common collector

Discussion in 'Homework Help' started by amangupta1219, Oct 30, 2012.

  1. amangupta1219

    Thread Starter New Member

    Oct 18, 2012
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    for the common collector configuration shown i need to find the voltage gain.
    The voltage gain formula is AV = (β+1)(roll RE)/(rπ+(β+1)(roll RE)
    But in the circuit RE is absent . So how do we consider it ??
    Nothing is given about the current source.

    from the circuit theory i know that the internal impedance of a current source is very high (am i correct ?).

    So in this case how do we relate it ?
     
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  2. panic mode

    Senior Member

    Oct 10, 2011
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    current source is considered high impedance (ideally infinite).
    i don't know what is (roll RE) and how you got it so I'm attaching my version.
    if something does not show up in the circuit, gets removed from equation.
     
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    Last edited: Oct 30, 2012
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  3. amangupta1219

    Thread Starter New Member

    Oct 18, 2012
    19
    0
    it is early resistance ro parallel with RE . I don't know how to do formatting here
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Use a small signal analysis and find the voltage gain.
    But for ideal case - ideal current source and R_load = ∞ the voltage gain is equal 1V/V
     
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  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The Early resistance acts as a parasitic load on the output, somewhat reducing the voltage gain, and possibly more significantly reducing the input impedance. More complex follower circuits may bootstrap the collector voltage in an attempt to reduce these effects.

    If you really have no information about the Early resistance RE, then I would think you should assume it to be infinite, and so simplify anywhere it appears in parallel to just the other element(s).
     
    Last edited: Oct 30, 2012
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