experiment 5

De La Salle UniversityCollege of EngineeringElectronics and Communications Engineering

Experiment Number: 5Experiment Title: RectificationDate Performed: February 17, 2010Date Submitted: February 24, 2010Instructor

: Engr. RoqueSubject/Section: LBYEC13Group Number: 2Group Leader

: Joseph Emil Aco Group Members: Derrick Lim

Carlota Elauria

Patrick Jawili

Richard Vergara

Presentation

:_______________________

Data and Results

:_______________________

Analysis and Conclusion :_______________________

Answers to Questions

:_______________________

Total

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Remarks :___________________________________________________________________

___________________________________________________________________

Instructors Signature:______________________________

I. OBJECTIVES

* To be familiar with AC to pulsating DC conversion.

* To be familiar with different types of rectifiers namely: half-wave rectifier, bridge-type full-wave rectifier and center-tapped transformer full-wave rectifier.

* To observe the effect of simple capacitor filter connected at the output of the rectified circuit.

THEORY

Power companies find it easier to mass-produce and distribute electricity in alternating current than direct current. Because of this, most electronic equipment gets their source from the AC power line.

The AC power line delivers a sinusoid, which is 22OVac and has a frequency of 60Hz. However, the average value or the direct current voltage produced by a sinusoid is zero because the positive alteration will

Figure 5.1 Sinusoid Signal just cancel out with the negative alteration.

Since most electronic equipment requires DC, then, they should be able to convert AC to DC. The process of converting AC to pulsating DC is called rectification.

This concept can be further visualized by considering

Figure 5.2 Half-Wave Figures 5.1, 5.2 and 5.3.

Rectified Signal

Notice that the average value or DC of the AC signal as shown in Figure 5.1 is equal to the sum of the area of both positive and negative alterations. Since both are of equal magnitude but opposite directions, then the net

Figure 5.3 Full-Wave sum is zero.

Rectified Signal

On the other hand, the signal shown in Figure 5.2 has its negative alteration removed. The average or DC voltage of this signal is now equal to the area of the positive alteration distributed over one period. The average voltage can be expressed as:

where: VM is the peak value of the rectified signalIf we can change the polarity of the negative alteration of the ac signal, then it will result in a fullwave rectified signal whose average voltage is twice that of the halfwave rectified signal. SHAPE \* MERGEFORMAT

where: VM is the peak value of the rectified signal

It is possible to reduce greatly the amplitudes of the ac components in the output voltage of a rectifier by means of one or two sections of a smoothing filter. Smoothing filters used with rectifiers hat supply power to electronic tubes and transistors are usually made of iron-core inductances and capacitors. In circuits where the load currents are very small, or if the load current need not be extremely smooth, only capacitors and resistors may be used in the filter. A filter usually takes one of three forms: (1) shunt capacitor, (2) Lsection, and (3) (section.

(a) (b)

Figure 5.4

When the load voltage has only a small amount of ac voltage, its variation about the average (DC) value, or the variation in the output voltage due to the charging and discharging, is called ripple voltage or simply ripple. The smaller the ripple, the better the filtering action. Ripple voltage can be m2asured graphically as shown in Figure 5.4-a for half-wave rectifier and Figure 5.4-b for full-wave rectifier.

A simple filter to remove the AC ripple components in the output of a rectifier is obtained by shunting a capacitor across the load resistance. If the value of capacitance is chosen so that its reactance at the fundamental frequency is much less than the value of load resistance, the AC components will have a low-reactance path around the load resistor. Only a small AC current flows through the load and produces a small ripple voltage. From Figure 5.4, the DC voltage across the capacitor filter can be obtained graphically as:

SHAPE \* MERGEFORMAT

where: Vm is the peak value of the filtered signal

Vr(p-p) is the ripple voltage across the capacitor filter in peak-to-peak

Ripple factor is an indication of the effectiveness of the filet and is defined as the ratio of the ripple voltage in RMS value to the DC (average) value of the filter output voltage. The ripple factor can be decreased by increasing the value of the filter capacitor.

SHAPE \* MERGEFORMAT

MATERIALS 1 -220:3-4.5-6-9-12 Vac Transformer

1 -Analog VOM

1 -Dual-Trace Oscilloscope

1 l00(F Capacitor

1 -330(F Capacitor

4 -1N4001 Diode

1 -lk( Resistor

1 -Breadboard

Connecting Wires

II. DATA AND RESULTS

Half-Wave Rectifier

3. Connect CH1 (the input channel) across the input source and CH2 (the output channel) across the 1k resistor. Set VERT MODE to DUAL. Set AC-GND-DC switch of CH1 to AC and AC-GND-DC switch of CH2 to DC. Measure the value of the peak output voltage.

VM = 6.0 volts

4. Draw the input and output waveforms as seen from the oscilloscope in Graph 5.1.

Graph 5.1

5. What can you say about the input and output waveforms?

It clips the negative cycle of the wave.

6. Connect the VOM across the 1k load resistor and measure the DC voltage.

VDC = 1.86 volts7. Calculate the average or DC signal content of the half-wave rectified signal using Equation 5.1. Then, compute for the percent error.

VDC = 1.98 volts% Error = 2.67%Half Wave Rectifier with Capacitor Filter

10. Connect CH1 (the input channel) across the input source and CH2 (the output channel) across the parallel combination of the 1k resistor and the 100F capacitor. Set VERT MODE to DUAL. Set AC-GND-DC switch of CH1 to AC and AC-GND-DC switch of CH2 to DC. Measure the value of the peak output voltage.

VM = 6.0 volts11. Draw the input and output waveforms as seen from the oscilloscope in Graph 5.2.

Graph 5.2

12. Compare the output waveforms in Graph 5.1 with the output waveform in Graph 5.2. What is your observation?

In graph 5.1 the negative half cycle of the output voltage is clipped while in the graph 5.2, the output voltage is clamped up and the voltage peak became smaller.

13. Measure the peak-to-peak ripple voltage using the oscilloscope.

Vr (p-p) = 1 volts14. Connect the VOM across the parallel combination of the 1k resistor and the 100F capacitor then measure the DC voltage.

VDC = 5.47 voltsCompare the DC voltage obtained from step 6 with the DC voltage obtained from step 14. Note down your observation.

The Vr (p-p) of the output voltage in graph 5.2 is smaller and clamped up compared to the output voltage peak in graph 5.1.

15. Calculate the average or DC signal content of the filtered signal using Equation 5.3. Then, compute for the percent error.

VDC = 5.5 volts% Error = 0.55%18. Connect CH1 (the input channel) across the input source and CH2 (the output channel) across the parallel combination of the 1k resistor and the 330F capacitor. Set VERT MODE to DUAL. Set AC-GND-DC switch of CH1 to AC and AC-GND-DC switch of CH2 to DC. Measure the value of the peak output voltage.

VM = 6 volts19. Draw the input and output waveforms as seen from the oscilloscope in Graph 5.3.

Graph 5.3

20. Compare the output ripple voltage in Graph 5.2 with the output ripple voltage in Graph 5.3. What is your observation? The ripple voltage of graph 5.3 is much smaller than the ripple voltage of graph 5.2.

Measure the peak-to-peak ripple voltage using the oscilloscope.

Vr (p-p) = 0.5 volts22. Connect the VOM across the parallel combination of the 1k resistor and the 330F capacitor. Then, measure the DC voltage.

VDC = 5.72 volts23. Calculate the average or the DC signal content of the filtered signal using Equation 5.3. Then, compute for the percent error.

VDC = 5.75 volts% Error = 0.52%Full-Wave Rectifier

26. Connect CH1 across the input AC source. Set VERT MODE to CH1. Set AC-GND-DC switch of CH1 to AC. Draw the input waveform in Graph 5.4.

Graph 5.4

NOTE: AT this point, you should only use one oscilloscope probe. You cannot connect the two probes of the oscilloscope, which has no common ground. At least one diode will burn if you try to connect the probe at the same time.

27. Now, connect CH1 across the 1k resistor. Set VERT MODE to CH1. Set AC-GND-DC switch of CH! to DC. Measure the value of the peak output voltage.

VM = 5.5 volts28. Draw the rectified output waveform in Graph 5.4.

29. What can you say about the input and output waveforms?

The output waveform has a smaller Vr (p-p).


VDC = 3.20 volts31. Calculate the DC voltage using Equation 5.2. Then, compute for the percent error.

VDC = 3.5 volts% Error = 9.38%Full-Wave Rectifier with Capacitor Filter

34. Connect CH1 across the input AC source. Set VERT MODE to CH1. Set AC-GND-DC switch of CH1 to AC. Draw the waveform in Graph 5.5.

Graph 5.5

35. Now, connect CH1 across the parallel combination of the 1k resistor and the 100F capacitor. Set VERT MODE to CH1. Set AC-GND-DC switch of the CH1 to DC. Measure the value of the peak output voltage.

VM = 6 volts36. Draw the output waveform in Graph 5.5.

37. Compare the output waveform in Graph 5.4 with the output waveform in Graph 5.5. What is your observation?

The output waveform of Graph 5.5 has a bigger Vr (p-p) value than in Graph 5.4.38. Measure the peak-to-peak ripple voltage using the oscilloscope.

Vr (p-p) = 1 volts39. Connect the VOM across the parallel combination of the 1k resistor and the 100F capacitor. Then, measure the DC voltage.

VDC = 5.12 voltsCompare the DC voltage obtained from step 30 with the DC voltage obtained from step 39. Note down your observation.

The output Vr (p-p) is smaller and clamped up.

40. Calculate the average or DC signal content of the filtered signal using Equation 5.3. Then, compute for the percent error.

VDC = 5.5 volts % Error = 7.42%43. Connect CH1 across the input AC source. Set VERT MODE to CH1. Set AC-GND-DC switch of CH1 to AC. Draw the waveform in Graph 5.6.

Graph 5.6

44. Now, connect CH1 across the parallel combination of the 1k resistor and 330F capacitor. Set VERT MODE to CH1. Set AC-GND-DC switch of CH1 to DC. Measure the value of the peak output voltage.

VM = 5.5 volts45. Draw the output waveform in Graph 5.6.

46. Compare the output ripple voltage in Graph 5.5 with the output ripple voltage in Graph 5.6. What is your observation?

The output Vr (p-p) is small and clamped up in the Graph 5.6.

47. Measure the peak-to-peak ripple voltage using the oscilloscope.

Vr (p-p) = 0. 5 volts48. Connect the VOM across the parallel combination of the 1k resistor and the 330F capacitor. Then, measure the DC voltage.

VDC = 5.12 volts49. Calculate the average or DC signal content of the filtered signal using Equation 5.3. Then, compute for the percent error.

VDC = 5. 25 volts% Error = 2.54% Computer Simluation

50. Turn on the computer.

51. Using Electronic Workbench MULTISM 8, simulate the full-wave bridge-type rectifier in Figure 5.7.

52. Set vertical resolution of CH1 of the oscilloscope to 5 volts/div. Set the horizontal resolution to 5 ms/div. Connect CH1 of the oscilloscope across the 1k resistor. Set AC-GND-DC switch of CH1 to AC and the AC-GND-DC switch of CH2 to DC. Set X-position and Y-position to 0. Measure the value of the peak output voltage.VM = 5 volts53. Draw the full-wave rectified signal as seen from the oscilloscope in Graph 5.7.


VDC= 2.949 volts

55. Compare the output waveform in Graph 5.4 with the output waveform in Graph 5.7. Note down your observation. They are very similar, both are graphs of full-wave rectifiers.56. Connect a 100F capacitor across the 1k resistor as shown in Figure 5.8 then simulate the circuit using Electronic Workbench MULTISM 8.

57. Set vertical resolution of CH1 of the oscilloscope to 5 volts/div. Set the horizontal resolution to 5 ms/div. Connect CH1 of the oscilloscope across the 1k resistor. Set AC-GND-DC switch of CH1 to AC and the AC-GND-DC switch of CH2 to DC. Set X-position and Y-position to 0. Measure the value of the peak output voltage.

VM = 5.008 volts58. Draw the filtered waveform as seen from the oscilloscope in Graph 5.8.

59. Measure the peak-to-peak ripple voltage using the oscilloscope.Vr(p-p) = 0 volts (approximately)60. Connect the VOM across the parallel combination of the 1k resistor and the 100F capacitor. Then, measure the DC voltage.VDC= 4.943 volts

61. Compare the output waveform in Graph 5.5 with the output waveform in Graph 5.8. Note down your observation. They are very similar, both are pulsating DC graphs.V. GUIDE QUESTIONS1. Discuss the importance of rectification.The first element of an electronics power supply that any incoming power will meet is the transformer and AC rectifier circuits. This element of any electronics power supply converts the incoming power to a form which can be accepted by the smoothing and regulator circuits.

When running from an AC source, a transformer is used to transform the incoming mains voltage to the correct value required for the power supply electronics circuitry. The resulting voltage waveform is an alternating current. This must be rectified to enable the power to be smoothed and regulated for use by electronics circuitry. To achieve this an AC rectifier circuit is used. While the rectifier circuit may appear to be very simple at first sight, there are several different forms of AC rectifier circuit that can be used. The choice of the actual AC rectifier circuit chosen will depend upon a number of factors, and it could impact upon the type of transformer used as well.2. Define ripple factor. What is the ripple factor for all the experimental circuits?Ripple factor () may be defined as the ratio of the root mean square (rms) value of the ripple voltage to the absolute value of the dc component of the output voltage, usually expressed as a percentage. However, ripple voltage is also commonly expressed as the peak-to-peak value. This is largely because peak-to-peak is both easier to measure on an oscilloscope and is simpler to calculate theoretically. Filter circuits intended for the reduction of ripple are usually called smoothing circuits.3. Discuss the importance of the bleeder resistor.A bleeder resistor is a resistor placed in parallel with a high-voltage supply for the purposes of discharging the energy stored in the power source's filter capacitors or other components that store electrical energy when the equipment is turned off.4. Discuss the importance of the shunt capacitor.Is a capacitor or group of capacitors which are placed across an Electric Power line or Electric Appliance to provide a voltage increase or to improve the power factor of the circuit.5. Discuss the effect and importance of an RC filter.

It is used to filter a signal waveform that changes the relative amounts of low-frequency and high-frequency information in their output signals relative to their input signals. It is used to select wanted signals and reject the signals which are not necessary.6. Compare the performance of a full-wave rectifier with that of a half-wave rectifier in five important respects. Discuss its advantages.

In half wave rectification, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one half of the input waveform reaches the output, it is very inefficient if used for power transfer. Half-wave rectification can be achieved with a single diode in a one phase supply, or with three diodes in a three-phase supply. Ripple factor is . Full-wave rectification converts both polarities of the input waveform to DC (direct current), and is more efficient. Ripple factor is /2.7. Discuss the operation of the half-wave and full-wave rectifier circuits when the shunt capacitor is connected.

During the positive half cycle of the ac input, the diode of the rectifier is forward biased and so it conducts. This quickly charges the capacitor C to peak value of the supply voltage VSmax because of almost zero charging time constant.After being fully charged, the capacitor holds the charge till input ac supply to the rectifier goes negative. During the negative half cycle, the diode gets reverse biased and so stops conduction. So the capacitor C discharges through load resistance RL and loses charge. Voltage across RL (VL) or across C (vc), both being equal, decreases exponentially with time constant CRL along the curve.In Full wave with shunt capacitor, capacitor discharges twice in one cycle. Because both the diodes conduct, non-conducting period has reduced. The result is that ripple voltage Vr has been reduced to half and Vdc has been increased relative to half wave rectifier. Voltage regulation in this case is better than that in half-wave rectifier.

8. Discuss briefly the applications of half-wave rectifiers.

Most half-wave rectifiers are suitable in audio circuits. It can be used to obtain the desired level of dc voltage (using step up or step down transformers). It provides isolation from the power line.9. Answer the computer simulation problems in Exercise 5: Design of Rectifier with Capacitor Filter.

Exercise 5

DESIGN OF RECTIFIER WITH CAPACITOR FILTER

A. Simulate the center-tapped transformer full-wave rectifier circuit shown in Figure 1 using MULTISIM 8 software and determine the following:

1. Sketch of the rectified signal across RL as seen from the oscilloscope.

2. Reading of the DC voltmeter connected across the RL.

-607.417 pA3. Reading of the DC ammeter connected in series with the RL.

-1.248 uV4. If a 330uF capacitor is connected across the R sketch the filtered signal as seen from the oscilloscope.

5. Reading of the DC voltmeter connected across the parallel combination of RL and 330(F capacitor.-1.32nV

B. Design a bridge-type full-wave rectifier with capacitor filter with an input of 22OVrms, 60Hz to the transformer and requires a maximum output voltage of 26V and a minimum output voltage of 22V across a 500( load. Determine the circuit configuration of the power supply, rating of the transformer, value of the filter capacitor and the PRV rating of the Silicon diode. Simulate your circuit using MULTISIM 8 software.

EQUATION 5.1

Vave(HWR) = VDC(HWR) = Vm / ( = 0.318 Vm

EQUATION 5.2

Vave(FWR) = VDC(FWR) = 2 X Vm / ( = 0.318 Vm

Vdc = Vm [Vr(p-p) / 2]

EQUATION 5.3

r = Vr(rms) / Vdc

EQUATION 5.4

experiment 5

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