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

The Correct Option is D

Provided values:
\[ V_s = 8V, \quad V_z = 5V, \quad R_L = 1k\Omega \]

Zener Diode Power Dissipation:

Given Zener power \(P_d = 10 \, \text{mW}\) and Zener voltage \(V_z = 5V\), the Zener current is calculated as \(I_z = \frac{P_d}{V_z} = \frac{10 \times 10^{-3}}{5} = 2 \, \text{mA}\).

Load Resistor Current:

The current through the load resistor is \(I_L = \frac{V_z}{R_L} = \frac{5V}{1k\Omega} = 5 \, \text{mA}\).

Maximum Zener Diode Current:

The maximum current through the Zener diode is \(I_{z_{max}} = 2 \, \text{mA}\).

Kirchhoff's Current Law (KCL) Application:

Applying KCL at the junction, the source current is \(I_s = I_L + I_z = 5 \, \text{mA} + 2 \, \text{mA} = 7 \, \text{mA}\).

Series Resistance \(R_s\) Calculation:

The series resistance \(R_s\) is determined by: \(R_s = \frac{V_s - V_z}{I_s} = \frac{8V - 5V}{7 \, \text{mA}} = \frac{3}{7} k\Omega \.

Minimum Current Analysis:

For minimum Zener diode current (to ensure regulation):

\[ I_{z_{min}} = 0 \, \text{mA} \]

The total circuit current in this case is equal to the load current:

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

The series resistance \(R_s\) corresponding to this minimum current is:

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

Conclusion:

The value of the series resistance \(R_s\) must fall within the following range:

\[ \frac{3}{7} k\Omega < R_s < \frac{3}{5} k\Omega \]

Therefore, the acceptable range for \(R_s\) is between \(\frac{3}{7} k\Omega\) and \(\frac{3}{5} k\Omega\).