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參數資料
| 型號： | TPS54040DGQ |
| 廠商： | TEXAS INSTRUMENTS INC |
| 元件分類： | 穩壓器 |
| 英文描述： | 0.5 A SWITCHING REGULATOR, 2500 kHz SWITCHING FREQ-MAX, PDSO10 |
| 封裝： | PLASTIC, MSOP-10 |
| 文件頁數： | 24/47頁 |
| 文件大小： | 2435K |
| 代理商： | TPS54040DGQ |
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Output Inductor Selection (LO)
IND
Vinmax
Vout
Lo min =
Io
K
Vinmax
sw
-
(28)
(
)
=
f
OUT
RIPPLE
O
SW
V
Vin m ax
- V
I
Vin m ax
L
(29)
( )
(
)
2
OUT
L(rms)
O
SW
V
Vinmax
- V
1
I
12
Vinmax
L
=
+
÷
÷
è
f
(30)
2
Iripple
Iout
ILpeak
+
=
(31)
Output Capacitor
SLVS918 – MARCH 2009................................................................................................................................................................................................... www.ti.com
To calculate the minimum value of the output inductor, use Equation 28.
KIND is a coefficient that represents the amount of inductor ripple current relative to the maximum output current.
The inductor ripple current will be filtered by the output capacitor. Therefore, choosing high inductor ripple
currents will impact the selection of the output capacitor since the output capacitor must have a ripple current
rating equal to or greater than the inductor ripple current. In general, the inductor ripple value is at the discretion
of the designer; however, the following guidelines may be used.
For designs using low ESR output capacitors such as ceramics, a value as high as KIND = 0.3 may be used.
When using higher ESR output capacitors, KIND = 0.2 yields better results. Since the inductor ripple current is
part of the PWM control system, the inductor ripple current should always be greater than 30 mA for dependable
operation. In a wide input voltage regulator, it is best to choose an inductor ripple current on the larger side. This
allows the inductor to still have a measurable ripple current with the input voltage at its minimum.
For this design example, use KIND = 0.3 and the minimum inductor value is calculated to be 42 H. For this
design, a nearest standard value was chosen: 47
H. For the output filter inductor, it is important that the RMS
current and saturation current ratings not be exceeded. The RMS and peak inductor current can be found from
For this design, the RMS inductor current is 0.501 A and the peak inductor current is 0.567 A. The chosen
inductor is a MSS1048-473ML. It has a saturation current rating of 1.44 A and an RMS current rating of 1.83A.
As the equation set demonstrates, lower ripple currents will reduce the output voltage ripple of the regulator but
will require a larger value of inductance. Selecting higher ripple currents will increase the output voltage ripple of
the regulator but allow for a lower inductance value.
The current flowing through the inductor is the inductor ripple current plus the output current. During power up,
faults or transient load conditions, the inductor current can increase above the calculated peak inductor current
level calculated above. In transient conditions, the inductor current can increase up to the switch current limit of
the device. For this reason, the most conservative approach is to specify an inductor with a saturation current
rating equal to or greater than the switch current limit rather than the peak inductor current.
There are three primary considerations for selecting the value of the output capacitor. The output capacitor will
determine the modulator pole, the output voltage ripple, and how the regulators responds to a large change in
load current. The output capacitance needs to be selected based on the more stringent of these three criteria.
The desired response to a large change in the load current is the first criteria. The output capacitor needs to
supply the load with current when the regulator can not. This situation would occur if there are desired hold-up
times for the regulator where the output capacitor must hold the output voltage above a certain level for a
specified amount of time after the input power is removed. The regulator also will temporarily not be able to
supply sufficient output current if there is a large, fast increase in the current needs of the load such as
transitioning from no load to a full load. The regulator usually needs two or more clock cycles for the control loop
30
Copyright 2009, Texas Instruments Incorporated
Product Folder Link(s): TPS54040
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相關代理商/技術參數 |
參數描述 |
|---|---|
| TPS54040DGQ | 制造商:Texas Instruments 功能描述:Power Management IC 制造商:Texas Instruments 功能描述:IC, STEP-DOWN CONVERTER, MSOP-10 |
| TPS54040DGQR | 功能描述:直流/直流開關轉換器 3.5V-42Vin,0.5A Step Down SWIFT Convrtr RoHS:否 制造商:STMicroelectronics 最大輸入電壓:4.5 V 開關頻率:1.5 MHz 輸出電壓:4.6 V 輸出電流:250 mA 輸出端數量:2 最大工作溫度:+ 85 C 安裝風格:SMD/SMT |
| TPS54040EVM-456 | 功能描述:電源管理IC開發工具 0.5A,12-42V In SWIFT Cnvrtr Eval Mod RoHS:否 制造商:Maxim Integrated 產品:Evaluation Kits 類型:Battery Management 工具用于評估:MAX17710GB 輸入電壓: 輸出電壓:1.8 V |
| TPS54040EVM-456 | 制造商:Texas Instruments 功能描述:EVALUATION MODULE FOR TPS54040 SWIFT(TM) |
| TPS54040-Q1 | 制造商:TI 制造商全稱:Texas Instruments 功能描述:0.5-A 42-V STEP-DOWN SWIFTa?￠ DC/DC CONVERTER WITH Eco-modea?￠ |
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