1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding impedance 'looking in'?

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    3. The attempt at a solution
    The impedance in question(Rib) is defined to be, exactly like Rin or any other impedance at a point:

    Rb = Vb/Ib

    I took the approach of first finding Ib. I did this by using the Beta value given and the emitter current. Using the relationship: Ib = Ie/(beta+1) = 10mA/131 = 76.24μA. Using Ib you can find Vb by making use of the fact that current is drawn into the base of the transistor for PNP devices. Vb = V2 - Ib*Rb = 736.6mV.

    Thus, Rib = Vb/Ib = 0.7366V/76.24uA = 9649Ω

    This solution is incorrect.
    How would one find Rin?

    Can someone more clearly explain to me the concept impedances at different part of the circuit? By this I mean, the impedance looking into a certain part of the circuit. I thought of it as the impedance at a certain point, if you will. As in, if you were to somehow 'see', or 'enter,' the base of this transistor, you would encounter Rib impedance. Maybe someone else can explain this concept better to me. I somewhat visualize this concept as one that must be defined by Ohm's Law. As such, if one measures the voltage at a given points as well as the associated current, then there must be a resistance seen by that point.

    Thank you.
    Last edited: Apr 21, 2013
  2. jcsd
  3. Apr 21, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    There is no such thing as "impedance at a point". It is always between two parts of the circuit.
    In this case, ##R_{ib}## is between ##v_{out}## and ground.

    It is easier to think of it in terms of DC resistance - which is what you get when you put the terminals of an Ohmmeter at the two points you want the resistance between.

    iirc. to get the output impedence here you have to short-circuit all the voltage sources and open circuit the current sources and model the transistor as a network of ideal components. It also helps to redraw the circuit with one ground rail instead of all those gnd symbols all over the place.
  4. Apr 22, 2013 #3
    By 'at a point' I meant 'looking in' from that point. The impedance I am looking for in this case is actually the impedance seen at the base of the transistor.
  5. Apr 22, 2013 #4
    Apparently, the solution turns out to be the rπ parameter associated with the use of the hybrid pi model for the BJT. Can someone explain this?
  6. Apr 22, 2013 #5
    Try replacing the pnp BJT with the hybrid pi model and have a look at all the paths looking in from the base to ground. One of them shorts out all the others (edit: Correction, the path to ground after r_π).
    Last edited: Apr 22, 2013
  7. Apr 22, 2013 #6
    Using the hybrid pi model, the emitter is grounded. So, the base terminal is connected to the pi parameter resistor then directly to ground. I suppose I did not think of it in this manner. I just have always been preached the equation I originally submitted in the first post.

    Rib = Vb/Ib
  8. Apr 22, 2013 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Well done ... just a niggle:
    There is no such thing as impedance 'looking in' from a point. Impedance requires two points by definition.
    i.e. the input impedance is the impedance "looking in" between the input terminals - there are two of them.

    This could trip you up more later.

    Anyway - you got the "model for the transistor" part :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted