ML4812CQX 查看數據表（PDF） - Fairchild Semiconductor

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ML4812CQX

Power Factor Controller

Fairchild Semiconductor 

PRODUCT SPECIFICATION

ML4812

The recommended maximum duty cycle is 95% at 100KHz

to allow time for the input inductor to dump its energy to

the output capacitors. For example, if: VOUT = 380V and

DON (max) = 0.95, then substituting in (3) yields VINDRY

= 20V. The effect of drying out is an increase in distortion at

low voltages.

For a given output power, the instantaneous value of the

input current is a function of the input sinusoidal voltage

waveform, i.e. as the input voltage sweeps from zero volts to

a maximum value equal to its peak so does the current.

The load of the power factor regulator is usually a switching

power supply which is essentially a constant power load. As

a result, an increase in the input voltage will be offset by a

decrease in the input current.

By combining the ideas set forth above, some ground rules

can be obtained for the selection and design of the input

inductor:

Step 1: Find minimum operating current.

IIN(min)PEAK = 1----.-4---1-V--4---I--×N----(P--m--I--N-a---(x--m--)---i--n----)

(4)

VIN(max) = 260V

PIN(min) = 50W

then:

IIN(min)PEAK = 0.272A

Gapped Ferrites, Molypermalloy, and Powdered Iron cores

are typical choices for core material. The core material

selected should have a high saturation point and acceptable

losses at the operating frequency.

One ferrite core that is suitable at around 200W is the

#4119PL00-3C8 made by Philips Components (Ferroxcube).

This ungapped core will require a total gap of 0.180" for this

application.

Oscillator Component Selection

The oscillator timing components can be calculated by using

the following expression:

fOSC = -R----T1---.-×-3---6-C----T-

(6)

For example:

Step 1: At 100kHz with 95% duty cycle TOFF = 500ns

calculate CT using the following formula:

CT

=

T----O----F---F----×-----I--D----I--S-

VOSC

=

1000pF

(7)

Step 2: Calculate the required value of the timing resistor.

(8)

RT = f---O----S-1--C-.--3-×--6---C----T-- = 1----0---0---K-----H--1--z-.--3-×--6---1---0---0----p---F-- = 13.6kΩ

choose RT = 14kΩ

Step 2: Choose a minimum current at which point the

inductor current will be on the verge of drying out. For this

example 40% of the peak current found in step 1 was chosen.

then:

ILDRY = 100mA

Step 3: The value of the inductance can now be found using

previously calculated data.

(5)

L1 = V-----I--N----ID--L-R--D--Y--R---×Y-----D×----O-f---ON---S-(--mC-----a---x----) = 1----0---0---m2---0--A-V-----×-×----1-0--0-.-9-0---5K-----H----z- = 2mH

The inductor can be allowed to decrease in value when the

current sweeps from minimum to maximum value. This

allows the use of smaller core sizes. The only requirement is

that the ramp compensation must be adequate for the lower

inductance value of the core so that there is adequate com-

pensation at high current.

Step 4: The presence of the ramp compensation will change

the dry out point, but the value found above can be consid-

ered a good starting point. Based on the amount of power

factor correction the above value of L1 can be optimized

after a few iterations.

Current Sense and Slope (Ramp)

Compensation Component Selection

Slope compensation in the ML4812 is provided internally.

Rather than adding slope to the noninverting input of the

PWM comparator, it is actually subtracted from the voltage

present at the inverting input of the PWM comparator. The

amount of slope compensation should be at least 50% of the

downslope of the inductor current during the off time, as

reflected to the inverting input of the PWM comparator. Note

that slope compensation is required only when the inductor

current is continuous and the duty cycle is more than 50%.

The downslope of the inductor current at the verge of

discontinuity can be found using the expression given below:

-d---i--L- = V-----O----U----T----–----V-----I--N----D---R----Y-- = 3----8---0---V------–----2---0----V-- = 0.18A ⁄ µs

(9)

dt

L

2mH

The downslope as reflected to the input of the PWM

comparator is given by:

SPWM

=

-V----O----U----T----–----V-----I--N----D---R----Y--

L

=

-R----S--

NC

(10)

SPWM

=

3----8---0---V------–----2---0----V--

2mH

×

1----0---0-

80

=

0.225V ⁄ µs

REV. 1.0.4 5/31/01

9

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