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Input Impedance of SIMPLE Circuit

Discussion in 'Electronic Basics' started by [email protected], Apr 26, 2005.

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  1. Guest

    Can anyone figure out the input impedance of the circuit attached here:
    http://physicsforums.com/showthread.php?t=73070 ?

    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)?
     
  2. Jim Thompson

    Jim Thompson Guest

    I don't respond to problem definitions posted on sites requiring
    registration. Sorry.

    ...Jim Thompson
     
  3. mike

    mike Guest

    Impedance is impedance. It defines the vector relationship between
    voltage and current at the port. Doesn't matter whether it's "input" or
    "output" or whether you've "turned off" sources. Turning on the sources
    won't affect the impedance, but it may require you to use different
    measurement techniques.
    mike

    --
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    with links. Delete this sig when replying.
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  4. Tim Wescott

    Tim Wescott Guest

    I second that. Jim's language is more diplomatic than what I had in
    mind, so I'll let it stand for me, too.
     
  5. Bob Monsen

    Bob Monsen Guest

    Post the gif/jpg on the newsgroup alt.binaries.schematics.electronic
     
  6. Jim Thompson

    Jim Thompson Guest

    WOW! It's been a long time since someone referred to me as
    "diplomatic" ;-)

    ...Jim Thompson
     
  7. G Patel

    G Patel Guest

    here:

    Correction: Let me draw you a text version of the circuit:

    100 ohms
    o---------/\/\/\----|
    |
    |
    -----
    | + | 20Vdc
    | - |
    Zin -> -----
    |
    |
    |
    o-------------------|




    That's a voltage source in the circuit. Of course this could be a more
    complex circuit with more voltage sources in different places, but this
    two element circuit serves to explain my confusion. I don't know what
    the input impedance is for circuits like this (for output impedance of
    complex circuits with sources I can simply short voltage sources or
    open current sources before calculating, but what would I do for input
    impedance?)
     
  8. Jim Thompson

    Jim Thompson Guest

    Impedance is generally defined incrementally, so a 1V CHANGE at the
    input will produce a 1/20 Amp change in current... thus the input
    impedance is 20 ohms, irrespective of the voltage source behind it ;-)

    ...Jim Thompson
     
  9. Guest

     
  10. Jim Thompson

    Jim Thompson Guest

     
  11. Jim Thompson wrote...
    Oops, time for another trip to the drugstore for stronger reading
    glasses. Or perhaps a trip to the kitchen so the good wife can
    give them a good cleaning. <sigh>
     
  12. Tim Wescott

    Tim Wescott Guest

    -- and this is why you short a voltage source and open a current source,
    on an input _or_ output. Change the voltage by 1V into a voltage source
    and the incremental current change is infinite. Change the voltage by
    1V into a current source and the incremental current change is zero.
     
  13. Ratch

    Ratch Guest

    Yes, the general definition of impedance is Z(s)=V(s)/I(s). Where Z(s),
    V(s), and I(s) are the Laplace transform of voltage, current and Impedance,
    If you can calculate the voltage and current Laplace transforms, then you
    can find the impedance provided the circuit is linear, which it usually is.
    This will work for alternating or DC sources. Ratch
     
  14. Ratch

    Ratch Guest


    OK, let's do the problem. Assume a constant DC voltage is applied to
    the terminals. V(s)=V/s. Writing the loop equation we get
    100*I(s)=V/s-20/s=(V-20)/s==>I(s)=(V-20)/100s . Z(s)=(V/s)/I(s)=100V/V-20 .
    Notice from the last equation, that if the 20 volt source is removed, the
    impedance value becomes 100 ohms resistive regardless of the input source
    voltage. Also from the last equation, it can be seen that if the input
    voltage is 20 volts, the impedance is infinite. This is because the 20
    volts of input voltage balances the 20 volt source voltage, so that no
    current exists in the circuit, thereby making the impedance infinite. Ratch
     
  15. The OP doesn't understand Thévenin and you want to sell him Laplace? Is
    this just an ego trip or do you really not understand the KISS
    Principle?
     
  16. Ratch

    Ratch Guest

    Did you read the OP's request? "Does anyone know a THOROUGH definition
    of input impedance? output impedance?" He did not ask for a KISS, and
    therefore incomplete definition. So I gave a complete, general, thorough,
    and correct definiton. In another post, I worked his example and explained
    the result to show him how it is done. Ratch
     
  17. G Patel

    G Patel Guest

    No not KISS. Didn't you read my OP? I wanted to go passed the simple
    cases that everyone talks about and delve deaper. I was taught the
    calculation method for input impedances and have used it before but
    then I started questioning how voltage and current sources would be
    dealt with. Then I realized that I could make circuits (like the
    simple one I posted) that would not have a contast Vin/Iin ratio. This
    is why I wanted a more STRICT definition of input impedance.

    Although I'm a Computer Science student, I do understand Thevenin and
    Laplace from my EE course. Thevenin applies to output impedances, and
    the method has specific treatments for voltage/current sources when
    looking BACK into a circuit. But I always wondered the method required
    to treat voltage and current sources for looking INTO a circuit (INPUT
    IMPEDANCE). V/I for this circuit is not constant, but other posters
    have noted that in such a case we only worry about the incremental
    effect on I for each Volt (which is the same result you get if you
    short voltage sources and open current sources - much like when
    calculating Thevenin output impedances).
     
  18. Ratch

    Ratch Guest

    No not KISS. Didn't you read my OP? I wanted to go passed the simple
    cases that everyone talks about and delve deaper. I was taught the
    calculation method for input impedances and have used it before but
    then I started questioning how voltage and current sources would be
    dealt with. Then I realized that I could make circuits (like the
    simple one I posted) that would not have a contast Vin/Iin ratio. This
    is why I wanted a more STRICT definition of input impedance.

    Although I'm a Computer Science student, I do understand Thevenin and
    Laplace from my EE course. Thevenin applies to output impedances, and
    the method has specific treatments for voltage/current sources when
    looking BACK into a circuit. But I always wondered the method required
    to treat voltage and current sources for looking INTO a circuit (INPUT
    IMPEDANCE). V/I for this circuit is not constant, but other posters
    have noted that in such a case we only worry about the incremental
    effect on I for each Volt (which is the same result you get if you
    short voltage sources and open current sources - much like when
    calculating Thevenin output impedances).

    I hope the Laplace definition of impedance and previous example have
    helped you. It is correct for input impedances, output impedances, and
    anything in between. It also works for active voltage/current sources, as
    the example illustrates. Ratch
     
  19. Jim Thompson

    Jim Thompson Guest

    I have to buy a new pair... dropped them on a concrete floor, and have
    a scratch right across the line of sight :-(

    ...Jim Thompson
     
  20. Ben Bradley

    Ben Bradley Guest

    Ahem. Did you see the bugmenot site in the earlier thread? And I
    didn't add this login, it was already there:

    http://bugmenot.com/view.php?url=http://physicsforums.com

    It was hardly worth going there. The first pic is two terminals
    that go to a 20V battery in series with a 100 ohm resistor. Just a
    SWAG, but I'd say the input impedance of that circuit is 100 ohms.
    Second pic, "Zin = V / I Formula works with Phasor Algebra also."
    But I'll have to ask Science Officer Spock how Phasor Algebra works. I
    know it has to do with the current through a Flux Capacitor.
     
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