Theory question on computing roll-off frequencies

[JonFrum]

Using the Easy Face as an example - how would you compute the rolloff freq due to the 1uF blocking cap at the output?  More specifically, what resistance/impedence would be plugged in to a R/C equation? It took me a few years to figure it out in tube amps, and now I have to start all over again.   

[Sir H C]

Fc = 1/(2*pi*r*c)

that should be the 3 dB rolloff point.

[JonFrum]

Fc = 1/(2*pi*r*c)

that should be the 3 dB rolloff point.

My question is, what would I use as "r" in the above application?

[brett]

Hi
IF there's an effect next in line, the resistance to ground will probably be less through the next effect than through the 500k pot, so R will be approximately equal to the resistance of the next effct.
IF there's a tube amp or buffer next in line, the 500k pot might be an equal or lesser impedance, so R will be at least 250k.

A good rule of thumb for choosing output caps is to estimate R in kohms for the lowest impedance thing that it's likely to drive (e.g. 50k).  The output cap is ideally more than 10 divided by that number (in this case 0.2 uF).  At a pinch, it can be 5 divided by that number (in this case, 0.1uF).

One reason why there's so many big output caps is that in the 60s and 70s, small electro caps of 1 or 2 uF were cheaper than greencaps of 0.2 or 0.5uF (hence that ridiculous 2.2uF input cap in fuzzfaces).  Another was that many pedals had low input impedance by today's standards.
cheers

[JonFrum]

Brett

Thanks for the answer - it went right to my issue. You don't see 1uF blocking caps in tube amps.

[WGTP]

By using a large output cap, you are effectively removing it's effect on the bass roll off.  Some FF's use .1uf or .01uf or less to make the tone less wooley.