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DFS04M 数据表(PDF) 20 Page - STMicroelectronics |
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DFS04M 数据表(HTML) 20 Page - STMicroelectronics |
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20 / 42 page ![]() AN1262 APPLICATION NOTE 20/42 13 SELF-SUPPLY CIRCUIT DESIGN To define the self-supply circuit it is necessary to select the bias rectifier and the supply capacitor (see fig. 7) since the turns number of the auxiliary winding has been defined already. The bias rectifier has to withstand a reverse voltage equal to: with an appropriate safety margin of 20-25%. The current rating is of little concern since the diode has to carry few mA. A popular 1N4148 (75V rating) or an UF4003 (200V rating) may be suitable choices. The supply capacitor has to be large enough to keep the device running during the time needed for the auxiliary winding to develop its correct voltage at start-up. A minimum value of 10 µF is recommended and any low cost electrolytic capacitor will do the job. The resistor Rs in series to D filters the voltage spike appearing on the pos- itive-going edge of the voltage generated by the self-supply winding that causes the voltage Vcc to increase with the converter's output load. The optimum value depends on the transformer's stray parameters (mainly the cou- pling between the auxiliary and the secondary winding) and can be found empirically once the transformer spec and construction have been frozen. A small and inexpensive axial inductor in the range of 1 to 10 µH may be used instead of RS, with even better results. Figure 7. Self-supply circuit 14 BROWNOUT PROTECTION DESIGN (L6590A AND L6590D ONLY) With reference to the schematic of fig. 8, the following relationships can be established for the ON (VinON) and OFF (VinOFF) thresholds of the input voltage: . Solving for R1 and R2: For a proper operation of this function, VinON must be less than VPKmin and VinOFF less than Vinmin (see the timing diagram of figure 8). V RE V V CC 1 V PK m ax V R -------------------- + ⋅ = L6590 L6590D L6590A Vcc D C Naux Rs V inO N R2 R1 R2 + ---------------------- ⋅ 2.5, = V inOF F 2.5 – R1 --------------------------------- 50 10 6 – ⋅ + 2.5 R2 -------- = R1 V inO N V inO FF – 50 10 6 – ⋅ ---------------------------------------- = R2 R1 2.5 V inON 2.5 – ------------------------------ ⋅ = |
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