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.
Best Answer
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.