LM3445MM datasheet(21/26 Pages) NSC

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capacitors are delivering power to the buck converter. When

the triac is firing at 135° the current through the LED string will

be small. Therefore the droop should be small at this point

and a 5% voltage droop should be a sufficient derating. With

this derating, the lowest voltage the buck converter will see is

about 42.5V in this example.

30060355

FIGURE 21. AC Line with Various Firing Angles

30060356

FIGURE 22. V

BUCK Waveforms with Various Firing Angles

To determine how many LEDs can be driven, take the mini-

mum voltage the buck converter will see (42.5V) and divide it

by the worst case forward voltage drop of a single LED.

Example: 42.5V/3.7V = 11.5 LEDs (11 LEDs with margin)

OUTPUT CAPACITOR

A capacitor placed in parallel with the LED or array of LEDs

can be used to reduce the LED current ripple while keeping

the same average current through both the inductor and the

LED array. With a buck topology the output inductance (L2)

can now be lowered, making the magnetics smaller and less

expensive. With a well designed converter, you can assume

that all of the ripple will be seen by the capacitor, and not the

LEDs. One must ensure that the capacitor you choose can

handle

the

RMS

current

the

inductor.

Refer

manufacture’s datasheets to ensure compliance. Usually an

X5R or X7R capacitor between 1 µF and 10 µF of the proper

voltage rating will be sufficient.

SWITCHING MOSFET

The main switching MOSFET should be chosen with efficien-

cy and robustness in mind. The maximum voltage across the

switching MOSFET will equal:

The average current rating should be greater than:

DS-MAX = ILED(-AVE)(DMAX)

RE-CIRCULATING DIODE

The LM3445 Buck converter requires a re-circulating diode

D10 (see the Typical Application circuit figure 4) to carry the

inductor current during the MOSFET Q2 off-time. The most

efficient choice for D10 is a diode with a low forward drop and

near-zero reverse recovery time that can withstand a reverse

voltage of the maximum voltage seen at V

BUCK. For a common

110V

AC ± 20% line, the reverse voltage could be as high as

190V.

The current rating must be at least:

D = 1 - (DMIN) x ILED(AVE)

Or:

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