AoE mathematical explanation of emitter follower impedance

Discussion in 'Math' started by BrianH, Sep 25, 2010.

  1. BrianH

    Thread Starter Active Member

    Mar 21, 2007
    43
    0
    Hi everyone,

    I have been doing some reading of the Art of Electronics and I have come across an explanation for the Emitter Follower configuration on page 66. The explanation serves to convince the reader that the emitter follower has a large input impedance compared to its output impedance. And it explains this with the help of some mathematics.

    Now, I intuitively understand that the emitter follower has a large input impedance compared to its output impedance because of the fact that an emitter follower has no voltage gain but it has current gain. However, whilst I understand and accept what the book is trying to say, I can't quite grasp the mathematics of it. Here's what the book says:

    \Delta IE = \DeltaVB / R. (where R is the load).

    I'm okay with that much. Since \DeltaVB = \DeltaVE it makes sense.

    The author then makes a leap from that statement to this one;

    \DeltaIB = \frac{1}{hfe + 1} * \DeltaIE.

    = \DeltaVB / R(hfe + 1)

    Now, I must confess that my understanding has been lost on that jump. Is another more mathematically adept member able to explain it a little more fully for me?

    Thanks very much in advance!

    Brian
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    There are two things to keep in mind

    1. Sum of the currents is zero, by KCL so: Ic + Ib + Ie = 0
    2. The definition of hfe is such that Ic = hfe*Ib
    If we adopt the convention that currents going into a node are positive we get:
    Code ( (Unknown Language)):
    1.  
    2. Ic + Ib - Ie = 0
    3. Ic + Ib = Ie
    4. hfe*Ib + Ib = Ie
    5. (hfe + 1)Ib = Ie
    6. Ib = Ie / (hfe + 1)
    7.  
    Taking differentials on both sides assuming hfe is constant over some range of Ib and Ie
    Code ( (Unknown Language)):
    1.  
    2. d(Ib) = (1 / (hfe + 1)) d(Ie)
    3.  
    Does that help?
     
  3. BrianH

    Thread Starter Active Member

    Mar 21, 2007
    43
    0
    Yes it does actually, thank you very much. Do you know, the part that was causing me confusion was actually very simple!

    I could not immediately see how a leap could be made from

    hfe*Ib + Ib = Ie

    to

    (hfe + 1)Ib = Ie

    However, having studied it for just a short moment, I have convinced myself that it is indeed true. Having convinced myself of that, and using this extra step that you have made but that the book didn't, I've been able to satisfy myself of the entire explanation on that page.

    Thanks very much!

    Brian.
     
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