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In a p-n-p transistor, 10⁸ electrons enter into the emitter in 10⁻⁸ s. If 1% electron is lost in the base, what will be the collector current and collector gain?

1 Answer

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Given:
Number of electrons entering the emitter = 10^8
Time taken for electrons to enter the emitter = 10^-8 s
Electrons lost in the base = 1%

To calculate the collector current and collector gain, we need to use the following equations:

Collector current = β × Base current
Collector gain (β) = Collector current / Base current

Let's first calculate the base current:

Base current = Electrons entering emitter - Electrons reaching collector
= 10^8 - 10^8 × 0.01
= 10^8 - 10^6
= 9.9 × 10^7 electrons

The time taken for the base current to flow is very small compared to the time taken for the electrons to enter the emitter, so we can assume that the base current flows instantly.

Now, let's calculate the collector current and collector gain:

Collector current = β × Base current
We do not know the value of β, so we cannot calculate the collector current directly.

However, we can use the law of conservation of charge to calculate the collector current. According to this law, the total current entering the emitter must be equal to the total current leaving the collector.

Total current entering emitter = Base current + Electrons reaching collector
= 9.9 × 10^7 + 10^8 × 0.99
= 1.0989 × 10^8 electrons

Since the time taken for the electrons to enter the emitter is 10^-8 s, the collector current can be calculated as follows:

Collector current = Total current leaving collector = Number of electrons reaching collector / Time taken
= 10^8 / 10^-8
= 10^16 electrons per second

Now, we can calculate the value of β as follows:

β = Collector current / Base current
= 10^16 / (9.9 × 10^7)
= 101.01

Therefore, the collector current is 10^16 electrons per second, and the collector gain is 101.01
User Sowmya
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