Minimum Series Resistance for Zener Diode Safety Circuit

To find the minimum series resistance required to ensure the safety of the Zener diode, we need to calculate the maximum safe current that can flow through the diode without exceeding its maximum power rating.

1. Given Data:

• Zener voltage of the diode, $V_Z = 10\text{ V}$
• Maximum power dissipation of the diode, $P_{\text{max}} = 0.5\text{ W}$
• Supply voltage, $V_s = 25\text{ V}$

2. Calculating the Maximum Safe Current ($I_{\text{max}}$): The power dissipated in the Zener diode when it is operating in breakdown region is given by: $P = V_Z \times I_Z$

To avoid damaging the diode, the maximum current $I_{\text{max}}$ passing through it must satisfy: $P_{\text{max}} = V_Z \times I_{\text{max}}$

Rearranging for $I_{\text{max}}$: $I_{\text{max}} = \frac{P_{\text{max}}}{V_Z} = \frac{0.5\text{ W}}{10\text{ V}} = 0.05\text{ A}$

3. Calculating the Voltage Drop Across the Series Resistance ($V_R$): The potential difference across the series resistor $R$ is the difference between the supply voltage and the Zener breakdown voltage: $V_R = V_s - V_Z$ $V_R = 25\text{ V} - 10\text{ V} = 15\text{ V}$

4. Determining the Minimum Series Resistance ($R_{\text{min}}$): By Ohm's law, the series resistance $R$ is given by: $R = \frac{V_R}{I_Z}$

For the current not to exceed $I_{\text{max}} = 0.05\text{ A}$, the resistance must be at least $R_{\text{min}}$: $R_{\text{min}} = \frac{V_R}{I_{\text{max}}} = \frac{15\text{ V}}{0.05\text{ A}} = 300\text{ }\Omega$

Thus, the minimum resistance to be used in series with the Zener diode for safety is $300\text{ }\Omega$.