Get Answers to all your Questions

header-bg qa

In a zener-regulated power supply a zener diode with v_2=6.0 \mathrm{~V} is used for regulation. The load current is to be 4 \mathrm{~mA} and the unregulated input is 10 \mathrm{~V}. What should be the value of series resistance R ?

Option: 1

140 \Omega


Option: 2

150 \Omega


Option: 3

162 \Omega


Option: 4

167 \Omega


Answers (1)

best_answer

Given I_L=4.0 \mathrm{~mA}
Let us take I_z to be fivetimes I_L or I_z=20 \mathrm{~mA}.

Total current I=I_z+I_L=24 \mathrm{~mA}.
Input voltage V_{m}=10 \mathrm{~V}
zener diode voltage V_z=6 \mathrm{~V}
Voltage drop across resistance
\begin{aligned} \Rightarrow & V_R=V_{\text {in }}-V_2=(10-6) \mathrm{V}=4 \mathrm{~V} \\ \Rightarrow & \frac{V_R}{I_R}=\frac{.4}{24 \times 10^{-3}} \Omega=167 \Omega \end{aligned}
 

Posted by

Rishi

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks