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DFS04M 数据表(PDF) 19 Page - STMicroelectronics |
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DFS04M 数据表(HTML) 19 Page - STMicroelectronics |
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19 / 42 page ![]() 19/42 AN1262 APPLICATION NOTE may be achieved by using paralleled capacitors. ESR, besides being responsible for capacitor heating, is what basically determines the switching frequency volt- age ripple superimposed on top of the DC value. This is true as long as the capacitive contribution to the ripple is negligible, that is if: (18) The specification on the maximum allowed output ripple is then translated into a requirement on the maximum ESR of the capacitor: (19) Anyway, once the specification on either the AC ripple current or the ESR is fulfilled, the resulting capacitance value definitely meets condition (18). If the requirement on ESR is very tight, there is an alternative to using a large number of output capacitors: it is possible to tolerate a higher ripple on Cout (provided the AC ripple requirement is met) and add an LC post filter, like the one shown in fig. 6, that attenuates the ripple to the desired level. Figure 6. Output post filter for ripple reduction The attenuation factor of such filter is approximately given by: which is the same for complementary duty cycles and minimum for D=0.5. Thus, to get the desired attenuation factor the following design equations can be applied: It is convenient to choose an off-the-shelf choke and then select a capacitor with an ESR low enough to get the desired attenuation level. For low output current (less than 1 A) ferrite beads may be used. At any rate, the DC current rating of the choke should be oversized to minimize DC voltage drop. In fact, the feedback should be connected upstream the post filter to avoid stability problems (see "Control loop compensation" section). C out > > 100 I out D x ⋅ V r% V out f sw ⋅⋅ -------------------------------------- ⋅ ESRx V r% 100 ---------- V ou t Is pk ----------- ⋅ = L Cout C' ESR ESR' ∆Vo ∆Vout Post filter Ka V out pp – ∆ ∆V Op p – ----------------------- D1 D – () ESR' f sw L ⋅ --------------- ⋅⋅ ≈ = Ka ESR' 4f sw L ⋅⋅ ----------------------- for Dx > 0.5 = Ka Dx 1 Dx – () ESR' f sw L ⋅ --------------- for Dx < 0.5 ⋅⋅ = |
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