Resistance of CB Amp

Discussion in 'Homework Help' started by Guitarras, Jun 30, 2012.

  1. Guitarras

    Thread Starter New Member

    Dec 10, 2010
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    Can someone show me that, if we don't neglect ro, the resistance seen from the collector is given by:

    R = ro*[1 + gm(RE//rpi)]
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What steps have you taken to try to solve it thus far?

    What is R (mathematically, what does it represent)?

    In general, how do you determine the resistance as seen from a particular viewpoint in a circuit?

    What is the small signal equivalent circuit that you are working with?
     
  3. Guitarras

    Thread Starter New Member

    Dec 10, 2010
    14
    0
    Thanks for the reply.
    [​IMG]
    So, I've tried already using the ∏ hibrid model (with ro) and I obtained that the resistance seen from the collector is:

    ro + (RE // [r∏/(hfe+1)])

    Knowing that r∏ = hfe/gm and other expressions i tried to obtain the expression

    ro*(1 + gm(RE//r∏))

    but I didn't succeed.

    -- // --

    I will answear your question about how to determine the resistance seen from a viewpoint with a practical example of a CB amplifier.
    [​IMG]

    Let's determine the resistance seen from the emitter terminal of the BJT (input resistance of the amplifier):
    Rin = [r∏/(hfe+1)] // [ro + RC//RB1].

    Let's determine the resistance seen from the output of the amplifier (output resistance of the amplifier):
    Rout = [ (r∏/(hfe+1))//[RE1//RE2 + RE3] + ro ] // RB1 // RC.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I believe the correct answer is actually ...

    R_{out}=r_\pi||R_E+r_o\[1+gm(r_\pi||R_E)\]
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    326
    I agree with t_n_k.

    I've been waiting for a day to see if anyone else would come up with this result.

    If the OP would post the details of his derivation, we could probably show him where he went wrong.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    I get the same thing. Rewriting it just a tad, we have:


    R_{out}=r_o\;+\;\left( R_E||r_\pi \right)\;+\;r_o g_m\left(R_E||r_\pi \right)

    The point here is just to emphasize that the transistor's output resistance sets the floor of the circuit's output resistance. For R_E significantly smaller than r_pi and for ß>>1, this simplifies to

    R_{out}=r_o \left( 1\;+\;\frac{R_E}{r_e} \right) \;+\;R_E

    If r_o >> R_E >> r_e, then this becomes:

    R_{out}=r_o \left(\frac{R_E}{r_e} \right)

    Now, seeing this is a bit troubling, because I would not expect R_e to improve things so much.

    But I can't help wondering how realistic this circuit is, it is shows no external collector resistance. For small R_E, the collector resistance, R_C, is basically in parallel with r_o. But for large R_E, that is not the case and so I have a feeling it would kill some of that factor.

    Worth playing with, but not right now.
     
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