L5980 datasheet(15/37 Pages) STMICROELECTRONICS

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Application informations

L5980

15/37

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

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

∆I

L=20% to 40% of the maximum

output current, the minimum inductance value can be calculated:

Equation 7

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

For example for VOUT=3.3V, VIN=12V, IO=0.7A and FSW=250kHz the minimum inductance

value to have

∆I

L=30% of IO is about 45µH.

The peak current through the inductor is given by:

Equation 8

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

Series

Inductor value (

µH)

Saturation current (A)

Wurth

TPC XL

10 to 33

1.3 to 2.5

PD3 L

33 to 68

1.35 to 1.8

Coilcraft

MSS1038

39 to 68

1.62 to 2.04

LPS6235

12 to 33

1.4 to 2.2

∆I

OUT

–

------------------------------ T

⋅

OUT

-------------- T

OFF

⋅

MIN

OUT

∆I

MAX

----------------------------

MIN

–

-----------------------

⋅

LPK

∆I

--------

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