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AN3142 数据表(PDF) 13 Page - STMicroelectronics

部件名 AN3142
功能描述  Solution for designing a 400 W fixed-off-time controlled
PDF  44 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

AN3142 数据表(HTML) 13 Page - STMicroelectronics

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AN3142
Designing a fixed-off-time PFC
Doc ID 17005 Rev 3
13/44
4.3
Power section design
4.3.1
Bridge rectifier
The input rectifier bridge can use standard slow recovery, low-cost devices.
Typically a 600 V device is selected in order to have good margin against mains surges. An
NTC resistor limiting the current at turn-on is required to avoid overstress to the diode
bridge.
The rectifier bridge power dissipation can be calculated using
Equation 18, Equation 19,
and
Equation 20. The threshold voltage (V
th) and dynamic resistance (Rdiode) of a single
diode of the bridge can be found in the component datasheet.
Equation 18
Equation 19
The power dissipated on the bridge is:
Equation 20
4.3.2
Input capacitor
The input filter capacitor (Cin) is placed across the diode bridge output. This capacitor must
smooth the high-frequency ripple and must sustain the maximum instantaneous input
voltage. In a typical application an EMI filter is placed between the mains and the PFC
circuit. In this application the EMI filter is reinforced by a differential mode Pi-filter after the
bridge to reject the differential noise coming from the whole switching circuit. The design of
the EMI filter (common mode and differential mode) is not described here. The value of the
input filter capacitor can be calculated as follows, simply considering the output power that
the PFC should deliver at full load:
Equation 21
The maximum value of this capacitor is limited to avoid line current distortion. The value
chosen for this design is 1µF.
A
53
.
3
2
A
99
.
4
2
2
I
2
I
in
inrms
=
=
=
A
25
.
2
A
99
.
4
2
I
2
I
in
avg
_
in
=
π
=
π
=
avg
_
in
th
inrms
2
diode
bridge
I
V
4
I
R
4
P
+
=
W
53
.
7
A
25
.
2
V
7
.
0
4
)
A
53
.
3
(
025
.
0
4
P
2
bridge
=
+
=
out
3
in
P
10
5
.
2
C
=
F
1
W
400
10
5
.
2
C
3
in
µ
=
=



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