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AN4501 Datasheet(PDF) 11 Page - STMicroelectronics |
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AN4501 Datasheet(HTML) 11 Page - STMicroelectronics |
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11 / 48 page  DocID026394 Rev 1 11/48 AN4501 Guideline for the selection of external components 47 Two 4.7 µF MLCC with a voltage capability of 50 V and with an ESR of 4 m Ω can be placed in parallel. The resulting ESR is 2 m Ω (ESR COUT). The resulting maximum output current ripple due to the bulk capacitance is: The additional current ripple caused by the ESR is: which is indeed negligible compared to IOUT,PP,MAX. The total output current ripple is 62 mA, fulfilling the target. Step 5: Selection of the input capacitor We assume that MLCC with a very low ESR is used. The contribution of the ESR to the input voltage ripple can be neglected as well, compared to the contribution of the charge and discharge of the input capacitor itself. The maximum inductor peak-to-peak current IL,PP,MAX is: The minimum input capacitor is calculated by: Using two 4.7 µF, 50 V input capacitor with an ESR of 4m Ω (ESR CIN), the maximum input voltage ripple caused by the bulk capacitance is: C OUT,MIN 1 Nb LED R DLED × -------------------------------------------- I OUT D MAX × I OUT,PP,MAX,TARGET F SW × -------------------------------------------------------------------------- × = C OUT,MIN 1 80.4 × ----------------- 1.0 0.791 × 0.075 450 10 3 ⋅ × --------------------------------------------- × 7.34µF == I OUT,PP,MAX 1 Nb LED R DLED × -------------------------------------------- I OUT D MAX × C OU T F SW × ------------------------------------ × == 1 80.4 × ----------------- 1.0 0.791 × 9.4 10 6 – ⋅ 450 10 3 ⋅ × ------------------------------------------------------- × 58.5mA == ESR COUT I L,PEAK,MAX × Nb LED R DLED × ------------------------------------------------------------------- 210 3 – 5.59 × ⋅ 80.4 × -------------------------------------- 4.7mA == I L,PP,MAX V OUT D 50 × 1D 50 – () × LF SW η × × --------------------------------------------------------------- 26 0.5 2 × 6.8 10 6 – 450 10 3 0.9 × ⋅ × ⋅ ---------------------------------------------------------------------- 2.36A == = C IN,MIN I L,PP,MAX 8F × SW V IN,PP,MAX,TARGET × -------------------------------------------------------------------------------- 2.36 8 450 10 3 0.100 × ⋅ × ------------------------------------------------------- 6.6µF == = I L,PP,MAX 8F × SW C IN × -------------------------------------- 2.36 8 450 10 3 9.4 10 6 – ⋅ × ⋅ × ----------------------------------------------------------------- 70mV == |
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