A zener diode of power rating 2W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of Rₛ for safe operation should be ___________ Ω.

Solution

Physics sir · 3 steps

Protect the zener at the worst moment: highest input, with all of the series current going into the diode.

1. Maximum safe current. The zener holds 10 V across itself and may dissipate at most 2 W. So the largest current it can carry is Iₘₐₓ = P/V_Z = 2/10 = 0.2 A.
2. Worst case. At 6 V the input is below the breakdown voltage, so the zener does not conduct and is in no danger. The risk is at 14 V, where Rₛ drops the extra 14 - 10 = 4 V. With no load attached, all of that current flows through the zener.

   (14 - 10)/Rₛ ≤ 0.2 A ⇒ Rₛ ≥ 4/0.2 = 20 Ω

3. So. The smallest series resistance that keeps the zener within its 2 W rating is Rₛ = 20 Ω.

The common slip: 70. Dividing the full 14 V by 0.2 A gives 70. Only the 4 V above breakdown appears across Rₛ, because the zener holds 10 V.

The idea to keep. For zener safety, find Iₘₐₓ = P/V_Z. Then use only the voltage dropped across Rₛ, which is V_(in,max) - V_Z, and assume the worst case where all of the current goes through the zener.

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