Q. 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
A) 10 $mA$
B) 40 $mA$
C) 80 $mA$
D) 160 $mA$
Explanation
(c) In space charge limited region, the plate current is given by Child’s law $i_{p} = KV_{p}^{3/2}$
Thus, $\frac{i_{p2}}{i_{p1}} = \left( \frac{V_{p2}}{V_{p1}} \right)^{3/2} = \left( \frac{600}{150} \right)^{3/2} = (4)^{3/2} = 8$
Or $i_{p2} = i_{p1} \times 8 = 10 \times 8\ mA = 80\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.