Question

In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance Rs to regulate the input unregulated supply is :

• A. 5kΩ
• B. 10Ω
• C. 1kΩ
• D. Non of these

Answer 1

Answer: (D)

Non of these

Answer 2

Given:
$V_s = 8V, \quad V_z = 5V, \quad R_L = 1k\Omega$

Power across the Zener diode:

$P_d = 10 \, \text{mW}, \quad V_z = 5V \implies I_z = \frac{P_d}{V_z} = \frac{10 \times 10^{-3}}{5} = 2 \, \text{mA}$

Current through the load resistor:

$I_L = \frac{V_z}{R_L} = \frac{5V}{1k\Omega} = 5 \, \text{mA}$

Maximum current through the Zener diode:

$I_{z_{max}} = 2 \, \text{mA}$

Applying KCL at the junction:

$I_s = I_L + I_z = 5 \, \text{mA} + 2 \, \text{mA} = 7 \, \text{mA}$

The series resistance $R_s$ is given by:

$R_s = \frac{V_s - V_z}{I_s} = \frac{8V - 5V}{7 \, \text{mA}} = \frac{3}{7} k\Omega$

Minimum current calculations:

For the minimum current through the Zener diode:

$I_{z_{min}} = 0 \, \text{mA} \quad \text{(to ensure Zener regulation)}$

Total current through the circuit:

$I_{s_{min}} = I_L = 5 \, \text{mA}$

The corresponding series resistance $R_s$ for minimum current is given by:

$R_{s_{min}} = \frac{V_s - V_z}{I_{s_{min}}} = \frac{8V - 5V}{5 \, \text{mA}} = \frac{3}{5} k\Omega$

Conclusion:

Thus, the value of the series resistance $R_s$ must satisfy:

$\frac{3}{7} k\Omega < R_s < \frac{3}{5} k\Omega$

Therefore, the suitable range for $R_s$ is between $\frac{3}{7} k\Omega$ and $\frac{3}{5} k\Omega$.