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LT1375 Datasheet(PDF) 11 Page - Linear Technology |
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LT1375 Datasheet(HTML) 11 Page - Linear Technology |
11 / 28 page 11 LT1375/LT1376 13756fd APPLICATIONS INFORMATION must withstand continuous fault conditions. If maxi- mum load current is 0.5A, for instance, a 0.5A inductor may not survive a continuous 1.5A overload condition. Dead shorts will actually be more gentle on the induc- tor because the LT1376 has foldback current limiting. 2. Calculate peak inductor current at full load current to ensure that the inductor will not saturate. Peak current can be significantly higher than output current, espe- cially with smaller inductors and lighter loads, so don’t omit this step. Powdered iron cores are forgiving because they saturate softly, whereas ferrite cores saturate abruptly. Other core materials fall in between somewhere. The following formula assumes continu- ous mode of operation, but it errs only slightly on the high side for discontinuous mode, so it can be used for all conditions. II VV V fL V PEAK OUT OUT IN OUT IN =+ − () ()( )( ) 2 VIN = Maximum input voltage f = Switching frequency, 500kHz 3. Decide if the design can tolerate an “open” core geom- etry like a rod or barrel, which have high magnetic field radiation, or whether it needs a closed core like a toroid to prevent EMI problems. One would not want an open core next to a magnetic storage media, for instance! This is a tough decision because the rods or barrels are temptingly cheap and small and there are no helpful guidelines to calculate when the magnetic field radia- tion will be a problem. 4. Start shopping for an inductor (see representative surface mount units in Table 2) which meets the re- quirements of core shape, peak current (to avoid satu- ration), average current (to limit heating), and fault current (if the inductor gets too hot, wire insulation will melt and cause turn-to-turn shorts). Keep in mind that all good things like high efficiency, low profile, and high temperature operation will increase cost, sometimes dramatically. Get a quote on the cheapest unit first to calibrate yourself on price, then ask for what you really want. Example: with L = 2 µH, VOUT = 5V, and VIN(MAX) = 15V, ImA OUT MAX ( ) − = () ⎛⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠() () − () = 1 5 500 10 2 10 15 2 5 15 5 338 2 36 .• • The main reason for using such a tiny inductor is that it is physically very small, but keep in mind that peak-to-peak inductor current will be very high. This will increase output ripple voltage. If the output capacitor has to be made larger to reduce ripple voltage, the overall circuit could actually wind up larger. CHOOSING THE INDUCTOR AND OUTPUT CAPACITOR For most applications the output inductor will fall in the range of 3 µH to 20µH. Lower values are chosen to reduce physical size of the inductor. Higher values allow more output current because they reduce peak current seen by the LT1376 switch, which has a 1.5A limit. Higher values also reduce output ripple voltage, and reduce core loss. Graphs in the Typical Performance Characteristics section show maximum output load current versus inductor size and input voltage. A second graph shows core loss versus inductor size for various core materials. When choosing an inductor you might have to consider maximum load current, core and copper losses, allowable component height, output voltage ripple, EMI, fault cur- rent in the inductor, saturation, and of course, cost. The following procedure is suggested as a way of handling these somewhat complicated and conflicting requirements. 1. Choose a value in microhenries from the graphs of maximum load current and core loss. Choosing a small inductor with lighter loads may result in discontinuous mode of operation, but the LT1376 is designed to work well in either mode. Keep in mind that lower core loss means higher cost, at least for closed core geometries like toroids. The core loss graphs show both absolute loss and percent loss for a 5W output, so actual percent losses must be calculated for each situation. Assume that the average inductor current is equal to load current and decide whether or not the inductor |
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