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In a Zener regulated power supply a Zener diode with $V_z=6 \mathrm{~V}$ is used for regulation. The load current is $4 \mathrm{~mA}$ and the unregulated input voltage is $10 \mathrm{~V}$. To get a Zener current
five times the load current the value of series resistor $R_S$ is nearly
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five times the load current the value of series resistor $R_S$ is nearly
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Verified Answer
The correct answer is:
$167 \Omega$
Given, $I_Z=5 I_L=20 \mathrm{~mA}$
Total current $I_s$ through $R_s$ is $I_s=I_Z+I_L$
$=20+4=24 \mathrm{~mA}$
As a potential drop of $6 \mathrm{~V}$ occurs across Zener diode and supply voltage is $10 \mathrm{~V}$.
So, potential drop across $R_s, V_s=10-6=4 \mathrm{~V}$ $\therefore$ Series resistor,
$R_s=\frac{V_s}{I_s}=\frac{4}{24 \times 10^{-3}} \approx 167 \Omega$
Total current $I_s$ through $R_s$ is $I_s=I_Z+I_L$
$=20+4=24 \mathrm{~mA}$
As a potential drop of $6 \mathrm{~V}$ occurs across Zener diode and supply voltage is $10 \mathrm{~V}$.
So, potential drop across $R_s, V_s=10-6=4 \mathrm{~V}$ $\therefore$ Series resistor,
$R_s=\frac{V_s}{I_s}=\frac{4}{24 \times 10^{-3}} \approx 167 \Omega$
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