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| 型号： | A8502KLPTR-T |
| 厂商： | Allegro Microsystems Inc |
| 文件页数： | 31/35页 |
| 文件大小： | 0K |
| 描述： | IC LED DVR WHITE MULTI 16-TSSOP |
| 标准包装： | 8,000 |
| 拓扑： | PWM，升压（升压） |
| 输出数： | 2 |
| 内部驱动器： | 是 |
| 类型 - 主要： | 车载，背光 |
| 类型 - 次要： | 白色 LED |
| 频率： | 580kHz ~ 2MHz |
| 电源电压： | 5 V ~ 40 V |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-TSSOP（0.173"，4.40mm）裸露焊盘 |
| 供应商设备封装： | 16-TSSOP-EP |
| 包装： | 带卷 (TR) |
| 工作温度： | -40°C ~ 125°C |
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A8502
Wide Input Voltage Range, High Efficiency
Fault Tolerant LED Driver
Step 4 Selecting the inductor. The inductor must be chosen such
that it can handle the necessary input current. In most applica-
inductor must be set such that the ? inductor ripple current is not
greater than the average minimum input current. As a first pass
tions, due to stringent EMI requirements, the system must operate
in continuous conduction mode throughout the whole input volt-
age range.
assume I ripple to be 30% of the maximum inductor current:
Δ I L = I IN (max) × I ripple
(39)
V OUT(OVP) + V d
V IN (min)
Step 4a Determining the duty cycle, calculated as follows:
D (max) =
V IN (min) + V OUT(OVP) + V d
(35)
then:
= 0.848 × 0.30 = 0.254 A
L = D (max)
Δ I L f SW
(40)
=
15.9 (V) + 0.4 (V)
5 (V) + 15.9 (V) + 0.4 (V)
= 76.5%
=
5 (V)
0.254 (A) 2 (MHz)
0.765 = 7.53 μ H
Step 4b Determining the maximum and minimum input current
to the system. The minimum input current will dictate the induc-
tor value. The maximum current rating will dictate the current
rating of the inductor. First, the maximum input current, given:
I OUT = # CHANNELS I LED (36)
Step 4d Double-check to make sure the ? current ripple is less
than I IN (min):
I IN (min) > 1 / 2 Δ I L (41)
0.265 A > 0.127 A
A good inductor value to use would be 10 μ H.
then:
= 2
0.120 (A) = 0.240 A
Step 4e Next insert the inductor value used in the design to deter-
mine the actual inductor ripple current:
I IN (max) =
=
V OUT(OVP) I OUT
V IN (min)
15.9 (V) 0.24 (A)
5 (V) 0.90
= 0.848 A
(37)
Δ I Lused =
=
V IN (min) D (max)
L used f SW
5 (V) 0.765
10 ( μ H) 2.0 (MHz)
= 0.191 A
(42)
where η is efficiency.
Step 4f Determining the inductor current rating. The inductor
I IN (min) =
Next, calculate minimum input current, as follows:
V OUT(OVP) I OUT
V IN (max)
(38)
current rating must be greater than the I IN (max) value plus half of
the ripple current Δ I L , calculated as follows:
L (min) = I IN (max) + 1 / 2 Δ I Lused (43)
=
15.9 (V)
16 (V)
0.24 (A)
0.90
= 0.265 A
= 0.848 (A) + 0.096 (A) = 0.944 A
Step 5 Determining the resistor value for a particular switching
A good approximation of efficiency, η , can be taken from the
efficiency curves located in the datasheet. A value of 90% is a
good starting approximation.
Step 4c Determining the inductor value. To ensure that the induc-
tor operates in continuous conduction mode, the value of the
frequency. Use the R FSET values shown in figure 7. For example,
a 10 k Ω resistor will result in a 2 MHz switching frequency.
Step 6 Choosing the proper switching diode. The switching diode
must be chosen for three characteristics when it is used in LED
lighting circuitry. The most obvious two are: current rating of the
diode and reverse voltage rating.
Allegro MicroSystems, LLC
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
31
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