Pedal output high pass calculation

[Unlikekurt]

Hi all,

Let's say you've got a coupling cap feeding the input of a volume control (voltage divider).  Does the series resistance of the voltage divider setting have any impact on the high pass corner frequency or is it purely based on the shunt resistance?

[antonis]

Any series resistance between cap and pot wiper should be considered part of HPF "upper" region..

For Lug 3 - wiper resistance = R1 and wiper - lug1 resistance = R2, corner frequency is f = 0.159 / [(C+R1)* R2]

[Rob Strand]

Let's say you've got a coupling cap feeding the input of a volume control (voltage divider).  Does the series resistance of the voltage divider setting have any impact on the high pass corner frequency or is it purely based on the shunt resistance?

If there is no load on the output side of the pot the high-pass frequency is unchanged  f3 = 1/(2*pi*C*R);  R = R pot. 

if there is a load on the pot of say RL  then
When the pot is on minimum clearly RL is output of the picture and you get same result as ignoring the load.

When the pot is on maximum the load resistance appears in parallel with the pot.   That reduces the resistance
seen by the cap and the cut-off rises.    So same formula but R = Rpot in parallel with RL.
If the pot is 100k and the load is 1M then R = 91k so not much different.  However, if the load is 10k  you get R = 9.1k which is a big difference.

You can do more elaborate calculation to see the effect at different pot positions.

[iainpunk]

you might want to bookmark this link if you are going to do simple highpass or lowpass calculations often, its a quick little rc freq. calculator.

cheers, Iain

[Unlikekurt]

Iain, I went there to start and then considered how the series resistance might come into play and posted my query.
Rob, gotcha on the load resistance as parallel to the shunt resistance.
Antonis,  I've tried to run the formula several times and keep getting bonkers results.  Can you show the work for 470pF, 1K series and 999k shunt?
I'm assuming the cap converts to 4.7e-10?
Thanks!

[iainpunk]

you only have to calculate the filter with the FULL resistance of the pot and the capacitor. if you turn it down, you change the level without filter change, since the resistance seen by the capacitor is still the same.

cheers

[Rob Strand]

Antonis,  I've tried to run the formula several times and keep getting bonkers results.  Can you show the work for 470pF, 1K series and 999k shunt?

Just a typo, should be,

f = 0.159 / [(C*(R1+ R2)]

Also, it's *exactly* the same as my formula Antonis has just split the pot into two R's

[Unlikekurt]

That makes sense now!  Its the overall resistance value of the divider.
Thanks for clearing that up in my head!

[anotherjim]

Well, I just make the volume pot value no more than 10k which won't have any significant loading effect feeding other pedals and amp channels. Then I also calculate the output cap considering some bright spark might plug into a 10k mixer line input, so that's 5k to calculate for. If that means the cap has to be electrolytic, then so be it!
If you hate the idea of electro caps in the audio path then pick the largest film cap you can and choose the lowest pot value that will work the bass end driving a 1M following load and hope that's all it gets used with!

With many vintage designs and some not so vintage amp simulator designs, the output pot can be very high, 500k to 1M. These don't have a snowflake in hells chance of driving a mixer, so you may as well calculate for a 1M amplifier input load as that is all they were intended to work into.

Anyway, you should do no harm making the cap too big.
It also proves why you should not use the output cap to force a bass-cut character. The only time you can do that is when you're certain of what the following load impedance is.

[iainpunk]

It also proves why you should not use the output cap to force a bass-cut character. The only time you can do that is when you're certain of what the following load impedance is.

the Shin-Ei Germanium FY-2 begs to differ! it uses only 2nf, to take out any and all low end...  (the Silicon version with its bridge tee filter doesn't do a better job with its output impedance tho)

for pedals meant to go in to other pedals or an amp, i use 100k pots,
when going in to other loads, like mixers, and , i prefer using a buffer after the volume pot, and using a 1k short circuit protection resistor.

cheers, Iain

[anotherjim]

Yeh, but Shin Ei were designing in a time when players used only one pedal and they assumed it was driving a guitar amplifier.