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| 型号： | A8502KLPTR-T |
| 厂商： | Allegro Microsystems Inc |
| 文件页数： | 27/35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 4a Determining the duty cycle, calculated as follows:
Step 4d Double-check to make sure the ? current ripple is less
V IN (min)
1 –
= = 72.04%
D (max) =1 –
V OUT(OVP) + V d
10 (V)
35.36 (V) + 0.4 (V)
(12)
than I IN (min):
I IN (min) > 1 / 2 Δ I L (18)
0.67 A > 0.19 A
A good inductor value to use would be 10 μ H.
Step 4e This step is used to verify that there is sufficient slope
Step 4b Determining the maximum and minimum input current
compensation for the inductor chosen. The slope compensation
Slope Compensation = = 3.6 A / μ s
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:
value is determined by the following formula:
3.6 f SW
2 10 6
(19)
I OUT = # CHANNELS I LED
= 2 0.120 (A) = 0.240 A
(13)
Next insert the inductor value used in the design:
V IN (min) D (max)
=
then:
I IN (max) =
=
V OUT(OVP) I OUT
V IN (min)
35.36 (V) 240 (mA)
10 (V) 0.90
= 0.94 A
(14)
Δ I Lused =
L used f SW
10 (V) 0.72
10 ( μ H) 2.0 (MHz)
Calculate the minimum required slope:
= 0.36 A
(20)
(1 – D (max))
where η is efficiency.
Next, calculate minimum input current, as follows:
Required Slope (min) =
Δ I Lused 1 10 – 6
1
f SW
(21)
= = 2.57 A/ μ s
(1 – 0.72)
I IN (min)=
=
V OUT(OVP) I OUT
V IN (max)
35.36 (V) 240 (mA)
14 (V) 0.90
= 0.67 A
(15)
0.36 (A) 1 10 – 6
1
2.0 (MHz)
If the minimum required slope is greater than the calculated slope
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
inductor must be set such that the ? inductor ripple current is not
greater than the average minimum input current. As a first pass
compensation, the inductor value must be increased.
Note: The slope compensation value is in A/ μ s, and 1 × 10 –6 is a
constant multiplier.
Step 4f Determining the inductor current rating. The inductor
current rating must be greater than the I IN (max) value plus half of
the ripple current Δ I L , calculated as follows:
assume I ripple to be 40% of the maximum inductor current:
L (min) = I IN (max) + 1 / 2 Δ I Lused
(22)
Δ I L = I IN (max) × I ripple
(16)
= 0.94 (A) + 0.36 (A) / 2 = 1.12 A
then:
= 0.94 (A) × 0.40 = 0.376 A
Step 5 Determining the resistor value for a particular switching
frequency. Use the R FSET values shown in figure 7. For example,
V IN (min)
L =
=
Δ I L f SW
D (max)
10 (V)
0.376 (A) 2 (MHz)
0.72 = 9.57 μ H
(17)
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
27
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