L5986 데이터 시트보기 (PDF) - STMicroelectronics

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L5986

2.5 A step-down switching regulator

STMicroelectronics 

L5986 - L5986A

Application information

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 continuos current mode (CCM), the inductance value can be calculated by the

following equation:

Equation 6

ΔIL

=

-V----I-N-----–-----V----O----U----T-

L

⋅

TON

=

V-----O----U---T-----+-----V----F-

L

⋅

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 previous section

to calculate minimum duty). So fixing ΔIL=20 % to 30 % of the maximum output current, the

minimum inductance value can be calculated:

Equation 7

LMIN

=

V-----O----U---T-----+-----V----F-

ΔIMAX

⋅

1-----–-----D----M-----I-N--

FSW

where FSW is the switching frequency, 1/(TON + TOFF).

For example for VOUT = 3.3 V, VIN = 12 V, IO = 2.5 A and FSW = 250 kHz the minimum

inductance value to have ΔIL = 30 % of IO is about 12 μH.

The peak current through the inductor is given by:

Equation 8

IL, PK

=

IO

+

Δ-----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 7. Inductors

Manufacturer

Coilcraft

Series

MSS1038

MSS1048

Inductor value (μH)

3.8 to 10

12 to 22

Saturation current (A)

3.9 to 6.5

3.84 to 5.34

15/39

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