Common collector output impedance

Discussion in 'General Electronics Chat' started by unseensoul, May 22, 2009.

  1. unseensoul

    Thread Starter Member

    Dec 13, 2008
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    Many textbooks derive the small signal output impedance of a common collector amplifier by short circuiting the small signal input and adding a source to the small signal output. Fair enough...

    However, I was trying to use a different approach based on Thevenin's theorem by setting all independent sources to zero and measuring the impedance seen at the output but it seems to not work. Why?

    Edit: I've attached the model I was using below...
     
    Last edited: May 22, 2009
  2. studiot

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    Are you trying to derive h parameters for common collector??
     
  3. unseensoul

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    Dec 13, 2008
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    I'm just trying to measure the output impedance of a common collector amplifier using Thevenin's approach instead of inserting a source to the output to measure it.
     
  4. The Electrician

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    Show a schematic of the model you're using, with all the sources, etc. Then maybe somebody could help you.
     
  5. unseensoul

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    Dec 13, 2008
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    Here's the model I'm using...

    [​IMG]
     
  6. PRS

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    Aug 24, 2008
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    I use this method: The resistance looking into the emitter is re. Where re=Vt/Ic and Vt is a constant dependent on the temperature (26 mV at room temp). So if you're Ic is 1 mA then re= 26 ohms.

    re is in parallel with RE, the emitter resistor. So a quick approximation is re//RE.

    But the source resistance and the base bias resistors reflect through from the base to the emitter and are in series with re. So a fuller equation would be:

    Zout=[(Rs//Rb) + re]//RE

    This has worked well for me.
     
  7. unseensoul

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    Dec 13, 2008
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    If you replace the ∏-model (in the image I provided) by the T-model your explanation fits quite well and Thevenin's Theorem works in measuring the resistance at the output terminals. Thanks ;)
    However, Thevenin's Theorem weirdly seems not to work if you're working with the ∏-model :confused:
     
  8. The Electrician

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    Shouldn't you divide Rs||Rb by (β+1)?
     
  9. The Electrician

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    The input source is already shorted (zero); if the source connected to the output for the textbook method is a current source, then setting it to zero means the output is open circuited.

    Show what you did next.
     
  10. unseensoul

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    Dec 13, 2008
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    I don't want and I'm not using any source attached to the output in order to calculate the output impedance. Actually, I'm trying to measure it using the thevenin equivalent resistance approach.

    However, that method doesn't work with the ∏-model as you'll get Rout = r_{∏} // Re if you short circuit the input (assuming the input source has zero internal resistance).

    But, if I replace the ∏-model (in the analysis) by the T-model I easily obtain Rout = re // Re (assuming the input source has zero internal resistance) which is the right answer.

    So why doesn't this work with the ∏-model? Perhaps I'm missing something...
     
  11. hgmjr

    Moderator

    Jan 28, 2005
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    The hybrid pi model that is contained in this wikipedia.org article shows the presense of the output resistance Ro in parallel with the output current source. Is it this output impedance that you are looking for?

    hgmjr
     
  12. steveb

    Senior Member

    Jul 3, 2008
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    I think what is happening is that you are missing the fact that the base side has to be reflected over to the emitter side using the current scaling. This needs to be done with either model.

    I think that The Electrician was pointing this out earlier when he said that the Rs//Rb combination needs to be divided by (beta+1)

    Looking back into the base side from the emitter side, you see hib (rb) because you are really saying that the resistor is hie, but the current scaling is (beta+1), so you use the hib value which is 1/(beta+1) times hie.

    With the conductance model you need to do the same thing which will bring gm into the equation for the output impedance. I think that 1/gm will be in parallel with Rpi, but I'd have to calculate to be sure.

    Anyway, in both cases, the current source parameter (either beta or gm) needs to show up in the final equation. So, basically to correct your T model, you need to put hie in place of hib in the equivalent circuit. Then when you reflect the hie using the current scaling, it will turn into rb (hib).

    The calculation of output impedance is always troublesome and confusing. There is something very nonintuitive about thinking about applying a source into the output. But, you can't just look backward and treat all resistors as they are. They are scaled, or reflected based on current gain. Because of this, I prefer to calculate by brute force to avoid errors. But, there is real value in being able to quickly determine the answer by inspection, which is possible with experience.
     
  13. The Electrician

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    In order to get Rout = r_{\pi} || Re, you apparently are assuming that the resistance looking back from the emitter to the base is r_{\pi}. How did you get that? I think it should be (with the input shorted)
    \frac{r_{\pi}}{\beta+1}   \mbox{~or~}   \frac{r_{\pi}}{g_{m} r_{\pi}+1}
     
  14. The Electrician

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    With the input not shorted, I get an output resistance of:

    \frac{R_{B}+r_{\pi}}{g_{m} r_{\pi}+1}\mbox{~||~}R_{E}

    alternatively:

    \frac{(R_{B}+r_{\pi})R_{E}}{(g_{m} r_{\pi}+1)R_{E}+R_{B}+r_{\pi}}

    If there is an input source resistance, Rs, just replace RB with RB||Rs.

    By setting RB to zero, we can get an expression for the output resistance, with the input shorted:

    \frac{r_{\pi}}{g_{m} r_{\pi}+1}\mbox{~||~}R_{E} \mbox{~or~} \frac{r_{\pi} R_{E}}{(g_{m} r_{\pi}+1)R_{E}+r_{\pi}}
     
    Last edited: May 23, 2009
  15. studiot

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    You have mentioned this several times. But then discuss theoretical analysis. Please clarify what you are actually doing.

    Ae you making physical on a real transistor or simulated measurements in a program, or are you trying to analyse and calculate?

    Greater detail of your methodology would be very useful.
     
  16. unseensoul

    Thread Starter Member

    Dec 13, 2008
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    I was indeed missing that. Although The Electrician has mentioned it before I was not getting it. However, I thought that current scaling would play no role in the solution once the input was short circuited (therefore the dependent current source would be 0) :confused:
     
  17. unseensoul

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    Dec 13, 2008
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    I think I've discussed what I was really doing above.
     
  18. unseensoul

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    Dec 13, 2008
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    Finally I got it! I think I was mixing things; that's why I was so confused :(

    Thank you very much you all for the help provided otherwise I'd still be stuck on it ;)
     
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