Chemistry, asked by Srishti57131, 11 months ago

When a p-n junction is reverse-biased, the current becomes almost constant at 25 µA. When it is forward-biased at 200 mV, a current of 75 µA is obtained. Find the magnitude of diffusion current when the diode is
(a) unbiased,
(b) reverse-biased at 200 mV and
(c) forward-biased at 200 mV.

Answers

Answered by bhuvna789456
4

Explanation:

It is given:

Under reverse bias, the drift current, i_{1}=25 \mu \mathrm{A}

Voltage that is forward bias, V = 200 mV

Under forward bias, the net current, i_{2}=75 \mu \mathrm{A}

(a) There will be no flow of current across the p-n junction when it is in unbiased condition, that is, diffusion current is equal to the drift current.

Therefore, diffusion current =  25 \mu A

(b) Under reverse bias, the applied voltage and the potential competes with the majority carriers’ motion across the junction.

So, there will be zero diffusion current.

(c) The motion of majority carriers under forward bias across the junction and it is supported by the voltage.

Let x be the actual current

So, Forward-biased current = (x − Drift current)

\begin{aligned}&\Rightarrow x-25 \mu A\\\\&\Rightarrow x=100 \mu A\end{aligned}

Answered by Satyamrajput
3

Answer:

Check a...bove ......................

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