参数资料
型号: TPS54260DRCR
厂商: Texas Instruments
文件页数: 29/52页
文件大小: 0K
描述: IC REG BUCK ADJ 2.5A 10SON
标准包装: 3,000
系列: SWIFT™, Eco-Mode™
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 0.8 V ~ 58 V
输入电压: 3.5 V ~ 60 V
PWM 型: 电流模式
频率 - 开关: 100kHz ~ 2.5MHz
电流 - 输出: 2.5A
同步整流器:
工作温度: -40°C ~ 150°C
安装类型: 表面贴装
封装/外壳: 10-VFDFN 裸露焊盘
包装: 带卷 (TR)
供应商设备封装: 10-SON 裸露焊盘(3x3)
www.ti.com
SLVSA86A – MARCH 2010 – REVISED DECEMBER 2010
Figure 49. 3.3V Output TPS54260 Design Example.
Output Inductor Selection (L O )
To calculate the minimum value of the output inductor, use Equation 28 .
K IND 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 K IND = 0.3 may be used.
When using higher ESR output capacitors, K IND = 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 150 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 K IND = 0.3 and the minimum inductor value is calculated to be 11 m H. For this
design, a nearest standard value was chosen: 10 m 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 2.51 A and the peak inductor current is 2.913 A. The chosen
inductor is a Coilcraft MSS1038-103NLB . It has a saturation current rating of 4.52 A and an RMS current rating
of 4.05 A.
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.
Lo min =
Vinmax - Vout Vout
Io ′ K IND Vinmax ′ ?sw
(28)
Copyright ? 2010, Texas Instruments Incorporated
Product Folder Link(s): TPS54260
29
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