Upload
monica-latoya-hepburn
View
213
Download
0
Embed Size (px)
Citation preview
7/27/2019 Rectification Lab
1/8
The University of Technology, Jamaica
Electronic Devices 1
Rectification and smoothing circuits
Monique Hepburn
1001296
7/27/2019 Rectification Lab
2/8
OBJECTIVE
To investigate rectification and smoothing circuits.
PRELAB/THEORY
Half-wave rectification
The diagram below shows a half wave rectifier. A rectifier is a circuit which converts ac to dc current.
The input signal will be a full sine wave whereas the output signal would be that as shown in the graph
below. The current will flow towards the anode of the diode and through it but in the opposite directionof current flow the diode blocks the current. This is the reason for the attained output graph.
7/27/2019 Rectification Lab
3/8
Full wave rectifier
In the diagram above the current in the forward direction will flow through the red diodes. The current
in the opposite direction will allow current through the blue diodes. This will produce a full wave
rectification as shown in the graph below.
7/27/2019 Rectification Lab
4/8
Capacitor Smoothing
When the voltage increases during the first half of the voltage peaks from the rectifier, the capacitor
charges up. Then as the voltage decreases to zero in the second half of the peaks, the capacitor issues its
stored energy the keep the output voltage constant. The diagram below shows the smoothing effect.
7/27/2019 Rectification Lab
5/8
EquipmentOscilloscope
Digital Multimeter
12V Centre tap transformer
(2) 2.2kohm resistor
(4) Silicon diode
10microFarad Capacitor
100microFarad Capacitor
Procedure
Half Wave rectification
The circuit of a half wave rectifier was constructed. A labelled sketch of the input and voltage was made.
The peak voltage value was recorded in the sketch to calculate the DC level of the signal. The DC output
was measured and recorded with the multimeter.
Full wave rectification
The circuit of a full wave rectifier was constructed.
7/27/2019 Rectification Lab
6/8
A sketch of the voltage at the points A, B and the output voltage was made. The average or equivalent
DC level of a full wave signal is 63.6 percent of the peak value. The peak voltage was recorded in the
sketch to calculate the DC level of the signal. The value of the DC output was measured and recorded.
Full wave rectifier and smoothing
A similar circuit was constructed as before with a transformer included. A labeled sketch of the input
voltage and output voltage was made. The peak voltage was recorded in the sketch to calculate the DClevel of the signal. The DC output was measured.
The 10 microfarad capacitor was placed across the output of the bridge rectifier. The oscilloscope was
used to view the output voltage. A sketch was made of the output.
The capacitor was replaced by the 100 micro farad capacitor and the oscilloscope was used to record
the output voltage. A sketch was made of this output. The 2.2 Kohm resistor was placed in parallel with
the one of the circuit and the effect of the output ripples was noted.
7/27/2019 Rectification Lab
7/8
Results(see attachment)
Replacing the 10 microfarad with a 100 microfarad gives a straight line
Volts against milliseconds
Placing the 2.2 kilo-ohm resistor with the 100 microfarad gives a straight line
Volts against milliseconds
7/27/2019 Rectification Lab
8/8
Analysis of Results
The DC level of the half wave rectification was less than the DC level of the full wave rectified circuit.
This was expected based on the formula used to calculate the DC levels of both circuits. The full wave
circuit dc voltage must obviously higher voltage level.
The measured and calculated values for the half wave rectifier were found to be similar: 2.64 volts for
the calculated value and the measured for the half wave was found to be 2.5.For the full wave, the DC
output was found to be 6.31V and the calculated value was 5.41V.For the half peak, the value of the
calculated voltage was found to be 5.06V. The DC voltage was found to be 1.18V.
When the voltage increases during the first half of the voltage peaks from the rectifier, the capacitor
charges up. Then as the voltage decreases to zero in the second half of the peaks, the capacitor issues its
stored energy the keep the output voltage constant. This is what accounts for the ripple.
Where a capacitor is placed in parallel with a load its voltage will be affected by the time constant T =
RC. The larger the capacitance the higher will be T, hence, the ripple factor will be lower.
ConclusionThe concept of full wave rectification and half wave rectification was appreciated as well as that for the
smoothing effect. It was understood how to find the DC voltage via calculation. The graphs obtained
match the theoretical concept behind half wave and full wave rectification.
References
Giancoli. D.C, (2000).Physics for Scientists and Engineers with Modern Physics 4thedition.
Pearson Education Inc, Prentice Hall.
Robert Boylestead, Louis Nashelsky(2005 updated ). Electronic Devices and Circuit Theory
seventh edition.