In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA will be __________ Ω.

Solution

Physics sir · 3 steps

The resistor drops everything the zener does not, so use 45 V across R, not the full 50 V.

1. Voltage across R. In breakdown the zener holds the output at V_z = 5 V. The series resistor R takes the rest: 50 - 5 = 45 V.
2. Worst case current. A smaller R lets a larger current I flow. The zener current I₁ = I - I₂ is largest when the load draws nothing (I₂ = 0), so then all of I goes through the zener. To keep the zener safe, I must not exceed 90 mA.

   Rₘᵢₙ = (Vᵢ - V_z)/I_(z,max) = (45 V)/(90 × 10⁻³ A)

3. So. Rₘᵢₙ = 45/0.09 = 500 Ω, so the answer is 500.

The common slip: 556. Putting the full 50 V across R gives 50/0.09 ≈ 556 Ω. The zener keeps 5 V for itself, so R sees only 45 V.

The idea to keep. In zener circuits, the series resistor always sees Vᵢ - V_z. Find the extreme current that R must allow or limit, then use Ohm's law on R alone.

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