- #1

Pushoam

- 962

- 52

## 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?

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##