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10
2005 Semtech Corp.
www.semtech.com
SC4809A/B/C
POWER MANAGEMENT
Application Information (Cont.)
The primary inductance can be calculated given an
acceptable current ripple,
I
.
I
was set to equal one-
half the peak primary current. For a CCM flyback design,
the peak primary current is calculated:
2
I
D
1
1
N
I
I
L
max
(max)
OUT
PEAK
+
=
Because the converter is operating in the continuous
mode, the maximum peak flux density B
, is limited by
the saturation flux density, B
. Taking all this into
consideration, the maximum core size is determined by.
31
.
MAX
4
RMS
B
PEAK
420
P
k
10
I
I
L
AP
=
where AP = the core area product in cm
4
,
k = winding factor,
B
MAX
≈
B
SAT
,
The result is compared to the product of the winding area,
Aw (cm
2
), and effective core area, Ae (cm
2
), listed in the
core manufacturer’s data sheet.
The minimum number of primary turns is determined by:
Ae
B
10
I
L
N
MAX
4
PEAK
P
P
=
Based upon this result and the predetermined turns ratio,
the number of secondary turns is established.
The energy stored in the flyback transformer is actually
stored in an air gap in the core. This is because the high
permeability of the ferrite material can’t store much
energy without saturating first. By adding an air gap, the
hysteresis curve of the magnetic material is actually tilted,
requiring a much higher field strength to saturate the
core. The length of the air gap is calculated by:
P
2
2
P
L
r
o
g
10
Ae
)
N
(
μ
μ
=
l
MOSFET Selection
The switching element in a flyback converter must have
a voltage rating high enough to handle the maximum input
voltage and the reflected secondary voltage, not to
mention any leakage inductance induced spike that is
inevitably present. Approximate the required voltage
rating of the MOSFET using.
(
)
(
)
3
V
V
N
N
V
V
V
D
O
S
P
L
(max)
IN
ds
+
+
+
=
where V
= the required drain to source voltage rating of
the MOSFET,
V
= the voltage spike due to the leakage inductance of
the transformer, estimated to be thirty percent of V
,
and the additions 1.3 factor includes an overall thirty
percent margin.
This FET will experience both switching and conduction
losses. The conduction losses will be equal to the I
2
R
losses, as shown by:
(
)
)
ON
(
DS
2
RMS
I
COND
P
R
=
Switching losses are the result of overlapping drain
current and drain to source voltage at turn on and turn
off.
The total switching losses are estimated based on
equation:
sw
ch
PEAK
I
DS
SW
2
DS
2
OSS
SW
f
t
V
f
)
V
(
C
P
+
=
where t
ch
:
)
th
(
gs
g
gd
CH
V
VDD
R
Q
t
=
Diode Selection
Schottky rectifiers have a lower forward voltage drop than
typical PN devices, making it the rectifier of choice when
considering reducing converter losses and improving
overall efficiency. Selecting the appropriate Schottky for
a specific application depends mainly on the working
peak reverse voltage rating and peak repetitive forward
current.