Bypass caps

Hiya,

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?

Steve

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Fat, sugar, salt, beer: the four essentials for a healthy diet.

Hi, Steve. Aah, the ever popular

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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. Chris

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.

John

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? Thanks,

Steve.

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Fat, sugar, salt, beer: the four essentials for a healthy diet.

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.

John

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

the

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.

John

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.

John

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

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Fat, sugar, salt, beer: the four essentials for a healthy diet.

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.

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.