L5983 查看數據表（PDF） - STMicroelectronics

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L5983

2A step-down switching regulator

STMicroelectronics 

Application informations

L5983

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=1.5A and FSW=250kHz the minimum inductance

value to have ∆IL=30% of IO is about 21µH.

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

MSS1038

MSS7341

Inductor Value (µH) Saturation Current (A)

15 to 22

3.3 to 6.8

15 to 18

3.3 to 6.2

2.3 to 3.25

2.5 to 2.75

3.2 to 3.6

2.5 to 3.5

16/37

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