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

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10 mA

40 mA

80 mA

160 mA

answer is C.

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Detailed Solution

In space charge limited region, the plate current is given by Child's law, $i_{p} = K V_{p}^{3 / 2}$

$T h u s, \frac{i_{p_{2}}}{i_{p_{1}}} = {(\frac{V_{p_{2}}}{V_{p_{1}}})}^{3 / 2} = {(\frac{600}{150})}^{3 / 2} = {(4)}^{3 / 2} = 8$

$o r i_{p_{2}} = i_{p_{1}} \times 8 = 10 \times 8 m A = 80 m A$

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