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 low power factor 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 cost and easy to implement 500W power factor 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 inductor 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 detailed derivations of all power, IC biasing and control component values and types 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 Virms = 88Vac to 264Vac, 60/50Hz

–DC regulated output voltage Vout = 400Vdc

– Full load output ripple voltage Vripple = ±8V

–Rated output power Pout = 500W

–Maximum output overvoltage Vomax = 450V

–Switching frequency fsw = 80kHz

– Maximum inductor current ripple IL = 23%

–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 Designing A High Power Factor Switching Preregulator With The L4981 Continuous Mode [1] and the corresponding Datasheet L4981A/B Power Factor Corrector [2].

November 2003

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AN827 APPLICATION NOTE

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 current, maximum surge current and thermal considerations.

Maximum peak reverse voltage:

Vprv = Virmsmax × 2 × 1.2(safety m arg in) = 264V × 2 × 1.2 = 448V

Therefore use a 600V rated diode.

Maximum forward average current:

Irmsmax	PO UT			=	500		=	6.31A
	= -----------------------	×---	--		------	0.9------
	Vrmsmin		n 88 ×
If ave =	Irmsmax × 2		=	6.31 ×		2	=	2.84A
	---------------p----------------			----------	-p-----	------

The thermal considerations require the Ifave rating to be significantly higher than the value calculated. The part chosen has a Ifave of 25A. Additionally, a small heatsink is required to keep the case temperature within specification.

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 combination. The input bridge diode’s maximum surge current rating must not be exceeded. This demoboard has a low cost and simple NTC for current inrush limiting. The efficiency can be improved 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 current condition or during normal operation. The fuse must have a current rating above the maximum 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 frequency 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

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AN827 APPLICATION NOTE

this input capacitor. However, the evaluation 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:

Cin >	Irms
	Kr 2-----×---p-----×---f--S----w----×---r----×---V-----rm-----s-----min-------
Cin > 0.25 ×	6.31	= 0.59mF
	2------×--p-----×---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 demoboard is 0.68μF.

Output Bulk Capacitor

The choice of the output bulk capacitor, C6, depends on the electrical parameters that affect the filter performance and also on the subsequent application.

Capacitance Value:

The value shall be chosen to limit the output voltage ripple according to the following formula:

Assume low ESR and Vripple= ±8V

Cout	=	Pout	Pout		= 207mF
		×---D----V----O-----×---V-----O--	= ------------------------	×----8----
	2p × 2f		2p × 120		× 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 to deliver the supply voltage for a specified time and for a specified dropout voltage, then the capacitor value will be determined by the following equation:

	2 × Pout × thold
	Cout = ---------------------------------------------	-
	Vo min2 – Vop min	2
Where:
Pout	is the maximum output power
Vomin	is the minimum output voltage at max. load
Vopmin	is the minimum operating voltage before "power fail" detection
thold	is the required hold-up time

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 NOTE

Vcap > Vout + Vripple + Vmargin = 400 + 8 + 40 = 448V

Where: Vout is the nominal regulated DC output voltage

Vripple is the ac voltage superimposed on the regulated DC output voltage

Vmargin is the allowance for tolerances in Vout and additional margin before OVP intervention

The capacitor chosen has a voltage rating of 450VDC. The overvoltage trip level of Pin 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:

BVdss > Vout + Vripple + Vmargin = 400 + 8 + 40 = 448V

The power mosfet chosen has a BVdss of 500V.

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 losses. The switching losses can be separated into two quantities, capacitive and crossover losses. The switching losses are dependent on the mosfet current di/dt. The maximum conduction (on-state) power losses can be calculated according to the following equations:

500

Pout

16 × 2 × Virms min

---------

16 × 2 × 88

IQ msm ax

=

× 2 –

=

0.9

×

2

–

h------×-------

2------V----irms------------min------

-------------3----p----×---V-----out------------------

2-----×---88----

-----3---p-----×---400-------------

IQrmsmax = 5.42A

Ponmax = IQrms 2max · R(DS)on max = 5.422 · 0.54 = 15.86W

Where:

IQrmsmax is the max. power mosfet rms current

Virmsmin is the min. specified rms input voltage

R(DS)on typ. = 0.27Ω at 25°C at 10A, VGS = 10V

R(DS)on max = 0.54Ω at 100°C

The capacitive switching losses at turn-on are calculated as follows:

P

= 3.3 × C

× V

1.5

1

× V

2

× f

= 2W

capaci tance

oss

out

+ --C

ext

sw

2

o ut

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AN827 APPLICATION NOTE

Where:

Coss = 650pF is the mosfet drain capacitance at 25V

Cext = 100pF is the equivalent stray capacitance of the layout and external parts

The estimated crossover switching losses (turnon and turn-off) are calculated as follows:

Pcrossover = Vout · IQrms · fsw · tcr + Prec = 400 · 5.42 · 80k · 40ns + 1.5 = 8.43W

Where:

tcr is the crossover time

Prec is the boost diode recovery power loss contribution

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 =	IQ1pk	× trise	=	8.92 × 40ns	=	892pF
		-------D----V-----o---ut---------			-----------400------------------

Therefore, use C11 = 820pF, 1000VDC rating

The resistors, R23-24, must dissipate the energy stored in the snubber capacitor upon turn-on of the power mosfet. The capacitor must fully discharge during the switching cycle.

The time constant of the RC combination is determined as follows:

R £

--1----

× ---

------

--1------------

= 1524

10

fsw

× C11

The power dissipated in the resistors, R23-24, is calculated as follows:

P

=

1

× V

2

× f

=

1

2

× 80k = 5.25W

diss

--C11

out

sw

-- × 820pF × 400

2

2

Therefore, use R23 = R24 = 510Ω, 3W rating.

The power mosfet chosen is the STMicoelectonics Part Number STW20NA50.

This part has a BVdss = 500V, RDSon = 0.27Ω, and is in a TO-247 package. In order to keep the junction temperature at a safe level, the mosfet is attached to an AAVID Heatsink Part Number 61085 with a thermal resis-

tance 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 (Vrrm), average forward current (Ifave), reverse recovery time (trr) and thermal considerations.

Voltage Rating:

The voltage rating of the boost diode is determined by the same equation as for the power mosfet. The value chosen is Vrrm = 600V.

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AN827 APPLICATION NOTE

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 (trr) and output voltage (Vout) . The switching losses are negligible compared to the conduction losses if a suitable ultra fast recovery diode is chosen. The conduction power losses can be calculated as follows:

	Io ut		Pout	=	500	= 1.25A
			= ----------		400---------
			Vout
IDrms	=	Pin		16 ×		2 × Vin rms min
		---------	-----------	------------------------------------------------- = 3.24A
	2V	in rms min			3	× p × Vout

Pcond = Vto × Iout + IDrms2 × Rd = 1.15 × 1.25 + 3.242 × 0.043 = 1.89W

Where:

Vto = 1.15V is the threshold voltage of the diode Rd = 0.043W is the diode differential resistance

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, trr, must be used. This circuit employs the STMicroelectronics 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 (Vrrm) of 600V, average forward current rating (Ifave) of 8A and reverse recovery time (trr) of 25ns.

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, Kr, can be considered between 10% to 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.

Vout		400
DILmax -4----×---f--SW----------×---L-	=	4-----×---80k------------×---0.5mH-----------------	= 2.50A

Occurs at Vinpk = Vout/2 = 200V; Vinrms = 141V

	DIL	=	Vinpk(Vout –		Vinpk)	For all other input voltages
			----------V----ou-------t--	×---f--sw-------	×---L------------

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