# Thread: cathode bypass calc. what am I doing wrong?

1. ## cathode bypass calc. what am I doing wrong?

The Single Ended Output Stage article on MerlinB's website gives the following formula for calculating the bypass capacitor value:

Ck=1/(2*pi*f*Rk)

using the values in his example circuit this is

Ck=1/(2*pi*10*220)

=72uF

72uF=0.072F, correct?

but when I do the calculation (using the calculator utility in Windows) I keep getting 7.2xxxxxxxxxx (rounded off to two places) rather than the expected 0.072

I am off by a factor of 100 and I can't see why.

Also, I solved the same equation for f and get

f=1/(2*pi*Rk*Ck)

Is this correct?

Thanks, it's been a loooong time since High School Algebra!

2. Originally Posted by Groover
72uF=0.072F, correct?

3. Originally Posted by Groover
but when I do the calculation (using the calculator utility in Windows) I keep getting 7.2xxxxxxxxxx
Are you sure there isn't an "e-5" at the end of those numbers? (Which means x10^(-5) )

4. Originally Posted by Merlinb
Are you sure there isn't an "e-5" at the end of those numbers? (Which means x10^(-5) )
Yes there was. I didn't know what it meant, and thought it was some sort of error message due to more places than the calculator could manage or something.

so one micro = 0.000,001
and one pico = 0.000,000,000,001
does one nano = 0.000,000,001 ?

Did I solve for f correctly?

5. think so,

1pF = 0.000000000001 F = 1e-12 F ( see why they use the e?)
1nF = 0.000000001 F = 1e-9 F
1uF = 0.000001 F = 1e-6 F
1mF = 0.001 F = 1e-3 F

some use mF to mean "micro" farad, or even mmF to be millimicro farad

micro is a greek "mu" and windows makes it an "m" as an english character...messes people up!

chemistry joke: 1.660538863zM = 1! (not factorial, although that works too)

6. Originally Posted by Groover
Did I solve for f correctly?
Yes.

f = 1/(2 pi R C)
is like Ohm's law for filters, commit it to memory!

7. ## one more question for you Merlin (or others)

That is a very simple formula compared to the formulae in the "Choosing Cathode Bypass Capacitors" article by you and David Ian James.

What neccessitates the higher complexity of the information and formulae in that article?

8. Originally Posted by Groover
That is a very simple formula compared to the formulae in the "Choosing Cathode Bypass Capacitors" article by you and David Ian James.

What neccessitates the higher complexity of the information and formulae in that article?
Mental masturbation

just do this ... it's an old ham radio trick for -3dB points in a simple RC circuit:

159160/ RC = Fz

example:
R = 1500 ohms
C = 10uF
1500 x 10 = 15000
159160/15000 = 10.67Hz

here's your circuit the ham radio way to find Fhz at -3db
10 x 220 = 2200
159160/2200 = 72.3Hz

9. Is this formula {f=1/(2piRC)} the same for all filter configurations?

The way I see it there are at least 3:

R in parallel with C

R in signal path and C to ground

C in signal path and R to ground

I suppose the answer is in the book, actually.

10. Originally Posted by Groover
Is this formula {f=1/(2piRC)} the same for all filter configurations?
Yes. It works for CR and RC. In the case where you have a resistor in parallel with C then you need to find out what the source resistance is, and you imagine this to also be in parallel with the cap.

That is a very simple formula compared to the formulae in the "Choosing Cathode Bypass Capacitors" article by you and David Ivan James.
That's because cathode bypassing is dealing with small amounts of negative feedback, which greatly complicates things. The simple formula becomes an approximation, but it's close enough for rock and roll.

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