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Input Impedance of SIMPLE Circuit

  1. Apr 25, 2005 #1
    Can anyone figure out the input impedance of the circuit I have attached?

    Does anyone know a THOROUGH definition of input impedance? output impedance?

    Output impedance is equivalent to the thevenin impedance, which mean we turn off all voltage/current sources. But for input impedance, what if the circuit has voltage/current sources, what do we do (like the circuit I have attached)?
     

    Attached Files:

    Last edited: Apr 25, 2005
  2. jcsd
  3. Apr 26, 2005 #2
    Still no answer :-(

    Still no answer from PhysicsForums :-( :frown:
     
  4. Apr 29, 2005 #3
    Damm.. i had a reply , but it dissapeared when i clicked on your attachment ..
    your first attachment , if you flip it over to the right ,and short the voltage source just like an output impedance ,what have you got??
     
  5. Apr 29, 2005 #4
    Your second attachment looks ok ..
     
  6. May 3, 2005 #5
    Convert the thevenin equivalent norton equivalent, that is a shunt resistor and a current source equal to the thevenin voltage divided by 100 ohms. The current source is supposedly infinit so doesn't affect the input impedance calculation.
    Clearly the input impedance is 100 ohms.
    Joe
     
  7. Jun 29, 2005 #6
    The circuit you have shown can be looked at several ways:

    (1) You can ignore the battery and replace it with an equivalent resistor representing it's internal resistance. Now you can calculate the 'impedance', and since there is no reactive component it is simply resistance with no phase shift.

    (2) However, you may be misleading yourself as to how the circuit actually behaves. While the theoretical basics are good for simple circuits, real circuits rarely conform, and the results will not be anything like what you learn in class:

    For example, suppose (since we are talking about impedance) that this is an audio application. Applying a D.C. voltage backwards into the output leads of an amplifier can have all kinds of disastrous effects on both the performance and the reliability of the circuit. So, although the 'impedance' we calculated is correct, the actual behaviour of the circuit could be bad. Even a small D.C. voltage applied backward might change the bias on a transistor or tube circuit, and result in the component operating in a non-linear part of its performance specs.

    Likewise, one rarely applies A.C. to a drycell, so, this could be just a representation of the input circuit of an op-amp or some other useful device. This means that although there is a 'load resistance' and D.C. voltage peeking out of the input leads, there must be more attached to this circuit than actually appears. Applying an A.C. signal for instance may or may not be able to penetrate other hidden components connected to the system, like capacitors or inductors. Obviously, the circuit will not behave as assumed by the diagram.
     
  8. Jul 5, 2005 #7

    berkeman

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    Staff: Mentor

    The input impedance is 100 Ohms plus the output impedance of the battery (frequency dependent). The impedance is deltaV/DeltaI, not V/I. If you are interested in the DC Zin, you need to measure the V and I at a couple of load currents, and use the load line to give you the Z. If you are interested in the Z at a particular frequency or range of frequencies, use a signal generator through a series resistor to find deltaV/deltaI.
     
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