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L5994

ADJUSTABLE TRIPLE OUTPUT POWER SUPPLY CONTROLLER

STMicroelectronics 

L5994 - L5994A

+3.3V Inductor

To define the inductor, it is necessary to determine firstly the inductance value. Its minimum value is given by:

L3MIN = V-----I-N--3--M-.--3--A---–X----(-⋅-V--f--S-I-N--W--M----⋅-A--I-X-O----U–----T-3--3-.--3-⋅--)-R-----F--

and a value L3 > L3MIN should be selected. Core geometry selection is connected to the requirements of the

specific application in terms of space utilization and other practical issues like ease of mounting, availability and

so on. As to the material, the choice should be directed towards ferrite, molypermalloy or KoolMµ, to achieve

high efficiency. These materials provide low core losses (ferrite in particular), so that the design can be concen-

trated on preventing saturation and limiting copper losses. Saturation must be avoided even at maximum peak

current:

IL3PK = IOUT3 + -23---.-⋅-3--f--S-⋅---(W--V----⋅-I-N-L---M-3----A⋅---VX----I-–-N----3M---.-A-3---X)-

To limit copper losses, the winding DC resistance, RL, should be as low as possible (in the range of mΩ). AC

losses can usually be neglected. A practical criterion to minimize DC resistance could be to use the largest wire

that fits the selected core.

Anyway the best solution, whenever possible, is to use an off-the-shelf inductor which meets the requirements

in terms of inductance and maximum DC current. Nowadays there is a broad range of products offered by man-

ufacturer, also for surface mount assemblies.

+5.1V Transformer

The primary winding carries the secondary power as well, thus the total primary average current is:

ITOT5

=

IOUT

5

+

-V----I--N---L---I--N-----⋅---I--O----U----T----1--2-

5.1

where VDRLIN is the voltage generated during the recirculation of the primary and fed into the input of the +12V

linear regulator. The turns ratio 1:n of the transformer is chosen so that VDRLIN is above 13V. To reduce the

turns ratio in order to minimize stray parameters, the secondary is referred to the 5.1V output, and therefore the

minimum value is given by:

ηMIN = V-----I-N----L---I--N---5--–--.--1-5---.--1----+-----V----f

where Vf is the forward drop across the rectifier (assume 1V to be conservative). Make sure the secondary is

connected with the proper polarity (see fig. 4). The minimum primary inductance value can be expressed as:

L5pmin

=

3--

4

⋅

-V----I-N-----⋅---f--S----W------⋅---[--I--T----O----T---5----⋅-5--R-.--1--F---⋅--⋅-(--(V---V--I-N-I--N--–---–--5--5-.--1-.-1--)--2)----–----η-----⋅---V----I--N-----⋅---I--O----U-----T---1---2---]

where RF, to get positive values for L5PMIN, must satisfy the inequality:

RF > I--T--η--O---⋅-T---V5----I⋅-N--(---V⋅----II-O-N----U-–--T--5--1--.-21----)-

and where VIN can be either VINMIN or VINMAX, whichever gives the higher value for L5PMIN. With a primary in-

ductance L5P > L5PMIN the primary peak current, which must not saturate the magnetic core, will be:

I5PK = I5TOT + -2--5--⋅-.--1f--S---⋅-W--(--V--⋅--I-L-N--5--M--P--A--⋅--X-V----–-I-N--5--M--.-1--A--)-X-- + η ⋅ IOUT12

As to the transformer realization, the considerations regarding to the +3.3V inductor can be here repeated.

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