A8519KETTR-R datasheet(29/35 Pages) ALLEGRO

29 / 35 page

Wide Input Voltage Range, High-Efficiency,

Fault-Tolerant LED Driver

A8519 and

A8519-1

Allegro MicroSystems, LLC

115 Northeast Cutoff

Worcester, Massachusetts 01615-0036 U.S.A.

1.508.853.5000; www.allegromicro.com

Step 4a: Determine the Duty Cycle.

D = 1 –

MAX

+ V )

OUT(ovp)

IN(min)

D = 1 –

MAX

(39.9 + 0.4)

= 0.75

Step 4b: Determine the maximum and minimum input current to

the system. The minimum input current will dictate the inductor

value. The maximum current rating will dictate the current rating

of the inductor.

IN(max)

× I

OUT(ovp)

OUT

IN(min)

I = #Channels × I

I = 4 × 0.060A = 0.240 A

OUT

LED

A good approximation of efficiency η can be taken from the

efficiency curves located on page 10. A value of 90% is a good

starting approximation.

IN(max)

10 V × 0.90

39.9 V × 240 mA = 1.06 A

IN(min)

14 V × 0.90

32.85 V × 240 mA = 0.625 A

V = 10 × 3.2 V + 0.85 V = 32.85 V

OUT

V × I

OUT

IN(max)

Step 4c: Determining the inductor value. To ensure that the

inductor operates in continuous conduction mode, the value of

the inductor needs to be set such that the ½ inductor ripple cur-

rent is not greater than the average minimum input current. A

first pass calculation for Kripple should be 30% of the maximum

inductor current.

I = I

× K

IN(max)

ripple

I = 1.06A × 0.3 = 0.318 A

L = 0.318 A × 2 MHz

10 V

× 0.75 = 11.79 µH

L =

I × f )

× D )

IN(min)

MAX

Double-check to make sure that ½ current ripple is less than

IIN(min).

IIN(min) > ½ DIL

0.625 A > 0.159 A

A good inductor value to use would be 10 µH.

Step 4d: This step is used to verify that there is sufficient slope

compensation for the inductor chosen. 6 A/µs slope compensation

value is applied inside the IC at 2 MHz switching frequency. The

slope compensation at any switching frequency can be deter-

mined by the following formula:

Slope Comp =

6 A/µs × f

2 × 10 6

Next, insert the inductor value used in the design:

ΔI

L(used)

× D

IN(min)

MAX

L(used) × f

ΔI

L(used)

= 0.375 A

10 V × 0.75

10 µH × 2 MHz

Required Min Slope =

ΔI

× ΔS × 10

L(used)

-6

× (1 – D )

MAX

where ΔS is taken from the following formula:

ΔS = 1 – 0.18

MAX

ΔS = 0.76

= 2.28 A/µs

Required Min Slope =

0.375 × 0.76× 10 -6

2 MHz × (1– 0.75)

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