A 8V Zener diode along with a series resistance R is connected across a 20 V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be _____ $$\Omega$$.

We need to find the minimum value of the series resistance $$R$$ required in the given Zener diode regulator circuit.

1. Identify the Given Circuit Parameters

• Supply Voltage ($$V_s$$) = $$20\text{ V}$$
• Zener Breakdown Voltage ($$V_z$$) = $$8\text{ V}$$
• Maximum Zener Current ($$I_{z(\text{max})}$$) = $$25\text{ mA} = 25 \times 10^{-3}\text{ A}$$

2. Understand the Condition for Minimum Resistance ($$R_{\text{min}}$$)

The series resistor $$R$$ drops the excess voltage between the source supply and the stabilized Zener voltage. The current flowing through $$R$$ splits into the Zener diode ($$I_z$$) and the load resistor ($$I_L$$):

$$I = I_z + I_L$$

For the resistance $$R$$ to be at its minimum allowable value, the total current $$I$$ flowing through it must be at its maximum safely sustainable limit. This worst-case scenario occurs when the load is disconnected (open-circuit condition, $$I_L = 0$$), forcing all of the circuit current to flow directly through the Zener diode ($$I = I_{z(\text{max})}$$).

3. Calculate the Voltage Drop Across Resistor $$R$$

The voltage drop ($$V_R$$) across the series resistor is the difference between the input supply voltage and the constant operating voltage maintained by the Zener diode:

$$V_R = V_s - V_z$$

$$V_R = 20\text{ V} - 8\text{ V} = 12\text{ V}$$

4. Calculate the Minimum Value of $$R$$

Using Ohm's law, we substitute the voltage drop and the maximum safe current limit:

$$R_{\text{min}} = \frac{V_R}{I_{z(\text{max})}}$$

$$R_{\text{min}} = \frac{12\text{ V}}{25 \times 10^{-3}\text{ A}} = \frac{12 \times 1000}{25} = 12 \times 40 = 480\ \Omega$$

Therefore, the minimum value of $$R$$ required to protect the diode from exceeding its maximum current rating is 480 Ω.