In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance Rs to regulate the input unregulated supply is :
- 5kΩ
- 10Ω
- 1kΩ
- Non of these
The Correct Option is D
Approach Solution - 1
Given:
\[ V_s = 8V, \quad V_z = 5V, \quad R_L = 1k\Omega \]
Power across the Zener diode:
\[ P_d = 10 \, \text{mW}, \quad V_z = 5V \implies I_z = \frac{P_d}{V_z} = \frac{10 \times 10^{-3}}{5} = 2 \, \text{mA} \]
Current through the load resistor:
\[ I_L = \frac{V_z}{R_L} = \frac{5V}{1k\Omega} = 5 \, \text{mA} \]
Maximum current through the Zener diode:
\[ I_{z_{max}} = 2 \, \text{mA} \]
Applying KCL at the junction:
\[ I_s = I_L + I_z = 5 \, \text{mA} + 2 \, \text{mA} = 7 \, \text{mA} \]
The series resistance \(R_s\) is given by:
\[ R_s = \frac{V_s - V_z}{I_s} = \frac{8V - 5V}{7 \, \text{mA}} = \frac{3}{7} k\Omega \]
Minimum current calculations:
For the minimum current through the Zener diode:
\[ I_{z_{min}} = 0 \, \text{mA} \quad \text{(to ensure Zener regulation)} \]
Total current through the circuit:
\[ I_{s_{min}} = I_L = 5 \, \text{mA} \]
The corresponding series resistance \(R_s\) for minimum current is given by:
\[ R_{s_{min}} = \frac{V_s - V_z}{I_{s_{min}}} = \frac{8V - 5V}{5 \, \text{mA}} = \frac{3}{5} k\Omega \]
Conclusion:
Thus, the value of the series resistance \(R_s\) must satisfy:
\[ \frac{3}{7} k\Omega < R_s < \frac{3}{5} k\Omega \]
Therefore, the suitable range for \(R_s\) is between \(\frac{3}{7} k\Omega\) and \(\frac{3}{5} k\Omega\).
Approach Solution -2
In the given Zener diode voltage regulator circuit, we are provided with the input voltage, Zener voltage, load resistance, and the maximum power rating of the Zener diode. The goal is to calculate the value of the series resistance Rs required for proper regulation.
Concept Used:
A Zener diode acts as a voltage regulator by maintaining a constant voltage (VZ) across its terminals when reverse-biased in its breakdown region. To protect the Zener diode from excessive current, a series resistor (Rs) is used to limit the current from the source.
The key equations for analyzing this circuit are:
- Load Current (IL): The current through the load resistor RL. \[ I_L = \frac{V_Z}{R_L} \]
- Maximum Zener Current (IZ(max)): The maximum current the Zener diode can handle is determined by its power rating PZ(max). \[ P_{Z(\text{max})} = V_Z \times I_{Z(\text{max})} \]
- Source Current (IS): The total current drawn from the input supply, which flows through Rs. By Kirchhoff's Current Law, it is the sum of the Zener current and load current. For safe operation, we design for the maximum Zener current. \[ I_S = I_{Z(\text{max})} + I_L \]
- Series Resistance (Rs): Calculated using Ohm's law for the voltage drop across it. \[ R_s = \frac{V_{in} - V_Z}{I_S} \]
Step-by-Step Solution:
Step 1: List the given parameters from the circuit diagram and problem statement.
\[ V_{in} = 8 \, \text{V} \] \[ V_Z = 5 \, \text{V} \] \[ R_L = 1 \, \text{k}\Omega = 1000 \, \Omega \] \[ P_{Z(\text{max})} = 10 \, \text{mW} = 10 \times 10^{-3} \, \text{W} \]
Step 2: Calculate the current flowing through the load resistor (IL). Since the load is parallel to the Zener diode, the voltage across it is VZ.
\[ I_L = \frac{V_Z}{R_L} = \frac{5 \, \text{V}}{1000 \, \Omega} = 0.005 \, \text{A} = 5 \, \text{mA} \]
Step 3: Calculate the maximum permissible current through the Zener diode (IZ(max)) using its power rating.
\[ I_{Z(\text{max})} = \frac{P_{Z(\text{max})}}{V_Z} = \frac{10 \times 10^{-3} \, \text{W}}{5 \, \text{V}} = 0.002 \, \text{A} = 2 \, \text{mA} \]
Step 4: The series resistor Rs must limit the total current from the source (IS) to ensure the Zener current does not exceed its maximum rating. The total current is the sum of the load current and the maximum Zener current.
\[ I_S = I_{Z(\text{max})} + I_L = 2 \, \text{mA} + 5 \, \text{mA} = 7 \, \text{mA} = 7 \times 10^{-3} \, \text{A} \]
Step 5: Calculate the value of the series resistance Rs using Ohm's law. The voltage drop across Rs is the difference between the input voltage and the Zener voltage.
\[ V_{R_s} = V_{in} - V_Z = 8 \, \text{V} - 5 \, \text{V} = 3 \, \text{V} \] \[ R_s = \frac{V_{R_s}}{I_S} = \frac{3 \, \text{V}}{7 \times 10^{-3} \, \text{A}} \]
Final Computation & Result:
Performing the final calculation:
\[ R_s = \frac{3}{0.007} \, \Omega \approx 428.57 \, \Omega \]
The value of the series resistance Rs is approximately 428.57 Ω.
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