Art of Electronics, Emitter Follower output impedance

Discussion in 'General Electronics Chat' started by raymondxloutlookcom, Nov 20, 2016.

  1. raymondxloutlookcom

    Thread Starter New Member

    Nov 20, 2016
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    Hi, I've been reading the Art of Electronics third version. On page 80 the author says that the output impedance of an emitter follower is Zsource/(1+beta). How can I derive that?

    Thanks very much in advance!

    Raymond
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Notice that Ie = Ic + Ib additional we know that Ic = Ib*β therefore Ie = Ib*β + Ib = Ib*(β + 1) and Ib = Ie/(β + 1)
    And because of this any base resistor will be seen at emitter as a RB/(β + 1) smaller resistor and any emitter resistor is seen at the base as a (β+1) larger.

    Rout2.png

    V1 = Ib*RB + Vbe + Ie*Re (1) and if we substitute Ib = Ie/(β + 1) we have

    V1 = Ie/(β + 1) * RB + Vbe + Ie*Re

    Ie/(β + 1) * RB + Ie*Re = V1 - Vbe

    Ie = (V1 - Vbe)/(RB/(β + 1) + Re)

    As you can see RB at emitter is seen as a RB/(β + 1) resistor.

    Or try this
    Rout.png

    Iin = Ie = Ib * (β + 1)

    Ib = Vin/RB

    Rout = Vin / Iin = ??
     
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  3. raymondxloutlookcom

    Thread Starter New Member

    Nov 20, 2016
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    The part that confuses me is ' be seen as '. What exactly does it mean?
    If we express the final equation using Ib, we have Ib= (V1 - Vbe)/(RB+ Re(β + 1) ), does it mean Re at base is seen as a Re(β + 1) resistor?

    Thanks!
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, exactly.
    RB resistor from emitter point of view (when we are looking into emitter) has a resistance RB/(β + 1)
     
  5. raymondxloutlookcom

    Thread Starter New Member

    Nov 20, 2016
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    Thanks very much! Now I understand it!
     
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