Output impedance of source follower

Discussion in 'Homework Help' started by ray242, Jan 2, 2017.

  1. ray242

    Thread Starter Member

    Nov 27, 2016
    Hi, I've been learning about JFET and I came across the source follower. My book gives me two ways for calculating the output impedance (as shown in the picture, by shorting the input etc.) but I don't understand why it is feasible. Could someone explains it to me in more detail?

    Thanks very much in advance.
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Hi, Have you ever tried to measure in real life an Internal resistance of a voltage source (battery)?
    You only have a one voltmeter and one given load resistance ?

    So first what we do is to measure a "open" source voltage ( Vsig = 9V ) next we connect a given load resistance (750Ω) to the battery terminals and measure the output voltage Vout = 8.88V.


    And from there we have to find the battery Internal resistance Rw. Do you know how to do this ?

    In your case we have almost exactly the same situation. Except the fact that now we do not know the value of a Load resistance, but we know Vsig and Rw (Internal resistance) values.
    Last edited: Jan 3, 2017
    ray242 likes this.
  3. MrAl

    AAC Fanatic!

    Jun 17, 2014

    The method in part A is very basic. If you know how to measure input impedance then you know how to measure output impedance as you just pretend the output is the input.
    However, that may be impractical for some circuits because you have to make sure you dont upset the biasing of the input or output of the circuit so they give you part B for an AC amplifier.

    If you apply a reasonable voltage you measure the current and then calculate the resistance using Ohm's Law:

    Sometimes a current source is used instead of a voltage source, and then you just measure the voltage and again use Ohm's Law:
    ray242 likes this.
  4. DickCappels


    Aug 21, 2008
  5. MrAl

    AAC Fanatic!

    Jun 17, 2014

    Hi there Dick,

    Yes the wording makes it sound like we are 'measuring' but really this method applies when calculating also. That's because in theory we can treat the circuit output as an input just like we can when doing an actual real life measurement, so in a theoretical calculation we can apply a voltage or current source just like we do when doing an actual measurement. It is still good to point that out though :)
    Yes there are others ways of doing it too.
    ray242 likes this.