- 2073 Shrawan · 4 marks
Zener current for two load values
Short answer
For RL=1.2 kΩ, open-circuit voltage 16×1.2/(1+1.2)=8.73 V < 10 V, so Zener off, Iz=0. For RL=3 kΩ, voltage 12 V > 10 V, Zener on: IR=(16−10)/1k=6 mA, IL=10/3k=3.33 mA, Iz=6−3.33=2.67 mA.
Find the zener current from the given circuit if (i) RL = 1.2 KΩ (ii) RL = 3 KΩ. [Figure: 16 V source in series with R = 1 KΩ; Zener diode (PzM = 80 mW, Vz = 10 V) in parallel with load RL]
Given: Source = 16 V, R = 1 kΩ, Zener Vz = 10 V with PZM = 80 mW, load RL in parallel with the Zener.
Step 1: Maximum Zener current
IZM = PZM/Vz = 80 mW / 10 V = 8 mA
Step 2: Check whether the Zener is ON (breakdown)
Remove the Zener and find the voltage across RL: VL = 16 × RL/(R + RL). The Zener conducts only if VL ≥ Vz = 10 V.
(i) RL = 1.2 kΩ
VL = 16 × 1.2/(1 + 1.2) = 16 × 0.5455 = 8.73 V
Since 8.73 V < 10 V, the Zener is OFF (not in breakdown).
Iz = 0
(The load voltage is then 8.73 V, and the load current IL = 8.73/1.2 kΩ = 7.27 mA.)
(ii) RL = 3 kΩ
VL = 16 × 3/(1 + 3) = 12 V
Since 12 V > 10 V, the Zener is ON, with Vo = 10 V.
IR = (16 − 10)/1 kΩ = 6 mA
IL = 10 V / 3 kΩ = 3.33 mA
Iz = IR − IL = 6 − 3.33 = 2.67 mA
This is less than IZM = 8 mA, so the Zener is safe.
Answer: (i) Iz = 0 (Zener off); (ii) Iz = 2.67 mA.
✓ Answer checked by Claude Opus 5.5 (high effort)
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