Question about input/output impedance

Discussion in 'General Electronics Chat' started by dsp_redux, May 7, 2009.

  1. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
    182
    5
    Hi,

    I know how to calculate the input/output impedance of a general circuit. Let's suppose it's a black box. I mount that on a breadboard. I know there is a technic to mesure the input/ouput impedance of that black box circuit, but I don't remember how. How do you mesure that, using a standard meter?
     
  2. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Put a variable resistor in series with the input or output and find the resistance at which the voltage is reduced by one half. Measure the variable resistor's value and that is the Zin or Zout of your black box.
     
  3. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
    182
    5
    That makes sense using the voltage divider principle. One of my electronic teacher told use about a method wich implied putting approximately the input impedance calculated at the input of the circuit, mesuring the input voltage, than shorting that impedance and mesuring again. Than there was a formula he had to calculate de real Zin... anyone have a clue?
     
  4. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    A picture is worth 1000 words in this case. But say you want to measure the input resistance of an amplifier. Theoretically, you guess an input resistance of 5kohms. So you put a 5k resistor in series between your source and the input to the amp.

    Use your scope or a voltmeter to determine the voltave at the source, Vs. Then put its probe on the other side of the test resistor, Rs. Call this voltage Vx. The input resistance is Rs*Vx/(Vs-Vx).
     
  5. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
    182
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    No problem with that. But the technique that particular teacher used was comparing Vx with Vx when RS was shorted and he had a formula to determine Zin and Zout...
    Edit: Nothing to do with the voltage divider technique.
     
  6. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I'm not familiar with that method. Sorry dsp redux. Now can you help me with my audio amp problem? See the post above or below this.
     
  7. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    First connect the Rs in series with the circuit input. Measure Vx1 at the circuit input. Based on the attenuator formula

    Vx1 = Vs * (Zin/(Zin+Rs))

    where Vs is the source voltage.

    Shortcircuit Rs. Measure Vx2 at the circuit input. Since there is no drop on Rs

    Vx2 = Vs.

    Replace Vs with Vx2 in the first formula.

    Vx1 = Vx2 * (Zin/(Zin+Rs)).

    This is an equation with one unknown: Zin. The result is:

    Zin = Rs (Vx1/(Vx2-Vx1))
     
  8. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
    182
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    Thank you... I guess that's the same thing since Vx2 = Vs. Why make it simple when it can be harder?
     
  9. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    This is the classical way of finding out the input impedance.
    I know is the same thing as PRS's post. But since you did not get it, and insisted on your teacher's method:

    I showed you the same method with the Rs shorted or not. It is not more complicated, it is shown in a different way.
     
  10. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    By the way, Zout is as follows:

    Connect a load, ZL. Measure the Vx1, the drop on ZL. Disconnect the load. Measure Vx2 at the black-box output. Zout is

    Zout = ZL * ((Vx2-Vx1)/Vx1)
     
    Last edited: May 8, 2009
  11. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
    182
    5
    Thanks a lot for the clarifications.
     
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