Physics HL
Sample Internal Assessment
6/7
2,071 Words
English
Free

# How does the ripple factor expressed as a ratio of output AC current and the output DC current of a bridge rectifier circuit with capacitive filter depends on the capacitance of the capacitive filter when varied over a range of 1mF to 10mF?

## Background information

### What is a bridge rectifier with a capacitive filter

A rectifier circuit is used to convert AC current to DC current. There are various types of rectifier circuits. Amongst them, the Bridge Rectifier Circuit is the most efficient and important rectifier circuit because it rectifies the AC current approximately 100% if proper circuit elements are used. It is a combination of four PN – Junction diodes which prevents the current to reach the output load in one half cycle of the AC signal and also rectifies the current of the other half cycle. The output terminal of the arrangement of the diodes is connected with the load resistance (load output) and a capacitor in parallel. This capacitor acts as a filter and prevents AC component to reach at the output load end.

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

### Video Course

• Figure 1 - Bridge Rectifier Circuit With Capacitive Filter

### How does bridge rectifier works

From figure 1, it could be observed that, during the positive half cycle of the supply AC current, diode D1 was in forward bias condition and diode D3 was in reverse bias condition. Hence, diode D1 would act as a closed-circuit path and allow current to flow through it and diode D3 would act as an open circuit path and would not allow current to flow through it. Similarly, during the negative half cycle of the supply AC current, diode D4 was in forward bias condition and diode D2 was in reverse bias condition. Hence, diode D4 would act as a closed-circuit path and allow current to flow through it and diode D2 would act as an open circuit path and would not allow current to flow through it.

In both the cycles, the direction of current delivered at the load output RL is same. Hence, the change of phase is not observed in the output end.

However, there is a change in the magnitude of current in the output. That is rectified using the capacitive filter which is explained in detail in Section 4.4.

### Reactance of a capacitor

Reactance might be defined as the opposition that was offered by a capacitor to flow current through it. Capacitive Reactance offered by a capacitor is expressed as the inverse of the product of angular frequency and the magnitude of capacitance of the capacitor.

$$X_C = \frac1{\omega C}$$

where,

X= Capacitive Reactance

ω = angular frequency of current component

C = capacitance of the capacitor

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ### What is the function of a capacitive filter in a bridge rectifier

The capacitive filter performs two major functions in the bridge rectifier circuit. Firstly, from the expression of reactance of the circuit, it could be said that a capacitor offers infinite reactance (or resistance) to DC current because frequency of a DC current is zero. Hence, angular frequency would also be zero resulting a value of infinite reactance. On the other hand, the reactance offered towards an AC source will be very less. For example, if an AC source of 50 Hz is fed through a capacitor with capacitance of 1mF, then the reactance offered by the capacitor would be:

$$X_{C} = \frac{1}{ωC}$$

$$X_{C} = \frac{1}{50×1×10^{-3}}$$

$$X_C = 20$$

If the load resistance is more than that of the reactance of the capacitor, then maximum AC component will pass through the capacitor and will not reach the load output resulting in reduction of AC component in the output.

Secondly, a capacitor acts as a battery (DC in nature). When AC current flows through it, a capacitor gets charged. During the first quarter cycle of AC, the magnitude of current increases. During this phase, the capacitor is charged as the current flows through the capacitor. During the second phase, the magnitude of AC current decreases and hence, the capacitor starts to discharge, supplying DC current across the load output. As a result, DC current is obtained in the output.

### Equivalent capacitance of capacitor when connected in parallel

Figure 2 - Capacitor Connected In Parallel Across AC Supply

From the definition of capacitance of the capacitor, it could be expressed:

q = CV

where,

q = charge stored in the capacitor

C = capacitance of the capacitor

V = voltage across capacitor

In figure 2, all the capacitors are connected in parallel across the AC supply. Hence, potential difference across every capacitor is same and equal to the voltage of the AC supply. Moreover, the it could be said that the charge dissociated by the AC supply is equal to the charge stored in every capacitor. Therefore:

qtotal = q+ q+…+ qn

Ceq × V = C× V+ C× V+ ... + Cn × Vn

where,

Ceq = equivalent capacitance of the circuit

C1, C2 ... Cn = capacitance of each capacitor connected in parallel

V1, V2, ... Vn = potential difference across each capacitor respectively

V = potential difference of the AC supply

Here, as all the capacitors are connected in parallel, the potential difference across every capacitor would be equal.

Ceq × V = C1 × V + C× V + ... + Cn × V

Ceq × V = V(C+ C2 + ... + Cn)

Ceq = C1 + C2 + ... + Cn

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ### Experimental methodology

The bridge rectifier circuit with capacitive filter was prepared. A step-down transformer was connected with household AC supply connect in the primary side of the transformer and the bridge rectifier circuit was connected in the secondary side of the transformer. The value of capacitance of the capacitive filter was varied throughout the experiment by connecting capacitors in parallel to the load output and the AC voltage (Vrms) and (Vdc) was measured using a multimeter across the load. The ripple factor has been calculated for each trial and the variation of ripple factor with an increase in capacitance of the capacitive filter was studied.

## Hypothesis

It was assumed that with an increase in capacitance of the capacitive filter in the bridge rectifier circuit, the ripple factor would decrease. This was assumed because with an increase in capacitance of the filter, more AC current will be bypassed through the path in which the capacitor was connected because capacitance offers extremely low resistance towards AC current. With an increase in capacitance of the capacitive filter, its reactance would decrease and hence, more AC current will flow through the filter, decreasing the amount of AC current component in the load.

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ## Variables

### Independent variable

Capacitance of the capacitive filter

The capacitance of the capacitive filter connected across the load of bridge rectifier circuit was the independent variable of this exploration. The magnitude of capacitance was varied over a range of 10mF from 1mF to 10mF at an interval of 1mF. The capacitance of the capacitive filter was increased by connecting identical capacitors of 1mF in parallel across the load resistor (output). The capacitance was increased over the above-mentioned range to understand the variation in AC component of voltage across the output over a wide range of capacitance to strength the correlation obtained. Moreover, the capacitance has been increased at a fixed interval to maintain a regularity in the observed magnitudes of output current.

### Dependent variable

Ripple Factor

Ripple factor is the dependent variable of the exploration. It is a measure of AC component present in a rectified AC signal. More the Ripple factor, less efficient the rectification methodology is. An ideal rectification circuit or an ideal AC to DC converter would have 0 ripple factor as there should not be any AC component in the output after rectification. Ripple Factor is the ratio of output AC current and the output DC current. It is also represented by the following formula:

$$\gamma = \sqrt{\frac{v^2_{rms}}{V^2_{dc}}\ -\ 1}$$

$$\gamma = Ripple \ \ Factor$$

$$v_{rms} = Output \ \ RMS \ \ Voltage$$

$$V_{dc} = Output \ \ DC \ \ Voltage$$

In the experiment, the rms output voltage and the dc output voltage has been obtained using a multimeter and by using the above-mentioned expression the ripple factor has been calculated.

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ### Controlled variable

Variable
Why it has been controlled?
How is the variable controlled?
If the load resistance was varied then it would have affected the potential drop across the load at the output. Hence, AC output voltage and the DC output voltage would have been changed affecting the ripple factor.
A fixed resistance of 1000 Ω was connected at the output for every trial of the experiment.
Input Voltage
If the input supply voltage was varied, then it would have varied the output voltage (both DC output voltage and AC output voltage) across the load. Consequently, the ripple factor would have been changed.
Input supply voltage was chosen to be the household supply AC voltage of 220 V, 50 Hz throughout the experiment.
Ratio of Turns of transformer
If the ratio of turns of primary and secondary coil would have changed, then the voltage across the rectifier circuit would be varied which would result in variation in DC output voltage and AC output voltage. As a result, the ripple factor would have been changed.
Same transformer has been used throughout the experiment.
Temperature
An increase in temperature would have increased resistance of the circuit affecting the output voltage.
The experiment has been done in a confined place with least possibility of change in temperature.
Figure 3 - Table On Controlled Variable

## Apparatus and materials required

### Apparatus table

Apparatus
Specification
Quantity
Least Count
Uncertainty (±)
AC Power Supply
220 V, 50 Hz
1
-
-
Transformer
Step Down (Turn Ratio equals to 44)
1
-
-
-
1
-
-
Multimeter
-
1
0.001 V
0.001V
PN – Junction Diode
Silicon
4
-
-
Resistor
1kΩ
1
-
-
Capacitor
Polar, 1000μF, 10V
10
-
-
Multimeter Probe
-
2
-
-
Figure 4 - Table On Apparatus Table

### Materials required

No materials are required to perform the experiment.

## Considerations

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

### Environmental considerations

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ## Data processing

### Raw data table

Figure 5 - Table On Variation In Magnitude Of AC Output Voltage And DC Output Voltage With Respect To Capacitance Of The Capacitive Filter

Sample Calculation

Mean AC Output Voltage for Capacitance equals 1.000mF

$$=\frac{3.195\,+\,3.191\,+\,3.196}{3}= 3.194V$$

Mean DC Output Voltage for Capacitance equals 1.000mF

$$=\frac{2.851\,+\,2.854\,+\,2.855}{3}= 2.853V$$

Standard Deviation of AC Output Voltage for Capacitance equals 1.000mF

$$\sqrt{\frac{(3.195\,-\,3.194)^{2}\,+\,(3.191\,-\,3.194)^{2}+\,(3.196\,-\,3.194)^{2}}3} = 0.003mF$$

Standard Deviation of AC Output Voltage for Capacitance equals 1.000mV

$$\sqrt{\frac{(2.851\,-\,2.853)^2\ +\ (2.854\ -\ 2.853)^2\ +\ (2.855\ -\ 2.853)^2}3} = 0.002mV$$

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ### Processed data table

Capacitance (±0.001mF)
Mean AC Output Voltage ±0.001V
Mean DC Output Voltage ±0.001V
Ripple Factor (±0.002)
1.000
3.194
2.853
0.503
2.000
3.126
2.815
0.483
3.000
3.054
2.773
0.461
4.000
2.984
2.735
0.436
5.000
2.915
2.691
0.416
6.000
2.843
2.626
0.415
7.000
2.772
2.586
0.386
8.000
2.705
2.545
0.360
9.000
2.625
2.494
0.328
10.000
2.564
2.414
0.358
Figure 7 - Table On Variation Of Ripple Factor With Respect To Capacitance (In mF)

Sample Calculation for Ripple Factor

Ripple Factor when Capacitance equals 1.000mF

$$\sqrt{\frac{v^2_{rms}}{v^{2}_{dc}}\ - 1}$$

$$\sqrt{\frac{3.194^2}{2.853^{2}}\ - 1}$$

= 0.503

Impact on uncertainty -

$$\gamma = \sqrt{\frac{V^2_{rms}}{v^{2}_{dc}}\ - 1}$$

Taking relative error in both the sides -

$$\frac{∆ \gamma}{ \gamma} = \frac{1}{2}\ ×\ \biggl[2\ × \frac{\Delta v_{rms}}{v_{rms}}\ +\ 2\ × \frac{\Delta V_{DC}}{V_{DC}}\biggl{]}$$

$$\frac{\Delta \gamma}{\gamma}=\frac{\Delta v_{rms}}{v_{rms}} \,+\, \frac{\Delta V_{DC}}{V_{DC}}$$

$$\frac{\Delta\ \gamma}{\gamma} = 0.001 + 0.001$$

$$\frac{\Delta\ \gamma}{\gamma} = 0.002$$

### Graphical analysis

Figure 8 - Variation Of Ripple Factor Versus Capacitance Of Capacitive Filter (In mF)

Analysis

In the graph shown above, variation of ripple factor with respect to the capacitance of the capacitive filter of a Bridge rectifier circuit has been obtained. Capacitance of the capacitive filter in mF has been plotted along the X – Axis, and the Ripple factor has been plotted along the Y – Axis. Ripple factor is a ratio and thus, it does not have any unit. It has been observed that with an increase in capacitance from 1mF to 10 mF, the ripple factor has decreased from 0.503 to 0.358. Thus, an overall negative correlation has been obtained between the ripple factor and the capacitance of the capacitive filter. The equation of trend obtained in the graph has been expressed as: y = - 0.0185x + 0.5161 where y represents the Ripple factor and x represents the capacitance of the capacitive filter. The ripple factor has decreased at a rate of 0.0185 with a unit increase of capacitance in mF.

From the X – intercept of the graph, the value of capacitance for an ideal rectifier could be found -

0 = - 0.0185x + 0.5161

0.0185 x = 0.5161

$$x = \frac{0.5161}{0.0185}$$

x = 27.897 mF

From the graph, it has been found that the ripple factor of a bridge rectifier circuit should be equal to zero if the capacitance of the capacitive filter is equal to 27.897 mF.

The data points are almost lying on the trendline and even if some are out of the trendline, they are lying very close to the curve of trend. Outliers are found on both the sides of the trendline, and hence, there is a negative and a positive deviation of the data points. However, as the value of the correlation coefficient shown in the graph using MS – Excel is 0.94 which is very close to 1, the correlation could be assumed to be reliable.

### Scientific justification

The decreasing trend between the ripple factor and the capacitance could be explained using two possible explanations:

Firstly, with an increase in capacitance of the capacitive filter, the capacitive reactance decreases as compared to the load resistance and by the current divider rule, the current AC current flowing through the capacitance increases. As a result, AC current across the output decreases and hence, the ripple factor decreases.

Secondly, with an increase in capacitance of the capacitive filter, the charge stored in the capacitor increases. As a result, in the even quarters of the AC cycle, when the magnitude of current decreases, the capacitor acts as a battery and discharges. As a result, DC current is supplied across the load. With an increase in stored charge, the DC current supplied across the load also increases also increases and hence the ripple factor decreases.

## Evaluation of hypothesis

The hypothesis that was assumed initially has been found to be true as there is a decreasing relationship between the ripple factor and the capacitance of the capacitive filter.

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ## Conclusion

How does the ripple factor expressed as a ratio of output AC current and the output DC current of a Bridge Rectifier Circuit with Capacitive filter depends on the Capacitance of the Capacitive filter when varied over a range of 1mF to 10mF?

The ripple factor of a bridge rectifier circuit with a capacitive filter decreases with an increase in capacitance of the capacitive filter.

• With an increase in capacitance from 1mF to 10 mF, the ripple factor has decreased from 0.503 to 0.358.
• The equation of trend obtained in the graph has been expressed as: y = - 0.0185x + 0.5161 where y represents the Ripple factor and x represents the capacitance of the capacitive filter.
• The ripple factor has decreased at a rate of 0.0185 with a unit increase of capacitance in mF.
• The ripple factor of a bridge rectifier circuit should be equal to zero if the capacitance of the capacitive filter is equal to 27.897 mF.
• The regression correlation coefficient of the trend obtained is equal to 0.94 which is very close to 1. It justifies that the correlation obtained is strong and reliable.

## Evaluation

### Strength

• For every value of capacitance (independent variable), three trial values of output voltage (both AC and DC) have been obtained and the mean value of the voltage has been calculated and used for calculation of ripple factor. This is known as method of triangulation. It reduces the error which might have been obtained due to experimental procedure or even human error. This method reduces the discrepancy (if any) obtained in any observation.
• The standard deviation has been calculated for all the observed values and the range of the standard deviation is very close to zero. It also determines that the presence of error or discrepancy in the raw data is negligible. It has made the exploration more coherent.
• The step-down transformer of 44:1 has been used in this exploration to convert the magnitude of the voltage from 220V to 5V. It improves the exploration as this transformer is used in any cell phone charger which has made the exploration more coherent.
• A wide range of capacitance has been used in this exploration. It has enabled the exploration to determine the variation of ripple factor over a wider range improving the reliability of the exploration.
• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

• ### Weakness

Random Error

No random errors are found in this exploration.

Systematic Error

Error
How would it affect the exploration?
How could it be improved?
Instrumentation error of Multimeter
It could increase the observed value of DC output voltage and AC output voltage. As a result, it would significantly affect the value of the ripple factor.
It could be improved by using Voltmeters with more significant figures.
Figure 9 - Table On Systematic Error

Methodological Limitations

• The frequency of the AC supply voltage is assumed to be constant and equal to 50 Hz. However, it cannot be controlled or verified throughout the exploration.
• Human error in observing the values of output voltage.
• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!

## Bibliography

• David McKay

Ivy league graduate, 4+ yrs of experience, vetted IB Examiner!