Determining DC impedance of transistor

Discussion in 'General Electronics Chat' started by phuzionz, Apr 24, 2009.

  1. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    How can i calculate the DC impedance of a normal VDB(voltage divider bias) transistor circuit.
    And how can i calculate this from a emitter follower.
    How can i know which load my input signal sees. Is this always the voltage diverder circuit?
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    In an emitter follower the input impedance seen by the source (signals the base of the transistor) is the load impedance times the gain of the transistor (plus any other resistors in series with the base).

    If you use a voltage divider you can say that the input impedance seen by the source is due to the voltage divider because the input impedance looking into the base of the transistor is very high.

    Be more specific to your question if you need more.
     
  3. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    During EMC testing we want to connect the real load to the cable harness.
    But in the scheme is that a transistor, i want to replace it with a resistor.
    Therefore i need to calculate the DC resistance, but also the AC resistance i think.

    The strange thing on this circuit is that the input signal is coupled to the collector, normally it is to the base.
    What infuence have that on the input impedance .

    Here you can found the link to the scheme:
    http://www.uploadarchief.net/files/download/emc1.jpg
    Pin 2 is connected with microcontroller
    Pin 1 is the input signal.
     
    Last edited: Apr 25, 2009
  4. mik3

    Senior Member

    Feb 4, 2008
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    I think your circuit controls when the speed signal appears on the output by switching on/off the transistor when apply a signal at point 2.

    I didn't understand what you want to do exactly.
     
  5. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
    47
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    Your analysis of the citcuit is correct. With pin 2 you can switch the transistor ON/OFF, when there is a signal on the collector, the base signal will appear on the emitter.

    Now I want to calculate the resistance (input impedance) for the speed signal (pin1). This is my real input sigal and not the base signal (pin2)
    What for resitance is signal line 1 seeing?
     
  6. mik3

    Senior Member

    Feb 4, 2008
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    What is connected on capacitor 0306?
     
  7. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Last edited: Apr 26, 2009
  8. mik3

    Senior Member

    Feb 4, 2008
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    The transistor while is conducting has a low impedance thus you can remove the transistor and connect your signal at the point where the emitter was connected.
    This is because the transistor here is like a switch which controls the signal.
    If you don't want control, you remove the switch (transistor).
     
  9. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    0
    Ok thank you.
    Thus if i understand you well, i can assume that the input impedance is 100 ohm for the speed signal (pin1). This is because the transistor has a low impedance in saturation mode.

    Is my explenation correct?
     
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