Voltage amplification done by a transistor amplifier

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The equivalent circuit is drawn below.
It is assumed that the potential drop across the pn junction in forward bias is 0 V(I can't take it 0.7 V as the input voltage is 0.5 V). In this case the p-n junction is equivalent to a connecting wire-piece.
Since $\alpha$ is nearly equal to 1, ##I_B# is taken to be 0.

So, the current flowing through $R_{in}$ is $I_E = \frac{500~mV}{20~\Omega} = 25 ~mA$
In fact, the above figure is wrong.Right?
It is the p-n junction in forward bias which should get replaced by R_in and np junction in reverse bias which should get replaced by R_out.
Then, the potential drop across the diode in forward bias is equal to the potential drop across the R_in i.e. 500 mV and the potential drop across the reverse bias is the potential drop across the R_out.
Potential drop across R_out = $I_C R_{out} = 2500 V$........(1)
Using Kirchoff's Voltage loop law, this voltage drop should be equal to the voltage drop across$R_C$.
But, due to Ohm's law, voltage drop across $R_C = I_C R_C = 25V$......(2)
So, there is a contradiction.
In (1), I have considered the n-p junction as a linear element. I think this is wrong. The n-p junction could not be replaced by R_out. Then , in this case, there is no need to know R_out. Right?
So, does it mean that , in general, in transistor, the forward bias p-n junction could be taken as a linear element, but the reverse bias p-n junction has to be taken as a non-linear element?
The voltage amplification is : $\frac{25V}{0.5V}=50$

 

[CWatters]

I'm a bit rusty but...

Potential drop across Rout = ICRout=2500VICRout=2500V I_C R_{out} = 2500 V ........(1)

No that's not correct. The output looks like a dependant current source with the 100K output impedance in parallel. So the 1k load resistor is in parallel with the 100K output impedance.

I agree that I_E = 25mA
I_B is small so I_C = I_E = 25mA
The output impedance is high compared to the load so the output impedance can be ignored.
So the voltage across RC is 25mA * 1K = 25V
So the voltage gain is 25/0.5 = 50.

 

[Pushoam]

No that's not correct. The output looks like a dependant current source with the 100K output impedance in parallel.

O.K.
So, in the equivalent circuit, the forward bias p-n junction gets replaced by an input resistance in series while the reverse bias junction gets replaced by an output resistance in parallel.
The output circuit acts as a current source with current $I_C$.

Is this right?

 

[CWatters]

No, I would simplify the transistor model to an amplifier model and draw it like this...

When the problem statement says the input resistance is 20R you shouldn't think of the circuit as a transistor and a separate 20R resistor. The 20R results from combining the effects of the transistor, it's bias circuit, feedback, base current (eg the whole thing) into one amplifier parameter for the input.

 

[Pushoam]

I got it.
The problem is with the connection at B in my circuit. The connection at B implies that $I_B >>I_C$.
Thank you.