Question

A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25V. For a 400 \( \Omega \) resistor connected in series, the zener current is found to be 4 times load current. The load current \( I_L \) and load resistance \( R_L \) are:

• A. \( I_L = 20 \, \text{mA}; \, R_L = 250 \, \Omega \)
• B. \( I_L = 10 \, \text{A}; \, R_L = 0.5 \, \Omega \)
• C. \( I_L = 0.02 \, \text{mA}; \, R_L = 250 \, \Omega \)
• D. \( I_L = 10 \, \text{mA}; \, R_L = 500 \, \Omega \)

Hint:

In zener diode circuits, the load current and load resistance can be found by using Ohm’s law and the given zener voltage.

Answer 1

The Correct Option is D

This problem requires the determination of load current (\( I_L \)) and load resistance (\( R_L \)) within a Zener diode voltage regulator circuit, based on provided parameters.

Concept Used:

Circuit operation relies on Zener diode characteristics and fundamental circuit laws.

1. Zener Diode Voltage Regulation: In reverse breakdown, a Zener diode maintains a constant voltage (\( V_Z \)) across its terminals. When \( R_L \) is in parallel with the Zener, the load voltage (\( V_L \)) equals \( V_Z \).
2. Ohm's Law: Governs the relationship between voltage (V), current (I), and resistance (R) via \( V = IR \).
3. Kirchhoff's Laws:
   • Kirchhoff's Voltage Law (KVL): Sum of voltages in a closed loop is zero. The input loop voltage drops across \( R_S \) and the Zener diode.
   • Kirchhoff's Current Law (KCL): Current entering a junction equals current leaving it. Source current (\( I_S \)) divides into Zener current (\( I_Z \)) and load current (\( I_L \)).

Key equations governing this circuit are:

\[ V_L = V_Z \] \[ I_S = \frac{V_{in} - V_Z}{R_S} \] \[ I_S = I_Z + I_L \]

Step-by-Step Solution:

Step 1: Identify given circuit parameters.

• Unregulated DC input voltage, \( V_{in} = 25 \, \text{V} \)
• Zener voltage, \( V_Z = 5 \, \text{V} \)
• Series resistance, \( R_S = 400 \, \Omega \)
• Condition: Zener current is 4 times load current, \( I_Z = 4 I_L \).

Step 2: Calculate total source current (\( I_S \)).

The Zener diode establishes a constant load voltage of 5 V. The voltage drop across \( R_S \) is the difference between input voltage and Zener voltage.

\[ V_{R_S} = V_{in} - V_Z = 25 \, \text{V} - 5 \, \text{V} = 20 \, \text{V} \]

Using Ohm's law for \( I_S \):

\[ I_S = \frac{V_{R_S}}{R_S} = \frac{20 \, \text{V}}{400 \, \Omega} = 0.05 \, \text{A} \]

Converted to milliamperes:

\[ I_S = 0.05 \, \text{A} \times 1000 \, \frac{\text{mA}}{\text{A}} = 50 \, \text{mA} \]

Step 3: Apply KCL and the given condition to find load current (\( I_L \)).

KCL states that \( I_S \) divides into \( I_Z \) and \( I_L \).

\[ I_S = I_Z + I_L \]

Substituting the condition \( I_Z = 4 I_L \):

\[ I_S = 4 I_L + I_L = 5 I_L \]

Solving for \( I_L \) with \( I_S = 50 \, \text{mA} \):

\[ 50 \, \text{mA} = 5 I_L \] \[ I_L = \frac{50 \, \text{mA}}{5} = 10 \, \text{mA} \]

Final Computation & Result:

Step 4: Determine load resistance (\( R_L \)).

Load voltage is \( V_L = V_Z = 5 \, \text{V} \). Ohm's law is used with the calculated load current.

Convert \( I_L \) to Amperes:

\[ I_L = 10 \, \text{mA} = 0.01 \, \text{A} \]

Calculate \( R_L \):

\[ R_L = \frac{V_L}{I_L} = \frac{5 \, \text{V}}{0.01 \, \text{A}} = 500 \, \Omega \]

The results for load current and load resistance are:

\( I_L = 10 \, \text{mA}; \, R_L = 500 \, \Omega \)