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AN966 Datasheet, PDF (11/21 Pages) STMicroelectronics – The front-end stage of conventional off-line converters
AN966 APPLICATION NOTE
As a result, neglecting fringing flux in the air gap region, the energy balance can be re-written as:
L ⋅ IL2pk ≈ ∆Hgap ⋅ ∆B ⋅ Ae ⋅ Igap
The flux density ∆B is the same throughout the core and the air gap and is related to the field strength
inside the air gap by the well-known relationship:
∆B = µ0 ⋅ ∆Hgap .
Then, considering Ampere’s law (applied to the air gap region only):
Igap ⋅ ∆Hgap ≈ N ⋅ ILpk ,
from the energy balance equation it is possible to obtain:
√ L
≈
µ0
⋅
N2 ⋅ Ae
Igap
⇒
N≈
L ⋅ lgap ,
µο ⋅ Ae
where N is the turn number of the winding.
As N is defined, it is recommended to check for the saturation of the core (see Pin 4 description). If the
check shows a result too close to the rated limit, an increase of lgap and a new calculation will be neces-
sary.
The wire gauge selection is based on limiting the copper losses at an acceptable value:
PCU
=
4
3
⋅
Ir2ms
⋅
RCU;
due to the high frequency ripple the effective wire resistance RCU is increased by skin and proximity ef-
fects. For this reason Litz wire or multi-wire solutions are recommended.
Finally, the space occupied by the winding will be evaluated and, if it does not fit the winding area of the
bobbin, a bigger core set will be considered and the winding calculation repeated.
It is now necessary to add an auxiliary winding to the inductor, in order for the ZCD pin to recognize
when the current through the inductor has gone to zero. It is anyway a low cost thin wire winding and the
turns number is the only parameter to be defined (see Pin 5 description).
POWER MOSFET
The choice of the MOSFET concerns mainly its RDSon, which depends on the output power, since the
breakdown voltage is fixed just by the output voltage, plus the overvoltage admitted and a safety margin.
The MOSFET’s power dissipation depends on conduction and switching losses.
The conduction losses are given by:
PON = IQ2 rms ⋅ RDSon
where:
√ IQrms = 2 ⋅ √2 ⋅ Irms ⋅
1
6
−
4 √2
9π
⋅
Virms
VO
.
The switching losses due to current-voltage cross occur only at turn-off because of the TM operation:
PCROSS = VO ⋅ Irms ⋅ tfall ⋅ ƒsw,
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