I have a standard common collector amp with RL parallel to RE in the output.
How do I calculate the output resistance?
I'm using the h-parameter model in AC analysis assuming that hrc=hoc=0...

As far as I know, Ro = Re / (hfc+1). Not sure about that though.

Circuit for example: http://upload.wikimedia.org/wikipedia/commons/archive/6/6e/20050626193503!Common_collector.png

 

I have a standard common collector amp with RL parallel to RE in the output.
How do I calculate the output resistance?
I'm using the h-parameter model in AC analysis assuming that hrc=hoc=0...

As far as I know, Ro = Re / (hfc+1). Not sure about that though.

Circuit for example: http://upload.wikimedia.org/wikipedia/commons/archive/6/6e/20050626193503!Common_collector.png

The circuit dont't seem to have a RL parallel to
RE.If ac is enabled,then RL is bypassed.I'am assumming that as far as the ac is concerned that would be a common collector with the fact that the out put is being taken from the emitter.I am not sure about this either but making a guest that hfc+1 is the gain. Dividing the gain into the input would not be output.

 

If we ignore the source resistance the output resistance of a CC amp is

Zo = [re + (R1//R2/B+1)]//RE and does not include the load resistor.

re is the emitter-base diode resistance (looking into the emitter), and is given by Vt/Ic where Vt=25mV at room temperature. So if you bias the transistor at 1mA this resistance is 25 ohms.

R1//R2 is reflected through the base by (Beta+1) and is in series with re, hence the addition. Beta is hfe, I think.

 

If we ignore the source resistance the output resistance of a CC amp is

Zo = [re + (R1//R2/B+1)]//RE and does not include the load resistor.

re is the emitter-base diode resistance (looking into the emitter), and is given by Vt/Ic where Vt=25mV at room temperature. So if you bias the transistor at 1mA this resistance is 25 ohms.

R1//R2 is reflected through the base by (Beta+1) and is in series with re, hence the addition. Beta is hfe, I think.

R1 and R2 are shorted out by the input source if the capacitor Xc<<(R1||R2).

 

Like I said, 'if' we ignore the source resistance. If we do not then the term R1//R2 should be changed to R1//R2//Rs. You're going to confuse this guy, Ron!

 

Like I said, 'if' we ignore the source resistance. If we do not then the term R1//R2 should be changed to R1//R2//Rs. You're going to confuse this guy, Ron!

Yeah, but your answer, and mine, are simplistic. The output impedance as a function of frequency is really a function of every component on the schematic, including all 3 caps, and Rsource (unless we assume it's zero).
In any case, I don't think ignoring the source is a good idea, because it will only be valid if Rsource=∞, which is highly unlikely.
Including the collector network would really complicate the issue.
I suspect that the instructor meant for the student to assume that X(Cin), X(Cout), X(Cc), and Rsource are all zero. He still needs an expression for deriving re as a function of R1, R2, and Re, and we have to assume that the transistor isn't saturated.

 

Ok. LOL! My message has to be ten characters, huh?

 

Ron H:

You said:"I don't think ignoring the source is a good idea, because it will only be valid if Rsource=∞, which is highly unlikely."

I agree with this point, but then you said:"I suspect that the instructor meant for the student to assume that X(Cin), X(Cout), X(Cc), and Rsource are all zero."

Why do you think the instructor meant for the student to assume that Rsource is zero? Your just said you didn't think it was a good idea, and that it is highly unlikely?

Besides, if one has a source impedance of zero, what do you need an emitter follower for?

I think the student should assume the capacitors are good AC shorts, and Rsource is finite. That's the typical case, and it's the case where an emitter follower is useful.

I wonder if our OP has gone away like they sometimes do.

Here's a heavy duty analysis of the common emitter stage and the common collector stage, including the effect of Rsource:

http://people.seas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_amps/bjt_amps.html

I've never seen this analysis carried out with quite so much algebra.

 

Electrician, you make some excellent points. It just seemed to me that the instructor would have added source impedance to the schematic, but maybe that's a leap he hoped the student would make.