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Diode Circuits or Uncontrolled Rectifier

Diode Circuits or Uncontrolled Rectifier. Rectification: The process of converting the alternating voltages and currents to direct currents. The main disadvantages of half wave rectifier are: . High ripple factor, Low rectification efficiency, Low transformer utilization factor, and,

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Diode Circuits or Uncontrolled Rectifier

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  1. Diode Circuits or Uncontrolled Rectifier Rectification: The process of converting the alternating voltages and currents to direct currents

  2. The main disadvantages of half wave rectifier are: High ripple factor, Low rectification efficiency, Low transformer utilization factor, and, DC saturation of transformer secondary winding.

  3. Performance Parameters rectification effeciency form factor ripple factor

  4. Single-phase half-wave diode rectifier with resistive load.

  5. the load and diode currents

  6. Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1. . (d) It is clear from Fig2.2 that the PIV is

  7. Half Wave Diode Rectifier With R-L Load Fig.2.3 Half Wave Diode Rectifier With R-L Load

  8. Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier

  9. PIV of each diode = Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.

  10. The PIV is

  11. Single-Phase Full Bridge Diode Rectifier With Resistive Load

  12. Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor. The PIV=300V Input power factor =

  13. Full Bridge Single-phase Diode Rectifier with DC Load Current

  14. Example 5 solve Example 4 if the load is 30 A pure DC Input Power factor=

  15. Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier.

  16. Example 6Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.

  17. Three-Phase Half Wave Rectifier

  18. ThePIV of the diodes is Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and the load  resistance is R=20. If the source inductance is negligible, determine (a) Rectification efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode.

  19. The PIV= Vm=650.54V

  20. Three-Phase Half Wave Rectifier With DC Load Current and zero source induct New axis

  21. Example 8 Solve example 7 if the load current is 100 A pure DC The PIV= Vm=650.54V

  22. Three-Phase Full Wave Rectifier With Resistive Load

  23. Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode .

  24. The PIV= Vm=650.54V

  25. Three-Phase Full Wave Rectifier With DC Load Current

  26. Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find; Commutation time and commutation angle. DC output voltage. Power factor. Total harmonic distortion of line current.

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