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AN3276 数据表(PDF) 7 Page - STMicroelectronics

部件名 AN3276
功能描述  ST solution for efficiency improvement in PFC applications
PDF  22 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

AN3276 数据表(HTML) 7 Page - STMicroelectronics

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AN3276
The new ST solution - BC2: energy recovery circuit
Doc ID 17975 Rev 1
7/22
2
The new ST solution - BC2: energy recovery circuit
The innovative circuit has been designed [see Section 5: References, 1.] to respect the five
soft switching criteria in Section 1.2. Figure 6 shows that two additional diodes D1 and D2
and two auxiliary windings NS1 and NS2 wound around the main boost inductor LB are
designed to reset the energy stored in the small inductor L.
Figure 6.
Novel energy recovery circuit: BC2
2.1
Concept description
The winding NS1 allows the IRM current from the boost diode DB to be recovered in the main
boost inductor when the transistor turns on. Since the mains input voltage modulates the LB
voltage, it also modulates the reflected voltage across NS1. This input voltage also
modulates the boost diode current IDB and its associated recovery current IRM. These
combined modulations allow the extra current IRM flowing in the inductor L to be reset into
the winding NS1 even in the worst case. The winding NS2 allows the extra current of L to be
injected into the output capacitor when the transistor turns off. The reflected voltage across
NS2 is also a function of the input voltage. This reflected voltage reaches its maximum when
the AC line voltage is low, corresponding to the maximum value of the inductor L current.
These combined variations allow the current flowing in the inductor L to be cancelled in the
bulk capacitor through the diode D2 even in the worst case. The benefits of these two
additional windings NS1 and NS2 are to switch off the diodes D1 and D2 with a low dI/dt
(about 10 A/µs) as in a discontinuous mode switching converter. Their recovery currents do
not affect the behavior of the BC2 circuit.
L
N
S2
m2 =NS2 /NP
DB
D
1
TR
m1 =NS1 /NP
D2
NS1
Np
LB
VOUT
Vmains



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