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Experiment 4

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Experiment 4

  1. 1. WESLEYAN UNIVERSITY – PHILIPPINES<br />Cabanatuan City<br />COLLEGE OF ENGINEERING<br />NAME: EDUARDO S. CUDIADATE:09 – 23 – 11 <br />SCHEDULE: Friday (3:00 – 6:00)<br />EXPERIMENT NO. 4<br />Half – Wave Rectifier<br />OBJECTIVES:<br />To learn how rectification is achieved using diodes in a circuit particularly half – wave rectification.<br />MATERIALS: <br />1 IN4001<br />1 resistor (value assigned by instructor)<br />1 unregulated power supply<br />1 multitester<br />1 oscilloscope<br />1 breadboard<br />Connecting Wires<br />DISCUSSIONS:<br />Rectification is the process of converting AC to DC. However, if we look at the output voltage of a rectifier, it does not look very much like a constant. How then does rectification produce DC.<br />A pure sinusoid has no DC component. It is because for each cycle of the sinusoid, there are much positive areas as there are negative areas, the total are therefore is zero.<br />It is very sure through that there is DC component as well as an AC component.<br />Half - wave rectification is the process of removing one – half of the input signal to establish a DC level. Since half of the input signal is removed by a half – wave rectifier, the DC value of a half – wave rectified signal is given by:<br />Half – wave Rectified signal, Vdc = Vp/π<br />Where : Vdc - is the DC value of a half wave rectified signal<br /> Vp – is the peak voltage of the signal<br />PROCEDURE:<br /><ul><li>Construct the circuit of Figure A as shown. Use the value of R assigned to you by your instructor.
  2. 2. Determine the output voltage of the rectifier circuit by connecting the multitester parallel to R. Note that the positive probe of the tester is connected at point A and the negative probe is connected at point B.
  3. 3. Determine the output voltage of the same circuit using the oscilloscope. Follow the procedure taught during the familiarization hours.
  4. 4. Draw and label the parts of the output wave form as viewed from the oscilloscope.</li></ul>DATA AND OBSERVATION:<br /><ul><li> Record the value of the output voltage as measured by the multitester in TABLE 1.
  5. 5. Record the peak to peak value of the output voltage as measured by the oscilloscope.
  6. 6. Determine manually the DC value of the measured output voltage using the multitester.
  7. 7. Determine manually the DC value of the measured output voltage using the oscilloscope.
  8. 8. Is the computed DC value for the measured output voltage using the multitester equal to DC value of the measured output voltage using the oscilloscope?
  9. 9. If not, what accounts for the difference?
  10. 10. Draw and label the parts of the output waveform as viewed from the oscilloscope.
  11. 11. Describe the viewed output waveform.
  12. 12. TABLE 1
  13. 13. Output VoltageMULTITESTEROSCILLOSCOPE8V7.63V</li></ul>FILL – IN QUESTION:<br /><ul><li>Rectification is the process of converting AC to DC.
  14. 14. A half – wave rectifier removes the ONE – HALF of the input signal.
  15. 15. A pure sinusoid has no DC component.
  16. 16. The output of a rectifier has both DC and AC component.
  17. 17. The DC value of a half – wave rectifier is equal to Vp/π.</li></ul>ANSWERS TO QUESTIONS:<br />5. The DC value measured with multitester is 8V while in the oscilloscope it is 7.6V. There is a <br />small difference of 0.4 in the measured value. The instruments used may account for that result maybe we have not properly calibrated the oscilloscope or maybe because of the presence of resistance that result for some changes with the measured voltage at the load resistor.<br /><ul><li>Vp </li></ul>21062951905000<br /> 0.5V/division<br />Vp = 4.8 (0.5 x 10) = 24V<br />Vrms = 0.707(24) = 7.63V <br /> ῳt<br /><ul><li>The viewed output waveform of the oscilloscope defines that the Vp is 24V and its Vrms is 7.63V. This gives us the idea that a half – wave rectifier gets only 180o of the input voltage and the rectifier circuit gets the positive cycle.</li></ul>CONCLUSION:<br />

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