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 objective is to determine the current traversing the Zener diode within the provided voltage regulator circuit.

Conceptual Framework:

The circuit functions as a Zener diode voltage regulator. When a Zener diode is subjected to reverse bias and the voltage across it meets or exceeds its breakdown voltage (\(V_Z\)), it enters the breakdown region, maintaining a stable voltage equal to \(V_Z\). Circuit current analysis employs Ohm's law and Kirchhoff's Current Law (KCL).

The solution methodology involves the following steps:

1. Ascertain that the Zener diode is operational within its breakdown region.
2. Upon confirmation, the voltage across the load resistor is established at the Zener voltage, \(V_Z\).
3. Compute the total current originating from the source and flowing through the series resistor.
4. Determine the current flowing through the load resistor.
5. Apply KCL at the junction to ascertain the current passing through the Zener diode.

Detailed Solution:

Step 1: Verify Zener diode breakdown mode.

For breakdown, the voltage across the diode must be at least \(10 \, \text{V}\). Assuming, hypothetically, the absence of the Zener diode, the open-circuit voltage (\(V_{OC}\)) across the \(500 \, \Omega\) resistor, calculated via the voltage divider rule, is:

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

As \(V_{OC} (14.3 \, \text{V})\) exceeds the Zener breakdown voltage \(V_Z (10 \, \text{V})\), the Zener diode is confirmed to be in its breakdown region, thus regulating the voltage across the parallel branch to \(10 \, \text{V}\).

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

With the voltage across the Zener diode and the \(500 \, \Omega\) resistor 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 from the source, flowing through this series resistor, is determined by 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 load resistor current (\(I_L\)).

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

The total source current divides between the Zener diode (\(I_Z\)) and the load resistor (\(I_L\)). KCL dictates:

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

Rearranging to solve for the Zener current:

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

Final Calculation and Outcome:

Substituting the computed values for total and load currents:

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

The current flowing through the Zener diode is 30 mA.