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Double load NPN amplifier

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faustoleali1974

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Dear Sirs, i'm trying to solve an exercise for an exam but i doesn't really understand how to do it

The circuit below:
**broken link removed**
LINK TO IMAGE


The exercise text: FIND THE APPROXIMATE VALUE OF THE AMPLIFIER IN THE FIGURE AT THE FREQUENCY fr:

Rcllector = 1250 Ohm
Remitter = 250 Ohm
L = 100 uH
C = 500 nF
Bo >>1
Bo * Remitter >> Rpi (resistance between base and emitter in the small signal model of the BJT)
fr = 1/(2*pi*sqrt(LC))

The solution of the exercise says:

The frequency fr is the resonance frequency of L-C parallel circuit. At this frequency the collector impedance is purely resistive (i think becouse Rcollector has a infinite impedance paralleled) SO THE GAIN IS THE ONE OF A DOUBLE LOAD CIRCUIT:

gain = Av = Vo/Vi = - Rcollector/Remitter = - 5


Can someone explain how to obtain this last formula?

---------- Post added at 03:16 ---------- Previous post was at 02:51 ----------

The formula is correct if I apply the small signal analysis that short circuit the supply with the ground but why must I obtain the gain with small signal analysis?
 

I don't want to cause confusion on your side, but - to be exact - the formula is not "correct".
It is an approximation, which in some cases may be accurate enough.
the following expression is much more exact:

gain=Rc/(g^-1 + Re) with transconductance g=h21/h11=Ic/Ut (thermal voltage app. Ut=26 mV)

You can decide by yourself if the approximation as mentioned by you is good enough;
that means: If Re>>1/g

Regarding your last question: "gain" is a small-signal parameter, since it is assumed that the input and ouput signal have the same shape (sinusoidal).
 
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