Bypass caps

S

Steve Evans

Guest
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
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.
 
On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

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?
Click to expand...
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
Click to expand...

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
 
On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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
Click to expand...
Hi, the old "rule of thumb" is to make the capacitor reactance one tenth of the
emitter resistor at the lowest frequency of interest.
 
On 18 Oct 2004 15:01:37 GMT, cbarn24050@aol.com (CBarn24050) wrote:

On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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

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

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
 
On Mon, 18 Oct 2004 13:22:51 -0700, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On 18 Oct 2004 15:01:37 GMT, cbarn24050@aol.com (CBarn24050) wrote:

On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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

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

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.
 
Subject: Re: Bypass caps
From: John Larkin jjlarkin@highSNIPlandTHIStechPLEASEnology.com
Date: 18/10/2004 21:22 GMT Standard Time
Message-id: <q898n0tacolpgmk7g57a3vbcd7j0q4pgam@4ax.com

On 18 Oct 2004 15:01:37 GMT, cbarn24050@aol.com (CBarn24050) wrote:

On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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

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
Click to expand...
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.
 
On 19 Oct 2004 01:09:14 GMT, cbarn24050@aol.com (CBarn24050) wrote:

Subject: Re: Bypass caps
From: John Larkin jjlarkin@highSNIPlandTHIStechPLEASEnology.com
Date: 18/10/2004 21:22 GMT Standard Time
Message-id: <q898n0tacolpgmk7g57a3vbcd7j0q4pgam@4ax.com

On 18 Oct 2004 15:01:37 GMT, cbarn24050@aol.com (CBarn24050) wrote:

On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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

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




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.
Click to expand...

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
 
Subject: Bypass caps
From: Steve Evans smevans@jif-lemon.co.mars
Date: 10/16/2004 5:23 PM Central Daylight Time
Message-id: <5083n0plp32l8haiud9cvlm6b0i7av2tdb@4ax.com

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
Click to expand...
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.

Good luck with your homework.
Chris
 
On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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.
Click to expand...
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.
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.
 
On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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.

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.
Click to expand...

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
 
On Mon, 18 Oct 2004 23:20:28 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

On Mon, 18 Oct 2004 13:22:51 -0700, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On 18 Oct 2004 15:01:37 GMT, cbarn24050@aol.com (CBarn24050) wrote:

On Sun, 17 Oct 2004 23:14:51 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

On Sat, 16 Oct 2004 16:45:38 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Sat, 16 Oct 2004 22:23:13 GMT, Steve Evans
smevans@jif-lemon.co.mars> wrote:

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

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.

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