1.

A zener diode has a breakdown voltage of 9.1 volts with a maximum power dissipation of 364 milliwatts. What is the maximum current the diode can handle?

Answer»

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Solution :`P_(MAX) = V_z I_z `( max)
` I_z ( max ) =(P_("max"))(/(V_z) = ( 364 xx 10^(-3))/( 9.1) = 40 mA `


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