In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance Rₛ to regulate the input unregulated supply is :

Solution

Physics sir · 4 steps

The Zener current must stay between zero and its safe maximum, so the current through Rₛ has a range.

1. Fixed voltages. When the Zener regulates, the load sees V_z = 5 V. So Rₛ always drops 8 - 5 = 3 V, and the load current is I_L = (5 V)/(1 kΩ) = 5 mA.
2. Zener current limits. The power rating caps the Zener current at I_(z,max) = (10 mW)/(5 V) = 2 mA. To keep regulating, the Zener must also carry more than zero current.
3. Current through Rs. The series current is I = I_L + I_z, so it must lie between 5 mA and 7 mA. Since Rₛ = (3 V)/I, a larger current needs a smaller resistance.

   (3 V)/(7 mA) < Rₛ < (3 V)/(5 mA) ⇒ 3/7 kΩ < Rₛ < 3/5 kΩ

4. So. Below 3/7 kΩ the Zener overheats, and above 3/5 kΩ it stops conducting. The answer is (d).

The idea to keep. For Zener design, fix the load current first. Then I_z runs from 0 to P/V_z, and Rₛ = (Vₛ - V_z)/(I_L + I_z) gives a range, not a single value.

Answer from the NTA final answer key. Solution by PrepWiser.