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LAB ACTIVITY 2:
HALF-WAVE RECTIFIER
I.Objectives
The objectives of this activity are to know what a rectifier is and to study and analyse theprocess of how a half-wave rectifier works.
II.Basic Concepts
Rectification is the process of turning an AC input into DC usually with the help of a diode.
During this process, the diode is referred to as a rectifier. (Boylestad, Nashelsky, 1998)
The Alternating Current
The alternating current (AC) is a stream of charges that periodically changes directionwhereas the direct current (DC) only goes in one direction (Andrew Rader Studios, 1997-2014).
From this, we can see that an AC voltage source creates current that changes polarity over a
period of time across a circuit. The most basic of all AC is the sinusoidal wave since all the other
types of waves are just derivatives of the sine wave using the Fourier transform. The period oftime where the AC voltage source completes one shift of polarity is called one full cycle and it is
divided into two parts, the positive half cycle, where the voltage input is positive, and the
negative half cycle, where the voltage input is negative. Other components to take note for anAC are its amplitude and its frequency. The amplitude is the range of voltage that an AC is
capable of making and the frequency determines how fast or, as its name suggests, how
frequently one full cycle occurs.
Half-wave Rectification
Half-wave rectification is a process where half of the full input AC cycle is used whenconverting AC to DC
[1]. A basic setup of this rectification process can be seen on Figure 2-1.
Figure 2-1. A Basic Circuit Setup for Half-wave Rectification.
The rectification process starts after an AC voltage input is applied across the given setup.
During the positive half cycle, the AC inputs polarity causes the given diode to become
forward-biased, thus allowing current to flow across the circuit. Ideally, this causes the loadresistance to have the same voltage as the AV input, but if the diode has a voltage drop, the
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loads voltage then becomes the input minus the voltage drop of the diode, that is, only for the
positive half cycle. Now during the negative half cycle, the polarity of the AC input reverses,
causing the given diode to become reverse-biased. As discussed in the previous activity, when adiode is in reverse-bias, it blocks all current going through it causing the load resistors voltageto become zero, thus effectively eliminating the negative half of the AC and causing the output
voltage to only have one polarity, which is the positive half in this case. This process can be seenon Figure 2-2.
Figure 2-2. The Half-wave Rectification Process[4]
The DC value of the rectified AC input is given by the equation: VDC = VL / , where VLis thevoltage across the load resistor. This is taken from the average of the net area of VLin one full
cycle[1]
.
Some characteristics and parameters of a half-wave rectifier that should be taken note of areits forward current, peak inverse voltage and ripple factor. The forward current and peak inverse
voltage are the current and voltage values that the rectifier can be able to withstand before
breaking. As for the ripple factor, it is defined as the amount of AC content in the DC output.
The less the ripple factor, the performance of the rectifier is more. (Administrator, 2013) To
lessen the ripple factor during rectification and to smoothen the output, a filter must be addedacross the load. This filter is usually in the form of a capacitor.
III.Materials and Equipment
Diode: 1N4001 (Table 2-1 for datasheet)
Resistors: 500 , 1000 , 10000 Connectors
Breadboard
Oscilloscope
Function Generator
IV.Procedure
Half-wave Rectification[2]
1. Construct the circuit shown in Figure 2-1.
2. Switch on the oscilloscope.3. Switch on the function generator and set the amplitude to 5Vp, 50Hz, sinusoidal.
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Figure 2-4. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of 500
Figure 2-5. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
100.
Figure 2-6. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
10000.
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Figure 2-7. Output of a 2 Vp, 50 Hz input at 5V/div and 5ms/div with a load resistor value of
1000
Figure 2-8. Output of a 10Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of1000
Figure 2-9. Output of a 5Vp, 50Hz at 5V/div a5ms/div with a load resistor value of 1000where
the diodes polarity is reversed.
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Figure 2-10. Output of a 5Vp, 25Hz input at 5V/div and 5ms/div with a load resistor value of
1000
Figure 2-11. Output of a 5Vp, 500Hz input at 5V/div and 5ms/div with a load resistor value of1000
VI(Peak), V Frequency, Hz Vo(Peak), V VI-Vo, V Resistor,
5 50 4.352 0.648 500
5 50 4.506 0.494 10000
5 50 4.388 0.612 1000
5 25 4.388 0.612 1000
4.982 500 4.370 0.612 1000
2 50 1.445 0.555 100010 50 9.348 0.652 1000
Table 2-2. Data of a Sinusoidal Wave Input
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Triangle Wave Input
Figure 2-12. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
500
Figure 2-13. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
1000.
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Figure 2-14. Output of a 5 Vp, 50 Hz input at 5V/div and 5ms/div with a load resistor value of
10000.
Figure 2-15. Output of a 2Vp, 50Hz input at 2V/div and 5ms/div with a load resistor value of
1000
Figure 2-16. Output of a 10Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
1000
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Figure 2-17. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
1000where the diodes polarity is reversed.
Figure 2-18. Output of a 5Vp, 25Hz input at 5V/div and 5ms/div with a load resistor value of
1000
Figure 2-19. Output of a 5Vp, 500 Hz input at 5V/div and 5ms/div with a load resistor value of
1000
VI(Peak), V Frequency, Hz Vo(Peak), V VI-Vo, V Resistor,
4.981 50 4.333 0.648 500
4.981 50 4.487 0.494 10000
4.981 50 4.369 0.612 1000
4.981 25 4.369 0.612 1000
4.735 500 4.127 0.608 1000
1.992 50 1.438 0.554 1000
9.962 50 9.311 0.651 1000
Table 2-3. Table of a Triangle Wave Input
Square Wave Input
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Figure 2-20. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
500
Figure 2-21. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of1000
Figure 2-22. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
10000
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Figure 2-23. Output of a 2Vp, 50Hz input at 2V/div and 5ms/div with a load resistor value of
1000
Figure 2-24. Output of a 10Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
1000
Figure 2-25. Output of a 5Vp, 50Hz input at 5V/div and 5ms/div with a load resistor value of
1000where the diodes polarity is reversed
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Figure 2-26. Output of a 5Vp, 25Hz input at 5V/div and 5ms/div with a load resistor value of
1000
Figure 2-27. Output of a 5Vp, 500Hz input at 5V/div and 5ms/div with a load resistor value of
1000
VI(Peak), V Frequency, Hz Vo(Peak), V VI-Vo, V Resistor,
5 50 4.352 0.648 500
5 50 4.506 0.494 10000
5 50 4.388 0.612 10005 25 4.388 0.612 1000
5 500 4.388 0.612 1000
2 50 1.445 0.555 1000
10 50 9.349 0.651 1000
Table 2-3. Data of a Square Wave Input
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VI.Conclusion
After the activity was conducted, it is concluded that the semiconductor diode, which is alsoknown as a rectifier, is a device used to convert AC to DC. During this experiment it was used as
a half-wave rectifier, in which it converted the given AC input into DC, through the use of the
rectifier diode, with only half of the given AC input. Only half was used because in AC, thepolarity changes over a period of time where one full alternation of its polarity is called a fullcycle. During half of this cycle, its polarity causes the diode to be in forward-bias, while on the
other half it causes the diode to be in reverse-bias. From this, we can see that on one half of the
cycle there is voltage on the load , while on the other half it is equal to zero since no currentpasses through it. This gives the load a voltage equivalent to the input voltage wherein its
polarity of causes the diode to be in forward bias minus the voltage drop across the diode. Since
the voltage across the load has only one polarity, we can now see that the AC input voltage has
been successfully converted into DC.From the data that was gathered, it can be seen that the difference between the input voltage
and the output voltage is across its peaks is the voltage drop of the diode. In this activity, the
diode that was used was a silicon based diode which has an average voltage drop of 0.7V whichcan also be seen in the data in the tables. However, the voltage difference can be affected if, the
input voltage is change or if the value of the load resistor is changed.
Another application of the rectifier diode is to be able to convert the full wave of the AC
input, that is, the Full-wave Rectification process, which will be further discussed in the nextlaboratory activity.
VII. References
1. Boylestad, R. &Nashelsky L. (1998).Electronic Devices and Circuit Theory.Prentince
Hall, Upper Saddle River, New Jersey.
2. Awad N., Al-Khawaldeh S.,Electronics Lab. Electrical Engineering Department,
University of Jordan.
3. Diodes Incorporated.1N4001 - 1N4007.http://www.diodes.com/datasheets/ds28002.pdf
4. Storr, Wayne. (2014).Power Diodes and Rectifiers. http://www.electronics-
tutorials.ws/diode/diode_5.html
5. Phyiscs4kids.com.Electricity & Magnetism Alternating Current.
http://www.physics4kids.com/files/elec_ac.html
6. Administrator. (Sept. 12, 2013). Characteristics and Working of Half-wave Rectifier.
http://www.electronicshub.org/characteristics-and-working-of-half-wave-rectifier/