain slowly drops to 1. So c5 is probably just there to prevent oscillation at ultrasonic frequencies.)

As a final note, I'll point out that the corner frequency calculations are symmetric with respect to R1, R2, C1, and C2. So most likely, rv1 and rv2 should a dual-gang pot rather than two separate pots, since the effect of both pots would be identical.