Application informations
L5981
5.2
Inductor selection
The inductance value fixes the current ripple flowing through the output capacitor. So the
minimum inductance value in order to have the expected current ripple has to be selected.
The rule to fix the current ripple value is to have a ripple at 20%-40% of the output current.
The inductance value can be calculated by the following equation:
Equation 6
∆IL
=
-V----I-N-----–-----V----O----U----T-
L
⋅
TON
=
V-----O----U---T--
L
⋅
TOFF
Where TON and TOFF are the on and off time of the internal power switch. The maximum
current ripple, at fixed Vout, is obtained at maximum TOFF that is at minimum duty cycle (see
previous section to calculate minimum duty). So fixing ∆IL=20% to 40% of the maximum
output current, the minimum inductance value can be calculated:
Equation 7
LMIN
=
V-----O----U---T-----+-----V----F-
∆IMAX
⋅
1-----–-----D----M-----I-N--
FSW
where FSW is the switching frequency, 1/(TON + TOFF).
For example for VOUT=3.3V, VIN=12V, IO=1A and FSW=250kHz the minimum inductance
value to have ∆IL=30% of IO is about 31µH.
The peak current through the inductor is given by:
Equation 8
IL, PK
=
IO
+
∆-----I-L-
2
So if the inductor value decreases, the peak current (that has to be lower than the current
limit of the device) increases. The higher is the inductor value, the higher is the average
output current that can be delivered, without reaching the current limit.
In the table below some inductor part numbers are listed.
Table 6. Inductors
Manufacturer
Wurth
Coilcraft
Series
TPC XLH
PD M
MSS1038
LPS6235
Inductor Value (µH) Saturation Current (A)
22 to 47
10 to 18
22 to 47
10 to 18
1.85 to 2.3
1.7 to 2.2
1.9 to 2.9
1.8 to 2.4
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