A zener diode of power rating 1.6 W is to be used as voltage regulator. If the zener diode has a breakdown of 8V and it has to regulate voltage fluctuating between 3 V and 10 V. The value of resistance Rₛ for safe operation of diode will be

Solution

Physics sir · 4 steps

Safety is decided at the worst moment: the highest input, with all the current pushed through the zener.

1. Maximum zener current. In breakdown the zener holds 8 V, so its power is P = V_Z I_Z. The largest current it can safely carry is I_(Z,max) = 1.6/8 = 0.2 A.
2. Worst case. The current through Rₛ is largest when the input is at its highest, 10 V. No load is connected, so all of that current flows through the zener. The 3 V end is below breakdown, so the zener carries almost no current there and is in no danger.

   Rₛ = (Vₘₐₓ - V_Z)/I_(Z,max) = (10 - 8)/0.2

3. Put in numbers. The drop across Rₛ is 10 - 8 = 2 V. So Rₛ = 2/0.2 = 10 Ω. This is the smallest resistance that keeps the zener current within 0.2 A.
4. So. The series resistance for safe operation is Rₛ = 10 Ω, which is option (c).

The idea to keep. For zener safety, take the maximum input voltage and find I_(Z,max) = P/V_Z. Then Rₛ = (Vₘₐₓ - V_Z)/I_(Z,max), because the series resistor drops only the voltage above breakdown.

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