A6211 ALLEGRO [Allegro MicroSystems], A6211 Datasheet - Page 10

• Manufacturer Part Number: A6211
• Description: Constant-Current 3-Ampere PWM Dimmable Buck Regulator LED Driver
• Manufacturer: ALLEGRO [Allegro MicroSystems]

A6211
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 R
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-
Figure 10. A6211 overcurrent protection tripped in the case of a fault
caused by the sense resistor pin shorted to ground; shows switch node,
V
i
LED
C1
C2
C3
SW
(ch3, 1 A/div.), t = 100 μs/div.
(ch1, 10 V/div.), output voltage, V
V
V
i
LED
OUT
SW
DS(on)
= 0.4 Ω and a duty cycle of 85%, then
t
OUT
(ch2, 10 V/div.), LED current,
Constant-Current 3-Ampere PWM Dimmable
Figure 11. Inductance selection based on I
V
eters for the inductor:
1. Determine the saturation current of the inductor. This can be
2. Determine the ripple current amplitude (peak-to-peak value). As
3. Calculate the inductance based on the following equations:
where
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 f
minimum inductance required is L = 10 μH in order to keep the
ripple current at 30% or lower. (Note: V
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.
OUT
D is the duty cycle,
T is the period 1/ f
V
done by simply adding 20% to the average LED current:
a general rule, ripple current should be kept between 10% and
30% of the average LED current:
D1
D1 (see figure 7).
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
= 12 V, ripple current = 30%
0
is the forward voltage drop of the Schottky diode
0.0
i
0.1 < i
L = (V
D = (V
SAT
Buck Regulator LED Driver
≥ i
RIPPLE(pk-pk)
IN
OUT
LED
0.5
– V
+ V
SW
× 1.2 .
OUT
, and
D1
) × D × T / i
1.0
) / ( V
115 Northeast Cutoff
1.508.853.5000; www.allegromicro.com
Allegro MicroSystems, Inc.
Worcester, Massachusetts 01615-0036 U.S.A.
/ i
L=47 μH
LED
LED Current, I
IN
L=33 μH
< 0.3 .
+ V
1.5
L=22 μH
RIPPLE
LED
D1
L=15 μH
OUT
) ,
LED
and f
SW
L=10 μH
2.0
=1 MHz, then the
= V
, and
(A)
SW
; V
IN
IN
/ 2 is the worst
= 24 V,
2.5
3.0
10