Question

A Zener diode of breakdown voltage 10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is

• A. 50 mA
• B. 0
• C. 30 mA
• D. 20 mA

Answer 1

The Correct Option is C

The problem asks for the current flowing through the Zener diode in the given voltage regulator circuit.

Concept Used:

The circuit is a Zener diode voltage regulator. When a Zener diode is reverse-biased and the voltage across it is greater than or equal to its breakdown voltage (\(V_Z\)), it enters the breakdown region and maintains a nearly constant voltage across its terminals, equal to \(V_Z\). The current in the circuit is then analyzed using Ohm's law and Kirchhoff's Current Law (KCL).

The steps to solve this are:

1. Verify that the Zener diode is operating in its breakdown region.
2. Once confirmed, the voltage across the load resistor is fixed at the Zener voltage, \(V_Z\).
3. Calculate the total current flowing from the source through the series resistor.
4. Calculate the current flowing through the load resistor.
5. Apply KCL at the junction to find the current flowing through the Zener diode.

Step-by-Step Solution:

Step 1: Confirm that the Zener diode is in breakdown mode.

For the diode to be in breakdown, the voltage across it must be at least \(10 \, \text{V}\). If we assume for a moment that the Zener diode is not present, the voltage across the \(500 \, \Omega\) resistor (the open-circuit voltage \(V_{OC}\)) would be given by the voltage divider rule:

\[ V_{OC} = V_{in} \times \frac{R_{load}}{R_{series} + R_{load}} = 20 \, \text{V} \times \frac{500 \, \Omega}{200 \, \Omega + 500 \, \Omega} = 20 \times \frac{500}{700} \approx 14.3 \, \text{V} \]

Since \(V_{OC} (14.3 \, \text{V})\) is greater than the Zener breakdown voltage \(V_Z (10 \, \text{V})\), the Zener diode is operating in its breakdown region. Therefore, it will regulate the voltage across the parallel branch to \(10 \, \text{V}\).

Step 2: Calculate the total current (\(I_{total}\)) flowing from the source.

Since the voltage across the Zener diode (and the parallel \(500 \, \Omega\) resistor) is fixed at \(10 \, \text{V}\), the voltage drop across the series resistor (\(R_S = 200 \, \Omega\)) is:

\[ V_{R_S} = V_{in} - V_Z = 20 \, \text{V} - 10 \, \text{V} = 10 \, \text{V} \]

The total current flowing from the source is the current through this series resistor, which can be found using Ohm's law:

\[ I_{total} = \frac{V_{R_S}}{R_S} = \frac{10 \, \text{V}}{200 \, \Omega} = 0.05 \, \text{A} = 50 \, \text{mA} \]

Step 3: Calculate the current (\(I_L\)) through the load resistor.

The voltage across the load resistor (\(R_L = 500 \, \Omega\)) is the Zener voltage, \(V_Z = 10 \, \text{V}\). The current through the load is:

\[ I_L = \frac{V_Z}{R_L} = \frac{10 \, \text{V}}{500 \, \Omega} = 0.02 \, \text{A} = 20 \, \text{mA} \]

Step 4: Apply Kirchhoff's Current Law (KCL) to find the Zener current (\(I_Z\)).

The total current from the source splits into two paths: one through the Zener diode (\(I_Z\)) and one through the load resistor (\(I_L\)). According to KCL:

\[ I_{total} = I_Z + I_L \]

Solving for the Zener current:

\[ I_Z = I_{total} - I_L \]

Final Computation & Result:

Substitute the calculated values for the total current and load current:

\[ I_Z = 50 \, \text{mA} - 20 \, \text{mA} = 30 \, \text{mA} \]

The current through the Zener diode is 30 mA.

Answer 2

Calculate Total Current \( I_S \):

The total resistance in the circuit is \( 200\Omega + 500\Omega = 700\Omega \). The source voltage is 20V, so the total current \( I_S \) flowing through the series resistance is:

\[ I_S = \frac{20}{700} = \frac{20}{700} \approx 28.6 \text{ mA} \]

Determine Voltage Across the 500 \( \Omega \) Resistor and Zener Diode:

Since the Zener diode is in breakdown mode (10V across it), the voltage drop across the 500 \( \Omega \) resistor is also 10V. The current \( I_1 \) through the 500 \( \Omega \) resistor is:

\[ I_1 = \frac{10}{500} = 20 \text{ mA} \]

Calculate Current Through the Zener Diode \( I_Z \):

The current \( I_Z \) through the Zener diode is:

\[ I_Z = I_S - I_1 = 28.6 - 20 \approx 30 \text{ mA} \]