Emitter follower related question
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 11th November 2009, 01:37 PM #1 diyAudio Member     Join Date: Feb 2008 Emitter follower related question Hello, When biasing an emitter-follower with a current source in the place of a typical emitter resistor, can the equivalent emitter resistor be calculated by V-emitter / I-emitter? I need this value in order to calculate input and output impedances. Thanks!
diyAudio Member

Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 6
Quote:
 Originally Posted by IG81 Hello, When biasing an emitter-follower with a current source in the place of a typical emitter resistor, can the equivalent emitter resistor be calculated by V-emitter / I-emitter? I need this value in order to calculate input and output impedances. Thanks!
The impedance from the current source can be calculated from delta-V/delta-I. That is much higher than Ve/Ie. So what you see from the base is that very high Isource impedance in parallel with the load, and most of the times it will be the load that dominates, meaning you can disregard the Isource.

The input impedance looking into the emitter is that Isource in parallel with whatever the impedance into the emitter itself is, and the latter will dominate, meaning you can disregard the Isource.

jd
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 11th November 2009, 02:21 PM #3 diyAudio Member     Join Date: Feb 2008 Thanks for a quick answer janneman. If I understand correctly, I can disregard the Isource impedance and simply use Rload in my calc. How about for the base? I also have a current source in place of the base-to-ground resistor in the "voltage divider", how does that go? ig
diyAudio Member

Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 6
Quote:
 Originally Posted by IG81 Thanks for a quick answer janneman. If I understand correctly, I can disregard the Isource impedance and simply use Rload in my calc. How about for the base? I also have a current source in place of the base-to-ground resistor in the "voltage divider", how does that go? ig
Well in any case the Isource is (very) high impedance. So the impedance of the base divider is the Isource in parallel with the resistor from base to supply (remember that for ac and signal, the supply is considered ground).

But do you have a circuit diagram? I suspect what you are trying to do at the base will not be thermally stable and might need individual adjustment for each device.

jd
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diyAudio Member

Join Date: Feb 2008
Here's what it looks like, don't look at component values, just topography.

ig
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diyAudio Member

Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 6
Quote:
 Originally Posted by IG81 Here's what it looks like, don't look at component values, just topography. ig
Can't find anything wrong here.

jd
__________________
/Yes! Its out: Linear Audio Vol 5!
I'm not an "accademic", just a plodder who loves a challenge - Ian Hegglun

diyAudio Member

Join Date: Feb 2008
Quote:
 Originally Posted by janneman Can't find anything wrong here. jd
So for the input Z of the circuit, can I also "neglect" Isource and just use the +rail-to-base resistor || with Zin-base?

Thanks,

ig

diyAudio Member

Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 6
Quote:
 Originally Posted by IG81 So for the input Z of the circuit, can I also "neglect" Isource and just use the +rail-to-base resistor || with Zin-base? Thanks, ig
Well, depends on the resistor values. If the Isource has 150k dynamic impedance and the other R is also 150k, clearly you can't ignore Isource: the parallel value is 75k.
If the R is 10k, the parallel value is about 9.3k. Are you familiar with calculating parallel impedances?

jd
__________________
/Yes! Its out: Linear Audio Vol 5!
I'm not an "accademic", just a plodder who loves a challenge - Ian Hegglun

diyAudio Member

Join Date: Feb 2008
Quote:
 Originally Posted by janneman Well, depends on the resistor values. If the Isource has 150k dynamic impedance and the other R is also 150k, clearly you can't ignore Isource: the parallel value is 75k. If the R is 10k, the parallel value is about 9.3k. Are you familiar with calculating parallel impedances? jd
Yeah, it's more how to translate a current source into an impedance that I was looking into here. I think I can see what you mean by "dynamic impedance" that should not be neglected if the lower Z it can take is not big enough to make the parallel impedance be close enough to the resistor above.

az

diyAudio Member

Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 6
Quote:
 Originally Posted by IG81 Yeah, it's more how to translate a current source into an impedance that I was looking into here. I think I can see what you mean by "dynamic impedance" that should not be neglected if the lower Z it can take is not big enough to make the parallel impedance be close enough to the resistor above. az
An Isource can be considered as an impedance but you can't measure it by its voltage over its current. You measure it by varying the voltage a bit and see how the current varies, then the impedance (the dynamic impedance) is delta-V/delta-I. Ideally, the current doesn't vary with voltage and then the dynamic impedance is infinite...

jd
__________________
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