ACT366
TYPICAL APPLICATION CONT’D
where
η
is the estimated circuit efficiency, f
L
is the
line frequency, t
C
is the estimated rectifier
conduction time, C
IN
is empirically selected to be
15µF + 10µF electrolytic capacitors based on the
2µF/W rule of thumb.
When the transistor is turned off, the voltage on the
transistor’s collector consists of the input voltage
and the reflected voltage from the transformer’s
secondary winding. There is a ringing on the rising
top edge of the flyback voltage due to the leakage
inductance of the transformer. This ringing is
clamped by a RCD network if it is used. Design this
clamped voltage as 50V below the breakdown of
the NPN transistor. The flyback voltage has to be
considered with selection of the maximum reverse
voltage rating of secondary rectifier diode. If a 100V
Schottky diode is used, then the flyback voltage can
be calculated:
V
RO
=
V
INDCMAX
×
(V
OUTCV
+
V
DS
) 375
×
(12
+
0.5 )
=
=
68.9V
V
DREV
-
V
OUTCV
100
×
0.8
-
12
N
P
=
L
P
=
A
LE
0 . 87 mH
80 nH / T
2
Rev 1, 14-Nov-12
=
110
(11)
The number of turns of secondary and auxiliary
windings can be derived when Np/Ns=7:
N
S
=
N
S
1
×
N
P
= ×
110
≈
14
N
P
7
(12)
(13)
N
A
=
N
A
×
N
S
=
1.1
×
14
=
16
N
S
The current sense resistance (R
CS
) determines the
current limit value based on the following equation:
R
CS
=
(
I
)
OUT
+
V
DS
)
OUTFL
+
I
OUTMAX
×
(V
η
system
L
P
×
f
SW
×
η
xfm
0.9
×
V
CSLIM
=
0.9
×
0.396
(
1.0
+
1.2
)
×
12.3
0.87
×
75
×
0.78
0.89
=
0.5R
(14)
selected
(15)
(5)
The voltage feedback resistors
according to below equation:
R
FB 1
=
are
where V
DS
is the Schottky diode forward voltage,
V
DREV
is the maximum reverse voltage rating of the
diode and V
OUTCV
is the output voltage.
The maximum duty cycle is set to be 50% at low
line voltage 85V
AC
and the circuit efficiency is
estimated to be 78%. Then the full load input
current is:
V
×
I
OUTPL
12
×
1
(6)
I
IN
=
OUTCV
=
=
170 . 9 mA
V
INDCMIN
×
η
90
×
78 %
The maximum input primary peak current at full
load base on duty of 50%:
I
PK
=
2
×
I
IN
2
×
170 .9
=
=
683 mA
D
50 %
V
INDCMIN
×
D
90
×
50 %
=
≈
0 . 87 mH
I
PK
×
f
SW
683 mA
×
75 kHz
16
0 .87
N
A
L
×
P
×
K
=
×
×
230000
≈
59 k
N
P
R
CS
110
0 .5
In actual application 59K is selected.
Where K is IC constant and K = 230000.
R
FB
2
=
( V
OUTCV
V
FB
+
V
DS
N
)
A
-
V
FB
N
S
R
FB 1
(16)
=
2 . 20
×
59 K
≈
11 k
( 12
+
0 . 45 )
×
1 . 1
-
2 . 20
(7)
(8)
The primary inductance of the transformer:
L
P
=
When selecting the output capacitor, a low ESR
electrolytic capacitor is recommended to minimize
ripple from the current ripple. The approximate
equation for the output capacitance value is given by:
C
OUT
=
I
OUTCC
×
D
1 .2
×
0 .5
=
=
200
μ
F
f
SW
×
△
V
RIPPLE
60 kHz
×
50 mV
(17)
ACT366 needs to work in DCM in all conditions,
thus N
P
/N
S
should meet
L
P
×
I
PK
+
V
INDCMIN
L
P
×
I
PK
(V
OUTCV
+
V
DS
)
×
N
P
N
S
<
0 .9 N
P
⇒
>
8
f
SW
N
S
A 600µF electrolytic capacitor is used to keep the
ripple small.
(9)
PCB Layout Guideline
Good PCB layout is critical to have optimal
performance. Decoupling capacitor (C4), current
sense resistor (R9) and feedback resistor (R5/R6)
should be placed close to V
DD
, CS and FB pins
respectively. There are two main power path loops.
One is formed by C1/C2, primary winding, NPN
transistor and the ACT366. The other is the
secondary winding, rectifier D8 and output
capacitors (C5,C6). Keep these loop areas as small
as possible. Connect high current ground returns,
-7-
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Copyright © 2012 Active-Semi, Inc.
The auxiliary to secondary turns ratio N
A
/N
S
:
N
A
V
DD
+
V
DA
+
V
R
13
+
0 . 25
+
1
=
=
≈
.1
1
N
S
V
OUTCV
+
V
DS
+
V
CORD
12
+
0 . 3
+
0 . 35
(10)
Where V
DA
is diode forward voltage of the auxiliary
side and V
R
is the resister voltage.
An EPC17 transformer gapped core with an
effective inductance A
LE
of 80nH/T
2
is selected.
The number of turns of the primary winding is:
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TM
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