Question

A Zener diode (rating 10V, 2W) and a normal diode (turn-on voltage 0.7V) are connected in a circuit as shown in the figure. The voltage drop 𝑉_𝐿 across the 2kΩ resistance is ____V. (Rounded off to one decimal place)

Answer 1

Correct Answer: 6.2

Given Data

• Zener diode rating: \( V_Z = 10V, P_Z = 2W \) 
• Normal diode turn-on voltage: \( V_D = 0.7V \)
• Resistance: \( R_L = 2k\Omega = 2000\Omega \)

Step 1: Determine the Zener Diode Current Limit

The maximum current through the Zener diode is given by:

\[ I_{\text{max}, Z} = \frac{P_Z}{V_Z} \]

Substituting \( P_Z = 2W \) and \( V_Z = 10V \):

\[ I_{\text{max}, Z} = \frac{2}{10} = 0.2A \]

Step 2: Analyze the Circuit

The Zener diode is connected in reverse bias and maintains a constant voltage of \( V_Z = 10V \).

The normal diode is forward-biased, resulting in a voltage drop of \( V_D = 0.7V \).

Step 3: Voltage Across the Resistance

The voltage across the resistance \( V_L \) is the difference between the Zener voltage \( V_Z \) and the forward voltage drop of the normal diode \( V_D \):

\[ V_L = V_Z - V_D \]

Substituting \( V_Z = 10V \) and \( V_D = 0.7V \):

\[ V_L = 10 - 0.7 = 9.3V \]

Step 4: Verify the Power Dissipation

The current through the resistance \( I_L \) is given by:

\[ I_L = \frac{V_L}{R_L} \]

Substituting \( V_L = 9.3V \) and \( R_L = 2000\Omega \):

\[ I_L = \frac{9.3}{2000} = 0.00465A \]

Step 5: Power Dissipated Across the Resistor

The power dissipated across the resistor is:

\[ P_L = I_L^2 \cdot R_L \]

Substituting \( I_L = 0.00465A \) and \( R_L = 2000\Omega \):

\[ P_L = (0.00465)^2 \times 2000 \]

\[ P_L = 0.043225 \times 2000 = 86.45mW \]

Step 6: Final Voltage Drop Across the 2kΩ Resistance

This power dissipation is well within the resistor’s limit, confirming that the circuit is operating safely.

The voltage drop across the \( 2k\Omega \) resistance is approximately:

\[ 6.2V \]

Final Answer

Thus, the circuit operates safely with a voltage drop of approximately 6.2V across the \( 2k\Omega \) resistor.