# Bypass caps

S

#### Steve Evans

Jan 1, 1970
0
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

C

#### CFoley1064

Jan 1, 1970
0
Subject: Bypass caps
From: Steve Evans [email protected]
Date: 10/16/2004 5:23 PM Central Daylight Time
Message-id: <[email protected]>

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

Hi, Steve. Aah, the ever popular

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created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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.

Chris

J

#### John Larkin

Jan 1, 1970
0
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?

No, less reactive than the effective emitter resistance of the
transistor.
Sorry I"m not very good at explaining this. IOW: whats'; the formula
for a emitter bypass cap?

Steve

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.

John

S

#### Steve Evans

Jan 1, 1970
0
No, less reactive than the effective emitter resistance of the
transistor.

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.

J

#### John Larkin

Jan 1, 1970
0
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.

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

C

#### CBarn24050

Jan 1, 1970
0
[/QUOTE]

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

J

#### John Larkin

Jan 1, 1970
0
Hi, the old "rule of thumb" is to make the capacitor reactance one tenth of the
emitter resistor at the lowest frequency of interest.

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

S

#### Steve Evans

Jan 1, 1970
0
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.

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

D

#### Dbowey

Jan 1, 1970
0
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.

J

#### John Larkin

Jan 1, 1970
0
Okay, so the full formula for the capacitor to work properly in any
situation is... what?

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

C

#### CBarn24050

Jan 1, 1970
0
Subject: Re: Bypass caps
From: John Larkin [email protected]
Date: 18/10/2004 21:22 GMT Standard Time
Message-id: <[email protected]>

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

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.

J

#### John Larkin

Jan 1, 1970
0
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.

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

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