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.
From figure 1, it could be observed that, during the positive half cycle of the supply AC current, diode D_{1} was in forward bias condition and diode D_{3} was in reverse bias condition. Hence, diode D_{1} would act as a closed-circuit path and allow current to flow through it and diode D_{3} 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 D_{4} was in forward bias condition and diode D_{2} was in reverse bias condition. Hence, diode D_{4} would act as a closed-circuit path and allow current to flow through it and diode D_{2} 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 R_{L} 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 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_{C }= Capacitive Reactance
ω = angular frequency of current component
C = capacitance of the capacitor
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.
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:
q_{total }= q_{1 }+ q_{2 }+…+ q_{n}
C_{eq }× V = C_{1 }× V_{1 }+ C_{2 }× V_{2 }+ ... + C_{n} × V_{n}
where,
C_{eq} = equivalent capacitance of the circuit
C_{1}, C_{2} ... C_{n} = capacitance of each capacitor connected in parallel
V_{1}, V_{2, ... }V_{n} = 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.
C_{eq} × V = C_{1} × V + C_{2 }× V + ... + C_{n} × V
C_{eq} × V = V(C_{1 }+ C_{2} + ... + C_{n})
C_{eq} = C_{1} + C_{2} + ... + C_{n}
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 (V_{rms}) and (V_{dc}) 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.
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.
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.
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.
No materials are required to perform the experiment.
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
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
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.
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.
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.
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.
Random Error
No random errors are found in this exploration.
Systematic Error
Methodological Limitations