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14.Semiconductor Electronics
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A Zener diode, having breakdown voltage equal to $15\,\,V,$ is used in a voltage regulator circuit shown in figure. The current through the diode is......$mA$
A
$5$
B
$10$
C
$15 $
D
$20$
(AIPMT-2011)
Solution
The voltage drop across $1 \mathrm{k} \Omega=V_{z}=15 \mathrm{V}$
The current through $1 \mathrm{k} \Omega$ is
$I^{\prime}=\frac{15 \mathrm{V}}{1 \times 10^{3} \Omega}=15 \times 10^{-3} \mathrm{A}=15 \mathrm{mA}$
The voltage drop across $250 \Omega=20 \mathrm{V}-15 \mathrm{V}=5 \mathrm{V}$
The current through $250 \Omega$ is
$I=\frac{5 \mathrm{V}}{250 \Omega}=0.02 \mathrm{A}=20 \mathrm{mA}$
The current through the zener diode is
$I_{2}=I-I^{\prime}=(20-15) \mathrm{mA}=5 \mathrm{mA}$
Standard 12
Physics
