1.

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 currentthe diode can handle?

Answer»

<P>40 mA
60 mA
50 mA
45 mA

Solution :The maximum permissible CURRENT is
`I_(Z_("max))=(P)/(V_(Z))=(364xx10^(-3))/(9.1)=40mA`


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