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 describes a Zener diode voltage regulator circuit. We need to determine the load current (\( I_L \)) and the load resistance (\( R_L \)) based on the given circuit parameters and conditions.

Concept Used:

The operation of this circuit is based on the properties of a Zener diode and fundamental circuit laws.

1. Zener Diode as a Voltage Regulator: When operated in the reverse breakdown region, a Zener diode maintains a nearly constant voltage across its terminals, known as the Zener voltage (\( V_Z \)). In this circuit, the load resistor \( R_L \) is connected in parallel with the Zener diode, so the voltage across the load is regulated to be equal to the Zener voltage, i.e., \( V_L = V_Z \).
2. Ohm's Law: This law relates voltage (V), current (I), and resistance (R) as \( V = IR \).
3. Kirchhoff's Laws:
   • Kirchhoff's Voltage Law (KVL): The sum of voltage drops around a closed loop is zero. For the input loop, the input voltage drops across the series resistor (\( R_S \)) and the Zener diode.
   • Kirchhoff's Current Law (KCL): The total current entering a junction is equal to the total current leaving it. The current from the source (\( I_S \)) splits into the Zener current (\( I_Z \)) and the load current (\( I_L \)).

The key equations 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: List the given circuit parameters.

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

Step 2: Calculate the total current (\( I_S \)) flowing from the source through the series resistor \( R_S \).

The Zener diode maintains a constant voltage of 5 V across the load. Therefore, the voltage drop across the series resistor \( R_S \) is the difference between the input voltage and the Zener voltage.

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

Using Ohm's law, we can find the total current \( I_S \):

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

Converting this to milliamperes (mA):

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

Step 3: Use Kirchhoff's Current Law (KCL) and the given condition to find the load current (\( I_L \)).

According to KCL, the total current \( I_S \) splits into the Zener current \( I_Z \) and the load current \( I_L \).

\[ I_S = I_Z + I_L \]

We are given the condition that \( I_Z = 4 I_L \). Substituting this into the KCL equation:

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

Now, we can solve for \( I_L \) using the value of \( I_S \) we calculated:

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

Final Computation & Result:

Step 4: Calculate the load resistance (\( R_L \)).

The voltage across the load resistance is the regulated Zener voltage, \( V_L = V_Z = 5 \, \text{V} \). We can find the load resistance using Ohm's law with the load current we just found.

First, convert the load current \( I_L \) back to Amperes:

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

Now, calculate \( R_L \):

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

Therefore, the load current and load resistance are:

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

Answer 2

From the circuit diagram, we have the following: \[ i = \frac{20}{400} \, \text{A} = 10 \, \text{mA} \quad \text{(Load current \( I_L \))} \] \[ V_L = 5V \quad \text{(Zener voltage)} \] Also, \[ R_L = \frac{V_L}{i} = \frac{5}{10 \times 10^{-3}} = 500 \, \Omega \]