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Input impedance BJT with input capacitance and bridge capacitance & pole

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AMSA84

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Hi guys,

I would like to know if someone can give me a help here to remember one detail of thevenin equivalent and BJT input impedance.

See the attached image.

1.jpg

I would like to compute the frequency pole at the point B1. From what I understood the solution is something like 1/[(rpi//Rs)(Cpi + 2Cµ)]

What I can't understand is how to get the (rpi // Rs). How's that possible? To get that result we should short-circuit the Vin source, but why, for what reason we should short the input (if that is the solution to get that parallel resistance).

Thank you in advance.
 

but why, for what reason we should short the input
Click to expand...
?? A voltage source has zero output impedance by definition, it's always acting as a short for other signals.
 

Sources never have 0 impedance in practice, so Rs is added for realistic model.
The pole is computed by the net // impedance of source (output Rs) and transistor input Zin.

So the input is not as you have shown. It is after Rs.
 

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