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
- 50 mA
- 0
- 30 mA
- 20 mA
The Correct Option is C
Approach Solution - 1
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:
- Verify that the Zener diode is operating in its breakdown region.
- Once confirmed, the voltage across the load resistor is fixed at the Zener voltage, \(V_Z\).
- Calculate the total current flowing from the source through the series resistor.
- Calculate the current flowing through the load resistor.
- 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.
Approach Solution -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:
\[ I_Z = I_S - I_1 = 28.6 - 20 \approx 30 \text{ mA} \]
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