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ADP2442ACPZ-R7 Datasheet(PDF) 24 Page - Analog Devices |
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ADP2442ACPZ-R7 Datasheet(HTML) 24 Page - Analog Devices |
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24 / 36 page ![]() ADP2442 Data Sheet Rev. B | Page 24 of 36 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 • 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 1 and Equation 2, 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 consideration of 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 58. For example, a 700 kHz, frequency is well within the maximum and minimum duty cycle limitations. Using Equation 3, SW FREQ f R 500 , 92 = RFREQ = 132 kΩ Inductor Selection Select the inductor by using Equation 7. 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 7, 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 6, ΔIL = 0.314 A. Although the maximum output current that is required is 1 A, the maximum peak current is 1.6 A. Therefore, the inductor must 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. Output Capacitor Selection Select the output capacitor by using Equation 12 and Equation 13 ) ( 8 ) ( ESR I V f I C L RIPPLE SW L MIN OUT × ∆ − ∆ × × ∆ ≅ Equation 12 is based on the output voltage ripple (ΔVRIPPLE), which is 1% of the output voltage. ∆ × ∆ ≅ DROOP SW STEP OUT MIN OUT V f I C 3 ) ( ) ( Equation 13 calculates the capacitor selection based on the transient load performance requirement of 2%. Perform these calculations, then use the equation that yields the larger capacitor size to select a capacitor. |
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