Question

The following Zener diode voltage regulator circuit is used to obtain 20 V regulated output at load resistance \( R_L \) from a 35 V dc power supply. Zener diodes are rated at 5W and 10V. The value of the resistance \( R \) is ......... \( \Omega \). 

Hint:

In voltage regulator circuits, ensure the power dissipation in the Zener diode is within the specified rating to avoid damage.

Answer 1

Correct Answer: 30

The power dissipation in the Zener diode is given by \[ P_{\text{Zener}} = \frac{V_Z^2}{R_L}. \] It is given that the power dissipation is 5 W and \( V_Z = 10\, \text{V} \). Therefore, \[ 5 = \frac{10^2}{R_L} \Rightarrow R_L = 20 \, \Omega. \] Now, to calculate \( R \), use Kirchhoff’s voltage law: \[ V_{\text{in}} = I R + V_Z, \] where \( V_{\text{in}} = 35\, \text{V} \) and \( V_Z = 10\, \text{V} \). The current \( I \) can be found using \[ I = \frac{V_Z}{R_L} = \frac{10}{20} = 0.5 \, \text{A}. \] Substituting into the equation for \( R \): \[ 35 = 0.5 \cdot R + 10 \Rightarrow R = \frac{25}{0.5} = 50 \, \Omega. \] Final Answer: The value of \( R \) is \( 50 \, \Omega \).