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ADP2441ACPZ-R7 数据表(PDF) 23 Page - Analog Devices |
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ADP2441ACPZ-R7 数据表(HTML) 23 Page - Analog Devices |
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23 / 32 page ![]() Data Sheet ADP2441 Rev. A | Page 23 of 32 DESIGN EXAMPLE Consider an application with the following specifications: VIN =24 V ± 10% VOUT = 5 V ± 1% Switching frequency = 700 kHz Load = 800 mA typical Maximum load current = 1 A Soft start time = 6 ms Overshoot ≤ 2% under all load transient conditions CONFIGURATION AND COMPONENTS SELECTION Resistor Divider The first step in selecting the external components is to calculate the resistance of the resistor divider that sets the output voltage. Using Equation 2 and Equation 3, kΩ 10 μA 60 6 . 0 STRING REF BOTTOM I V R REF REF OUT BOTTOM TOP V V V R R kΩ 3 . 73 V 6 . 0 V 6 . 0 V 5 kΩ 10 TOP R Switching Frequency Choosing the switching frequency involves considering the trade-off between efficiency and component size. Low frequency improves the efficiency by reducing the gate losses but requires a large inductor. The choice of high frequency is limited by the minimum and maximum duty cycle. Table 11. Duty Cycle VIN Duty Cycle 24 V (Nominal) DNOMINAL = 20.8% 26 V (10% Above Nominal) DMIN = 19% 22 V (10% Less than Nominal) DMAX = 23% Based on the estimated duty cycle range, choose the switching frequency according to the minimum and maximum duty cycle limitations, as shown in Figure 55. For example, a 700 kHz, frequency is well within the maximum and minimum duty cycle limitations. Using Equation 4, SW FREQ f R 500 , 92 RFREQ = 132 kΩ Soft Start Capacitor For a given soft start time, the soft start capacitor can be calculated using Equation 5, SS SS SS REF C I t V REF SS SS SS V t I C nF 10 V 6 . 0 ms 6 μA 1 SS C Inductor Selection Select the inductor by using Equation 9. SW IN OUT IN OUT IDEAL f V V V V L ) ( 3 . 3 μH 3 . 18 μH 66 . 18 kHz 700 V 24 V ) 5 24 ( V 5 3 . 3 IDEAL L In Equation 9, VIN = 24 V, VOUT = 5 V, ILOAD(MAX) = 1 A, and fSW = 700 kHz, which results in L = 18.66 μH. When L = 18 μH (the closest standard value) in Equation 8, ΔIL = 0.314 A. Although the maximum output current required is 1 A, the maximum peak current is 1.6 A. Therefore, the inductor should be rated for higher than 1.6 A current. Input Capacitor Selection The input filter consists of a small 0.1 μF ceramic capacitor placed as close as possible to the IC. The minimum input capacitance required for a particular load is SW PP OUT MIN IN f V D D I C ) 1 ( _ where: VPP = 50 mV. IOUT = 1 A. D = 0.23. fSW = 700 kHz. Therefore, μF 9 . 4 kHz 700 V 05 . 0 ) 22 . 0 1 ( 22 . 0 A 1 _ MIN IN C Choosing an input capacitor of 10 μF with a voltage rating of 50 V ensures sufficient capacitance over voltage and temperature. |
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