Confusion regarding Transistors

Question

In a transistor, we know that current amplification factor alpha (DC) for CB Configuration is given by :

Alpha = Ic / Ie

Where, Ic = collector current; Ie = Emitter current

This implies that :

        Ic = alpha × Ie ....................(1)

Also, the total current is given by :

  Ic = alpha × Ie + Icbo ................(2)

Where, Icbo = collector base current with open Emitter (leakage current)

From 1 and 2,

Ic = Ic + Icbo
Icbo = 0 ...................................(3)

This means for any numerical values of alpha, Ic and Ie, the leakage current is always going to be 0. But practically, this is not the case. A small current of the order of micro/nano amps flows as Leakage current. This contradicts the above equation. Does this mean to say that the above equation is faulty? Please explain.

check out this image from my textbook... This is where i got the second equation which I have highlighted in yellow. Also have a look at the image at the top right.

Answer 1

Yes any real transistor does have some leakage. Usually a resistance Ro is added in the model between the collector and emitter to account for leakage.

The equation Ic = alpha * Ie is only an approximation based on a simplified model of a transistor.

It is no more faulty than say Newtons laws of motion, even though we know that they become inaccurate at relativistic speeds. In the same way the equation in your book gets very close to the real answer most of the time.

Most text books start with the simplified model, and then gradually build up their model of a transistor adding additional parts to more closely approximate the real behavior. Starting with the full model would probably hinder understanding for most people.