Datasheet

ripple
Lpeak OUT
I
I = I +
2
2
2
OUT IN OUT
Lrms OUT
IN
V (V max V )
1
I = I +
12 V max L1 sw
æ ö
´ -
´
ç ÷
´ ´ ¦
è ø
IN OUT OUT
ripple
IN
V max V V
I =
L1 V max sw
-
´
´ ¦
IN OUT OUT
OUT ind IN
V max V V
L1 =
I K V max sw
-
´
´ ´ ¦
TPS54318
SLVS975A –SEPTEMBER 2009–REVISED SEPTEMBER 2013 www.ti.com
For this design example, use K
IND
= 0.3 and the inductor value is calculated to be 1.4 μH. For this design, a
nearest standard value was chosen: 1.5 μH. 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 is found from Equation 20
and Equation 21.
For this design, the RMS inductor current is 3.01 A and the peak inductor current is 3.42 A. The chosen inductor
is a Coilcraft XPL7030-152ML_. The inductor has an RMS current rating of 12.3 A and a saturation current rating
of 20.2 A. The current ratings for this exceed the requirement, but the inductor was chosen for small physical
size and low series resistance for high efficiency.
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 increases 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.
(18)
(19)
(20)
(21)
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 must 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 must supply
the load with current when the regulator cannot. This situation occurs 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 requires 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 22 shows the minimum output capacitance necessary to accomplish this.
For this example, the transient load response is specified as a 3% change in V
OUT
for a load step from 1.25 A to
2.75 A For this example, ΔI
OUT
= 2.75 – 1.25 = 1.5 A and ΔV
OUT
= 0.03 × 1.8 = 0.054 V. Using these numbers
gives a minimum capacitance of 56 μF. This value does not take the ESR of the output capacitor into account in
the output voltage change. For ceramic capacitors, the ESR is usually small enough to ignore in this calculation.
Equation 23 calculates the minimum output capacitance needed to meet the output voltage ripple specification.
Where ƒsw is the switching frequency, V
ripple
is the maximum allowable output voltage ripple, and I
ripple
is the
inductor ripple current. In this case, the maximum output voltage ripple is 30 mV. Under this requirement,
Equation 23 yields 3.2 μF.
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