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Am I confusing input & output impedance?

  1. May 23, 2015 #1
    1. The problem statement, all variables and given/known data

    design a common-emitter amplifier with an output impedance of 4.7kOhms and again of-10 using a transistor with a beta=200 powered by a 12V DC supply.

    2. Relevant equations

    v=ir
    3. The attempt at a solution

    ------------------------------------------------------
    rest of the problem i get; just this one thing bugging me.
    the solution starts out with solving for the collector current. I understand why the output is taken at the collector, but i don't get why they set R_c equal to the given output impedance. i thought the output impedance was the equivalent resistance as seen from where ever the "output" was being taken (in this case the collector).
     
  2. jcsd
  3. May 23, 2015 #2

    CWatters

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    Perhaps think of a amplifier as a voltage source that has some internal resistance (capacitance or inductance). The output impedance is a measure of how near the output is to an ideal voltage source. If the output impedance is very low then the load (the input impedance of the next stage) will have little effect.

    The output impedance of a common emitter amp is roughly equal to the collector resistor. The required output resistance was specified as 4.7K so they set the collector resistor at that value.
     
  4. May 23, 2015 #3

    LvW

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    ...because the exact output resistance is ro||Rc. And the internal BJT output resistance ro is very large (20...50 kohms).
     
  5. May 23, 2015 #4
    but if we were in saturation mode would we take into account r0?
     
  6. May 23, 2015 #5

    CWatters

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    Why would the transistor be saturated? It should be in a linear mode.
     
  7. May 23, 2015 #6
    high power switch?
     
  8. May 23, 2015 #7

    CWatters

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    Well yes in a switch the transistor might be saturated but normally you would hope it's not saturated in an amplifier. What happens to the output if it does?
     
  9. May 23, 2015 #8

    LvW

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    In your question, you spoke about a common-emitter amplifer, didn`t you?
     
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