AN1028

Application Note 1028
Basic Steps to Design a PSR Flyback Converter Using AP3706/08N
Prepared by Su Qinghua and Dong Wenhui
System Engineering Department
Introduction
Vdri is a simplified driving signal of primary transistor.
IP is the primary side current.
IS is the secondary side current.
VS is the voltage of secondary side.
AP3706/08N uses Pulse Frequency Modulation
(PFM) method to realize Discontinuous Conduction
Mode (DCM) operation for flyback power supplies;
uses Primary Side Regulation (PSR) to provide Constant Voltage (CV) and Constant Current (CC) regulation. The principle of PFM is different from that of
Pulse Width Modulation (PWM), so the design steps of
the transformer for PFM operation is also different
from that for PWM operation. In the other side,
AP3706/08N uses PSR to provide CV/CC regulation
without requiring opto-coupler and loop compensation, which makes the design of transformer more critical than that with PWM. The following design steps
focus on the transformer design. Some design guides
to select diodes and transistor are also included.
Figure 2 shows the relatively ideal operation waveforms to illustrate some parameters used in the following design steps. The nomenclature of the parameters
in Figure 2 are listed as below:
TSW is the period of switching frequency
Tonp is the time of primary side “ON”
Tons is the time of secondary side “ON”
Toff is the discontinuous time
Ipk is the peak current of primary side
Ipks is the peak current of secondary side
VS equals the sum of VO and forward voltage of rectification diode
Figure 1 is a simplified flyback converter controlled
by AP3706 with a 3-winding transformer: Primary
winding (NP), Secondary winding (NS) and Auxiliary
winding (NAUX). The AP3706 senses the Auxiliary
winding feedback voltage at FB pin and obtains power
supply at VCC pin. In the circuit:
The only difference between AP3706 and AP3708N
is cable compensation. AP3708N has built-in it, while
AP3706 has not. Their design steps are quite similar.
Vg
VS
VIN
+
+
CS
OUT
AP3706
BIAS
NP
NS
LM
LS
Vd
IS
IO
-
+
+
VO
-
VAux
Vdri
VCC
+
NAux
iP
FB
GND COMP
VDD
RCS
Figure 1. Simplified Flyback Converter Using AP3706
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Application Note 1028
Figure 2. Ideal Operation Waveforms for a Simplified Flyback Converter
Basic Design Step
Tonp = Ipk ×
Step 1: Select a Reasonable Ipk of Flyback Converter Using AP3706/08N
LM
Vg
where
LM is the inductance of primary winding.
1-1. Calculate the Maximum Turn Ratio of the
Transformer
The maximum turn ration of the transformer should
be calculated first to ensure the system working in
DCM under all working conditions, especially at the
minimum input voltage and full load.
Vg is the rectified DC voltage of input.
When Vg is at the minimum value, the maximum
Tonp can be obtained. So,
Tonp _ max = Ipk ×
If the system meets equation (1) at minimum input
voltage and full load, the circuit can work in DCM in
all working conditions.
TSW ≥ Tonp + Tons
LM
Vg _ min
......(2)
For the secondary side current,
......(1)
Tons = Ipks ×
For the primary side current,
LS
VS
......(3)
where
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Application Note 1028
LS is the inductance of secondary winding.
VS=VO+Vd, Vd is the forward voltage of secondary
diode.
PIN =
At the full load, the system will work in the
boundary of CC regulation. IO is determined by:
In Flyback converter, the energy stored in the
magnetizing inductance LM when the primary
transistor turns ON. So the power transferring from
the input to the output is given by:
1 Tons
× Ipks
IO = ×
2 TSW
Then, Ipks can be defined as:
......(4)
......(10)
Ipks = k × I O
then,
TSW =
......(9)
η
where η is the system efficiency.
For (3), in CV regulation, the VS is a constant
voltage, so Tons is a constant value with different
input voltage.
1
PI N = × LM × Ipk 2 × f SW
2
VO × I O
In the design of AP3706/08N,
1
f SW
=
LM × Ipk 2
2 × PIN
k=
Substitute TSW, Tonp and Tons in equation (1) with
(4), (2) and (3):
2 × TSW
= 3 .5
Tons
With (8), (9) and (10), it can be obtained:
N PS ≤Vg _ min× (
LM × Ipk
L
LM
≥Ipks × S + Ipk ×
2 × PIN
Vs
Vg _ min
2
N PS _ MAX = Vg _ min× (
Because the peak current and inductance of primary
side and secondary side have the following
relationship:
......(12)
......(6)
Because above calculations are all based on ideal
conditions without considering power loss, k is given
an approximately value 4 instead of the real value 3.5.
1-2. Calculate the Peak Current of Primary
Side and Current Sense Resistor
Ipk can be calculated by the output current:
......(7)
From (5), (6) and (7):
Ipk
1
1
≥
+
2 × PIN VS × N PS Vg
1
k ×η
)
−
2 × VO VO + Vd
(k ≈4)
here, NPS=NP/NS is the turn ratio of primary to
secondary sides, then
L
LS = M2
N PS
......(11)
So, the maximum turn ration of primary and
secondary side can be obtained:
......(5)
Ipks = N PS × Ipk
1
k ×η
− )
2 × VO VS
Ipk =
......(8)
Ipks
N PS _ MAX
=
k × IO
N PS _ MAX
......(13)
here, k=4.
In AP3706/08N, 0.5V is an internal reference
voltage. If the sensed voltage VCS reaches to 0.5V, the
Because
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Application Note 1028
The turns of primary winding is:
power transistor (APT13003 for this application) will
be shut down and Tonp will be ended. So, RCS can be
calculated as:
RCS =
0.5V
Ipk
NP =
......(14)
NS =
N AUX =
2-1. Calculate the Inductance of Primary Side--LM
The primary side inductance LM is relative with the
stored energy. LM should be big enough to store
enough energy, so that PO_max can be obtained from
this system.
......(15)
Vdr = VO +
......(16)
Vg _ max× N S
NP
......(21)
Maximum reverse voltage of Auxiliary side is:
Vdar = VAUX +
2-2. Re-calculate the Turn Ratio of Primary to
Secondary Side---NPS
From formula (13), the turn ratio of primary and
secondary side NPS should be re-calculated.
k × IO
(k ≈4)
Ipk
......(20)
3-1. Select Diodes of Secondary and Auxiliary
Sides
Maximum reverse voltage of secondary side is:
To achieve good overall system performance, the
optimum switching frequency fSW is recommended to
be 50 to 60kHz under full load.
N PS =
N S × VAUX
VS
Step 3: Select Diode and Primary
Transistor
Then, LM is:
2 × PO
Ipk × f SW ×η
......(19)
here,
VAUX can be set to its typical value 15V.
Vs is equal to VO+Vd.
Ae can be obtained automatically after core-type is
selected.
From equation (9), the output power can be
obtained from:
2
NP
N PS
The turns of Auxiliary winding is:
Step 2: Design Transformer
LM =
......(18)
The turns of secondary winding is:
Then, select RCS with a real value from the standard
resistor series. After RCS determined, Ipk should be
modified according to the selected RCS.
1
PO = × LM × Ipk 2 × f SW ×η
2
LM × Ipk ×108
Ae × ∆B
Vg _ max× N AUX
NP
Vdc _ max = Vdc _ spike + Vg _ max +
......(22)
VS × N P
NS
......(23)
In (21) and (22), the maximum DC input voltage
should be used.
......(17)
2-3. Calculate the Turns of Primary,
Secondary and Auxiliary Windings
First, select the reasonable core-type and ∆B . Then,
decide the turns of 3-winding transformer
respectively.
3-2. Select the Primary Side Transistor
It should be noted that the value of Vdc_spike will
be different in circuits with different snubber.
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Application Note 1028
Design Example
RCS =
Design Specification
0.5V
Ipk
......(14)
RCS ≈ 2.1Ω
Input Voltage
85 to 265 VAC
Output Voltage
VO=5.5V
Re-calculate peak current of primary side,
Output Current
IO=0.5A
Ipk = 238mA
Efficiency
75%
Step 2: Design Transformer
Settings Selected by Users
fSW=55kHz
Switching Frequency
Forward
Diode
Voltage
of
Secondary
Vd=0.4V
Forward Voltage of Auxiliary Diode
Vda=1V
Feedback
Winding
VAUX=15V
Voltage
2-1. Calculate the Inductance of Primary Side--LM
of
Auxiliary
Core_type
EE16
(Ae=19.2mm2)
∆B
∆B=2850GS
Vdc_spike
200V
LM =
......(16)
LM = 2.35mH
2-2. Re-calculate the Turn Ratio of Primary to
Secondary Side---NPS
N PS =
Design Step
k × IO
(k ≈4)
Ipk
......(17)
N PS = 8.4
2-3. Calculate the Turns of
Secondary and Auxiliary Windings
Step 1: Select a Reasonable Ipk for
the Flyback Converter Using AP3706
1-1. Calculate the Maximum Turn Ratio of the
Transformer
N PS _ MAX = Vg _ min× (
2 × PO
Ipk 2 × f SW ×η
1
k ×η
−
)
2 × Vo Vo + Vd
NP =
LM × Ipk ×108
Ae × ∆B
Primary,
......(18)
N P = 102T
......(12)
(k ≈4)
NS =
Vg _ min = VIN _ min× 2 − 40
NP
N PS
......(19)
N S = 12T
N PS _ MAX = 8.259
N AUX =
1-2. Calculate the Peak Current of Primary
Side and Current Sense Resistor
N S × VAUX
VS
......(20)
N AUX = 31T
Ipk =
k × IO
Ipks
=
N PS _ MAX N PS _ MAX
......(13)
Ipk = 242mA
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Application Note 1028
Step 3: Select Diode and Primary
Transistor
Maximum reverse voltage of secondary side.
Vdar = V AUX +
3-1. Select Diodes of Secondary and Auxiliary
Sides
Maximum reverse voltage of secondary side.
Vdr = VO +
Vg _ max× N S
NP
Vg _ max × N AUX
NP
......(22)
Vdar = 129V
3-2. Select the Primary Side Transistor
......(21)
Vdc _ max = Vdc _ spike + Vg _ max +
Vg _ max = VIN _ max× 2
VS × N P
NS
......(23)
Vdc _ max = 625V
Vdr = 50V
Design Results Summary
1. Calculate the Maximum Peak Current of Primary Side and RCS
Peak Current of Primary Side
Ipk=238mA
Current Sense Resistor
RCS=2.1Ω
2. Design Transformer
Inductance of Primary Side
LM=2.35mH(+/-8%)
Turn Ratio of Primary to Secondary Windings
NPS=8.4
Turns of Primary Side
NP=102T
Turns of Secondary Side
NS=12T
Turns of Auxiliary Side
NAUX=31T
3. Select Diode and Primary Transistor
Maximum Reverse Voltage of Secondary Diode
Vdr=50V
Maximum Reverse Voltage of Auxiliary Diode
Vdar=129V
Voltage Stress of Primary Transistor
Vdc_max=625V
Sep. 2009 Rev. 1.1
BCD Semiconductor Manufacturing Limited
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