APW7128 Просмотр технического описания (PDF) - Anpec Electronics

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APW7128

30V, 1.2MHz, White LED Driver

Anpec Electronics 

APW7128

Application Information

Connecting More LED Strings

The APW7128 can drive 8 LED strings in parallel and up

to 8 LEDs per string (VF<3.5V). Each string must have the

same number of LEDs. In the applications that have the

same total number of LEDs, more strings and less LEDs

in series are more efficiency than less strings and more

LEDs in series.

Brightness Control

The method for dimming the LEDs is to apply a PWM

voltage through an RC filter into the FB pin.

The RC filter is used to convert the PWM voltage into an

analog voltage. The values of the R and C depend upon

the frequency of the PWM voltage and the amount of al-

lowable ripple voltage on FB pin. The LED current is pro-

portional to the PWM duty cycle. 0 % duty delivers maxi-

mum LED current and 100% duty delivers minimum LED

current. The values of R1 and R2 are calculated by the

following equations:

R2 =

V ×(I ×R3 + V −I ×R3 − V ) FB LED(max)

ADJ(low) LED(min)

ADJ(high)

V ×I + V ×I − V ×I − V ×I ADJ(high) LED(max)

FB LED(min)

ADJ(low) LED(min)

FB LED(max)

V FB × (1 + R 2 ) − R 2 × V ) ADJ (low

R1 =

R3 R3

I LED (max)

where:

ILED(max) is the maximum LED current

ILED(min) is the minimum LED current

VADJ(high) is the maximum PWM voltage level

VADJ(low) is the minimum PWM voltage level

V is the FB pin Voltage

FB

APW7128

FB

1

R2

R1

VADJ(high)

R3

PWM

Voltage

R4

C4

VADJ(low)

Figure 3. Dimming with the PWM Voltage

Inductor Selection

A larger value of inductor will reduce the peak inductor

current, resulting in smaller input ripple current, higher

efficiency and reducing stress on the internal MOSFET.

However, the larger value of inductor has a large

dimension, lower saturation current, and higher series

resistance.

A good rule for determining the inductance is to allow the

peak-to-peak ripple current to be approximately 30% to

50% of the maximum input current. Calculate the required

inductance value by the equation:

L

=

(V − V )× V OUT

IN

IN

∆I ×F × V L

SW

OUT

∆IL = (30% − 50%)×IIN

IIN = VOUT×ILOAD

VIN×η

It is necessary to choose an inductor that ensures the

inductor saturation current rating to exceed the peak

inductor current for the application.

To make sure that the peak inductor current is below the

current-limit 2.5A. Calculating the peak inductor current

by the following equation:

IPEAK = IIN + 0.5 × ∆IL

IPEAK = VOUT × ILOAD + (VOUT − VIN) × VIN

VIN × η

2 × L × FSW × VOUT

where

η is the efficiency

Schottky Diode Selection

A fast recovery time and low forward voltage Schottky di-

ode is necessary for optimum efficiency. Ensure that the

diode’s average and peak current rating exceed the aver-

age output current and peak inductor current. In addition,

the diode’s reverse voltage must exceed output voltage.

Capacitor Selection

An input capacitor is required to supply the ripple current

to the inductor and stabilize the input voltage. Larger

input capacitor values and lower ESR provide smaller

input voltage ripple and noise. The typical value for input

capacitor is 2.2µF to 10µF.

Copyright © ANPEC Electronics Corp.

10

Rev. A.2 - Aug., 2010

www.anpec.com.tw

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