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In the circuit shown below, the knee current of the ideal Zener diode is 10 mA. To maintain 5 V across RL, the minimum value of RL in Ω and the minimum power rating of the Zener diode in mW respectively are
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- 125 and 125
- 125 and 250
- 250 and 125
- 250 and 250
Correct Option: B
Is = Iz + IL
Is – Iz = IL 
Two extreme condition :
If Iz (min),then IL (max)
If Iz (max) then IL (min) = 0
| Iz (max) = Is = | = 50 mA | |
| 10 |
Iz (min) = Is – IL (max)
IL (max) = Is – Iz (min) = Is – Iz = (50 – 10) = 40mA
| RL (min) = | = | K = 125 Ω | ||
| IL (max) | 40 |
Pz = Vz × Iz (max) = 5 × 50 mA = 250 mw
