Transistor as an amplifier (doubt)

AI Thread Summary
The npn transistor operates as an amplifier in its active state, where the emitter-base is forward biased and the collector-base is reverse biased. Amplification occurs because the collector current (cc) is approximately equal to the emitter current (ec), leading to a higher output power due to the greater resistance of the collector-base junction. The discussion clarifies that the output power should consider the external load resistance (RC) rather than the internal collector-base resistance (Rcb). Additionally, for high effective collector resistances, the internal output resistance (ro) of the transistor may also need to be factored into calculations for accurate gain determination. Understanding these resistances is crucial for analyzing the performance of common emitter amplifiers.
latencymech
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??
 
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latencymech said:
So why are we considering the resistance in output power to be R(collector base)?
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.
 
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Likes latencymech
Thank you.
 
NascentOxygen said:
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.
.

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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