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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier –Uncontrolled:

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Presentation on theme: "Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier –Uncontrolled:"— Presentation transcript:

1 Power Electronics and Drives (Version ), Dr. Zainal Salam, Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier –Uncontrolled: R load, R-L load, R-C load –Controlled –Free wheeling diode Single-phase, full wave rectifier –Uncontrolled: R load, R-L load, –Controlled –Continuous and discontinuous current mode Three-phase rectifier –uncontrolled –controlled

2 Power Electronics and Drives (Version ), Dr. Zainal Salam, Rectifiers DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches. Basic block diagram Input can be single or multi-phase (e.g. 3- phase). Output can be made fixed or variable Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC AC inputDC output

3 Power Electronics and Drives (Version ), Dr. Zainal Salam, Single-phase, half-wave, R-load +vs_+vs_ +vo_+vo_ vovo ioio vsvs  t  

4 Power Electronics and Drives (Version ), Dr. Zainal Salam, Half-wave with R-L load +vs_+vs_ +vo_+vo_ +vR_+vR_ +vL_+vL_ i

5 Power Electronics and Drives (Version ), Dr. Zainal Salam, R-L load

6 Power Electronics and Drives (Version ), Dr. Zainal Salam, R-L waveform  t vovo vs,vs,   ioio  vRvR vLvL   

7 Power Electronics and Drives (Version ), Dr. Zainal Salam, Extinction angle

8 Power Electronics and Drives (Version ), Dr. Zainal Salam, RMS current, Power

9 Power Electronics and Drives (Version ), Dr. Zainal Salam, Half wave rectifier, R-C Load +vs_+vs_ +vo_+vo_ iDiD       VmVm V max vsvs vovo V min  iDiD   VoVo

10 Power Electronics and Drives (Version ), Dr. Zainal Salam, Operation Let C initially uncharged. Circuit is energised at  t=0 Diode becomes forward biased as the source become positive When diode is ON the output is the same as source voltage. C charges until V m After  t=  /2, C discharges into load (R). The source becomes less than the output voltage Diode reverse biased; isolating the load from source. The output voltage decays exponentially.

11 Power Electronics and Drives (Version ), Dr. Zainal Salam, Estimation of 

12 Power Electronics and Drives (Version ), Dr. Zainal Salam, Estimation of 

13 Power Electronics and Drives (Version ), Dr. Zainal Salam, Ripple Voltage

14 Power Electronics and Drives (Version ), Dr. Zainal Salam, Capacitor Current

15 Power Electronics and Drives (Version ), Dr. Zainal Salam, Peak Diode Current

16 Power Electronics and Drives (Version ), Dr. Zainal Salam, Example A half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume  and  are calculated as 48 and 93 degrees respectively. Determine (a) Expression for output voltage (b) peak-to peak ripple (c) capacitor current (d) peak diode current.     VmVm V max vsvs vovo V min  iDiD     VoVo

17 Power Electronics and Drives (Version ), Dr. Zainal Salam, Example (cont’)

18 Power Electronics and Drives (Version ), Dr. Zainal Salam, Controlled half-wave +vo_+vo_ +vs_+vs_ igig iaia  t v vovo  i g  t  t v s

19 Power Electronics and Drives (Version ), Dr. Zainal Salam, Controlled h/w, R-L load +vs_+vs_ i +vo_+vo_ +vR_+vR_ +vL_+vL_  t vsvs   vovo  ioio 

20 Power Electronics and Drives (Version ), Dr. Zainal Salam, Controlled R-L load

21 Power Electronics and Drives (Version ), Dr. Zainal Salam, Examples 1.A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(  t), (b) average current, (c) the power absorbed by the load. 2. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance.

22 Power Electronics and Drives (Version ), Dr. Zainal Salam, Freewheeling diode (FWD) Note that for single-phase, half wave rectifier with R-L load, the load (output) current is NOT continuos. A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos +vs_+vs_ ioio +vo_+vo_ +vR_+vR_ +vL_+vL_ +vs_+vs_ +vo_+vo_ D 1 is on, D 2 is off ioio v o = v s ioio +vo_+vo_ ioio D 2 is on, D 1 is off v o = 0

23 Power Electronics and Drives (Version ), Dr. Zainal Salam, Operation of FWD Note that both D 1 and D 2 cannot be turned on at the same time. For a positive cycle voltage source, –D 1 is on, D 2 is off –The equivalent circuit is shown in Figure (b) –The voltage across the R-L load is the same as the source voltage. For a negative cycle voltage source, –D 1 is off, D 2 is on –The equivalent circuit is shown in Figure (c) –The voltage across the R-L load is zero. –However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path. –But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop. –Hence the “negative part” of v o as shown in the normal half-wave disappear.

24 Power Electronics and Drives (Version ), Dr. Zainal Salam, The inclusion of FWD results in continuos load current, as shown below. Note also the output voltage has no negative part. FWD- Continuous load current     i D1 ioio output Diode current i D2 vovo   t

25 Power Electronics and Drives (Version ), Dr. Zainal Salam, Full wave rectifier Center-tapped D1D1 isis +vs_+vs_  v o + i D1 i D2 ioio + v s1 _ + v s2 _ D2D2 + v D1  + v D2  Center-tapped (CT) rectifier requires center-tap transformer. Full Bridge (FB) does not. CT: 2 diodes FB: 4 diodes. Hence, CT experienced only one diode volt-drop per half-cycle Conduction losses for CT is half. Diodes ratings for CT is twice than FB +vs_+vs_ isis i D1 +vo_+vo_ ioio Full Bridge D1D1 D2D2 D4D4 D3D3

26 Power Electronics and Drives (Version ), Dr. Zainal Salam, Bridge waveforms +vs_+vs_ isis i D1 +vo_+vo_ ioio Full Bridge D1D1 D2D2 D4D4 D3D3     VmVm VmVm -V m -Vm-Vm vsvs vovo v D1 v D2 v D3 v D4 ioio i D1 i D2 i D3 i D4 isis

27 Power Electronics and Drives (Version ), Dr. Zainal Salam, Center-tapped waveforms Center-tapped D1D1 isis +vs_+vs_  v o + i D1 i D2 ioio + v s1 _ + v s2 _ D2D2 + v D1  + v D2      VmVm VmVm -2V m vsvs vovo v D1 v D2 ioio i D1 i D2 isis

28 Power Electronics and Drives (Version ), Dr. Zainal Salam, Full wave bridge, R-L load +vs_+vs_ isis i D1 +vo_+vo_ ioio +vR_+vR_ +vL_+vL_ vovo vsvs ioio, i D2 i D3,i D4 isis  t  

29 Power Electronics and Drives (Version ), Dr. Zainal Salam, Approximation with large L

30 Power Electronics and Drives (Version ), Dr. Zainal Salam, R-L load approximation vovo vsvs ioio i D1, i D2 i D3,i D4 isis  t  

31 Power Electronics and Drives (Version ), Dr. Zainal Salam, Examples Given a bridge rectifier has an AC source V m =100V at 50Hz, and R-L load with R=100ohm, L=10mH a)determine the average current in the load b)determine the first two higher order harmonics of the load current c)determine the power absorbed by the load

32 Power Electronics and Drives (Version ), Dr. Zainal Salam, Controlled full wave, R load +vs_+vs_ isis i D1 +vo_+vo_ ioio T1T1 T4T4 T2T2 T3T3

33 Power Electronics and Drives (Version ), Dr. Zainal Salam, vs_+vs_ isis i D1 +vo_+vo_ ioio +vR_+vR_ +vL_+vL_ Controlled, R-L load   vovo Discontinuous mode     ioio   vovo Continuous mode ++   ioio

34 Power Electronics and Drives (Version ), Dr. Zainal Salam, Discontinuous mode

35 Power Electronics and Drives (Version ), Dr. Zainal Salam, Continuous mode

36 Power Electronics and Drives (Version ), Dr. Zainal Salam, Single-phase diode groups In the top group (D 1, D 3 ), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) i d. For example, when v s is ( +), D 1 conducts i d and D 3 reverses (by taking loop around v s, D 1 and D 3 ). When v s is (-), D 3 conducts, D 1 reverses. In the bottom group, the anodes of the two diodes are at common potential. Therefore the diode with its cathode at the lowest potential conducts i d. For example, when v s (+), D 2 carry i d. D 4 reverses. When v s is (-), D 4 carry i d. D 2 reverses. +vs_+vs_ +vo_+vo_ vpvp vnvn ioio D1D1 D3D3 D4D4 D2D2 v o =v p  v n

37 Power Electronics and Drives (Version ), Dr. Zainal Salam, Three-phase rectifiers D1D1 v o =v p  v n +vo_+vo_ v pn v nn ioio D3D3 D2D2 D6D6 + v cn - n + v bn - + v an - D5D5 D4D4    VmVm VmVm v an v bn v cn vnvn vpvp v o =v p - v n 

38 Power Electronics and Drives (Version ), Dr. Zainal Salam, Three-phase waveforms Top group: diode with its anode at the highest potential will conduct. The other two will be reversed. Bottom group: diode with the its cathode at the lowest potential will conduct. The other two will be reversed. For example, if D 1 (of the top group) conducts, v p is connected to v an.. If D 6 (of the bottom group) conducts, v n connects to v bn. All other diodes are off. The resulting output waveform is given as: v o =v p -v n For peak of the output voltage is equal to the peak of the line to line voltage v ab.

39 Power Electronics and Drives (Version ), Dr. Zainal Salam, Three-phase, average voltage vovo    V m, L-L vovo 

40 Power Electronics and Drives (Version ), Dr. Zainal Salam, Controlled, three-phase VmVm v an v bn v cn T1T1 +vo_+vo_ v pn v nn ioio T3T3 T2T2 T6T6 + v cn - n + v bn - + v an - T5T5 T4T4  vovo

41 Power Electronics and Drives (Version ), Dr. Zainal Salam, Output voltage of controlled three phase rectifier EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz. The load R=10 ohm. Determine the delay angle required to produce current of 50A.


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