参数资料
型号: TPS54478RTET
厂商: Texas Instruments
文件页数: 23/38页
文件大小: 0K
描述: IC REG BUCK SYNC ADJ 4A 16WQFN
标准包装: 1
系列: SWIFT™
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 0.6 V ~ 4.5 V
输入电压: 2.95 V ~ 6 V
PWM 型: 电流模式
频率 - 开关: 500kHz
电流 - 输出: 4A
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: 16-WFQFN 裸露焊盘
包装: 标准包装
供应商设备封装: 16-WQFN 裸露焊盘(3x3)
其它名称: 296-29064-6
SLVSAS2 – JUNE 2011
OUTPUT INDUCTOR SELECTION
The inductor selected works for the entire TPS54478 input voltage range. To calculate the value of the output
inductor, use Equation 19 . K IND is a coefficient that represents the amount of inductor ripple current relative to the
maximum output current. The inductor ripple current is filtered by the output capacitor. Therefore, choosing high
inductor ripple currents impacts 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, K IND is normally from 0.1 to 0.3 for the majority of applications.
For this design example, use K IND = 0.3 and the inductor value is calculated to be 1.05 μ H. For this design, a
nearest standard value was chosen: 1.2 μ 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 4.01 A and the peak inductor current is 4.53 A. The chosen inductor
is a Coilcraft XAL5030-122ME. It has a saturation current rating 0f 11.8 A (20% inductance loss) and a RMS
current rating of 8.7 A (20 ° C. temperature rise). The series resistance is 6.78 m Ω typical.
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.
L1 =
Vinmax - Vout Vout
Io ′ Kind Vinmax ′ | sw
(19)
vertical spacer
Iripple =
vertical spacer
Vinmax - Vout Vout
L1 Vinmax ′ | sw
(20)
Io +
′ ? ÷
ILrms =
2
1 ? Vo (Vinmax - Vo) ?
12 è Vinmax ′ L1 ′ | sw ?
2
(21)
vertical spacer
ILpeak = Iout +
Iripple
2
(22)
OUTPUT CAPACITOR
There are three primary considerations for selecting the value of the output capacitor. The output capacitor
determines the modulator pole, the output voltage ripple, and how the regulator 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 is temporarily not 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 to see the
change in load current and output voltage and adjust the duty cycle to react to the change. The output capacitor
must be sized to supply the extra current to the load until the control loop responds to the load change. The
output capacitance must be large enough to supply the difference in current for 2 clock cycles while only allowing
a tolerable amount of droop in the output voltage. Equation 23 shows the minimum output capacitance necessary
to accomplish this.
Copyright ? 2011, Texas Instruments Incorporated
23
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