# Question regarding collector leakage current with open emitter

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

$$\alpha = I_{C} / I_{E}$$

Where, $I_{C}$ = collector current; $I_{E}$ = Emitter current

This implies that:

$$I_{C} = \alpha * I_{E} \space \space (1)$$

Also, the total current is given by:

$$I_{C} = \alpha * I_{E} + I_{CBo} \space \space (2)$$

Where, $I_{CBo}$ = collector base current with open Emitter (leakage current)

From 1 and 2,

$$I_{C} = I_{C} + I_{CBo}$$ $$I_{CBo} = 0$$

This means for any numerical values of alpha, $I_{C}$ and $I_{E}$, 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.

• I think equation two should read Ic=Ie+Ib, but maybe ask on the electronics stack exchange. – George Herold Nov 3 '16 at 16:34
• That's a general case...but the 2nd equation is when emitter is open – Aditya DS Nov 3 '16 at 17:11

The reason for this contradiction is that different definitions for the common base current gain are used in equations (1) and (2). In equation (1), $\alpha$ is the DC current gain defined as the ratio of the DC collector to emitter current $\frac{I_C}{I_E}$ as measured at the terminals. In equation (2), $\alpha$ is defined as the the ratio of the change of collector current divided by the change in emitter current at constant collector-base voltage $V_{CB}$. It corresponds to the small signal common base current gain $\partial I_C/\partial I_B$ at constant $V_{CB}$. For practical purposes, the DC current gain (1) is often used to approximate the low frequency AC current gain defined by $\alpha=\partial I_C/\partial I_B$ because in good transistors their values are usually very close to each other.