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In space charge limited region, the plate current in a diode is \mathrm{10\ mA} for plate voltage \mathrm{150\ V}. If the plate voltage is increased to \mathrm{600\ V}, then the plate current will be

Option: 1

\mathrm{10\ mA}


Option: 2

\mathrm{40\ mA}


Option: 3

\mathrm{80\ mA}


Option: 4

\mathrm{160\ mA}


Answers (1)

best_answer

In space charge limited region, the plate current is given by Child’s law \mathrm{i_p=K V_p^{3 / 2}}
\mathrm{Thus,\ \frac{i_{p 2}}{i_{p 1}}=\left(\frac{V_{p 2}}{V_{p 1}}\right)^{3 / 2}=\left(\frac{600}{150}\right)^{3 / 2}=(4)^{3 / 2}=8}\\ \mathrm{ Or\ i_{p 2}=i_{p 1} \times 8=10 \times 8 \mathrm{~mA}=80 \mathrm{~mA}}

Posted by

Ritika Kankaria

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