Question

Question 3

3. 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 ?

Answer 1

$$\begin{gathered} Currentincreaseswithinceaseinvoltage.Power=voltage\times current.Whenvoltageismaximumthencurrentwillbemaximum.Thereforemaximumpowerdissipationoccuursatmaximumvoltage. \\ maximumvoltage,V_{max}=breakdownvoltage=9.1V \\ {P}_{max}=V_{max}{I}_{max} \\ {P}_{max}=364mW=0.364W \\ {V}_{max}=9.1V \\ 0.364=9.1I_{max} \\ {I}_{max}=\frac{0.364}{9.1}=0.04A=40mA \end{gathered}$$