FAIRCHILD FAN6747

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AN-6747
Applying FAN6747 to Control a Flyback Power Supply with
Peak Current Output
1. Introduction
Highly integrated PWM controller, FAN6747, is optimized
for applications with motor load, such as printers and
scanners, that inherently impose some kind of overload
condition on the power supply during acceleration mode.
FAN6747 provides a two-level OCP function that allows the
SMPS to stably deliver peak power during the motor
acceleration without causing premature shutdown, while
protecting the SMPS from overload condition.
Green-mode and burst-mode functions with a low operating
current maximize the light-load efficiency so that the power
supply can meet stringent standby power regulations.
Figure 1.
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
The frequency-hopping function reduces electro-magnetic
interference (EMI) of a power supply by spreading the
energy over a wider frequency range. The constant power
limit function minimizes the component stress in abnormal
condition and helps optimize the power stage. Protection
functions such as OCP, OLP, OVP, and OTP are fully
integrated into FAN6747, which improves the SMPS
reliability without increasing system cost.
This application note presents design considerations to
apply FAN6747 to a flyback power supply with peak load
current profile. It covers designing the transformer, selecting
the components, and closing the feedback loop. Figure 1
shows a typical application circuit using FAN6747.
Typical Application
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AN-6747
APPLICATION NOTE
(Design Example) The specifications of the target
system are:
VLINEMIN =90VRMS, VLINEMAX=264VRMS
Line frequency (fL) = 60Hz
Nominal output power (PNO) = 20W (32V/0.625A)
Peak output power (PPO) = 70W (32V/2.187A)
Peak load duration (tPO) < 100ms
Estimated efficiency: ηN = 0.87 and ηP = 0.83
2. Design Considerations
Flyback converters have two operation modes; continuous
conduction mode (CCM) and discontinuous conduction
mode (DCM). CCM and DCM each have advantages and
disadvantages. In general, DCM provides better switching
conditions for the rectifier diodes, since the diodes are
operating at zero current just before becoming reverse
biased and the reverse recovery loss is minimized. The
transformer size can be reduced using DCM because the
average energy storage is low compared to CCM. However,
DCM causes high RMS current, which increases the
conduction loss of the MOSFET severely for low line
condition. Thus, especially for applications with peak load
profile, such as printer and scanner; it is typical to design
the converter such that the converter operates in CCM for
low line and peak load condition to maximize efficiency.
PINP =
PPO
70
=
= 84 W
ηP
0.83
PINN =
PNO
20
=
= 23 W
ηN
0.87
FAN6747 can be used for this application because the
peak load duration is less than the OCP delay time of
220ms.
In this section, a design procedure is presented using the
schematic of Figure 1 as a reference. An offline SMPS with
20W/32V nominal output power and 70W/32V peak output
power has been selected as a design example.
[STEP-2] Determine the Input Capacitor (CIN) and
the Input Voltage Range
It is typical to select the input capacitor as 1.5~2μF per watt
of peak input power for universal input range (85-265VRMS)
and 0.7~0.8μF per watt of peak input power for European
input range (195V-265VRMS). With the input capacitor
chosen, the minimum input capacitor voltage at peak load
condition is obtained as:
[STEP-1] Define the System Specifications
Designing a power supply with peak load current profile,
the following specifications should be determined first:
„ Line voltage range (VLINEMIN and VLINEMAX)
„ Line frequency (fL)
(
„ Nominal output power (PNO)
VINPMIN = 2 • VLINE MIN
„ Peak output power (PPO) and its duration (tPO)
„ Estimated efficiencies for nominal load (ηN) and peak
load (ηP).
)
2
−
PINP • (1 − D CH )
CIN • fL
( 3)
The minimum input capacitor voltage at nominal load
condition is obtained as:
(
The power conversion efficiency must be estimated to
calculate the input powers for each condition. Typically,
the efficiency at peak load condition is lower than that
of nominal load since most of the components of power
supply are selected for nominal load condition.
If no reference data is available, set ηN = 0.7~0.75 and
ηP = 0.65~0.7 for low-voltage output applications and
ηN = 0.8~0.85 and ηP = 0.75~0.8 for high-voltage
output applications.
VINNMIN = 2 • VLINE MIN
)
2
−
PINN • (1 − D CH )
CIN • fL
( 4)
where DCH is the input capacitor charging duty ratio defined
as shown in Figure 2, which is typically about 0.2.
The maximum input capacitor voltage is given as:
VINMAX = 2 VLINEMAX
( 5)
With the estimated efficiency, the input power for peak
load condition is given by:
PINP =
PPO
ηP
(1)
The input power for nominal load condition is given
by:
PINN =
PNO
ηN
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
(2)
Figure 2.
Input Capacitor Voltage Waveform
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AN-6747
APPLICATION NOTE
As can be seen in Equation (7), the voltage stress across the
MOSFET can be reduced by reducing VRO; however, this
increases the voltage stresses on the rectifier diodes in the
secondary side. Therefore, VRO should be determined by a
trade-off between the voltage stresses of MOSFET and
diode. Because the actual drain voltage rises above the
nominal MOSFET voltage due to the leakage inductance of
the transformer, as shown in Figure 3, it is typical to set VRO
around 70~100V so that VDSNOM is 430~450V for 600V
MOSFET (73~78% of MOSFET voltage rating).
(Design Example) By choosing a 120μF capacitor for
the input capacitor, the minimum input voltages for peak
and nominal load are obtained, respectively, as:
(
VINPMIN = 2 • VLINE MIN
= 2 • (90 )2 −
2
−
PINP • (1 − D CH )
CIN • fL
84 • (1 − 0.2)
120 × 10 − 6 • 60
(
VINNMIN = 2 • VLINE MIN
= 2 • (90)2 −
)
)
2
−
= 83 V
PINN • (1 − D CH )
CIN • fL
23 • (1 − 0.2)
120 × 10
−6
• 60
(Design Example) By determining VRO as 100V:
VINMAX = 2 • VLINEMAX = 2 • 264 = 373V
VRO + VINP MIN
VDSNOM = VINMAX + VRO
=
100
= 0.55
100 + 83
[STEP-4] Determine the Transformer Primary-Side
Inductance (LM)
The transformer primary-side inductance is determined for
the minimum input voltage and peak load condition. With
the DMAX from step 3, the primary-side inductance (LM) of
the transformer is obtained as:
[STEP-3] Determine the Reflected Output Voltage
(VRO)
When the MOSFET is turned off, the input voltage (VIN),
together with the output voltage reflected to the primary,
(VRO) are imposed across the MOSFET, as shown in Figure
3. With a given VRO, the maximum duty cycle (DMAX) and
the maximum nominal MOSFET voltage (VDSNOM) are
obtained as:
DMAX =
VRO + VINP MIN
VDSNOM = VINMAX + VRO = 373 + 100 = 473V
The maximum input voltage is obtained as:
VRO
VRO
DMAX =
= 117 V
LM =
(V
MIN
• D MAX
2PINP fSW K RF
INP
)
2
( 8)
where fSW is the switching frequency and KRF is the ripple
factor at peak load and minimum input voltage condition, as
shown in Figure 4.
( 6)
The ripple factor is closely related to the transformer size
and the RMS value of the MOSFET current. Even though
the conduction loss in the MOSFET can be reduced by
reducing the ripple factor, too small a ripple factor forces an
increase in transformer size. From a practical point of view,
it is reasonable to set KRF = 0.3~0.6 for the universal input
range and KRF = 0.4~0.8 for the European input range.
( 7)
Once LM is calculated by determining KRF from Equation
(8), the peak current and RMS current of the MOSFET for
minimum input voltage and peak load condition are
obtained as:
IDS PK = IEDC +
I DS
RMS
where
and
Figure 3.
=
ΔI
2
( 9)
2
⎡
⎛ ΔI ⎞ ⎤ D
2
⎢3(I EDC ) + ⎜ ⎟ ⎥ MAX
⎝ 2 ⎠ ⎥⎦ 3
⎢⎣
IEDC =
ΔI =
VINP
PINP
MIN
• DMAX
VINP MINDMAX
L M fSW
(10)
(11)
(12)
Output Voltage Reflected to the Primary
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
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AN-6747
APPLICATION NOTE
K RF
The peak drain current at minimum input voltage and peak
load condition was obtained from Equation (9) in step 4.
The peak drain current at minimum input voltage and
nominal load condition is given as:
ΔI
=
2 I EDC
CCM:
ΔI
(
PINN • VINMIN + VRO
IDS.NPK =
I DS PK
VINNMIN • VRO
)+
VINNMIN • VRO
(
2L M fSW • VINNMIN + VRO
)
(13)
Figure 4.
MOSFET Current and Ripple Factor (KRF)
DCM:
(Design Example) Determining the ripple factor as 0.375:
LM =
(V
IEDC =
ΔI =
MIN
• DMAX
2PINP fSW K RF
INP
V INP
PINP
MIN
)
•DMAX
2
=
=
(83 • 0.55)2
2 • 84 • 65 × 10 3 • 0.375
84
= 1.84A
83 • 0.55
CCM:
[3(1.84)
2
ΔI
= 1.84 + 0.69 = 2.53
2
+ (0.69 )2
2PINNL M f SW •
R CS <
R CS <
MIN
+ VRO
MIN
• VRO
MIN
+ VRO
INN
VINN
(V
INN
VINNMIN • VRO
)>1
(15)
)<1
(16)
0.48
(17)
IDS.NPK
0.825
(18)
IDS.NPK
(Design Example) For minimum input voltage and
nominal load condition, the operation mode is DCM as:
[STEP-5] Determine the Sensing Resistor Value
The current sensing resistor value should be determined
considering the over-current protection threshold and the
pulse-by-pulse current limit threshold, as shown in Figure
5. The peak value of current sensing voltage (VCS) should
be lower than the pulse-by-pulse current limit level for peak
load condition. It should be lower than the OCP threshold
for nominal load conditions to prevent false triggering of
OCP protection during normal operation.
2PINNL M fSW •
=
(V
MIN
+ VRO
MIN
• VRO
INN
VINN
)
2 • 23 • 508 × 10 − 6 • 65 × 10 3 •
(117
+ 100
117 • 100
)<1
The peak drain current at minimum input voltage and
nominal power condition is given as:
Pulse-by-Pulse Current Limit Threshold
IDS.NPK =
0.825V
Peak Power
Condition
2 • PINN
=
fSW • L M
2 • 23
65 × 10 • 508 × 10 − 6
3
= 1.18 A
The conditions for the sensing resistor are given as:
R CS <
0.48V
Nominal Power
Condition
R CS <
0.48
IDS.N
PK
0.825
IDS.P
PK
=
0.48
= 0.41Ω
1.18
=
0.825
= 0.33Ω
2.53
A 0.33Ω resistor is selected for the current-sensing resistor.
VCS = I DS ⋅ RCS
Figure 5.
(V
The condition for the sensing resistor is given as:
]0.355 = 1.4A
OCP Threshold
2PINNL M f SW •
DCM:
2
⎡
⎛ ΔI ⎞ ⎤ D
IDS RMS = ⎢3(IEDC )2 + ⎜ ⎟ ⎥ MAX
⎢
⎝ 2 ⎠ ⎥⎦ 3
⎣
=
(14)
Whether the converter operates in CCM or DCM at
minimum input voltage and nominal load condition is
determined by:
= 508μH
VINPMINDMAX
83 • 0.55
=
= 1.38 A
L M fSW
508 × 10 − 6 • 65 × 10 3
IDS PK = IEDC +
2 • PINN
fSW • L M
IDS.NPK =
Determining Current Sensing Resistor
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
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AN-6747
APPLICATION NOTE
[STEP-6] Determine the Minimum Primary Turns
With a given core, the minimum number of turns for the
transformer primary side to avoid the core saturation is
given by:
N P MIN =
L • 0 . 825 / R CS
L M ILIM
× 10 6 = M
× 10 6
B SAT A e
B SAT A e
n=
VRO
NP
=
NS VO + VF
where NP and NS are the number of turns for primary side
and secondary side, respectively, VO is the output voltage;
and VF is the diode (DO) forward-voltage drop.
(19)
Determine the proper integer for NS such that the resulting
NP is larger than NPmin obtained from Equation (19).
where Ae is the cross-sectional area of the core in mm2, ILIM
is the pulse-by-pulse current limit level determined by
0.825V threshold, RCS is current sensing resistor, and BSAT
is the saturation flux density in Tesla.
The number of turns for the auxiliary winding for VDD
supply is determined as:
The pulse-by-pulse current limit level is included in
Equation (19) because the inductor current reaches the
pulse-by-pulse current limit level during the load transient
or overload condition. Figure 6 shows the typical
characteristics of ferrite core from TDK (PC40). Since the
saturation flux density (BSAT) decreases as the temperature
rises, the high-temperature characteristics should be
considered. If there is no reference data, use BMAX =0.3T.
NA =
VDD * + VFA
• NS
VO + VF
(21)
where VDD is the nominal value of the supply voltage and
VFA is the forward-voltage drop of DDD as defined in Figure
7. Since VDD increases as the output load increases, it is
proper to set VDD at 3~5V higher than VDD UVLO level (9V)
to avoid the over-voltage protection condition during the
peak load operation.
Figure 7.
Figure 6. Typical B-H Characteristics of Ferrite Core
(TDK/PC40)
Simplified Transformer Diagram
(Design Example) Assuming the diode forward-
voltage drop is 1V, the turn ratio is obtained as:
(Design Example) An EF25/13/11 core is selected
2
with effective cross-sectional area of 78mm . Choosing
the saturation flux density as 0.27T, the minimum
number of turns for the primary side is obtained as:
NP MIN
(20)
n=
VRO
NP
100
=
=
= 3.03
NS VO + VF 32 + 1
Then, determine the proper integer for NS such that the
resulting NP is larger than NPmin as:
L • 0.825 / R CS
508 × 10 −6 • 0 .825 / 0.33
= M
× 10 6 =
× 10 6 = 60
B SAT A e
0.27 • 78
NS = 20,NP = n • NS = 61 > NPMIN
Setting VDD* as 13V, the number of turns for the
auxiliary winding is obtained as:
[STEP-7] Determine the Number of Turns for Each
Winding
Figure 7 shows a simplified diagram of the transformer.
First, calculate the turn ratio (n) between the primary side
and the secondary side from the reflected output voltage
determined in step 3 as:
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
NA =
VDD * + VFA
13 + 1
• 20 = 9
• NS =
32 + 1
VO + VF
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AN-6747
APPLICATION NOTE
(Design Example) The diode voltage and current are
[STEP-8] Determine the Wire Diameter for Each
Winding Based on the RMS Current of Winding
The maximum RMS current of the secondary winding is
obtained as:
1 − DMAX
DMAX
ISECRMS = n • IDSRMS
calculated as:
VDO = VO +
(22)
IDO RMS = n • IDSRMS
The current density is typically 6~10A/mm2 when the wire
is long (>1m). When the wire is short with a small number
of turns, a current density of 8~14A/mm2 is also acceptable.
These current densities are based on the peak load condition
and therefore almost twice conventional power supply
design. Avoid using wire with a diameter larger than 1mm
to avoid severe eddy current losses and to make winding
easier. For high current output, use parallel windings with
multiple strands of thinner wire to minimize skin effect.
= 3 . 03 • 1 . 4
1 − 0 . 55
= 3 . 84 A
0 . 55
[STEP-10] Feedback Circuit Configuration
The FAN6747 employs peak-current-mode control, as
shown in Figure 8. A current-to-voltage conversion is
accomplished externally with current-sense resistor RCS.
Under normal operation, the FB level controls the peak
inductor current as:
side winding is obtained from step 4 as 1.4A. The RMS
current of the secondary-side winding is calculated as:
= 3 . 03 • 1 . 4
1 − DMAX
DMAX
10A and 200V diode is selected, assuming a very small
heat-sink is used for the diode.
(Design Example) The RMS current of the primary-
ISECRMS = n • IDSRMS
VINMAX
373
= 32 +
= 155 V
3.03
n
1 − DMAX
DMAX
IDS • R CS + VSLOPE = IDS • R CS + 0.35 • D =
VFB − 0.6
(27)
4
1 − 0 . 55
= 3 . 84 A
0 . 55
where VFB is the voltage of FB pin, VSLOPE is synchronized
positive-going ramp, and D is duty cycle ratio.
0.45mm (8A/mm2) and 0.55mm (12A/mm2) diameter
wires are selected for primary and secondary windings,
respectively.
[STEP-9] Choose the Rectifier Diode in the
Secondary-Side Based on Voltage and Current
Ratings
The maximum reverse voltage and the RMS current of the
rectifier diode are obtained as:
VDO = VO +
VINMAX
n
IDORMS = n • IDSRMS
(23)
1 − DMAX
DMAX
(24)
Figure 8.
The typical voltage and current margins for the rectifier
diode are:
VRRM > 1.3 • VDO
(25)
IF > 1.5 • IDO RMS
(26)
Figure 9 is a typical feedback circuit mainly consisting of a
shunt regulator and a photo-coupler. R1 and R2 form a
voltage divider for output voltage regulation. RF and CF are
adjusted for control-loop compensation. A small-value RC
filter (e.g. RFB= 100Ω, CFB= 1nF) placed from the FB pin to
GND can increase stability substantially. The maximum
source current of the FB pin is about 325μA. The
phototransistor must be capable of sinking this current to
pull the FB level down at no load. The value of the biasing
resistor, RBIAS, is determined as:
where VRRM is the maximum reverse voltage and IF is the
current rating of the diode.
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
Peak Current Mode Circuit
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AN-6747
APPLICATION NOTE
VO − VOPD − VKA
• CTR > 325 × 10 − 6
R BIAS
A two-stage hold-up capacitor configuration (CDD1 and
CDD2) is typically used to increase the hold-up time while
minimizing startup time. Initially, the FAN6747 HV startup
circuit is enabled before it begins normal switching
operation. Therefore, the current supplied by the HV pin
can charge capacitor CDD1 while supplying the startup
current to FAN6747. When VDD reaches the turn-on voltage
of 16.5V (VDD-ON), FAN6747 begins switching operation and
the HV startup circuit is disabled. Then the current required
by FAN6747 is supplied from the auxiliary winding of
transformer.
(28)
where VOPD is the drop voltage of photodiode, about 1.2V;
VKA is the minimum cathode to anode voltage of shunt
regulator (2.5V); and CTR is the current transfer rate of the
opto-coupler.
vO
5.2V
iD
R BIAS
It is typical to use a 150~250kΩ resistor for the HV pin to
improve the immunity against line surge.
vFB
3R
RDB
C FB
R1
R
CF
RF
R2
Figure 9.
Feedback Circuit
The feedback compensation network transfer function of
Figure 9 is obtained as:
νˆ FB
ω 1 + s / ω ZC
=− I •
νˆ O
s 1 + s / ω PC
where ωI =
(29)
RB
1
1
; ω ZC =
;ω =
(RF + R1 )CO PC RB CFB
R1R DB CF
Figure 10.
RB is the internal feedback bias resistor; and R1, RD, RF, CF,
and CFB are shown in Figure 9.
(Design Example) Assuming CTR is 100%;
Leading-Edge Blanking (LEB)
Each time the power MOSFET is switched on, a turn-on
spike occurs across the sense resistor, caused by primaryside capacitance and secondary-side rectifier reverse
recovery. To avoid premature termination of the switching
pulse, a leading-edge blanking time is built in. During this
blanking period (270ns), the PWM comparator is disabled
and cannot switch off the gate driver. Thus, an RC filter
with a small RC time constant is enough for current sensing
(e.g. 100Ω + 470pF). A non-inductive resistor is
recommended for RCS.
VO − VOPD − VKA
• CTR > 325 × 10 − 6
R BIAS
R BIAS <
VO − VOPD − VKA
325 × 10 − 6
=
32 − 1.2 − 2.5
325 × 10 − 6
Startup Circuit
= 87kΩ
5.1kΩ resistor is selected for RDB.
The voltage divider resistors for VO sensing are
selected as 120kΩ and 10kΩ.
[STEP-11] Design the Startup Circuit
Figure 10 shows the typical startup circuit for FAN6747.
HV pin has an internal high-voltage startup circuit that is
disabled when VDD reaches its turn-on threshold. Since HV
pin is also used to obtain line voltage information for
brownout protection and power limit line compensation, it
is typical to connect the HV pin to the AC line through a
resistor and diode.
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
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AN-6747
APPLICATION NOTE
NTC thermister decreases and RT pin voltage drops. When
the voltage of the RT pin is less than 1.05V but over 0.7V,
the PWM turns off after 16ms (tD_OTP-LATCH). When RT pin
voltage is less than 0.7V, OTP is triggered after the 185μs
(tD_OTP2-LATCH) debounce time.
If the RT pin is not connected to the NTC resistor for overtemperature protection, a 100KW resistor to ground to
prevent noise interference is recommended. This pin is
limited by the internal clamping circuit.
Figure 11.
Current Sensing
Thermal Protection
Figure 12 shows the internal blocks for thermal protection.
A constant current, IRT, of 100μA is provided from the RT
pin. For over-temperature protection, an NTC thermistor in
series with a resistor can be connected between the RT and
GND pins. As temperature increases, the impedance of
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
Figure 12.
Thermal Protection Circuit
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AN-6747
APPLICATION NOTE
Printed Circuit Board (PCB) Layout
Two suggestions with different advantages
disadvantages for ground connections are offered:
PCB layout is a very important design issue for highfrequency switching current/voltage application. Good PCB
layout minimizes excessive EMI and helps the power supply
survive during surge / ESD tests.
Guidelines:
ƒ To get better EMI performance and reduce line
frequency ripples, the output of the bridge rectifier
should be connected to capacitor C1 first, then to the
switching circuits.
ƒ
The high-frequency current loop is in C1 –
transformer – MOSFET – RS – C1. The area enclosed
by this current loop should be as small as possible.
Keep the traces (especially 4 → 1) short, direct, and
wide. High-voltage traces related to the drain of
MOSFET and RCD snubber should be kept far way
from control circuits to prevent unnecessary
interference. If a heatsink is used for the MOSFET,
connect this heatsink to ground.
ƒ
As indicated by 3, the ground of control circuits should
be connected first, then to other circuitry.
ƒ
As indicated by 2, the area enclosed by transformer
auxiliary winding, D1, C2, D2, and C3 should also be
kept small. Place C3 close to the FAN6747 for good
decoupling.
Figure 13.
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
and
ƒ
GND3 → 2 → 4 → 1: This could avoid common
impedance interference for sense signal.
ƒ
GND3 → 2 → 1 → 4: This could be better for ESD
testing where the earth ground is not available on the
power supply. Regarding the ESD discharge path, the
charges go from secondary through the transformer
stray capacitance to GND2 first. The charges then go
from GND2 to GND1 and back to the mains. Control
circuits should not be placed on the discharge path.
Point discharge for common choke can decrease highfrequency impedance and increase ESD immunity.
ƒ
Should a Y-cap between primary and secondary be
required, connect this Y-cap to the positive terminal of
C1. If this Y-cap is connected to the primary GND, it
should be connected to the negative terminal of C1
(GND1) directly. Point discharge of this Y-cap also
helps for ESD. However, the creepage between these
two pointed ends should be large enough to satisfy the
requirements of applicable standards.
Layout Considerations
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AN-6747
APPLICATION NOTE
Design Summary
Figure 14 shows the final schematic of the 20W (70W peak) power supply of the design example.
Figure 14.
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
Final Schematic of Design Example
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AN-6747
APPLICATION NOTE
Transformer Specification
4
JP3(fly line)
N3
3
1
5
10
6
N1
N2
5
1
N4
9
2
Bottom View
Figure 15.
Transformer Specification
Winding Specification
Pin
5Æ3
Insulation Tape
Shielding Lead to Pin 4
Insulation Tape
N2
JP3 Æ 9
Insulation Tape
Shielding Lead to Pin 4
Insulation Tape
N3
3Æ4
Insulation Tape
N4
1Æ2
N1
Diameter / Thickness
Turns
0.45mm
30
3
65
3
20
3
65
3
30
6
9
0.55mm
0.45mm
0.2mm
Insulation Tape
3
Core: EF25/13/11 (Ae=78 mm2)
Bobbin: EF25/13/11
Inductance: 508μH
© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
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11
AN-6747
APPLICATION NOTE
Related Datasheets
FAN6747 — Highly Integrated Green-Mode PWM Controller for Peak Power Management
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1.
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© 2010 Fairchild Semiconductor Corporation
Rev. 1.0.1 • 9/16/10
2.
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