ST AN827 Application note
AN827
APPLICATION NOTE
A 500W HIGH POWER FACTOR WITH THE L4981A
CONTINUOUS MODE IC
The widespread use of passive AC/DC off-line converters causes l ow power fac tor and high line current
harmonic distortion. To reduce these phenomena and to comply with relevant regulatory agency requirements , designers are employing active power factor correction in their off-line SMPS applications. This
paper describes a practical, low c ost and easy to i mpl ement 500W power fac tor corrected application
that employs the L4981A Continuous Mode PFC IC.
INTRODUCTION
Reduction of line current harmonic distortion and improvement of power factor is of great concern to many designers of off-line switched mode power supplies. This concern has been motivated by present and impending
regulatory requirements regarding line current harmonics. The reasons for improving power factor and reducing
line current harmonic distortion are well known and understood. Active power factor correction using the boost
topology and operating in the continuous i nductor current control mode is an excellent method to comply with
these requirements and is well accepted in the industry.
This paper will present a practical power factor corrected design for a 500 Watt output and universal mains input
application. The detail ed derivations of al l power, I C biasing and control component values and t ypes will be
shown. The evaluation results from an actual working demoboard will be presented as well as several relevant
oscillograms.
DESIGN SPECIFICATIONS
The design specifications given below are realized by the implementation of a functional demoboard.
The design target specifications are as follows:
– Universal mains input AC voltage V
– DC regulated output voltage V
– Full load output ripple voltage ∆V
– Rated output power P
– Maximum output overvoltage V
– Switching frequency f
– Maximum inductor current ripple ∆I
– Input power factor PF > 0.99
– Input line current total harmonic distortion <5%
To meet these specifications, the selection of component values and material types is very important. The next
sections will describe the component selection criteria along with some critical derivations. For detailed explanations on the controller operation and pin description, refer to Application
Note AN628
the corresponding Datasheet
Designing A High Power Factor Switching Preregulator With The L4981 Continuous Mode [1]
out
sw
L4981A/B Power Factor Corrector [2].
out
= 500W
omax
= 80kH z
= 88Vac to 264Vac, 60/50Hz
irms
= 400Vdc
= ±8V
ripple
= 450V
= 23%
L
and
November 2003
1/20
AN827 APPLICATION NO TE
POWER COMPONENTS SELECTION
The power component values and types are derived and selected in the next section. Please refer to Figure 2,
500 Watt Demoboard Schematic.
Input Diode Bridge
The input diode bridge, D1, can be a standard slow-recovery type. The selection criteria include the maximum
peak reverse breakdown voltage, maximum forward average c urrent, maximum surge current and thermal considerations.
Maximum peak reverse voltage:
V
prvVirmsmax
2 1.2 safety m inarg()264V 2 1.2 448V=⋅⋅=⋅⋅=
Therefore use a 600V rated diode.
Maximum forward average current:
I
rmsmax
I
fave
The thermal considerations require the I
chosen has a I
of 25A. Additionally, a small heatsink is required to keep the case temperature within speci-
fave
P
OUT
-----------------------------
V
rmsm in
I
rmsmax
--------------------------------
rating to be significantly higher than the value calculated. The part
fave
2⋅
π
500
------------------- 6.31A===
n⋅
88 0.9⋅
6.31 2⋅
----------------------- 2.84A===
π
fication.
Maximum surge current:
There is a significant inrush current at start-up due to the large value bulk capacitor, C6, at the output. There is
minimal impedance from the mains to this capacitor, thus at the peak of the input voltage waveform a large inrush current exists. This inrush current can be significantly reduced by some means of current limiting such as
an NTC or triac/resistor combinati on. The input bridge diode’ s maxi mum surge current rating must not be exceeded. This demoboard has a low cost and simple NTC for current inrush limit ing. The effi ciency c an be i mproved by using the triac/resistor scheme, however the cost and complexity increases.
Input Fuse
The input fuse, F1, must open during severe current overloads without tripping during the transient inrush cur-
rent condition or during normal operation. The fuse must have a current rating above the m aximum continuous
current (6.3Arms) that occurs at the low line voltage (88V). The fuse chosen for this demoboard has a continuous current rating of 10A/250VAC.
Input Filter Capacitor
The input filter capacitor, C3, is placed across the diode bridge output. This capacitor must smooth the high fre-
quency ripple and must sustain the maximum instantaneous input voltage. In a typical application an EMI filter
will be placed between the mains and the PFC circuit. This demoboard does not have the EMI filter except for
2/20
AN827 APPLICATION NOTE
this input capacitor. However, the eval uat i on results listed in Table 1 were made with an EMI filter placed between the mains input and the PFC circuit. The design of the EMI filter is not described here. The value of the
input filter capacitor can be calculated as follows:
I
Cin Kr
Cin 0.25
---------------------------------------------------------->
2 π fSwrV
⋅⋅ ⋅⋅
---------------------------------------------------- - 0.59µF=⋅>
2 π 80k 0.06 88⋅⋅ ⋅ ⋅
Where:
Kr is the current ripple coefficient r = 0.02 to 0.08
The maximum value of this capacitor is limited to avoid line current distortion. The value chosen for this demo-
board is 0.68
µ
F.
Output Bulk Capacitor
The choice of the output bulk capacitor, C6, depends on the electri cal parameters that affect the fi lter perf or-
mance and also on the subsequent application.
Capacitance Value:
rms
rms m in
6.31
The value shall be chosen to limit the output voltage ripple according to the following formula:
Assume low ESR and
∆
V
= ±8V
ripple
P
C
--------------------------------------------- -
out
2π 2f ∆VOV
out
⋅⋅ ⋅
O
P
out
------------------------------------------- 207 µF===
2π 120 8 400⋅⋅⋅
The value chosen is 330uf to ensure that the maximum specified voltage ripple is not exceeded.
Although the ESR does not normally affect the voltage ripple, it has to be considered for the power losses due
to the line and switching frequency ripple currents. It is important to verify that the low and high frequency ripple
currents do not exceed the manufacturer’s specified ratings at the operating case temperature. Capacitors may
be connected in parallel to decrease the equivalent ESR and to increase the ripple current handling capability.
If a specific hold-up time is required, that is the capacitor has t o deliver the suppl y voltage for a specifi ed time
and for a specified dropout voltage, then the capacitor value will be determined by the following equation:
2P
⋅⋅
C
out
----------------------------------------------=
V
o min
outthold
2
V
–
op min
2
Where:
P
is the maximum output power
out
V
is the minimum output voltage at max. load
omin
V
is the minimum operating voltage before "power fail" detection
opmin
t
is the required hold-up time
hold
Voltage Rating:
The capacitor output voltage rating should not be exceeded under worst case conditions. The minimum voltage
rating is calculated as follows:
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AN827 APPLICATION NO TE
V
> V
cap
out
+ ∆V
ripple
+ V
= 400 + 8 + 40 = 448V
margin
Where: V
is the nominal regulated DC output voltage
out
∆
V
is the ac voltage superimposed on the regulated DC output voltage
ripple
∆
V
is the allowance for tolerances in V
margin
and additional margin before OVP intervention
out
The capacitor chosen has a voltage rating of 450VDC. The overvoltage trip level of P in 3 (OVP) must be set
below 450VDC.
Power Mosfet
The power mosfet, Q1, is used as the active switch due to its high frequency capability, ability to be driven directly from the controller and availability. The main criteria for its selection include the drain to source breakdown
voltage (BVdss), delivered power and temperature considerations.
Voltage Rating:
The power mosfet has to sustain the maximum boosted output dc voltage according to the following equation:
> V
BV
dss
The power mosfet chosen has a BV
+ ∆V
out
of 500V.
dss
ripple
+ V
= 400 + 8 + 40 = 448V
margin
Power Rating:
The main parameters to consider are Rdson and the thermal characteristics of the package and heatsink. The
main losses in the power mosfet are the conduction and switching l osses. The swi tching l osses can be separated into two quantities , c apaciti ve and c rossover l osses. The swit ching l osses are dependent on t he mosfet
current di/dt. The maximum conduction (on-state) power losses can be calculated according to the following
equations:
500
--------- -
0.9
------------------ 2
288⋅
16 2 88⋅⋅
-----------------------------–⋅=⋅=
3π400⋅
I
Qrmsmax
P
onmax
= 5.42A
= I
Qrms
I
Qmsmax
2max · R
P
out
----------------------------------------- 2
η 2 V
⋅
(DS)on max
= 5.422 · 0.54 = 15.86W
irms min
16 2 V
⋅⋅
-----------------------------------------------–
⋅
3πV
irms min
out
Where:
I
max is the max. power mosfet rms current
Qrms
V
min is the min. specified rms input voltage
irms
R
on typ. = 0.27Ω at 25°C at 10 A, VGS = 10V
(DS)
R(
on max = 0.54Ω at 100°C
DS)
The capacitive switching losses at turn-on are calculated as follows:
P
capaci c etan
1.5
3.3 C
ossVout
1
-- - C
extVout
2
2
⋅+⋅
f
2W=⋅⋅=
sw
4/20
AN827 APPLICATION NOTE
Where:
C
= 650pF is the mosfet drain capacitance at 25V
oss
C
= 100pF is the equivalent stray capacitance of the layout and external parts
ext
The estimated crossover switching losses (turnon and turn-off) are calculated as follows:
P
crossover
= V
out
· I
Qrms
· fsw · t
+ Prec = 400 · 5.42 · 80k · 40ns + 1.5 = 8.43W
cr
Where:
is the crossove r tim e
t
cr
P
is the boost diode recovery power loss contribution
rec
To reduce the turn-off losses in the mosfet, an RCD turn-off snubber has been employed. The capacitor value
is calculated as follows:
C11
I
Q1pktrise
-----------------------------
∆V
out
8.92 40 ns⋅
----------------------------- - 892pF===
400
⋅
Therefore, use C11 = 820pF, 1000VDC rating
The resistors, R23-24, must dissipate the energy stored in the snubber capacitor upon t urn-on of the power
mosfet. The capacitor must fully disc harg e during the sw itching cycl e.
The time constant of the RC combination is determined as follows:
1
1
------
10
----------------------- - 1524=⋅≤
C11⋅
f
sw
R
The power dissipated in the resistors, R23-24, is calculated as follows:
P
diss
1
-- - C11 V
2
out
2
1
-- - 820p F 400
f
sw
2
2
80 k 5.25W=⋅⋅⋅=⋅⋅=
Therefore, use R23 = R24 = 510
Ω
, 3W rating.
The power mosfet chosen is the STMicoelectonics Part Number STW20NA50.
This part has a BV
= 500V, R
dss
= 0.27Ω, and is in a TO-247 package. In order to keep the junction tem-
DSon
perature at a safe level, the mosfet is attached to an AAVID Heatsink Part Num ber 61085 with a thermal resistance of 3.0°C/W. This will keep the mosfet junction temperature at a safe level at worst case conditions, lowline input voltage (88V) and full load (500W).
The thermal resistance of the heatsink may need to decrease depending upon the ambient temperature, type
of enclosure (vented or non-vented) and the method of cooling (natural or forced convection).
Boost Diode
The main criteria for the selection of the boost diode, D2, include the repetitive peak reverse breakdown voltage
(V
), average forward current (I
rrm
), reverse recovery time (trr) and thermal considerations.
fave
Voltage Rating:
The voltage rating of the boost diode i s determined by the same equati on as for t he power mosfet. T he value
chosen is V
= 600V.
rrm
5/20
AN827 APPLICATION NO TE
Current Rating:
The power losses in the boost diode consist of the conduction and switching losses. The switching losses are
a function of the reverse recovery ime (t
pared to the conduction losses if a suitable ultra fast recovery diode is chosen. The conduction power losses
can be calculated as follows:
) and output voltage (V
rr
) . The switching losses are negligible com-
out
I
I
Drms
P
condVtoIoutIDrms
----------------------------------
2V
out
P
in
in rms min
2
P
-----------
V
out
out
Rd 1.15 1.25 3.2420.043 1.89W=⋅+⋅=⋅+⋅=
500
--------- - 1.25A===
400
⋅⋅
16 2 V
------------------------------------------------- 3.24A==
3πV
⋅⋅
in rms min
out
Where:
V
= 1.15V is the threshold voltage of the diode Rd = 0.043W is the diode differential resistance
to
The diode must sustain the average output current and also keep the power losses to a minimum in order to
keep the diode junction temperature within acceptable limits. The switching losses can be significantly reduced
if an ultra-fast diode is employed. Since this circuit operates in the continuous current mode, the mosfet has to
recover the boost diode minority carrier charge at turn-on.
Thus, a diode with a small reverse recover time, t
, must be used. This circuit employs the STMicroelectronics
rr
Turboswitch Diode Part Number STTA806D. This part offers the best solution for the continuous current mode
operation due to its very fast reverse recovery time, 25ns typical. This part has a breakdown voltage rating (V
of 600V, average forward current rating (I
) of 8A and reverse recovery time (trr) of 25ns.
fave
rrm
The diode is attached to the same heatsink as the power mosfet, Q1. The STTA806D is non-isolated thus requiring a thermal insulator with good heat transfer characteristics. The STTA806DI is an isolated package and
can be attached directly to the heatsink. Silicone thermal grease may be applied to improve the thermal contact
between the diode and heatsink.
)
Boost Inductor
The boost inductor, T1, design starts with defining the minimum inductance value, L, to limit the high frequency
current ripple,
∆
IL. The next step is to define the number of turns, air gap length, ferrite core geometry, size and
type for the specified power level. Finally, the wire size and type are determined.
In the continuous mode approach, the acceptable current ripple factor, K
, can be considered between 10% to
r
35%. For this design, the maximum specified current ripple factor is 23%. The maximum current ripple occurs
when the peak of the input voltage is equal to Vout/2.
Occurs at V
6/20
inpk
= V
= 200V; V
out/2
V
∆I
∆I
L
-------------------------
Lmax
4fSWL⋅⋅
= 141V
inrms
V
inpkVoutVinpk
--------------------------------------------------=
V
outfsw
out
–()
L⋅⋅
400
---------------------------------------- - 2.50A==
4 80k 0.5mH⋅⋅
For all other input voltages
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