AN-303: IX9908 Design Considerations (European)

Application Note: AN-303
INTEGRATED CIRCUITS DIVISION
IX9908
Design Considerations
(European Voltage Version)
AN-303-R01
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AN-303
INTEGRATED CIRCUITS DIVISION
1
Introduction
This application note provides general guidelines for
designing an off-line LED driver using IXYS Integrated
Circuits Division’s IX9908. The IX9908 can be
configured to drive an external MOSFET device in a
quasi-resonant flyback converter power stage that
provides a constant current output to a LED string
while maintaining a high power factor.
Figure 1
This driver features a high voltage start-up circuit that
eliminates the need for a VCC resistor, thereby
improving overall power dissipation. In addition, there
are multiple safety features such as under-voltage
lockout, over-voltage protection, digital soft-start,
foldback correction, and cycle-by-cycle peak current
limiting. The IX9908 is an excellent choice for many
phase-cut dimming and high power-factor correction
LED lighting applications.
IX9908 Block Diagram
Vcc
ZCV
Ringing
Suppress
Blanking
High Voltage
Startup
HV
Gate
Control
GD
Vcc
Vcc Monitor
Over - Under
Voltage Lockout
Reference Voltage
Generator
Over-Voltage Protection
Leading
Edge
Blanking
Foldback
Sense Amp
Control
Logic
Over-Temp
Sensor
Soft-Start
Control
Foldback
Correction
VR
Shorted Winding Detection
GND
Leading
Edge
Blanking
CS
Leading
Edge
Blanking
2
LPF
Analog
Mux
PFC
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INTEGRATED CIRCUITS DIVISION
Figure 2
IX9908 Application Circuit
DVCC
RZCV1
AC
-
RSN
DOUT
COUT
DSN
CZCV
CVCC
+
T1
CSN
Aux
CIN
RZCV2
VCC
Q1
ZCV HV GD
LEDs
IX9908
RIN1
RIN2
1.1
VR
GND CS
CC
RCS
DVR
CVR
LED Driver Specifications
The following equations and component selections are
based on the following LED driver specifications:
Parameter
AC Input Voltage
Minimum Voltage
Symbol
Rating
VAC_min
190
Maximum Voltage
VAC_max
265
AC Input Frequency
fAC
50
Units
Vrms
Hz
Auxiliary Voltage
Vaux_max
18
V
Auxiliary Current
Iaux_max
30
mA
LED String Voltage
VLEDstring
40
V
LED String Current
ILED_max
300
mA
Estimated Efficiency
Oscillator Frequency

fS
85
100
%
kHz
Power Factor
Maximum Duty Cycle
PF
Dmax
98
50
%
%
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AN-303
INTEGRATED CIRCUITS DIVISION
2
Component Selection
2.1
DC Bulk Voltage at Low and High Line
V DC_bulk_min =
2  V AC_min =
2  190V AC  269V
V DC_bulk_max =
2  V AC_max =
2  265V AC  375V
2.2
Output Power Calculation
P out = V LEDstring  I LED_max
The breakdown voltage for this bridge is based on the
maximum input voltage.
V bridge =
P out
P in = --------------  PF
12W
= --------------------------  14.41W
0.85  0.98
The rectifier bridge is exposed to high surge currents,
so select a bridge that can handle at least five times
the Irms input current. In this case, a 0.6A bridge would
be sufficient.
Primary Inductor Currents Calculation
I pri_peak
2.8
V DS = V DC_bulk_max + V ro + V spike
Assume that Vspike = 50V.
V DS = 375V + 269V + 50V  694V
Due to the voltage spikes resulting from leakage
inductance, the MOSFET should be selected with an
adequate margin. A good choice would be between
800V and 900V.
D max
I rms = I pri_peak  ----------3
= 0.297A peak  0.5
-------  0.121A rms
3
2.9
Note that duty cycle above 50% will result in converter
stability issues such as sub-harmonic oscillations.
Reflected Flyback Voltage Calculation
V LEDstring + V f
40V + 0.7V
- = ---------------------------  269V
V ro = ---------------------------------0.1515
N
 ------S-
 N P
See 3.2 Determine Transformer Turns Ratio for
NS/NP ratio.
2.6
Calculate MOSFET Voltage Rating
269V  0.5
= ------------------------------------------------  0.297A peak
4.535mH  100kHz
For Lpri, see 3.1 Primary Inductance Calculation.
I rms
2  265V AC  1.5  562V
Select the next higher standard voltage, 600V.
2  V AC_min  D max
I pri_peak = -------------------------------------------------L pri  f S
2.5
Input Rectifier Bridge
Input Power Calculation
P in
2.4
A small, 0.1F, 630VDC polyester film filter capacitor
should be adequate for this application.
2.7
P out = 40V  300mA = 12W
2.3
follows the rectified AC line voltage, enabling a very
high power factor of up to 98% with low total harmonic
distortion (THD).
Input Capacitor, CIN
Design RCD Snubber
In a flyback topology, a snubber circuit is required to
clamp the voltage caused by the leakage inductance,
which is present in all transformers. Leakage
inductance is highly dependent on the transformer
construction, so care should be taken to keep it less
than 2% of primary inductance. When no leakage
inductance is known, a leakage inductance value of
90H (2% of 4.535mH) can be used as a starting point
to calculate the snubber’s resistor, capacitor, and diode
values.
DSN is selected as 1N4007GP (1000V, 1A axial lead).
Energy stored in the leakage inductance Le:
2
2
W1 = 0.5L e  I rms = 0.5  90H  0.121  0.66J
The IX9908 is designed to operate without the need of
a large bulk capacitor. This operating method enables
the input current to form a triangular shape that closely
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Average power transferred from W1 to the snubber:
Figure 3
P1 = W1  f S = 0.66J  100kHz  66mW
V max = V DC_bulk_max + V ro + V spike
1.05
V max = 375V + 269V + 50V = 694V
0.95
VCS vs. IZCV
1.00
0.90
VCS (V)
• Vspike = 50V
• Vro = 269V
0.85
0.80
0.75
Limiting the voltage to 694V.
0.70
V SN = 694V – 269V = 425V
0.65
2
2
V SN
425V
- = ----------------  2.7M
R SN = ----------66mW
P1
TS
10s
C SN = --------- = -----------------  3.7pF
2.7M
R SN
TS
1
1 - = ------------------------------------------  3.7pF
= ------------------C SN » --------100kHz  2.7M
R SN
f S  R SN
0.60
200
For margin, CSN =1nF is selected. This value is a
starting point, and might have to be adjusted to get the
desired voltage spike suppression.
The current equation is given by:
I ZCV
2200
R ZCV1  V ZCVOVP
R ZCV2 = -------------------------------------------V oovpth – V ZCVOVP
18.7k  3.7V
R ZCV2 = -----------------------------------  1.494k
50V – 3.7V
The delay capacitor can be approximated:
R ZCV1 + R ZCV2
C ZCV = t RC  ------------------------------------R ZC1  R ZC2
18.7k + 1.494k
C ZCV = 1s   ----------------------------------------------  723pF
 18.7k  1.494k 
Where tRC = 1s (see IX9908 data sheet).
2.11 Primary Peak Current Control
V CSmax
R CS = ----------------I pri_peak
Solving for RZCV1:
0.75V
R CS = --------------------------  2.53
0.297A peak
V DC_bulk_min N A
R ZCV1 = ------------------------------  ------I ZCV
NP
From the VCS vs. IZCV graph below we select
IZCV = 1000A as the recommended current.
For NA/NP value, see 3.2 Determine Transformer
Turns Ratio.
R01
1800
The value of the current sense resistor (RCS) can be
selected by using following equation:
V DC_bulk_min  N A
= -----------------------------------------R ZCV1  N P
269V
R ZCV1 = -------------------  0.0696  18.7k
1000A
1000
1400
IZCV (µA)
The over voltage detection can be programmed by
RZCV2 resistor. The output, VLEDstring, is 40V, so select
an over voltage protection of 50V, and from the
datasheet, VZCVOVP = 3.7V
2.10 Over-Voltage Protection & Zero-Crossing
Detection
The application schematic in Figure 2 IX9908
Application Circuit shows that the voltage from the
auxiliary winding is connected to the zero crossing pin
(ZCV), via RC network resistor RZCV1, RZCV2 and
CZCV. The circuit provides a delay so that switch-on can
occur at the voltage valley, thus enhancing efficiency.
The line voltage is sensed indirectly through the current
in RZCV1.
600
The RIN1 and RIN2 resistor values can be selected to
scale the input voltage at the VR
pin. The RIN1 resistor is selected with consideration of
losses and high power factor correction. In this
example we select RIN1 = 2040k. Select two standard
resistors, 1.00M and 1.04M, package size 1206.
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INTEGRATED CIRCUITS DIVISION
See 2.4 Primary Inductor Currents Calculation. For
adequate margin, select a 3Arms diode, such as
STPS3150V.
The RIN2 value can be calculated:
R IN1  R CS  G PWM  I pri_peak
R IN2 = -----------------------------------------------------------------------------------------------V DC_bulk_min –  G PWM  I pri_peak  R CS 
RIN2 = 17.3k. Select the next standard value, 17.4k,
where:
PWM-Op gain, GPWM = 3 (from the IX9908 data sheet).
Select the CVR capacitor to be 2.2nF and the DVR
diode can be BAS16, 100V, or equivalent.
The output capacitor can be selected based on the
desired output voltage ripple. The dynamic resistance
of the LED driven at a given current should also be
considered, as this will determine the current ripple
through the LED string.
In this application, the LEDs have a dynamic resistance
of 1 measured at 300mA operating current. 12 LEDs
result in 12 of dynamic resistance.
2.12 Selection of VCC Capacitor and Auxiliary
Blocking Diode
In this application, a simple and inexpensive power
supply for the IC is made from the auxiliary winding by
using a blocking diode and a capacitor.
V out_rip = I LED_max  R dynamic
V out_rip = 300mA  12 = 3.6V PP
The output capacitor can be approximated:
2  I LED_max
C out = -------------------------------------------------------V out_rip  2    100Hz
The auxiliary VCC diode blocking voltage can be
calculated:
2  300mA
C out = --------------------------------------------------------  265F
3.6V PP  2    100Hz
NA
V DVcc   2  V AC_max  ------- + V aux_max

N P
V DVcc   375V  0.0696  + 18V  44V
In this case, BAS16, 100V, or equivalent, would be
selected.
Select one 270F capacitor, EEV-FR1H271L. This
capacitor is rated at 105°C for 10,000 hours with
100kHz frequency.
For a dimming application a 22F capacitor would be
required: if there is no dimming requirement, then a
10F capacitor will suffice.
2.13 Output Schottky Diode and Capacitor
The Schottky output diode is exposed to large currents
when the converter is operated in critical conduction
mode, CCM. Care should be taken to ensure adequate
margins for the voltage and current ratings.
The required blocking voltage for DOUT:
NS
V d_out   2  265V AC_max  ------- + V LEDstring

N P
V d_out   375V  0.1515  + 40V  97V
See 3.2 Determine Transformer Turns Ratio for the
NS/NP ratio. Select a Schottky diode with a 150V rating.
The output diode is exposed to large peak currents.
V ro
1 – D max
I d_rms = I rms  --------------------  ----------------------------------V LEDstring + V f
D max
I d_rms  0.121A rms  1  6.609  0.8A rms
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INTEGRATED CIRCUITS DIVISION
3
Transformer Design
A step-by-step guide for transformer design will be
presented here. Primary inductance and turns ratio will
be calculated, and the appropriate wire size will be
selected. The transformer core and coil former will be
selected to support design power requirements, and a
general guideline will be presented for transformer
construction to achieve the best efficiency, and
avoiding transformer saturation at higher temperatures.
3.1
Primary Inductance Calculation
2
L pri
 V DC_bulk_min    D max 
= ----------------------------------------------------------------2  P IN  F S
3.3
The first step is to find the RMS current through the
primary winding, which can be approximated with the
following equation:
D max
I rms = I pri_peak  ----------3
I rms = 0.297A peak  0.5
-------  0.121A rms
3
The primary winding wire area can be calculated:
I rms 0.121A rms
-  0.0202mm 2
A w_pri = ----------  ----------------------J max 6A/mm 2
2
 269V  0.85  0.5 
L pri = ----------------------------------------------------  4.535mH
2  14.41W  100kHz
3.2
Determine Transformer Turns Ratio
D max
NS
V LEDstring
----------------------- = -------  --------------------N P 1 – D max
V IN
Where Jmax is the current density of a wire, which is
stated by the wire manufacturer to be 6A/mm2.
The diameter is determined by the following formula:
AW
- =
Wire Diameter = 2  ------
Secondary to primary turns ratio:
N
40V + 0.7V
------S- = ---------------------------  0.1515
NP
2  190V
The auxiliary to primary turns ratio:
V aux_max + V f
NA
------- = ----------------------------------------------------------NP
1 – D max
2  V AC_min  --------------------D max
NA
18V
+
0.7V
------- = ------------------------------------------------  0.0696
1 – 0.5
NP
2  190V  ---------------0.5
The auxiliary to secondary turns ratio:
V aux_max + V f
NA
------- = ---------------------------------NS
V LEDstring + V f
AW
AW
---------  --------------4
0.7854
2
N
V LEDstring + V f
------S- = ---------------------------------NP
2  V AC_min
Where Vf is the voltage drop of the output diode.
Determine Primary Winding Wire Size
D w_pri  0.0202mm
----------------------------  0.1604mm
0.7854
From Table 1: Wire Gauge Table this is converted to
37 SWG.
3.4
Determine Secondary Winding Wire Size
The peak current of secondary winding can be
determined by the following formula:
2  I LEDmax
 0.3A- = 1.2A
I sec_peak = --------------------------- = 2------------------peak
1 – 0.5
1 – D max
The RMS current can be found:
1 – D max
-  1.2A peak  0.408  0.490A rms
I sec_rms = I sec_peak  -------------------3
Calculate the secondary wire size:
I sec_rms 0.490A rms
2
A w_sec = -----------------  ------------------------  0.0817mm
6
6
N
18V + 0.7V
------A- = ---------------------------  0.4595
40V + 0.7V
NS
Calculate the secondary wire diameter:
2
----------------------------  0.323mm
D w_sec  0.0817mm
0.7854
Convert using Table 1: Wire Gauge Table to 29 SWG.
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AN-303
INTEGRATED CIRCUITS DIVISION
3.5
Determine Auxiliary Winding Wire Size
The maximum current in the auxiliary winding can be
calculated using:
2  I aux_max
 30mA- = 0.12A
I aux_peak = --------------------------- = 2---------------------peak
1 – 0.5
1 – D max
Table 1:
Standard Wire
Gauge
(SWG)
Equivalent
Wire Diameter
(mm)
Equivalent
Wire Area
(mm2)
15
1.829
2.6268
16
1.626
2.0755
17
1.422
1.5890
The RMS current can be calculated:
I aux_rms
1 – D max
= I aux_peak  --------------------  0.049A rms
3
The required wire area is given:
I aux_rms
2
A w_aux = ------------------  0.0082mm
6
Wire diameter:
2
----------------------------  0.102mm
D w_aux  0.0082mm
0.7854
Convert using Table 1: Wire Gauge Table to SWG 42.
8
Wire Gauge Table
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18
1.219
1.1675
19
1.016
0.8107
20
0.914
0.6567
21
0.813
0.5189
22
0.711
0.3973
23
0.610
0.2919
24
0.559
0.2452
25
0.5080
0.2027
26
0.4572
0.1642
27
0.4166
0.1363
28
0.3759
0.1110
29
0.3454
0.0937
30
0.3150
0.0779
31
0.2946
0.0682
32
0.2743
0.0591
33
0.2540
0.0507
34
0.2337
0.0429
35
0.2134
0.0358
36
0.1930
0.0293
37
0.1727
0.0234
38
0.1524
0.0182
39
0.1321
0.0137
40
0.1219
0.0117
41
0.1118
0.0098
42
0.1016
0.0081
43
0.0914
0.0066
44
0.0813
0.0052
45
0.0711
0.0040
46
0.0610
0.0029
47
0.0508
0.0020
48
0.0406
0.0013
49
0.0305
0.0007
50
0.0254
0.0005
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3.6
Calculate number of turns for the primary winding:
Select Inductor Core & Calculate Turns
The Ferroxcube catalog lists the following
specifications for the E20/10/5 core set.
L pri
2
4.535mH
--------  -----------------------  turn  193 turns
121.99nH
AL
N pri =
• (le/Ae) core factor = 1.37 mm-1
• Ve effective volume = 1340 mm3
• le effective length = 42.8 mm
Calculate the maximum flux density:
AL
B max = N pri  I pri_peak  -----Ae
• Ae effective area = 31.2 mm2
The air gap, G, of 0.3mm may be used for initial
calculation. The effective permeability of the core can
be calculated:
B max 
i
 e = ----------------------G  i
1 + --------------le
31.2mm
2
 0.224T
Note that Bmax for 1P2400, MnZn ferrite, is 0.36T at
100°C.
Calculate number of turns for secondary and auxiliary
windings:
Where i = 2000 (for N27 material or 1P2400) and
G = 0.3mm.
NS
N sec = N pri  -------  193turns  0.1515  29 turns
NP
NA
N aux = N pri  -------  193turns  0.0696  13 turns
NP
2000
 e = --------------------------------------------  133
 20001 + 0.3mm
--------------------------------42.8mm
Calculate Inductance Factor AL for this core:
Calculate to make sure windings will fit the winding
area of coil former. The Ferroxcube data sheet
provides information for 8-pin coil former and specifies
winding area of 27mm2.
o  e
A L = ---------------le
---- Ae
–7
121.99nH193turn  0.297A peak  ----------------------2
turn ----------------------------------------------------------------------------------
–3
10
4  10 H  133 ----------  121.99nH
----------------------A L  -----------------------------------------
2
2
1.37
m  turn
turn
The fill factor has to be taken into consideration as this
will affect winding area for this coil former. In general
we can use fill factor Ku of 0.3 to 0.7 as a starting point.
In this case 0.3 will be selected.
Where 0 = vacuum permeability = 4 •10-7H/m.
1
2
Total Winding Area = -------    N pri  A w_pri  +  N sec  A w_sec  +  N aux  A w_aux    27mm
Ku
2
2
2
Total Winding Area  3.33    193  0.0234mm  +  29  0.0937mm  +  13  0.0081mm    27mm
2
Total Winding Area  24.44mm  27mm
2
2
Note: The total winding area should not be exceeded due to the selection of approximate standard wire gauges
from Table 1: Wire Gauge Table.
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INTEGRATED CIRCUITS DIVISION
3.7
Flyback Transformer Construction
The transformer block diagram shows the primary split
into two sections, the first with 96 turns, the second
with 97 turns. The secondary and auxiliary are wound
between the two primary windings. The transformer
stack-up has three layers of insulation tape between
the primary and the secondary side. This method
enables the transformer to pass the safety standard for
1
+
Primary (NO0)
0.1727mm
37 SWG
96T
6
2
Primary (NO1)
0.1727mm
37 SWG
+
97T
Secondary (N2)
0.3453mm
29 SWG
3
4
+
Auxiliary (N1)
0.0081mm
42 SWG
Primary (NO1)
Basic Insulation
Auxiliary (N1)
Basic Insulation
Triple Insulation
Secondary (N2)
+
29T
electrical strength requirement. In the previous step,
we selected an air gap of 0.3mm as our starting point.
The air gap is critical because it allows the transformer
to extend its maximum saturation; however, if the gap is
larger, then it could contribute to higher leakage
inductance. In this example, it is possible to achieve a
leakage inductance less than 2% of primary.
Triple Insulation
Primary (NO0)
7
Bobbin
13T
5
Bobbin
Primary (NO0)
Coils wound
on bobbin &
insulated with
XFMR tape
Installation of
ferrite core
halves
Core halves
secured with
XFMR tape
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Specification: AN-303-R01
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