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1. In a p-n-p transistor 10^8 electrons enters in emitter in 10^-8 second.if 1% of the electrons are wasted in the base, then what will be the value of COLLECTOR CURRENT and the CURRENT AMPLIFICATION FACTOR?
2. (Original post by Miraz Noor)
In a p-n-p transistor 10^8 electrons enters in emitter in 10^-8 second.if 1% of the electrons are wasted in the base, then what will be the value of COLLECTOR CURRENT and the CURRENT AMPLIFICATION FACTOR?
For a PNP bipolar semiconductor, conventional current enters the Emitter and then gets divided between the Base and Collector.

IE = rate at which charge enters the emitter

IE = dqE / dt

dqE = (ne x qe) = 1x108 x 1.60x10-19 = 1.60x10-11 Coulombs

IE = 1.60x10-11 / 1x10-8 = 1.60 mA

Current divided in the ratio: 1% base, (100% - 1%) collector = 99% collector

IB = 0.01 x 1.60x10-3 = 16x10-6 A

IC = 0.99 x 1.60x10-3 = 1.59x10-3 A

Because the question does not state the circuit configuration (hence hybrid parameters cannot be invoked at this stage):

Current gain IC / IB

1.59x10-3 / 16x10-6

99

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