Datasheet

'i
L
=
V
IN
x D
f
SW
x L
L
2-VIN(MAX)
=
0.6 x 0.4 x 16
0.5 x 0.5
= 15.4 PH
L
1-VIN(MAX)
=
16 x 0.6
0.5 x 0.5
= 38.4 PH
L
2-VIN(MIN)
=
0.78 x 0.22 x 9
0.5 x 0.5
= 6.2 PH
L
1-VIN(MIN)
=
9 x 0.78
0.5 x 0.92
= 15.3 PH
L
2
=
D(1-D) x V
IN
I
O
x f
SW
LM5022
www.ti.com SNVS480G –JANUARY 2007–REVISED DECEMBER 2013
(9)
By calculating in terms of volts, amps and megahertz the inductance value will come out in µH.
For this design Δi
L
will be set to 40% of the maximum I
L
. Duty cycle is evaluated first at V
IN(MIN)
and at V
IN(MAX)
.
Second, the average inductor current is evaluated at the two input voltages. Third, the inductor ripple current is
determined. Finally, the inductance can be calculated, and a standard inductor value selected that meets all the
criteria.
Inductance for Minimum Input Voltage
D
VIN(MIN)
= (40 – 9.0 + 0.5) / (40 + 0.5) = 78% I
L-VIN(MIN)
= 0.5 / (1 – 0.78) = 2.3A Δi
L
= 0.4 x 2.3A = 0.92A (10)
(11)
(12)
Inductance for Maximum Input Voltage
D
VIN(MAX)
= (40 - 16 + 0.5) / (40 + 0.5) = 60% I
L-VIN(MIAX)
= 0.5 / (1 – 0.6) = 1.25A Δi
L
= 0.4 x 1.25A = 0.5A (13)
(14)
(15)
Maximum average inductor current occurs at V
IN(MIN)
, and the corresponding inductor ripple current is 0.92A
P-P
.
Selecting an inductance that exceeds the ripple current requirement at V
IN(MIN)
and the requirement to stay in
CCM for V
IN(MAX)
provides a tradeoff that allows smaller magnetics at the cost of higher ripple current at
maximum input voltage. For this example, a 33 µH inductor will satisfy these requirements.
The second criterion for selecting an inductor is the peak current carrying capability. This is the level above
which the inductor will saturate. In saturation the inductance can drop off severely, resulting in higher peak
current that may overheat the inductor or push the converter into current limit. In a boost converter, peak current,
I
PK
, is equal to the maximum average inductor current plus one half of the ripple current. First, the current ripple
must be determined under the conditions that give maximum average inductor current:
(16)
Maximum average inductor current occurs at V
IN(MIN)
. Using the selected inductance of 33 µH yields the
following:
Δi
L
= (9 x 0.78) / (0.5 x 33) = 425 mA
P-P
(17)
The highest peak inductor current over all operating conditions is therefore:
I
PK
= I
L
+ 0.5 x Δi
L
= 2.3 + 0.213 = 2.51A (18)
Copyright © 2007–2013, Texas Instruments Incorporated Submit Documentation Feedback 15
Product Folder Links: LM5022