rectification lab

7/27/2019 Rectification Lab

1/8

The University of Technology, Jamaica

Electronic Devices 1

Rectification and smoothing circuits

Monique Hepburn

1001296


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.


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.


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.


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.


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.


7/8

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


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.

rectification lab

Documents