Datasheet

R13
(BUCK)
=
1.25V
10 x I
PEAK
I
OUT
I
1(PEAK)
=
0.8
+
I
RIPPLE
2
Buck Mode
Buck-Boost Mode
(V
OUT
+ Vin)I
OUT
I
2(PEAK)
=
0.8 x V
IN2
+
I
RIPPLE
2
V
OUT
(V
IN1
- V
OUT
)
V
IN1
x f x I
RIPPLE
L1 =
Buck Mode
V
IN2
(V
OUT
)
(V
OUT
+ V
IN2
) x f x I
RIPPLE
L1 =
Buck-Boost Mode
LM5118, LM5118-Q1
SNVS566G –APRIL 2008–REVISED FEBRUARY 2013 www.ti.com
where
• V
OUT
is the output voltage
• V
IN1
is the maximum input voltage
• f is the switching frequency
• I
RIPPLE
is the selected inductor peak to peak ripple current (1.2 A selected for this example)
• V
IN2
is the minimum input voltage
• (10)
The resulting inductor values are:
L1 = 28 µH, Buck Mode
L1 = 9.8 µH Buck-Boost mode
A 10 µH inductor was selected which is a compromise between these values, while favoring the buck-boost
mode. As will be illustrated in the compensation section below, the inductor value should be as low as possible to
move the buck-boost right-half-plane zero to a higher frequency. The ripple current is then rechecked with the
selected inductor value using the equations above,
I
RIPPLE(BUCK)
= 3.36A
I
RIPPLE(BUCK-BOOST)
= 1.17A
Because the inductor selected is lower than calculated for the Buck mode, the minimum load current for CCM in
buck mode is 1.68 A at maximum VIN.
With a 10 µH inductor, the worst case peak inductor currents can be estimated for each case, assuming a 20%
inductor value tolerance.
(11)
For this example, the two equations yield:
I
1(PEAK)
= 5.43A
I
2(PEAK)
= 13.34A
An acceptable current limit setting would be 6.7A for buck mode since the LM5118 automatically doubles the
current limit threshold in buck-boost mode. The selected inductor must have a saturation current rating at least
as high as the buck-boost mode cycle-by-cycle current limit threshold, in this case at least 13.5A. A 10 µH 15
amp inductor was chosen for this application.
R13 = R
SENSE
To select the current sense resistor value, begin by calculating the value of R
SENSE
for both modes of operation.
(12)
R13
(BUCK)
= 23 mΩ
For the buck-boost mode, R
SENSE
is given by:
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