ML4819CP データシートの表示（PDF） - Fairchild Semiconductor

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ML4819CP

Power Factor and PWM Controller “Combo”

Fairchild Semiconductor 

The value of R9 (pin 3) depends on the selection of R2

(pin 6).

R2

=

VIN(MAX)PEAK

ISINE (PEAK)

=

260 × 1.414

0.72mA

= 510kΩ

(12)

R9

>

VCLAMP × R2

VIN(MIN)PEAK

=

4.8 × 510kΩ

80 × 1.414

= 22kΩ

(13)

Choose R9 = 27kΩ

The peak of the inductor current can be found

approximately by:

ILPEAK

= 1.414 × POUT

VIN (MIN) RMS

= 1.414 × 200

90

= 3.14A

(14)

Next select NC, which depends on the maximum switch

current. Assume 4A for this example. NC is 80 turns.

R11 =

VCLAMP × NC

ILPEAK

=

4.9 × 80

4

= 100Ω

(15)

Where R11 is the sense resistor, and VCLAMP is the current

clamp at the inverting input of the PWM comparator. This

clamp is internally set to 5V. In actual application it is a

good idea to assume a value less than 5V to avoid

unwanted current limiting action due to component

tolerances. In this application VCLAMP was chosen as 4.8V.

Having calculated R11 the value SPWM and of R18 can

now be calculated:

SPWM

=

380V − 20

2mH

×

100

80

=

0.225V

/

µs

R18

=

ASC

×

2.5 × R9

SPWM × RT

×

CT

R18

=

0.7

2.5 × 28.8k

× (0.225 × 106) × 14kΩ

× 1nF

≅

30kΩ

(16)

Choose R18 = 33kΩ

The following values were used in the calculation:

R9 = 27kΩ

RT = 14kΩ

ASC = 0.7

CT = 1nF

ML4819

VOLTAGE REGULATION COMPONENTS

The values of the voltage regulation loop components are

calculated based on the operating output voltage. Note

that voltage safety regulations require the use of sense

resistors that have adequate voltage rating. As a rule of

thumb if 1/4W through-hole resistors are used, two of

them should be put in series. The input bias current of the

error amplifier is approximately 0.5µA, therefore the

current available from the voltage sense resistors should

be significantly higher than this value. Since two 1/4W

resistors have to be used the total power rating is 1/2W.

The operating power is set to be 0.4W then with 380V

output voltage the value can be calculated as follows:

R5 = (380V)2 / 0.4W = 360kΩ

(17)

Choose two 178kΩ, 1% connected in series.

Then R6 can be calculated using the formula below:

R6

=

VREF × R5

VB − VREF

=

5V × 356kΩ

380V − 5V

=

4.747kΩ

(18)

Choose 4.75kΩ, 1%. One more critical component in the

voltage regulation loop is the feedback capacitor for the

error amplifier. The voltage loop bandwidth should be set

such that it rejects the 120Hz ripple which is present at

the output. If this ripple is not adequately attenuated it

will cause distortion on the input current waveform.

Typical bandwidths range anywhere from a few Hertz to

15Hz. The main compromise is between transient

response and distortion. The feedback capacitor can be

calculated using the following formula:

C8

=

1

3.142 × R5

× BW

C8

=

1

3.142 × 356kΩ

×

2Hz

=

0.44µF

(19)

REV. 1.0 10/10/2000

11

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