Full Wave Bridge Rectifier with Capacitor Filter Circuit
1. Aim
To study the construction, working, and characteristics of a Full Wave Bridge Rectifier with Capacitor Filter circuit, and to analyze its performance in terms of ripple factor, efficiency, and regulation.
2. Apparatus Used
| S.No. | Component/Equipment | Specifications | Quantity |
|---|---|---|---|
| 1 | Step-down Transformer | 230V AC / 12-0-12V, 500mA | 1 |
| 2 | Diodes | 1N4007 (1A, 1000V) | 4 |
| 3 | Filter Capacitor | 1000μF, 25V | 1 |
| 4 | Load Resistor (Variable) | 1kΩ, 2W | 1 |
| 5 | Digital Multimeter | - | 2 |
| 6 | Dual Channel Oscilloscope | 20MHz | 1 |
| 7 | Connecting Wires | - | As required |
| 8 | Breadboard | - | 1 |
3. Diagram
Fig 1: Full Wave Bridge Rectifier with Capacitor Filter Circuit Diagram
4. Theory
A Full Wave Bridge Rectifier is a circuit that converts alternating current (AC) to direct current (DC) by utilizing all halves of the input AC waveform. It consists of four diodes arranged in a bridge configuration to provide full-wave rectification.
During the positive half-cycle of the input AC signal:
- Diodes D1 and D3 are forward biased (conduct)
- Diodes D2 and D4 are reverse biased (do not conduct)
- Current flows through D1, the load, D3, and back to the source
During the negative half-cycle of the input AC signal:
- Diodes D2 and D4 are forward biased (conduct)
- Diodes D1 and D3 are reverse biased (do not conduct)
- Current flows through D2, the load, D4, and back to the source
In both cycles, current flows through the load in the same direction, effectively converting AC to pulsating DC.
Capacitor Filter: The pulsating DC output from the rectifier still contains significant AC components (ripple). A capacitor filter is added parallel to the load to reduce this ripple. The capacitor charges to the peak voltage during the conduction period and discharges slowly through the load when the rectified voltage falls below the capacitor voltage.
The effectiveness of the capacitor filter depends on:
- The capacitance value
- The load resistance
- The frequency of the input AC signal
Higher capacitance, higher load resistance, and higher frequency all lead to better filtering (lower ripple).
5. Formula
1. DC Output Voltage:
$V_{DC} = \frac{2V_m}{\pi} \approx 0.637 \times V_m$ (without filter)
$V_{DC} \approx V_m - \frac{I_{DC}}{2fC}$ (with capacitor filter)
Where:
$V_m$ = Peak value of the input voltage
$I_{DC}$ = DC load current
$f$ = Frequency of the input AC signal
$C$ = Capacitance of the filter capacitor
2. Ripple Voltage:
$V_{ripple} \approx \frac{I_{DC}}{fC}$
3. Ripple Factor:
$r = \frac{V_{ripple(rms)}}{V_{DC}} \approx \frac{1}{2\sqrt{3}fCR_L}$
Where $R_L$ is the load resistance
4. Rectifier Efficiency:
$\eta = \frac{P_{DC}}{P_{AC}} \times 100\%$
For bridge rectifier: $\eta_{max} \approx 81.2\%$
5. Voltage Regulation:
$Regulation = \frac{V_{NL} - V_{FL}}{V_{FL}} \times 100\%$
Where:
$V_{NL}$ = No-load output voltage
$V_{FL}$ = Full-load output voltage
6. Form Factor:
$Form Factor = \frac{V_{rms}}{V_{DC}}$
7. Peak Inverse Voltage (PIV):
$PIV = V_m$ (for each diode in bridge rectifier)
6. Procedure
- Connect the circuit as per the circuit diagram, with the bridge rectifier formed by four diodes D1, D2, D3, and D4.
- Initially, do not connect the filter capacitor to observe the unfiltered output.
- Connect the primary side of the transformer to the AC mains supply (230V, 50Hz).
- Set the variable resistor to its maximum value (to minimize load current).
- Switch on the power supply and observe the input AC waveform on Channel 1 of the oscilloscope.
- Observe the unfiltered output waveform across the load resistor on Channel 2 of the oscilloscope.
- Measure and record the following values:
- RMS value of input AC voltage (Vin(rms))
- RMS value of output voltage (Vo(rms))
- DC value of output voltage (Vo(DC))
- Peak value of output voltage (Vo(peak))
- Now connect the filter capacitor across the load resistor.
- Observe the filtered output waveform on the oscilloscope.
- Measure and record the following values:
- DC value of filtered output voltage (VDC)
- Ripple voltage (Vripple)
- Vary the load resistance and record the output DC voltage and ripple voltage for different load currents.
- Switch off the power supply.
- Calculate the ripple factor, efficiency, form factor, and regulation.
- Plot the relevant characteristics of the rectifier with and without filter.
7. Observation Table
Table 1: Rectifier without Filter
| S.No. | Load Resistance RL (Ω) | Input Voltage Vin(rms) (V) | Output Voltage Vo(DC) (V) | Output Voltage Vo(rms) (V) | Ripple Factor |
|---|---|---|---|---|---|
| 1 | |||||
| 2 | |||||
| 3 | |||||
| 4 | |||||
| 5 |
Table 2: Rectifier with Capacitor Filter
| S.No. | Load Resistance RL (Ω) | Input Voltage Vin(rms) (V) | Output Voltage Vo(DC) (V) | Ripple Voltage Vripple (V) | Ripple Factor |
|---|---|---|---|---|---|
| 1 | |||||
| 2 | |||||
| 3 | |||||
| 4 | |||||
| 5 |
8. Calculations
For each observation, calculate the following parameters:
1. Ripple Factor:
$r = \frac{V_{ripple(rms)}}{V_{DC}}$
Where:
$V_{ripple(rms)} = \sqrt{V_{o(rms)}^2 - V_{o(DC)}^2}$
2. Form Factor:
$Form Factor = \frac{V_{o(rms)}}{V_{o(DC)}}$
3. Rectifier Efficiency:
$\eta = \frac{P_{DC}}{P_{AC}} \times 100\%$
$\eta = \frac{V_{o(DC)}^2/R_L}{V_{o(rms)}^2/R_L} \times 100\%$
$\eta = \frac{V_{o(DC)}^2}{V_{o(rms)}^2} \times 100\%$
4. Voltage Regulation:
$Regulation = \frac{V_{NL} - V_{FL}}{V_{FL}} \times 100\%$
Where:
$V_{NL}$ = No-load output voltage (highest RL)
$V_{FL}$ = Full-load output voltage (lowest RL)
Sample Calculation:
For example, if:
Vo(DC) = 10V
Vo(rms) = 10.5V
Then:
Vripple(rms) = $\sqrt{10.5^2 - 10^2}$ = $\sqrt{110.25 - 100}$ = $\sqrt{10.25}$ = 3.2V
Ripple Factor = $\frac{3.2}{10}$ = 0.32
Form Factor = $\frac{10.5}{10}$ = 1.05
Efficiency = $\frac{10^2}{10.5^2} \times 100\%$ = $\frac{100}{110.25} \times 100\%$ = 90.7%
9. Result
From the experiment conducted on a Full Wave Bridge Rectifier with Capacitor Filter, the following results were obtained:
- The bridge rectifier successfully converted AC voltage to pulsating DC voltage.
- The addition of the capacitor filter significantly reduced the ripple in the output voltage.
- The average DC output voltage of the rectifier circuit (without filter) was found to be approximately 0.637 times the peak voltage of the AC input.
- With the capacitor filter, the DC output voltage increased to approximately the peak voltage of the AC input minus the voltage drop due to the ripple.
- The ripple factor was found to be approximately ______ without filter and ______ with filter.
- The efficiency of the bridge rectifier was approximately ______ %.
- The voltage regulation of the rectifier with capacitor filter was approximately ______ %.
- We observed that as the load current increased, the output voltage decreased and the ripple voltage increased.
These findings confirm the theoretical behavior of full wave bridge rectifiers and the effectiveness of capacitor filters in reducing ripple voltage.
10. Precautions
- Ensure all connections are tight and proper before switching on the power supply.
- Verify that the rating of components (diodes, capacitors, resistors) matches the circuit requirements.
- Observe correct polarity when connecting diodes and capacitors in the circuit.
- Do not exceed the rated current and voltage values of the components.
- Use appropriate scales on the oscilloscope to observe the waveforms clearly.
- Ensure the ground connections of oscilloscope probes are connected correctly.
- Discharge the capacitor before making any circuit modifications.
- Handle the components carefully to avoid damage due to static electricity.
- Start with the highest value of load resistance and then decrease it gradually.
- Switch off the power supply immediately if any component gets excessively hot.
- Keep the breadboard clean and organized to avoid short circuits.
- Avoid touching the circuit components when the power is ON.
11. Viva Voice Questions
Q1: What is the main advantage of a bridge rectifier over a center-tapped full-wave rectifier?
A bridge rectifier doesn't require a center-tapped transformer, making it more economical. It also utilizes both halves of the transformer secondary voltage, resulting in a higher output voltage. However, it requires four diodes instead of two.
Q2: Why is the ripple frequency in a full-wave bridge rectifier twice that of a half-wave rectifier?
In a full-wave bridge rectifier, both positive and negative half-cycles of the input AC are converted to DC, resulting in two pulses per cycle. In contrast, a half-wave rectifier only utilizes one half-cycle, producing only one pulse per cycle. Therefore, the ripple frequency in a full-wave rectifier is twice that of a half-wave rectifier.
Q3: How does a capacitor filter improve the output of a rectifier?
A capacitor filter charges to the peak value of the rectified voltage and discharges slowly through the load resistance when the rectified voltage falls below the capacitor voltage. This process helps in maintaining a more constant DC output by reducing the ripple voltage. The capacitor essentially acts as a temporary storage device, supplying energy to the load during the periods when the rectified voltage is low.
Q4: What happens to the ripple voltage and DC output voltage when the load current increases?
When the load current increases (load resistance decreases): 1. The DC output voltage decreases due to increased voltage drop across the diode's internal resistance and faster discharge of the filter capacitor. 2. The ripple voltage increases because the capacitor discharges more quickly through the lower load resistance between consecutive peaks.
Q5: What is the Peak Inverse Voltage (PIV) rating of diodes in a bridge rectifier, and why is it important?
In a bridge rectifier, the PIV rating of each diode should be at least equal to the peak value of the input AC voltage (Vm). This is important because during each half cycle, two diodes conduct while the other two are reverse biased with a voltage equal to the peak input voltage. If the PIV rating is inadequate, the diodes may break down under reverse bias conditions, leading to circuit failure and possibly component damage.
Q6: How would you calculate the optimal value of the filter capacitor for a given load and acceptable ripple?
The optimal value of the filter capacitor can be calculated using the formula: $C = \frac{I_{DC}}{2f \times V_{ripple}}$ where IDC is the load current, f is the frequency of the input AC (for full-wave rectifier, it's twice the input frequency), and Vripple is the acceptable peak-to-peak ripple voltage. A larger capacitance will result in smaller ripple, but practical limitations like cost, size, and leakage current need to be considered.
Q7: What is the maximum theoretical efficiency of a full-wave bridge rectifier?
The maximum theoretical efficiency of a full-wave bridge rectifier is approximately 81.2%. This is higher than a half-wave rectifier (40.6%) because the full-wave rectifier utilizes both half-cycles of the input AC waveform. The efficiency is limited because the output waveform, even after full-wave rectification, is not a perfect DC waveform and still contains AC components.
Q8: Explain the difference between voltage regulation in a rectifier with and without a filter capacitor.
Without a filter capacitor, the voltage regulation in a rectifier is relatively better (lower percentage) because the output voltage variation between no-load and full-load conditions is smaller. With a filter capacitor, the no-load voltage is close to the peak value of the input, while the full-load voltage drops significantly due to the discharge of the capacitor. This results in poorer voltage regulation (higher percentage). However, the filtered output provides a much more stable DC voltage with lower ripple, which is generally more important than voltage regulation for most applications.
Q9: Why are four diodes used in a bridge rectifier instead of just two?
Four diodes are used in a bridge rectifier to form a complete bridge network that enables full-wave rectification without requiring a center-tapped transformer. The arrangement allows current to flow through the load in the same direction during both positive and negative half-cycles of the input AC signal. Two diodes conduct during the positive half-cycle, and the other two conduct during the negative half-cycle, ensuring that the load always receives current in the same direction.
Q10: What other types of filters can be used with rectifiers, and how do they compare to capacitor filters?
Besides capacitor filters, other types include: 1. Inductor (choke) filters: Use an inductor in series with the load to oppose changes in current, reducing ripple but causing voltage drop. 2. LC filters: Combine inductors and capacitors for better filtering, especially at high currents. 3. π-filters: Use two capacitors and an inductor for very effective filtering but are bulkier and more expensive. 4. RC filters: Use resistors and capacitors but cause significant voltage drop. Compared to simple capacitor filters, these alternatives can provide better filtering (lower ripple factor) but may be more expensive, bulkier, or cause greater voltage drop across the filter components.