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Nuclei

Question
CBSEENPH12047983

A zener diode is specified as having a breakdown voltage of 9.1 V, with a maximum power dissipation of 364 mW. What is the maximum current the diode can handle?

  • 40 mA

  • 60 mA

  • 50 mA

  • 45 mA

Solution

A.

40 mA

Current increases with increase in voltage.

           Power = voltage × current

When voltage is maximum then current will be maximum.

Therefore maximum ower dissipation occurs at maximum voltage

           Pmax = Vmax IZmax

            Pmax = 364 mW

                     = 364  × 10-3 W

          Vmax = 9.1 V

The maximum permissible current is

            IZmaxPmaxVmax

                     = 364 × 10-39.1

               IZmax = 40 mA