Implementing Power Factor Correction in AC/DC Converters
4.2 DC/DC Converterized Rectifiers
A full-wave diode rectifier with R load has a high PF. If this rectifier supplies an R–C load, the PF is poor. Using a DC/DC converter in this circuit will improve the PF. The PFC rectifier circuit is shown in Figure 4.1.
The resistor emulation of the PFC rectifier is carried out by the DC/DC converter. The input to the DC/DC converter is a fully rectified sinusoidal voltage waveform. A con-stant DC voltage is maintained at the output of the PFC rectifier. The DC/DC converter is switched at a switching frequency fsthat is many times higher than the line frequency f . The input current waveform into the diode bridge is modified to contain a strong fundamental sinusoid at the line frequency but with harmonics at a frequency several times higher than the line frequency.
Since the switching frequency fsis very high in comparison with the line frequency f , the input and output voltages of the PFC rectifier may be considered as constant throughout the switching period. Thus, the PFC rectifier can be analyzed like a regular DC/DC converter:
vs= Vssinθ,
v1= Vs| sin θ| with θ = 2πft. (4.1)
The voltage transfer ratio of the PFC rectifier is required to vary with angleθ in a half supply period. The voltage transfer ratio of the DC/DC converter is
Tvv(θ) = VDC
V1(θ) = VDC
Vssinθ with fs f , (4.2)
where VDCis the local average DC output voltage.
Tvv in a supply period is shown in Figure 4.2. The high voltage transfer ratio in the vicinity ofωt = 0◦and 180◦can be achieved by using converters such as boost, buck–boost, or fly-back converters.
R D1
– +
D4
D3
D2
VDC V1
iO i1
is Vs
AC DC/DC
converter C
FIGURE 4.1 PFC rectifier.
Implementing Power Factor Correction in AC/DC Converters 97
p/2
p 0
0
0 IO 2IO iO Tvv v1
3p/2
2p p 2p
q q q
FIGURE 4.2 DC/DC converter output current required.
To prove this technique, a full-wave diode rectifier with R–C load (R= 100 Ω and C = 100μF) plus a buck–boost converter is investigated. Before applying any converter, the input voltage and current waveforms are shown in Figure 4.3. The fundamental harmonic of the input current delays the input voltage by an angleθ = 33.45◦.
The harmonics (FFT spectrum) of the input current are shown in Figure 4.4.
The harmonics, values are listed in Table 4.1.
1000.00 –10.00
–5.00 0.0 5.00 10.00 –400.00 –200.00 0.0 200.0 400.0
Iin
1010.00 1020.00
Time (ms)
1040.00
Time 0.0018588 Frequency 537.981
Vin 172.308
Iin Measure
6.67783
Vin
FIGURE 4.3 Input voltage and current waveforms.
Frequency (kHz)
1.00 0.80
0.60 0.40
0.20 0.0
0.0 1.00 2.00 3.00 4.00 5.00
Iin
(49.9, 4.546)
(150.1, 2.645)
(549.6, 0.523) (649.6, 0.473)
(750.67, 0.288) (848.9, 0.295) (949.9, 0.316) (449.5, 0.738)
(349.4, 0.746) (250.2, 0.833)
FIGURE 4.4 FFT spectrum of the input current.
The THD of the input current is obtained as THD2=α
i=2(Ii/I1)2= 0.4625 and the DPF is obtained as cos(33.45◦)= 0.834. Therefore,
PF= DPF
1+ THD2 = cos 33.45
√1+ 0.4625 = 0.689. (4.3)
A buck–boost converter (see Figure 5.7 in Chapter 5) is used for this purpose. The circuit diagram is shown in Figure 4.5.
The input voltage is 311 V (peak)/50 Hz. The duty ratio k is calculated as 20 chopping periods for a half-cycle. For one cycle, there are 40 chopping periods (maintain the same duty ratio) corresponding to its frequency of 2 kHz. The inductance value was set as L= 0.6 mH and the capacitance value as 800μF to maintain the output voltage at 200 V. The duty ratio k was calculated to set a constant DC output voltage of 200 V (see Table 4.2).
The duty ratio k waveform in two-and-a-half cycles is shown in Figure 4.6 and the switch (transistor) turn-on and turn-off in a half-cycle are shown in Figure 4.7.
TABLE 4.1
Harmonic Current Values of Normal AC to DC Converter
Current Frequency (Hz) Fourier Component
I1 50 4.546
I3 150 2.645
I5 250 0.833
I7 350 0.746
I9 450 0.738
I11 550 0.523
I13 650 0.473
I15 750 0.288
I17 850 0.295
I19 950 0.316
Implementing Power Factor Correction in AC/DC Converters 99
311Vp
Vduty_ra1 Vramp
V
~ +
–
L = 0.6 mH
C = 100 m R = 100
FIGURE 4.5 Buck–boost converter used for PFC with R–C load.
The input voltage and current waveforms are shown in Figure 4.8. From the waveform we can see that the fundamental harmonic delay angleθ is about 3.21◦. The output voltage of the buck–boost converter is 200 V, as shown in Figure 4.9.
The FFT spectrum of the input current is shown in Figure 4.10 and the harmonic components are shown in Table 4.3.
TABLE 4.2
Duty Ratio k in the 20 Chopping Periods in a Half-Cycle ωt (deg) Input Voltage= 311 sin(ωt) (V) k
9 48.65 0.804
18 96.1 0.676
27 141.2 0.586
36 182.8 0.522
45 219.9 0.476
54 251.6 0.443
63 277.1 0.419
72 295.8 0.403
81 307.2 0.394
90 311 0.391
99 307.2 0.394
108 295.8 0.403
117 277.1 0.419
126 251.6 0.443
135 219.9 0.476
144 182.8 0.522
153 141.2 0.586
162 96.1 0.676
171 48.65 0.804
180 0 ∞
1.00 0.80 0.60 0.40 0.20 0.0
1000.00 1010.00 1020.00 1030.00
Time (ms)
1040.00 1050.00
Vduty_ratio
FIGURE 4.6 Duty ratio k waveform in two-and-a-half cycles.
Vswitch
1.00 0.80 0.60 0.40 0.20 0.0
1000.00 1000.10 1000.20 1000.30
Time (ms)
1000.40 1000.50
FIGURE 4.7 Switch turn-on and turn-off waveform in a half-cycle.
From the data in Table 4.3, the THD of the input current is obtained as THD2=
α
i=2(Ii/I1)2= 0.110062 and DPF is obtained as cos(3.21◦)= 0.998431. Therefore,
PF= DPF
1+ THD2 = cos 3.21
√1+ 0.110062 = 0.95. (4.4)
Using this technique, PF is significantly improved from 0.689 to 0.95.
VinIin
1000.00 1010.00 1020.00
Time (ms)
1030.00 1040.00
15.00
–15.00 –10.00 –5.00 10.00 5.00 0.0 400.00 200.00
–200.00 –400.00 0.0
FIGURE 4.8 Input voltage and current waveforms.
Implementing Power Factor Correction in AC/DC Converters 101
100.00
Vload
0.0 –100.00 –200.00 –300.00 –400.00
0.0 0.50 1.00
Time (s)
1.50 2.00
FIGURE 4.9 Output voltage of the buck–boost converter.
3.00
2.50
2.00
1.50
1.00
0.50 0.0
–0.50
0.0 0.20 0.40
Frequency (kHz)
0.60 0.80
(949.062, 0.011) (850.6, 0.1) (750.7, 0.01)
(549.7, 0.077) (451.26, 0.295) (149.5, 0.664)
(50, 2680)
Iin
(349.5, 0.379) (250.2, 0.313)
(649.5, 0.071)
FIGURE 4.10 FFT spectrum of the input current.
TABLE 4.3
Harmonic Components of the Input Current
Current Frequency (Hz) Fourier Component
I1 50 2.680
I3 150 0.664
I5 250 0.313
I7 350 0.379
I9 450 0.295
I11 550 0.077
I13 650 0.071
I15 750 0.010
I17 850 0.100
I19 950 0.011
From the above investigation, we know that using a buck–boost converter to implement PFC can be successful, but the output voltage has a negative polarity. If a P/O Luo-converter or SEPIC or a P/O buck–boost converter is used, we can obtain the P/O voltage.
Example 4.1
A P/O Luo-converter (see Figure 5.11 in Chapter 5) is used to implement PFC in a single-phase diode rectifier with an R–C load. The AC supply voltage is 240 V/50 Hz and the required output voltage is 200 V. The switching frequency is 4 kHz. Determine the duty cycle k in a half supply period (10 ms). Other component values for reference are the following: R= 100 Ω, C = CO= 20μF, and L1= L2= 10 mH.
SOLUTION
Since the supply frequency is 50 Hz and the switching frequency is 4 kHz, there are 40 switching periods in a half supply period (10 ms). The voltage transfer gain of the P/O Luo-converter is
VO= k 1− kVin, k= VO
VO+ Vin = 200 200+ 240√
2 sinωt. Duty cycle k is listed in Table 4.4.