The diode used in the circuit has barrier voltage 0.5 V and maximum power rating is 10 mW. What should be the minimum value of resistance R connected in series with the diode

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400 Ω

200 Ω

100 Ω

50 Ω

answer is C.

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Detailed Solution

PDmax=IDmax×VD10×10−3=IDmax×0.5IDmax=10×10−30.5=20×10−3A=20mARmin=V−V0Imax=2.5−0.520×10−3=100Ω

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