A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance Rₚ is .................... Ω.

Solution

Physics sir · 4 steps

The power rating fixes the largest safe current; R_p must drop the extra 12 V at exactly that current.

1. Worst case. The n side of the zener goes to the positive terminal, so it is reverse biased. When the slider sits at the far end, the full 20 V appears across Rₚ and the zener together. This is the case Rₚ must protect against.
2. Maximum safe current. In breakdown the zener holds 8 V, and its power is P = V_z I. So the largest current it can carry is Iₘₐₓ = 0.5/8 = 1/16 A = 62.5 mA.

   Rₚ = (V - V_z)/Iₘₐₓ = (20 - 8)/(1/16) Ω

3. Put in numbers. Rₚ must drop the remaining 20 - 8 = 12 V, so Rₚ = 12 × 16 = 192 Ω.
4. So. The protective resistance is 192 Ω.

The common slip: 320. Dividing the full 20 V by the current, without subtracting the 8 V held by the zener, gives 320 instead of 192.

The idea to keep. For a zener, Iₘₐₓ = P/V_z. The series resistor drops whatever the supply gives beyond V_z, so R = (V - V_z)/Iₘₐₓ. Always take the largest supply voltage the circuit can apply.

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