L5987 View Datasheet(PDF) - STMicroelectronics
Part Name
Description
MFG CO.
'L5987' PDF : 40 Pages View PDF
Application information
L5987
6.2
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.
In the continuous current mode (CCM), the inductance value can be calculated by the
following equation:
Equation 10
IL = -V----I-N-----–---L--V----O----U----T- TON = V-----O----U---T-L----+-----V----F- TOFF
Where TON is the conduction time of the internal high-side switch and TOFF is the
conduction time of the external diode [in CCM, FSW = 1 / (TON + TOFF)]. The maximum
current ripple, at fixed VOUT, is obtained at maximum TOFF that is at minimum duty cycle
(see Section 6.1 to calculate minimum duty). So fixing IL = 20% to 30% of the maximum
output current, the minimum inductance value can be calculated:
Equation 11
LMIN = V-----O----U-I--MT----A-+---X--V----F- 1-----–-F----DS----WM-----I-N--
where FSW is the switching frequency, 1 / (TON + TOFF).
For example for VOUT = 3.3 V, VIN = 12 V, IO = 3 A and FSW = 250 kHz the minimum
inductance value to have IL = 30% of IO is about 10 H.
The peak current through the inductor is given by:
Equation 12
IL PK = IO + ---2--I-L-
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 Table 7 some inductor part numbers are listed.
Manufacturer
Coilcraft
Wurth
SUMIDA
Table 7. Inductors
Series
Inductor value (H)
MSS1038
3.8 to 10
MSS1048
12 to 22
PD Type L
8.2 to 15
PD Type M
2.2 to 4.7
CDRH6D226/HP
1.5 to 3.3
CDR10D48MN
6.6 to 12
Saturation current (A)
3.9 to 6.5
3.84 to 5.34
3.75 to 6.25
4 to 6
3.6 to 5.2
4.1 to 5.7
18/40
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