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Let me be clear: By amplification I am talking about the process of boosting the strength of an AC signal. Thus, where the output power to input power ratio is greater than 1.

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Now suppose we have a simple Common Emitter amplifier with a voltage divider base(and a positive collector supply voltage).A capacitor is an AC short and a DC open. Now when we add an AC signal to the base, the capacitors experience a change in voltage and began to conduct, causing a surge of current to flow from points E' to C' on the circuit diagram. There is also current flowing through C(BC'). Thus small changes on the base voltage(due to input) control larger changes in the emitter-collector current. And that's how transistors can be used to boost AC signal power.

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  • \$\begingroup\$ You are mixing two ways of analyzing a circuit: the term "AC voltage/current" implies a steady state (sinusoidal/phasor) way of looking at it. That's ok as long as everything is behaving linearly (large signal behaviour of diodes is not linear BTW). But on the other hand "experience a change in voltage and began to conduct" implies a transient/differential equation way of looking at it. That's always possible because it's the most general way of looking at it. But mixing both ways together is not useful and just confusing. \$\endgroup\$
    – Curd
    Jul 16, 2018 at 20:34
  • \$\begingroup\$ But small changes on any resistor's voltage cause large changes in its current. You're just describing capacitor reactance for large uF values, not describing amplification. The amplification is created by current source I(t), so AC produced by I(t) must not be ignored. \$\endgroup\$
    – wbeaty
    Jul 20, 2018 at 19:39

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You say AC and then immediately involve the capacitors and that's not strictly correct and not also needed.

Power amplification in a transistor means that a change at the input can result in a change at the output and the changes are such that the change (which is a signal) has greater power at the output than at the input. These changes are modeled in the GP model and "work" even when all the capacitors have a value of 0 (zero).

What I describe above is small signal behavior and that has by itself nothing to do with the capacitors. The capacitors are present in the model to model the capacitors that are actually present in the transistor and their effect is that they limit the bandwidth of the transistor circuit.

At high frequencies the power gain will be smaller or even non-existent (power loss) due to these capacitors.

If what you say is true and the capacitors are involved in power amplification then gain at DC would be impossible. A non-AC coupled common emitter circuit has a DC gain of about gm*Rc.

But back to your question, can the model be used?

Yes of course it can, as long as it models the change in Ic due to a change in Vbe (or Ib) it can. And that behavior is included in the model. If is was not included then the Gummel-Poon model would be "silly" as it would not model the main function of a BJT: power gain.

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  • \$\begingroup\$ From the Gummel-Poon schematic, there is no path from the emitter to the collector in this equivalent circuit except through the capacitors and the I(T)(saturation current lead). So this model does not show the DC current path. It does show though how current can go into the emitter and out the base through the forward biased diodes when there is a base voltage > the barrier potential. \$\endgroup\$
    – Mr X
    Jul 20, 2018 at 19:06
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    \$\begingroup\$ As explained in my answer, if what you write above in your comment is true then what use is the GP model? If it does not model active mode where the collector-to-emitter current is determined by the base-emitter current then what is the point of the model? It would be pretty useless or do you disagree? Fact is, that (DC) current is modeled by the controlled current source \$I_T\$. \$\endgroup\$ Jul 20, 2018 at 21:26
  • \$\begingroup\$ I agree. And It is my understanding that transistor amplifiers have an upper cutoff frequency where there is unity gain. \$\endgroup\$
    – Mr X
    Jul 21, 2018 at 21:28
  • \$\begingroup\$ @Bimpelrekkie, with reference to your above comment - are you aware that in the GP model the collector-to-emitter current is NOT determined by the base-emitter current, but - of course - by the base-emitter voltage (in accordance with the physical reality)? (Sorry, just now I have seen that this is a very old thread). \$\endgroup\$
    – LvW
    Aug 27, 2021 at 15:48
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    \$\begingroup\$ @LvW Sure but the BE voltage causes a current (through the BE diode). So I'd say voltage and current are related. So wether Ic is determined by Ib or Vbe doesn't make a difference (in my opinion) as long as the relation Ib <=> Vbe is clear (which I think it is, since there's a diode). There are many endless discussions wether BJTs are current or voltage controlled. My opinion is that they're both and you can pick whatever suits you best. \$\endgroup\$ Aug 27, 2021 at 15:59

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