RL1632R-R150-F datasheet(10/17 Pages) ALLEGRO

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Constant-Current 3-Ampere PWM Dimmable

Buck Regulator LED Driver

A6211

Allegro MicroSystems, Inc.

115 Northeast Cutoff

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

1.508.853.5000; www.allegromicro.com

Thermal Budgeting

The A6211 is capable of supplying a 3 A current through its

high-side switch. However, depending on the duty cycle, the

conduction loss in the high-side switch may cause the package to

overheat. Therefore care must be taken to ensure the total power

loss of package is within budget. For example, if the maximum

temperature rise allowed is ∆T = 50 K at the device case surface,

then the maximum power dissipation of the IC is 1.4 W. Assum-

ing the maximum RDS(on) = 0.4 Ω and a duty cycle of 85%, then

the maximum LED current is limited to 2 A approximately. At a

lower duty cycle, the LED current can be higher.

Fault Handling

The A6211 is designed to handle the following faults:

• Pin-to-ground short

• Pin-to-neighboring pin short

• Pin open

• External component open or short

• Output short to GND

The waveform in figure 10 illustrates how the A6211 responds

in the case in which the current sense resistor or the CS pin is

shorted to GND. Note that the SW pin overcurrent protection is

tripped at around 3.75 A, and the part shuts down immediately.

The part then goes through startup retry after approximately

380 μs of cool-down period.

Component Selections

The inductor is often the most critical component in a buck con-

verter. Follow the procedure below to derive the correct param-

eters for the inductor:

1. Determine the saturation current of the inductor. This can be

done by simply adding 20% to the average LED current:

iSAT ≥ iLED × 1.2 .

2. Determine the ripple current amplitude (peak-to-peak value).As

a general rule, ripple current should be kept between 10% and

30% of the average LED current:

0.1 < iRIPPLE(pk-pk) / iLED < 0.3 .

3. Calculate the inductance based on the following equations:

L = (VIN – VOUT) × D × T / iRIPPLE , and

D = (VOUT + VD1) / ( VIN + VD1 ) ,

where

D is the duty cycle,

T is the period 1/ fSW, and

VD1 is the forward voltage drop of the Schottky diode

D1 (see figure 7).

Inductor Selection Chart

The chart in figure 11 summarizes the relationship between LED

current, switching frequency, and inductor value. Based on this

chart: Assuming LED current = 2 A and fSW =1 MHz, then the

minimum inductance required is L = 10 μH in order to keep the

ripple current at 30% or lower. (Note: VOUT = VIN / 2 is the worst

case for ripple current). If the switching frequency is lower, then

either a larger inductance must be used, or the ripple current

requirement has to be relaxed.

Figure 11. Inductance selection based on ILED and fSW; VIN = 24 V,

VOUT = 12 V, ripple current = 30%

Figure 10. A6211 overcurrent protection tripped in the case of a fault

caused by the sense resistor pin shorted to ground; shows switch node,

VSW (ch1, 10 V/div.), output voltage, VOUT (ch2, 10 V/div.), LED current,

iLED (ch3, 1 A/div.), t = 100 μs/div.

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

LED Current, ILED

(A)

L=15 μH

L=22 μH

L=33 μH

L=47 μH

L=10 μH

VSW

VOUT

iLED

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