Search any question & find its solution
Question:
Answered & Verified by Expert
In space charge limited region, the plate current ina diode is $10 \mathrm{~mA}$ for plate voltage $150 \mathrm{~V}$. If the plate voltagte is increased to $600 \mathrm{~V}$, then the plate current will be
Options:
Solution:
2427 Upvotes
Verified Answer
The correct answer is:
$80 \mathrm{~mA}$
In space charge limited region, the plate current is given by Child's law $i_{p}=K V_{p}^{3 / 2}$ Thus,
$$
\frac{i_{p 2}}{i_{p 1}}=\left(\frac{V_{p 2}}{V_{p 1}}\right)^{3 / 2}=\left(\frac{600}{150}\right)^{3 / 2}=(4)^{3 / 2}=8
$$
or, $i_{p 2}=i_{p 1} \times 8=10 \times 8 \mathrm{~mA}=80 \mathrm{~mA}$...(i)
$$
\frac{i_{p 2}}{i_{p 1}}=\left(\frac{V_{p 2}}{V_{p 1}}\right)^{3 / 2}=\left(\frac{600}{150}\right)^{3 / 2}=(4)^{3 / 2}=8
$$
or, $i_{p 2}=i_{p 1} \times 8=10 \times 8 \mathrm{~mA}=80 \mathrm{~mA}$...(i)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.