Saved Bookmarks
| 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» <P> Solution :`P_(MAX) = V_z I_z `( max)` I_z ( max ) =(P_("max"))(/(V_z) = ( 364 xx 10^(-3))/( 9.1) = 40 mA ` |
|