Question: What is the source of the leakage current in a transistor? If the emitter current of a transistor is 8 MA and 1B is 1/100 of 1c, determine the levels of 1C and 1B.
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• 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

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

renu • 257 views
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 modified 10 weeks ago  • written 3 months ago by RB ♦♦ 110
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