If the emitter current of a transistor is 8MA and 1B is $(\frac{1}{100})$ of 1c

Determine the levels of 1C and 1B.

Solution: Source of leakage current in a transistor.

In BJT, as temperature increases covalent bonds in semiconductor lattice breaks and minority charge carriers are generated. these minority charge carriers are source for leakage current.

The leakage current doubles for every 10 degree Celsius rise in temperature.

Given: $1E$ = 8 MA

$1_B = \frac{1_c}{100}$

$1_C = ?$

$1_B = ?$

$\because$ $1_C = B 1_B$ Here $1_C = 100 1_B$

$\therefore$ B = 100

$\because$ $1_E = (1 + B) 1_B$

8MA = 101 $1_B$

$1_B = 79.207 \mu \ A$

$1_C = b 1_B$

$1_C = 100 \times 79.207 \ MA$

$1_C = 7.92 \ MA$