• The main source of the leakage current in a transistor are thermally generated minority carrier.

consider common emitter configuration.

• when the switch is open, the emitter base junction is an open circuit and so the value of input or base current is zero.

• But a leakage current 1CEO flows between collector and emitter. this leakage current 1CEO is not only due to the thermally generated minority carriers (1 CBO) across the C - B junction but also due to the movement of holes which flow across the base emitter junction.

Given:

1E = 8 MA

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

1c = ?

1B = ?

$\because$ 1E = 1B + 1C

8 MA = 1B + 1C

8MA = 1B + 1C

8 MA = $\frac{1}{100} 1c + 1c$

8 MA = $(\frac{1}{100} + 1) 1_c$

8 MA = $\frac{101}{100} 1_c$

$Ic = \frac{100 \times 8 MA}{101} =$

IC = 7.92 MA

$\because$ $IB = \frac{1}{100} IC = \frac{1}{100} \times 7.92 MA$