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Bypass caps

Discussion in 'Electronic Basics' started by Steve Evans, Oct 16, 2004.

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  1. Steve Evans

    Steve Evans Guest


    How do you go about working out the proper value for an emmiter bypass
    cpacitor in a common emitter amplifier stage? (the cap thats' in
    paralell iwth the emtier resitor). AIUI, it has to form an AC ground
    at the signal frequency at the emitter so dows that imply that it must
    be less reactive than the equivalent interstage coupling cap would be?
    Sorry I"m not very good at explaining this. IOW: whats'; the formula
    for a emitter bypass cap?

  2. CFoley1064

    CFoley1064 Guest

    Subject: Bypass caps
    Hi, Steve. Aah, the ever popular

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    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    To find the impedance Xc of a cap C at a frequency f, you use the formula

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

    Just determine where you want the cut. To begin with, calculate the frequency
    where the impedance of the cap has the same magnitude as the emitter resistor.

    For audio applications, that should be somewhere below audio frequency (10 to
    30 Hz is a good start). It more or less depends on what kind of low end
    rolloff you want.

    Good luck with your homework.
  3. John Larkin

    John Larkin Guest

    No, less reactive than the effective emitter resistance of the

    The effective emitter resistance in ohms is about Re = 25/Ie, where Ie
    is transistor emitter current in milliamperes. Assuming the base drive
    is low impedance, pick a low-frequency cutoff point Fc, 20 Hz or
    whatever, and calculate C = 1 / (2*pi*Fc*Re) C in farads.

    If the base drive is not low impedance, it's a little more complex,
    and you can tolerate a smaller cap.

  4. Steve Evans

    Steve Evans Guest

    Perhaps I didn't explain it properly. Let me clarify:
    In choosing the apporpirate value for the bypayss cap., do I only need
    to consider making its reactance much less than that of RE at the
    lowest frequency of operation? I meaN,, is RE the only other factor
    I'm up against or do I have to take into account anyything else as
    well? Like is the impedence of the source signal and /or the imput
    imedence of the transistor relevant to this calculation as well?

  5. John Larkin

    John Larkin Guest

    To quote myself,

    "If the base drive is not low impedance, it's a little more complex,
    and you can tolerate a smaller cap."

    Rather than hassle the math, just use the calculated value I
    suggested, or a bit more. Or use an opamp.

  6. CBarn24050

    CBarn24050 Guest


    Hi, the old "rule of thumb" is to make the capacitor reactance one tenth of the
    emitter resistor at the lowest frequency of interest.
  7. John Larkin

    John Larkin Guest

    That won't work right if the effective emitter impedance (the 25/Ie
    thing) is much lower than the external resistor. For, say, Ie = 2 mA,
    Re is only 12 ohms, but the external resistor might be a k or so.

  8. Steve Evans

    Steve Evans Guest

    Okay, so the full formula for the capacitor to work properly in any
    situation is... what?
  9. Dbowey

    Dbowey Guest

    re: "Okay, so the full formula for the capacitor to work properly in any
    situation is... what?"

    Buy a decent book and study a while. Quit looking for simple answers until you
    are able to understand more.
  10. John Larkin

    John Larkin Guest

    Uh, my introductory no-price consulting offer has just expired, and a
    general solution would be a nuisance. Crack a textbook and do a heap
    of algebra if you want the gory details.

  11. CBarn24050

    CBarn24050 Guest

    Subject: Re: Bypass caps
    Well John it depends on what you mean by won't work right. It will effectively
    bypass the emitter resistor. There is no way I know of to bypass Re.
  12. John Larkin

    John Larkin Guest

    I was referring to the value of capacitance needed, emitter to ground,
    to get a desired 3dB low frequency roll point. Gain is max when the
    emitter is effectively grounded, and 100% of the input signal voltage
    appears across the inaccessible Rb. Gain is then about Rl / Rb which
    is equivalent to Gm * Rl. Any added impedance from emitter to ground
    reduces gain.

    This is basic stuff, in all the books.

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