Datasheet

( )
= +
RIPPLE
OUT
L peak
I
I I
2
( )
( )
( )
(
)
( )
2
OUT OUT
IN max
2
OUT
L rms
O SW
IN max
V V V
1
I I
12 V L f
æ ö
´ -
ç ÷
= + ´
ç ÷
´ ´
ç ÷
è ø
( )
( )
OUT OUT
IN max
RIPPLE
O SW
IN max
V (V V )
I
V L f
´ -
=
´ ´
( )
( )
( )
OUT
IN max
OUT
O min
OUT IND SW
IN max
V V
V
L
I K V f
-
= ´
´ ´
TPS54140A
SLVSB55B –MAY 2012–REVISED JANUARY 2014 www.ti.com
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 100 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
= 0.2 and the minimum inductor value is calculated to be 7.6μH. For this
design, a nearest standard value was chosen: 10μH. 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
Equation 30 and Equation 31.
For this design, the RMS inductor current is 1.506 A and the peak inductor current is 1.62 A. The chosen
inductor is a MSS6132-103. It has a saturation current rating of 1.64 A and an RMS current rating of 1.9 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.
(28)
(29)
(30)
(31)
Output Capacitor
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
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