A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode as shown in the diagram. The value of protective resistance $$R_p$$ is ___ $$\Omega$$.

We need to determine the value of the protective resistance $$R_p$$ connected in series with a Zener diode under the maximum potential drop arrangement.

1. Extract Given Parameters from the Problem

• Zener Voltage ($$V_z$$): $$8\text{ V}$$
• Maximum Power Dissipation Rating ($$P_{\text{max}}$$): $$0.5\text{ W}$$
• Total Supply Voltage ($$V_s$$): $$20\text{ V}$$

2. Find the Maximum Safe Zener Current ($$I_{z\text{,max}}$$)

The maximum power dissipation inside a Zener diode occurs when it carries its peak breakdown current while maintaining its rated Zener voltage. The relation is expressed as:

$$P_{\text{max}} = V_z \times I_{z\text{,max}}$$

Isolating the maximum current ($$I_{z\text{,max}}$$):

$$I_{z\text{,max}} = \frac{P_{\text{max}}}{V_z} = \frac{0.5\text{ W}}{8\text{ V}} = \frac{1}{16}\text{ A} = 0.0625\text{ A}$$

3. Determine the Voltage Across the Protective Resistor ($$V_{Rp}$$)

When the potential divider is set such that the maximum potential drop is applied across the Zener system, the Zener diode operates in its breakdown region, fixing the voltage across itself and any parallel load at exactly $$V_z = 8\text{ V}$$.

By applying Kirchhoff's Voltage Law (KVL) around the main input loop, the voltage drop across the series protective resistance ($$R_p$$) is the remaining portion of the supply voltage:

$$V_{Rp} = V_s - V_z$$

$$V_{Rp} = 20\text{ V} - 8\text{ V} = 12\text{ V}$$

4. Calculate the Value of Protective Resistance ($$R_p$$)

To ensure the Zener diode is safely protected and does not burn out under maximum potential conditions, the series resistor must limit the loop current to exactly the maximum safe Zener current ($$I_{z\text{,max}}$$). Using Ohm's law:

$$R_p = \frac{V_{Rp}}{I_{z\text{,max}}}$$

Substituting the values calculated above:

$$R_p = \frac{12\text{ V}}{\left(\frac{1}{16}\text{ A}\right)} = 12 \times 16 = 192\ \Omega$$

Final Answer: $$192$$