Help me understand a bright cap with a resistor in series

[framos]

Hi everyone, here´s a beginner question.

I added a resistor in series with my Plexi´s 500pf bright cap on the gain pot in order to 'reduce' a bit the ammount of boost. I seems to work the way I intended, but I just want to make sure I really understand how it works.

Here´s is how I see it: say R1 is the resistance offered by the pot, and R2 is the resistance added in series with the bright cap.

Then the higher frequencies that bypasses the cap 'sees' the parallel R1//R2 resistance while the other (lowet) frequencies 'see' R1 only.

Is this understanding correct?

Thanks!

[martin manning]

I would explain it this way:
The volume pot is a voltage divider with R1 being the resistance from the input to the wiper and R2 being the resistance from the wiper to ground.
The output voltage (ignoring source and load impedances) is Vin * R2/(R1+R2).
The bright cap is across R1. At high frequencies the cap looks like a short, so the output voltage is then Vin * R2/((R1//0)+R2) = R2/R2 =1 (no attenuation).
If you put a resistor R3 in series with the cap, at high frequencies the output voltage is Vin * R2/((R1//R3)+R2).
Since R1//R3 is > 0, there will be some attenuation of high frequencies, as you observed.

[Winder]

Great explanation. Also:

https://en.wikipedia.org/wiki/RC_circuit

[framos]

I would explain it this way:
The volume pot is a voltage divider with R1 being the resistance from the input to the wiper and R2 being the resistance from the wiper to ground.
The output voltage (ignoring source and load impedances) is Vin * R2/(R1+R2).
The bright cap is across R1. At high frequencies the cap looks like a short, so the output voltage is then Vin * R2/((R1//0)+R2) = R2/R2 =1 (no attenuation).
If you put a resistor R3 in series with the cap, at high frequencies the output voltage is Vin * R2/((R1//R3)+R2).
Since R1//R3 is > 0, there will be some attenuation of high frequencies, as you observed.

It´s clear!

High frequencies Vin * R2/((R1//R3)+R2)
Other frequencies Vin * R2/(R1+R2)

So yes, only the high frequencies would see the parallel resistance introduced by R3, that was the key point I wanted to confirm.

Thanks!