Treble bleed frequency cutoff

Question

I built an attenuator for my guitar amplifier. I took the circuit from here but left out the 4 ohm toggle since my amplifiers all have 8 ohm outputs. Below is a simplified schematic of the circuit (with the L-pad set to 20 dB attenuation):

The 4.7u capacitor is supposed to be a treble bleed. My intuitive understanding is that the capacitor allows the high frequencies to bypass the L-pad while the low frequencies are still attenuated. However, I don't notice any difference with the switch open or closed while playing.

I tried to measure the frequency response of the whole attenuator with a signal generator and a scope. The amplitude of the output signal did not change when toggling the capacitor for any frequency up to 20kHz. However I realise that the signal generator and scope have very different output and input impedance compared to the amplifier and speaker, so I don't really trust these results.

My next idea was to calculate the expected cutoff of the treble bleed, but I am not sure how exactly the high-pass filter equation applies here. My understanding is that the speaker is part of the filter, so

$$ R = \frac{1}{\frac{1}{0.89}+\frac{1}{8}} $$ $$ f_c = \frac{1}{2\pi\times4.7\text{u}\times R} = 42281 $$

42 kHz is of course too high for audio, so now I am not sure if my analysis is correct.

• What is the correct way to analytically determine the treble bleed cutoff frequency?
• How can I reliably verify this using the signal generator and scope?

Answer 1

The amp is 0 Ohms and the speaker may be neglected. But the ESR of the capacitor must be <0.1 Ohm which rules 99% of electrolytics, so a metal filter cap from a microwave oven or store must be used.

In theory, the cap is not a LPF but a 4kHz HPF The -20dB pad rises to -17 dB at 4kHz

Simply use 7.2 with C to compute breakpoint.

But to make a LPF is harder with a pad as R becomes 0.89 and needs a cap 8x bigger as a shunt.