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LM3445MM データシート(PDF) 21 Page - National Semiconductor (TI) |
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LM3445MM データシート(HTML) 21 Page - National Semiconductor (TI) |
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21 / 26 page ![]() 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 of the inductor. Refer to 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: I 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: I D = 1 - (DMIN) x ILED(AVE) Or: 21 www.national.com |
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