JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2026
- Easy
- MCQ
The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at \(V _{ Z }=5 V\) and the desired current in load is 5 mA . The unregulated voltage source can supply upto 25 V . Considering the Zener diode can withstand four times of the load current, the value of resistor \(R _{ S }\) (shown in circuit) should be ___________ \(\Omega\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 800 \( \Omega \)
Step-by-step Solution
Detailed explanation
Given \(V_z=5 V\), desired load current \(I_L=5 mA\).
Maximum Zener current \(I_Z=4 I_L=4 \times 5 mA=20 mA\).
Total current through series resistor \(I=I_L+I_Z=5 mA+20 mA= 25 mA\).
Maximum input voltage \(V_{\text {in }}=25 V\).
Voltage across resistor \(R_S=(25-5) V =20 V\).
Therefore \(R_S=\frac{20}{25 \times 10^{-3}}=800 \Omega\).
Maximum Zener current \(I_Z=4 I_L=4 \times 5 mA=20 mA\).
Total current through series resistor \(I=I_L+I_Z=5 mA+20 mA= 25 mA\).
Maximum input voltage \(V_{\text {in }}=25 V\).
Voltage across resistor \(R_S=(25-5) V =20 V\).
Therefore \(R_S=\frac{20}{25 \times 10^{-3}}=800 \Omega\).
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