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A Zener diode is specified having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle.A. 40 mAB. 60 mAC. 50 mAD. 45 mA

Answer» Correct Answer - A
The maximum permissible current is
`I_(Z_("max"))=P/V_z=(364xx10^(-3))/9.1=40 mA`


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