- Asked 4/15
- 2071 Magh · 5 marks
Zener current: 16 V, 1 kΩ, RL 3 kΩ
Short answer
The unloaded divider gives 12V > VZ=10V, so the Zener conducts and clamps Vo=10V. IR = (16-10)/1k = 6mA, IL = 10/3k ≈ 3.33mA. IZ = IR - IL = 2.67mA.
Find the zener current from the given zener diode network when RL = 3 kΩ and Vo = 10 V. [Figure: Vi = 16 V source, series resistor R = 1 kΩ, Zener diode in parallel with RL; Vo = 10 V.]
Zener regulator with Vi = 16 V, R = 1 kΩ, RL = 3 kΩ and Vo = 10 V.
Check whether the Zener is ON: with the Zener removed, the voltage across RL would be 16 × 3/(1 + 3) = 12 V, which is above 10 V, so the Zener breaks down and holds Vo = 10 V.
- Current through R: IR = (Vi − Vo)/R = (16 − 10)/1 kΩ = 6 mA
- Load current: IL = Vo/RL = 10/3 kΩ = 3.33 mA
- By KCL at the output node: IZ = IR − IL = 6 − 3.33
16 V o--[1k]--+-------+--o Vo=10 V
| |
Z RL 3k
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0 V o--------+-------+
Answer: IZ = 2.67 mA
✓ Answer checked by Claude Opus 5.5 (high effort)
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