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LT1766IGN 数据表(PDF) 13 Page - Linear Technology |
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LT1766IGN 数据表(HTML) 13 Page - Linear Technology |
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13 / 30 page ![]() LT1766/LT1766-5 13 1766fc APPLICATIONS INFORMATION the high side for discontinuous mode, so it can be used for all conditions. II I I VV V Vf L PEAK OUT LP P OUT OUT IN OUT IN =+ =+ ()( ) ()( )( )() () – - 2 2 EMI Decide if the design can tolerate an open core geometry 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. 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 radiation will be a problem. Additional Considerations After making an initial choice, consider additional factors such as core losses and second sourcing, etc. Use the experts in Linear Technology’s Applications department if you feel uncertain about the final choice. They have experience with a wide range of inductor types and can tell you about the latest developments in low profile, surface mounting, etc. Maximum Output Load Current Maximum load current for a buck converter is limited by the maximum switch current rating (IP). The current rating for the LT1766 is 1.5A. Unlike most current mode converters, the LT1766 maximum switch current limit does not fall off at high duty cycles. Most current mode converters suffer a drop off of peak switch current for duty cycles above 50%. This is due to the effects of slope compensation required to prevent subharmonic oscilla- tions in current mode converters. (For detailed analysis, see Application Note 19.) The LT1766 is able to maintain peak switch current limit over the full duty cycle range by using patented circuitry* to cancel the effects of slope compensation on peak switch current without affecting the frequency compensation it provides. Maximum load current would be equal to maximum switch current for an infinitely large inductor, but with finite inductor size, maximum load current is reduced by one- half peak-to-peak inductor current (ILP-P). The following formula assumes continuous mode operation, implying that the term on the right is less than one-half of IP. IOUT(MAX) = Continuous Mode I– I 2 = I P LP-P P − + () − () ()( )( ) VV V V V Lf V OUT F IN OUT F IN – 2 For VOUT = 5V, VIN = 8V, VF(D1) = 0.63V, f = 200kHz and L = 20μH: I A OUT MAX () − =− + () − () ()( )() =− = 15 5 0 63 8 5 0 63 2 20 10 200 10 8 15 0 21 1 29 63 . .– . •• .. . Note that there is less load current available at the higher input voltage because inductor ripple current increases. At VIN = 15V, duty cycle is 33% and for the same set of conditions: I A OUT MAX () . .– . •• .. . =− + () − () ()( )() =− = − 15 5 0 63 15 5 0 63 2 20 10 200 10 15 15 044 1 06 63 To calculate actual peak switch current with a given set of conditions, use: II I VV V V V Lf V SW PEAK OUT P OUT OUT F IN OUT F IN () =+ =+ +− () ()( )( ) I 2 L-P () – 2 Reduced Inductor Value and Discontinuous Mode If the smallest inductor value is of most importance to a converter design, in order to reduce inductor size/cost, discontinuous mode may yield the smallest inductor solu- tion. The maximum output load current in discontinuous mode, however, must be calculated and is defined later in this section. *Patent # 6, 498, 466 |
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