4) Consider the half-wave rectifier circuit shown below. The ac input voltage to the circuit is a 60 Hz signal from a power transformer. MR752 Vout +6.3 V (rms) 47000 µF R. a) What is the peak output voltage across the load resistor, assuming RL = 10 2? Account for the voltage drop across the diode (look it up!). b) Estimate the voltage ripple, assuming the same load resistance.
4) Consider the half-wave rectifier circuit shown below. The ac input voltage to the circuit is a 60 Hz signal from a power transformer. MR752 Vout +6.3 V (rms) 47000 µF R. a) What is the peak output voltage across the load resistor, assuming RL = 10 2? Account for the voltage drop across the diode (look it up!). b) Estimate the voltage ripple, assuming the same load resistance.
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![### Half-Wave Rectifier Circuit Analysis
Consider the half-wave rectifier circuit shown below. The AC input voltage to the circuit is a 60 Hz signal from a power transformer.
![Circuit Diagram]
- **AC Source**: Provides a 6.3 V (RMS) input.
- **Diode (MR752)**: Rectifies the AC signal.
- **Capacitor**: 47000 µF
- **Load Resistor (\( R_L \))**: 10 Ω
- **Output Voltage (\( V_{out} \))**
#### Questions:
a) **What is the peak output voltage across the load resistor, assuming \( R_L = 10 \, \Omega \)?**
Account for the voltage drop across the diode (look it up!).
b) **Estimate the voltage ripple, assuming the same load resistance.**
#### Diagram Explanation:
The diagram depicts a half-wave rectifier circuit where an AC input signal is fed into a diode, specifically an MR752. The diode allows only half of the AC wave to pass through to the load resistor (\( R_L \)), converting it to a DC signal with some ripple. A capacitor is connected in parallel with the output to reduce the ripple effect by temporarily storing charge and releasing it when needed.
#### Analysis Steps:
1. **Calculate the Peak Voltage**: Convert the RMS voltage to peak voltage using the formula \( V_{peak} = V_{rms} \times \sqrt{2} \).
2. **Diode Voltage Drop**: Consider the typical forward voltage drop (\( V_{f} \)) of the MR752 diode. Subtract this from the peak voltage to find the actual peak output voltage across \( R_L \).
3. **Estimate Voltage Ripple**: Use the formula for ripple voltage \( V_{ripple} = \dfrac{I}{f \cdot C} \), where \( I \) is the current, \( f \) is the frequency (60 Hz in this case), and \( C \) is the capacitance (47000 µF).
This setup provides a basic understanding of how half-wave rectification works and the factors influencing the output voltage and ripple in such a circuit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b2dbe0f-1b57-45b1-b136-e64e5f1cae33%2Fc779dfe6-fdcb-4ceb-a538-779cb4722aa5%2F4mnyx3b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Half-Wave Rectifier Circuit Analysis
Consider the half-wave rectifier circuit shown below. The AC input voltage to the circuit is a 60 Hz signal from a power transformer.
![Circuit Diagram]
- **AC Source**: Provides a 6.3 V (RMS) input.
- **Diode (MR752)**: Rectifies the AC signal.
- **Capacitor**: 47000 µF
- **Load Resistor (\( R_L \))**: 10 Ω
- **Output Voltage (\( V_{out} \))**
#### Questions:
a) **What is the peak output voltage across the load resistor, assuming \( R_L = 10 \, \Omega \)?**
Account for the voltage drop across the diode (look it up!).
b) **Estimate the voltage ripple, assuming the same load resistance.**
#### Diagram Explanation:
The diagram depicts a half-wave rectifier circuit where an AC input signal is fed into a diode, specifically an MR752. The diode allows only half of the AC wave to pass through to the load resistor (\( R_L \)), converting it to a DC signal with some ripple. A capacitor is connected in parallel with the output to reduce the ripple effect by temporarily storing charge and releasing it when needed.
#### Analysis Steps:
1. **Calculate the Peak Voltage**: Convert the RMS voltage to peak voltage using the formula \( V_{peak} = V_{rms} \times \sqrt{2} \).
2. **Diode Voltage Drop**: Consider the typical forward voltage drop (\( V_{f} \)) of the MR752 diode. Subtract this from the peak voltage to find the actual peak output voltage across \( R_L \).
3. **Estimate Voltage Ripple**: Use the formula for ripple voltage \( V_{ripple} = \dfrac{I}{f \cdot C} \), where \( I \) is the current, \( f \) is the frequency (60 Hz in this case), and \( C \) is the capacitance (47000 µF).
This setup provides a basic understanding of how half-wave rectification works and the factors influencing the output voltage and ripple in such a circuit.
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