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.