A diode has Zener voltage of 10 text{ V} and | vedclass

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(A) The maximum current $I_{max}$ that the Zener diode can handle is given by $P = V_z I_{max}$.Therefore, $I_{max} = \frac{P}{V_z} = \frac{0.5 	ext{

Vedclass · Accepted Answer

$300$

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(A) The maximum current $I_{max}$ that the Zener diode can handle is given by $P = V_z I_{max}$.Therefore, $I_{max} = \frac{P}{V_z} = \frac{0.5 	ext{ W}}{10 	ext{ V}} = 0.05 	ext{ A}$.The voltage across the series resistor $R$ is $V_R = V_{supply} - V_z = 25 	ext{ V} - 10 	ext{ V} = 15 	ext{ V}$.For safety, the minimum resistance $R_{min}$ is calculated using Ohm's law: $R_{min} = \frac{V_R}{I_{max}} = \frac{15 	ext{ V}}{0.05 	ext{ A}} = 300 	ext{ } \Omega$.

$A$ diode has Zener voltage of $10 \text{ V}$ and maximum power dissipation of $0.5 \text{ W}$. The minimum resistance to be used in series with this diode for safety when it is connected to a $25 \text{ V}$ power supply is . . . . . . $\Omega$.

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