ac analysis emitter follower

Discussion in 'General Electronics Chat' started by soedar, Apr 7, 2011.

  1. soedar

    Thread Starter New Member

    Apr 7, 2011
    I'm confused :eek:. How did they get to the following equation for the output impedance of an emitter follower? It's on page 270 of "Electronic devices and circuit theory", Robert L.Boylestad and Louis Nashelsky, 9th edition.

    Equation 5.64: Zo=ro//RE//(βre/(β+1))

    I would only get to Zo=ro//RE//βre

    I have attached the ac equivalent circuit.
  2. saqib altaf

    New Member

    Apr 7, 2011
    Dear u have written the wrong value of Z0 its like

    1 can be ignored as B is a very high value so


    as ro is very greater than re than we hav


    for example ro//re if ro=2ok ohm and re=12 ohmthan re//ro is nearly =re thats y we ignore our resultant ro is

  3. soedar

    Thread Starter New Member

    Apr 7, 2011
    Hello Saqib,

    I have attached a scan of the page. Check it out...
  4. soedar

    Thread Starter New Member

    Apr 7, 2011
    The question is: How did they get to equation 5.64?

    In theory to establish the output impedance Zo, the Vi should be set to zero, the base current Ib will then be zero. Further more the current source should be replaced by an open circuit.

    So if I now look into the circuit from the output side, after doing all the things described above, I would get Zo=ro//RE//βre.

    But they get Zo=ro//RE//(βre/(β+1)). How did they get the βre/(β+1)???
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Ib will not gonna to be 0A
    Because in general we can calculate output impedance by applying a voltage source Vin to the output node.
    Zo = Vin/Iin


    So I hope that now is is clear to see that
    Ib = Vin/(βre+RB)

    And also remember that Ie = Ib + Ic = (β+1)*Ib

    Zo = Vin/Iin

    Iin = Iro+IRe + Ie
    Last edited: Apr 7, 2011