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AN2728 bảng dữ liệu(PDF) 8 Page - STMicroelectronics |
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AN2728 bảng dữ liệu(HTML) 8 Page - STMicroelectronics |
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8 / 20 page ![]() Selecting components for your applications AN2728 8/20 You must choose the resistor divider according to the following equation: Equation 1 with VFB = 0.6 V. Figure 9 shows the feedback voltage versus temperature. We suggest using a resistor with values in the range of 10 k Ω to 100 kΩ. Lower values are suitable as well but increase current consumption. For output voltages close to the feedback voltage, we suggest adding a very small capacitor in parallel with R1 in the range of 10 pF. As an alternative, we suggest increasing the current in the resistor divider by decreasing the R1 and R2 value. 2.2 Input capacitor The input capacitor must be able to support the maximum input operating voltage and the maximum RMS input current. Since step-down converters draw current from the input in pulses, the input current is squared and the height of each pulse is equal to the output current. The input capacitor has to absorb switching current that can be as high as the load current divided by two (worst case, with duty cycle of 50 %). For this reason, the quality of these capacitors has to be very high to minimize its power dissipation generated by the internal ESR, thus improving the system reliability and efficiency. The critical parameter is usually the RMS current rating, which must be higher than the RMS input current. The maximum RMS input current (flowing through the input capacitor) is: Equation 2 where η is the expected system efficiency, D is the duty cycle and IOUT the output DC current. This function reaches its maximum value at D = 0.5 and the equivalent RMS current is equal to IOUT divided by 2 (considering η= 1). + = 2 1 FB OUT R R 1 V V [] + = 2 1 FB OUT R R 1 V V [] Figure 9. Feedback voltage vs. temperature 550 570 590 610 630 650 -75 -50 -25 0 25 50 75 100 125 150 175 T [°C] V IN = 3.6 V, IOUT = 10 mA 550 570 590 610 630 650 -75 -50 -25 0 25 50 75 100 125 150 175 T [°C] V IN = 3.6 V, IOUT = 10 mA η D η D 2 D I I 2 2 OUT RMS + × - × = η D η D 2 D I I 2 2 OUT RMS + × - × = |
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