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最新版功率因数校正(PFC)手册(中).pdf
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更新于2023-05-22
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最新版功率因数校正(PFC)手册是第一版PFC手册的升级版。通过一步一步的设计指导和系统级的对比,PFC手册评估了300W电源应用中
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ON Semiconductor
http://onsemi.com
85
CHAPTER 7
Bridgeless PFC
Introduction
The need for higher efficiencies from the PFC stage has led the circuit designers to look closely at all sections of the circuit
and develop possible lower loss alternatives. One section that contributes significantly to the losses is the input bridge rectifier.
As a result, the alternatives to eliminate the diode bridge or convert it into a dual-use circuit have been explored for many years.
This elimination/conversion of diode bridge brings about its own set of challenges. This chapter provides a more in-depth look
at the bridgeless techniques and works through a design example for an 800-W bridgeless PFC converter.
Why remove the bridge?
EMI
Filter
AC
Line
PFC
Stage
D
1
D
2
D
3
D
4
Figure 7−1. The Input Current Flows Through Two Diodes
Figure 7−1 portrays the diode bridge that is usually inserted between the EMI filter and the PFC stage. This bridge rectifies
the line voltage to feed the PFC stage with a rectified sinusoid input voltage. It is well known that as a result of this structure,
the input current must flow through two diodes before being processed by the PFC boost stage:
♦
For one line half-wave, D1 and D4 conduct (red arrows of Figure 7−1)
♦
For the other one, D2 and D3 convey the current (blue arrows of Figure 7−1)
As a matter of fact, two diodes of the bridge are permanently inserted in the current path. Unfortunately, these components
exhibit a forward voltage that leads to conduction losses.
The mean value of the current seen by the bridge is the line current averaged over one half−line cycle. Hence we can write
the following equation:
ǂ
I
bridge
ǃ
T
line
2
+
ǂ
I
in
(t)
ǃ
T
line
2
+
22
Ǹ
p
@ I
in(rms)
(eq. 1)
The line rms current can be easily expressed as a function of the power and of the line voltage:
I
in(rms)
+
P
out
h @ V
in(rms)
(eq. 2)
Where:
♦
P
out
is the output power
♦
h is the efficiency
♦
V
in(rms)
is the rms line voltage
Since two diodes permanently see the average input current, the bridge consumes a power that can be computed as follows:
P
bridge
+ 2 @ V
f
@
ǂ
I
bridge
ǃ
T
line
2
^ 2 @ V
f
@
22
Ǹ
@ P
out
h @ p @ V
in(rms)
(eq. 3)















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