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参数资料
| 型号： | A6210GEUTR-T |
| 厂商： | Allegro Microsystems Inc |
| 文件页数： | 6/12页 |
| 文件大小： | 0K |
| 描述： | IC LED DRIVER HIGH BRIGHT 16-QFN |
| 特色产品： | A6210 Buck-Regulating LED Driver |
| 标准包装： | 1 |
| 拓扑： | PWM，降压（降压），切换式电容器（充电泵） |
| 输出数： | 1 |
| 内部驱动器： | 是 |
| 类型 - 主要： | 通用 |
| 类型 - 次要： | 高亮度 LED（HBLED） |
| 频率： | 2MHz |
| 电源电压： | 9 V ~ 46 V |
| 输出电压： | 46V |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-WSQFN 裸露焊盘 |
| 供应商设备封装： | 16-QFN-EP（4x4） |
| 包装： | 标准包装 |
| 工作温度： | -40°C ~ 105°C |
| 产品目录页面： | 1141 (CN2011-ZH PDF) |
| 其它名称： | 620-1281-6 |
A6210
3 A, 2 MHz Buck-Regulating LED Driver
of physically small, low value inductors. An output capacitor is
not necessary either to reduce the ripple current or to close the
control loop.
High efficiencies are achieved via drive circuits optimized to
minimize switching losses and the current sense voltage has a
typical voltage drop of only 183 mV. The current in the LED
string can be pulse width modulated (PWM) via the DIS (Dis-
able/Enable) pin. See figure 4.
The actual current control is maintained on the valley of the cur-
rent ripple. The average LED current is the valley level plus half
the inductor ripple current, as shown in figure 2.
To avoid potential mistriggering issues, it is recommended that
the ripple current that flows through the sense resistor (R2) does
Note: V f is the forward voltage drop of the recirculation diode
and sense resistor (R2). The valley current is determined by the
sense voltage (183 mV) divided by the sense resistor.
Worked example
This example uses the brief specification outlined in the typical
application circuit on page 1. The following information is used
as a starting point:
V IN = 24 V ,
3 LEDs producing V LED = 12 V ,
I LED = 500 mA, and
I av = I VALLEY + ,
not develop a ripple voltage of less than 20 mV.
The average LED current can be found from:
I RIPPLE
2
substituting values:
(4)
LED ripple current, I RIPPLE = 60 mA .
The duty cycle can be found initially. Assume the forward voltage
drop of the re-circulation diode is 400 mV, and that the sense
resistor is 183 mV. Then:
+ ? ?
?
?
where:
I av =
183 mV
R2
? 1
2
×
V IN – V LED
L
× t on
?
?
?
,
(5)
D =
V LED + V f 12 + 0.58
V IN + V f = 24 + 0.58 =
0.39
.
(7)
t on
V LED + V f
= V IN + V f ×
1
f SW
.
(6)
One of the objectives is to maximize the switching frequency to
minimize the inductor value. When driving at very high switching
frequencies, the duty cycle may be limited due to the minimum
2000
1800
t on + t off = 1/ f SW
1 / I
1600
1400
1 LED
2 LEDs
3 LEDs
4 LED s
t on
t off
2 RIPPLE
5 LEDs
1200
1000
800
600
0
Valley Current
Average
LED Current
400
10 4
10 5
10 6
Time
Resistor, R1 (k Ω )
Figure 1. Switching frequency versus value of external resistor R1 on the
TON pin.
Figure 2. Current control
Allegro MicroSystems, LLC
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
6
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