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Homework Help: Transistor as an amplifier (doubt)

  1. Mar 6, 2015 #1
    The npn transistor is said to function as an amplifier only in its active state (emitter base is forward biased and collector base is reverse biased). It is considered that majority of the electrons reaching the base layer is dragged into the collector layer. So the emitter current (ec) is approximately equal to the collector current (cc).

    The reason for amplification is given by the equation

    P(input) = I^2 (ec) * R(emitter base)
    P(output) = I^2(cc) * R(collector base)

    Since R(collector base) >>> R(emitter base) & I(ec) = I(cc), P(output) > P(input)... [because the collector base junction is reverse biased and it offers high resistance to the flow of majority charge carriers across the junction]

    My question:-

    The resistance offered by the collector base junction is only to the flow of electron holes from base (p type) to collector (n type) or the flow of electrons from collector (n type) to base (p type). So why are we considering the resistance in output power to be R(collector base)? This resistance does not oppose the motion of electrons from the base to the collector right??
  2. jcsd
  3. Mar 6, 2015 #2


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    Staff: Mentor

    Hi, welcome to the PF. :smile:

    I don't think we are considering the device's Rcb to be the output. You should be using the external resistance in series with the collector as the load being driven. It is denoted RC.

    You need to indicate the circuit under consideration, though I can surmize it to be the common emitter amplifier.

    Pin is (base current)^2 x rbe,
    there may be a factor beta in there, too, depending on the terms you use.
  4. Mar 7, 2015 #3
    Thank you.
  5. Mar 7, 2015 #4


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    Yes - that is the normal procedure for calculating the signal output voltage.
    However, if the effective collector resistance Rc,eff (including the input resistance of the next stage - if existent) is rather large (tens of kohms) it may be necessary to take into account also the BJT`s internal output resistance r,o because the transistor is not an ideal current source. The value of r,o is, typically, in the range (20....50)kohms and is identical to the inverse slope of the Ic=f(Vce) characteristics. In this case, the gain determining resistance is (R,eff||r,o).
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