A zener of breakdown voltage V z = 8 V and maximum zener current, I ZM = 10 mA is subjectd to an input voltage Vᵢ = 10 V with series resistance R = 100 Ω. In the given circuit R L represents the variable load resistance. The ratio of maximum and minimum value of R L is _____________.

Solution

Physics sir · 4 steps

The series resistor fixes the total current at 20 mA; the zener and the load just share it.

1. Current through R. While the zener is in breakdown, the load voltage stays at 8 V. So R drops 10 - 8 = 2 V, and I = 2/100 = 20 mA whatever R_L is.
2. Split the current. This current divides as I = I_Z + I_L. The zener current can range from 0 to I_(ZM) = 10 mA, so I_L ranges from 10 mA to 20 mA.

   R_L = V_Z/I_L ⇒ R_(L,max) = 8/(10 × 10⁻³) = 800 Ω, R_(L,min) = 8/(20 × 10⁻³) = 400 Ω

3. Match the limits. The largest R_L draws the least load current, so the zener carries its maximum 10 mA. The smallest R_L draws all 20 mA, and the zener current drops to zero.
4. So. R_(L,max)/R_(L,min) = 800/400 = 2.

The common slip: 0.5. Dividing the minimum resistance by the maximum gives 0.5. The question asks for maximum over minimum, 800/400.

The idea to keep. In a zener regulator, find the fixed current through the series resistor first. Then give the zener anywhere from zero to its maximum current, and the load gets the rest.

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