A 8 V Zener diode along with a series resistance R is connected across a 20 V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be _______ Ω.

Solution

Physics sir · 3 steps

The Zener holds 8 V, so R takes the other 12 V. Keep the R current within 25 mA even with no load.

1. Voltage across R. The Zener sits in parallel with R_L and fixes the voltage there at 8 V. The series resistor drops the rest: 20 - 8 = 12 V.
2. Worst case. A smaller R lets more current through it. The Zener is in the most danger when the load draws nothing (R_L removed). Then all the current through R flows through the Zener, and it must not exceed 25 mA.

   Rₘᵢₙ = (Vₛ - V_Z)/I_(Z,max) = (12 V)/(25 × 10⁻³ A)

3. So. Rₘᵢₙ = 480 Ω, so the answer is 480.

The common slip: 800. Using the full 20 V supply instead of the 12 V drop across R gives 800, not 480.

The idea to keep. In a Zener regulator, the series resistor always carries supply voltage minus Zener voltage. Size it for the worst case, when the load draws no current and the Zener takes all of it.

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