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

  1. #1
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    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. #2
    Capacitater Steve Conner's Avatar
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    Quote Originally Posted by Groover View Post
    72uF=0.072F, correct?
    No. A microfarad is one millionth of a farad, so 72uF = 0.000072 farad
    "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

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    Quote Originally Posted by Groover View Post
    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) )

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    Quote Originally Posted by Merlinb View Post
    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?

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    Senior Member tedmich's Avatar
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    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!

    oh, a zeptofarad is e-24!
    chemistry joke: 1.660538863zM = 1! (not factorial, although that works too)
    Last edited by tedmich; 04-06-2010 at 05:12 PM.

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    Quote Originally Posted by Groover View Post
    Did I solve for f correctly?
    Yes.

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

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    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?

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    Senior Hollow State Tech Bruce / Mission Amps's Avatar
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    Quote Originally Posted by Groover View Post
    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
    Last edited by Bruce / Mission Amps; 04-07-2010 at 01:32 AM.
    Bruce

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    303-955-2412

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    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. #10
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    Quote Originally Posted by Groover View Post
    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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