Question

A Zener diode is specified to have break-down voltage of $9.1\text{V}$ with a maximum power dissipation of $364\text{mW}.$ What is the maximum current that the diode can handle ?

• A. $30\text{mA}$
• B. $40\text{mA}$
• C. $60\text{mA}$
• D. $70\text{mA}$

Answer 1

The correct option is B $40\text{mA}$


As,

$\text{Power = voltage}\times \text{current}$

$\Rightarrow 364\times {10}^{-3}\text{W}=9.1\text{volt}\times I_Z\left(\text{in A}\right)$

$\Rightarrow I_z=\frac{364}{9.1}\times {10}^{-3}\text{A}=\frac{364}{9.1}\text{mA}$

$\therefore I_Z=40\text{mA}$

Hence, $\left(\text{B}\right)$ is the correct answer.