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| 型号： | LM3495MTCX/NOPB |
| 厂商： | National Semiconductor |
| 文件页数： | 21/36页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK SYNC ADJ 10A 16TSSOP |
| 标准包装： | 1 |
| 系列： | PowerWise® |
| 类型： | 降压（降压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 0.6 V ~ 5.5 V |
| 输入电压： | 2.9 V ~ 18 V |
| PWM 型： | 电流模式，混合 |
| 频率 - 开关： | 200kHz ~ 1.5MHz |
| 电流 - 输出： | 10A |
| 同步整流器： | 是 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-TSSOP（0.173"，4.40mm 宽） |
| 包装： | 标准包装 |
| 供应商设备封装： | 16-TSSOP |
| 产品目录页面： | 1300 (CN2011-ZH PDF) |
| 配用： | LM3495EVAL-ND - BOARD EVALUATION LM3495 |
| 其它名称： | LM3495MTCXDKR |
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SNVS410F – FEBRUARY 2006 – REVISED APRIL 2013
By calculating in terms of amperes, volts, and megahertz, the inductance value will come out in micro henries.
The second minimum inductance equation specific to the LM3495 is:
L MIN2 =
64 x (R DSON-LO + R SNS )
f SW
x
V IN
V IN + 2
(17)
By calculating in terms of milliohms and kilohertz the inductance value will come out in micro henries.
For this design:
L MIN1 =
13.2V - 1.2V
500 kHz x 3A
x 0.1 = 0.8 P H
L MIN2 =
64 x 3.4 m :
500 kHz
x
12V
12V + 2
= 0.4 P H
(18)
Whichever equation gives the higher value for inductance is the one which should be followed.
The second criterion for selecting an inductor is the peak current carrying capability. This is the level above
which the inductor will saturate. In saturation the inductance drops off severely, often to 20% to 30% of the rated
value. In a buck converter, peak current, I PK , is equal to the maximum load current plus one half of the ripple
current. For this example:
I PK = 10A + 1.5A = 11.5A
(19)
Hence an inductor must be selected that has a peak current rating greater than 11.5A and an average current
rating greater than 10A. To ensure a robust design, the inductor selected should maintain approximately 50% of
its rated inductance during the worst-case peak current from an output short circuit. For a low-side current limit
the peak current during an output short circuit can be estimated as I CL plus Δ i (O-MAX) . Δ i (O-MAX) is calculated by
substituting zero for output voltage in the expression for Δ i O . Inductor core materials with soft saturation
characteristics are preferred. One inductor that meets the peak current guidelines is an off-the-shelf 1.0 μH
component that can handle a peak current of 18A and an average current of 14A. The inductor current ripple and
peak inductor current should be recalculated for the selected inductance value, L ACTUAL , by rearranging the
equation for minimum inductance:
' i O =
V IN(max) - V O
f SW x L ACTUAL
xD
' i O =
13.2V - 1.2V
0.5 MHz x 1 P H
x 0.1 = 2.4A P-P
I PK = 10A + 2.4A/2 = 11.2A
(20)
OUTPUT CAPACITOR
The output capacitor in a switching regulator is selected on the basis of capacitance, equivalent series resistance
(ESR), size, and cost. An important specification in switching converters is the output ripple voltage, Δ v O . At 500
kHz the impedance of most capacitors is very small compared to ESR, hence ESR becomes the main selection
guide. In this design the load requires a 1% ripple, which results in a Δ v O of 10 mV P-P . Maximum ESR is then:
ESR MAX =
' V O
' i O
(21)
ESR MAX is 10 m ? . Multi-layer ceramic, aluminum electrolytic, tantalum, solid aluminum, organic, and niobium
capacitors are all popular in switching converters. Generally, by the time enough capacitors have been paralleled
to obtain the desired ESR, the bulk capacitance is more than enough to supply the load current during a transient
from no-load to full load. In this example the load could transition quickly from 0A to 5A, (or from 5A to 0A), so
moderate bulk capacitance is needed. Two MLCC capacitors rated 100 μF, 6.3V each with ESR of 1.5 m ? will
work well.
Copyright ? 2006–2013, Texas Instruments Incorporated
Product Folder Links: LM3495
21
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