Discussion in 'Pickups & Electronics [BG]' started by papalo, Apr 28, 2003.

1. ### papalo

Apr 16, 2003
I understand the way capacitors work, allowing only only certain frequenicies to pass through them etc.
I have seen anywhere from .22mfd to .80mfd on basses, so my question is:
Does anyone know the actual frequency that is being allowed to pass through depending on the cap?

For instance if you use a .50 cap for your passive tone control, what are the frequencies that are passing through? Etc. Etc. Or which ones are being cut?
Is there some sort of formula for it? thank you.

2. ### l0calh05t

Oct 14, 2001
Cottbus, Germany
there is a formula to calculate how high the impedance (frequency dependant resistance) of a capacitor is.

Zc = 1/(2*Pi*Freq*C) (if I remember correctly)

3. ### papalo

Apr 16, 2003
Thank you. I'm not certain that it answers my question but I appreciate the response.

Thanks.

4. ### papalo

Apr 16, 2003
Bump. Anyone? thanks

5. ### throbbinnut

It depends on the rest of the circuit, specifically, the resistance that the capacitor sees in parallel with itself determines the effect.

The RC time constant can be used to figure out the -3dB corner frequency.

So, here's some figures from a spreadsheet I made. The columns will be goofed up, but hey...

Enter resistance (ohms) Enter capacitor (uF) Calculated Frequency (Hz)
33000 0.000047 102666.45
47000 0.0047 720.85
180 100 8.85
1500 25 4.25
100000 0.00022 7237.98
22000 0.000047 153999.68
150 25 42.46
68000 0.33 7.10
130 47 26.06

So you really need to know the R that is in circuit, which depending on things may include the pickups, tone control resistance, etc. One way to measure this would be to take an ohmmeter and put its leads across the cap in question to see what the resistance that the cap is seeing (this will change as the Tone control is rotated, as this is what makes the frequency cut change.)

f = 1 / (2*pi*R*C)

Plug in R and C and calculate f in Hz. Remember that .1uF should be entered as .1 x 10^-6.

Chris

6. ### papalo

Apr 16, 2003
Thank you very much Chris.
I am confused about one part of your post. How does the value of the capacitor have any relavence on the resistance?
I didn't realize that a capacitor in a circuit changes resistance, actually I am pretty certain it doesn't.
The resistance will change as I turn the tone knob because the tone pot is a variable resister.
Either way the forumla you presented, if indeed is for real, is way over my head.
thank you

7. ### l0calh05t

Oct 14, 2001
Cottbus, Germany
the capacitor itself has a impedance (frequency dependant resistance) ,the total impedance is Z = sqrt( Rohm² + Xcap²) (if wired in series)

8. ### Joris

Calculating RC filters (that's what you're doing) requires quite some knowledge about complex numbers and mathematics in general.

The simplest approach to RC filter design really is what everyone else said: apply those equations. If you can't make heads or tails from them, have someone do it for you. We'll be glad to help, if you post your specific circuit, with all the component values.

Regards,
Joris.

9. ### geshel

Oct 2, 2001
Seattle
I believe what he is saying is not that the capacitor changes the resistance, but that you have to measure the resistance across the capacitor due to other parts of the circuit (eg, the tone knob) to calculate the frequency.

f = 1 / (2*pi*R*C) shouldn't be "way over your head". It's just a simple equation: multiply two times pi times the resistance (mentioned above) times the capacitance. Divide 1 by that number. That's the filter frequency. Since most capacitors are listed in uF (microfarads), divide the value listed by one million first (0.1 uF = 0.1 / 1,000,000 F).

10. ### Paul A

Dec 13, 1999
Hertfordshire U.K!
11. ### papalo

Apr 16, 2003
Ah yes of course, I write this stuff down and think about it all day Sorry, consider me stupid but It is over my head. I was looking for a simple answer which there seems not to be one. I am a musician, not a 'tech' of any kind. thanks for all who helped