LTC3854EMSE datasheet(21/28 Pages) LINER

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LTC3854

3854fa

applicaTions inForMaTion

However, the amount of capacitance needed is determined

not only by the allowed ripple in steady state but by the

maximum energy stored in the inductor. The capacitance

must be sufficient in value to absorb the change in induc-

tor current when a high current to low current transient

occurs. The minimum capacitance to assure the inductor’s

energy is adequately absorbed during a 5A load step for

a maximum overshoot of 2% is:

COUT ≥

L •

∆IL2

2 •

∆VOUT • VOUT

COUT ≥

0.56µH •(5A)2

0.02 • 1.2V

COUT ≥583µF

A maximum overshoot or undershoot of 2% for a 5A load

step will require an ESR of:

ESR

< 0.02 •

VOUT

∆ILOAD

= 0.02 •

1.2V

≤ 5mΩ

Several quality capacitors are available with low enough

ESR.

Multilayer ceramic capacitors tend to have very low ESR

values.ItisalsoagoodpracticetoreducetheESLbyputting

several capacitors in parallel on the output (a parallel bank

of larger and smaller capacitors will improve performance

in both a DC and a transient condition).

To keep ripple very low and design for any possible large

excursions in current 2x 330µF (tantalum or polymer

surface) and 1x 47µF polymer low ESR type were con-

nected in parallel.

Choosing FB Resistors (See Figure 3)

VOUT = 0.8 1+









RB = 0.5RA

Using 1% 10.0k for RA gives 1% 4.99k for RB.

Choosing CIN Capacitors

CINischosenforaRMScurrentratingofatleastIOUT(MAX)/2

= 6A. Again, keeping ESR low will improve performance

and reduce power loss (several capacitors in parallel is

once again a good choice). We will use an 180µF 25V

electrolytic with 2x 10µF 25V low ESR ceramic capacitors

connected in parallel.

Choosing MOSFETs

The power dissipation in the main and synchronous FETs

canbeeasilyestimated.ChoosingaRenesasRJK0305DPB

for the main FET results in the following parameters:

BVDSS = 30V

RDS(ON) = 13mΩ maximum at 25°C, VGS = 4.5V

QGD=1.5nCatVDS,test10VresultsinCMILLER=1.5nC/10V

= 150pF

QG = 8nC, typical, at VGS = 4.5V

VMILLER = 2.8V

At VIN = 20V, IOUT = 15A, estimated TJ = 100°C for the

top FET and given

VINTVCC = 5.0V

RDR,PULLUP = 2.6Ω

RDR,PULLDOWN = 1.5Ω

the total losses in the main FET will be:

PMAIN =

1.2V

20V

•(15A)2 • 1

+ 0.005 • 100°C – 25°C

(

)

(

)

• 13m

Ω + 20V

( )2 • 15A

• 150pF

•

2.5

Ω

− 2.8V

1.2

Ω

2.8V









• fSW

PMAIN = 0.55W

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