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    				| 1. | A Zener diode has a Zener voltage of 2.4 V and a 500 mW power rating. What should be the maxi-mum current through the diode if you design conservatively with a safety factor of 2? | 
| Answer» Data : Vz = 2.4 V, PZM = 500 mW A conservative design includes a safety factor to keep the power dissipation well below the rated maximum power. Thus, with a safety factor of 2, the operating power of the Zener diode is PZ = \(\cfrac12\) PZM = \(\cfrac{500}2\) = 250 m W Therefore, the Zener current, IZ = \(\cfrac{p_z}{V_z}\) = \(\cfrac{250mW}{2.4\,V}\) = 104.2 mA | |