1. 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,
Continuous and discontinuous current mode
Three-phase rectifier
uncontrolled
controlled
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2. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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 input
DC output
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3. Single-phase, half-wave, R-load+ vs _ vo vo io vs w t p 2p
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4. Half-wave with R-L load+ vs _ vo vR vL i
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5. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
R-L load
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6. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
R-L waveform w t vo
vs, p 2p io b vR vL 3p 4p
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7. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Extinction angle
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8. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
RMS current, Power
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9. Half wave rectifier, R-C Load+ vs _ vo iD p 2p 3p 4p Vm
Vmax vs vo
Vmin
p /2 iD
3p /2 D Vo a q
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10. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Operation
Let C initially uncharged. Circuit is energised at wt=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 Vm
After wt=p/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.
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11. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Estimation of q
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12. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Estimation of a
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13. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Ripple Voltage
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Capacitor Current
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Peak Diode Current
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16. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Example
A half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume a and q 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. p 2p 3p 4p Vm
Vmax vs vo
Vmin
p /2 iD
3p /2 a q D Vo
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17. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Example (cont’)
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18. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Controlled half-wave + vs _ ig ia w t v vo a i g s + vo _
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19. Controlled h/w, R-L load+ vs _ i vo vR vL w t vs p 2p vo a io b
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20. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Controlled R-L load
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21. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Examples
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(wt), (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.
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22. 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 _ io vo vR vL io
vo= vs io + vo _
D2 is on, D1 is off
vo= 0 + vs _ vo
D1 is on, D2 is off
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23. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Operation of FWD
Note that both D1 and D2 cannot be turned on at the same time.
For a positive cycle voltage source,
D1 is on, D2 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,
D1 is off, D2 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 vo as shown in the normal half-wave disappear.
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24. FWD- Continuous load current
The inclusion of FWD results in continuos load current, as shown below.
Note also the output voltage has no negative part. p 2p 3p 4p
iD1 io
output
Diode
current
iD2 vo w t
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25. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Full wave rectifier + vs _ is
iD1 vo io
Full Bridge D1 D2 D4 D3
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
Center-tapped D1 is + vs _
- vo
iD1
iD2 io
vs1
vs2 D2
+ vD1 -
+ vD2 -
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26. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Bridge waveforms + vs _ is
iD1 vo io
Full Bridge D1 D2 D4 D3 p 2p 3p 4p Vm
-Vm vs vo
vD1 vD2
vD3 vD4 io
iD1 iD2
iD3 iD4 is
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27. Center-tapped waveformsis + vs _
- vo
iD1
iD2 io
vs1
vs2 D2
+ vD1 -
+ vD2 - p 2p 3p 4p Vm
-2Vm vs vo
vD1
vD2 io
iD1
iD2 is
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28. Full wave bridge, R-L load+ vs _ is
iD1 vo io vR vL vo vs io
iD1
, iD2
iD3
,iD4 is w t p 2p
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29. Approximation with large L
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30. R-L load approximationvo vs io
iD1
, iD2
iD3
,iD4 is w t p 2p
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31. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Examples
Given a bridge rectifier has an AC source Vm=100V at
50Hz, and R-L load with R=100ohm, L=10mH
determine the average current in the load
determine the first two higher order harmonics of the load current
determine the power absorbed by the load
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32. Controlled full wave, R load+ vs _ is
iD1 vo io T1 T4 T2 T3
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33. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Controlled, R-L load + vs _ is
iD1 vo io vR vL p 2p vo
Discontinuous mode b a +a io p 2p vo
Continuous mode
p+a a b io
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34. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
Discontinuous mode
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Continuous mode
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36. Single-phase diode groups+ vs _ vo vp vn io D1 D3 D4 D2
vo =vp -vn
In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id.
For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3). When vs is (-), D3 conducts, D1 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 id.
For example, when vs (+), D2 carry id. D4 reverses. When vs is (-), D4 carry id. D2 reverses.
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37. Three-phase rectifiersD1 io
+ van D3 D5
vpn
+ vbn n + vo _ D2
+ vcn
vnn
vo =vp -vn D6 D4 p 2p 4p Vm
van
vbn
vcn vn vp
vo =vp - vn 3p
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38. 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 D1 (of the top group) conducts, vp is connected to van.. If D6 (of the bottom group) conducts, vn connects to vbn . All other diodes are off.
The resulting output waveform is given as: vo=vp-vn
For peak of the output voltage is equal to the peak of the line to line voltage vab .
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39. Three-phase, average voltage
p/3
2p/3
Vm, L-L
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40. Controlled, three-phase+ vo _
vpn
vnn io T3 T2 T6
+ vcn n
+ vbn
+ van T5 T4 a
van
vbn
vcn Vm vo
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41. 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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