ACT366 Ver la hoja de datos (PDF) - Active-Semi, Inc
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ACT366 Datasheet PDF : 11 Pages
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ACT366
Rev 1, 14-Nov-12
TYPICAL APPLICATION CONT’D
where η is the estimated circuit efficiency, fL is the
line frequency, tC is the estimated rectifier
conduction time, CIN is empirically selected to be
15µF + 10µF electrolytic capacitors based on the
2µF/W rule of thumb.
NP =
LP =
A LE
0 .87 mH
80 nH / T 2 = 110
(11)
The number of turns of secondary and auxiliary
windings can be derived when Np/Ns=7:
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:
NS
=
NS
NP
× NP
1
=
7
× 110
≈14
(12)
NA
=
NA
NS
×
NS
= 1.1 ×14
=16
(13)
The current sense resistance (RCS) determines the
current limit value based on the following equation:
RCS =
( ) 0.9×VCSLIM
=
IOUTFL+IOUTMAX×(VOUT +VDS)
LP
×fSW
×
ηsystem
ηxfm
0.9×0.396
(1.0+1.2)×12.3 =0.5R
0.78
0.87×75× 0.89
(14)
VRO
=
VINDCMAX×(VOUTCV +VDS
VDREV -VOUTCV
)
=
375×(12 +0.5
100×0.8 -12
)
=
68.9V
(5)
where VDS is the Schottky diode forward voltage,
VDREV is the maximum reverse voltage rating of the
diode and VOUTCV is the output voltage.
The maximum duty cycle is set to be 50% at low
line voltage 85VAC and the circuit efficiency is
estimated to be 78%. Then the full load input
current is:
IIN
= VOUTCV × IOUTPL
VINDCMIN × η
12 × 1
=
= 170 .9 mA
90 × 78 %
(6)
The maximum input primary peak current at full
load base on duty of 50%:
IPK
=
2 × IIN
D
=
2 ×170 .9
50%
= 683 mA
(7)
The primary inductance of the transformer:
The voltage feedback resistors are selected
according to below equation:
RFB1
=
NA
NP
×
LP
RCS
16 0.87
×K = ×
× 230000
110 0.5
≈59 k
(15)
In actual application 59K is selected.
Where K is IC constant and K = 230000.
R FB 2 =
( V OUTCV
V FB
+ V DS
)NA
NS
- V FB
R FB 1
2 . 20
=
× 59 K ≈11 k
( 12 + 0 . 45 ) × 1 . 1 - 2 . 20
(16)
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:
LP
= VINDCMIN × D
IPK × fSW
=
90 × 50 %
683 mA × 75 kHz
≈0 .87 mH
(8)
COUT
= IOUTCC × D
fSW ×△VRIPPLE
1.2 × 0.5
=
60 kHz × 50 mV
= 200 μF
(17)
ACT366 needs to work in DCM in all conditions, A 600µF electrolytic capacitor is used to keep the
thus NP/NS should meet
ripple small.
LP × IPK
VINDCMIN
+
LP × IPK
(VOUTCV + VDS
)
×
NP
NS
< 0.9 ⇒NP
fSW NS
>8
(9)
The auxiliary to secondary turns ratio NA/NS:
NA =
VDD + VDA + VR
= 13 + 0.25 + 1 ≈1.1
NS VOUTCV + VDS + VCORD 12 + 0.3 + 0.35
(10)
Where VDA is diode forward voltage of the auxiliary
side and VR is the resister voltage.
An EPC17 transformer gapped core with an
effective inductance ALE of 80nH/T2 is selected.
The number of turns of the primary winding is:
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 VDD, 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,
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