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ADP2389ACPZ-R7 数据表(PDF) 17 Page - Analog Devices |
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ADP2389ACPZ-R7 数据表(HTML) 17 Page - Analog Devices |
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17 / 23 page ![]() Data Sheet ADP2389/ADP2390 DESIGN EXAMPLE This section provides the procedures of selecting the external components based on the example specifications listed in Table 7. The schematic of this design example is shown in Figure 32. Table 7. Step-Down DC-to-DC Regulator Requirements Parameter Specification Input Voltage 12.0 V ± 10% Output Voltage 1.2 V Output Current 12 A Output Voltage Ripple 12 mV Load Transient ±5%, 3 A to 9 A, 2 A/µs Switching Frequency 500 kHz OUTPUT VOLTAGE SETTING Select a 10 kΩ resistor as the top feedback resistor (RTOP) and calculate the bottom feedback resistor (RBOT) using the following equation: 0.6 0.6 BOT TOP OUT R R V = × − To set the output voltage to 1.2 V, the resistors values are RTOP = 10 kΩ and RBOT = 10 kΩ. FREQUENCY SETTING Use the following equation to calculate the value of RT: 67,000 (k ) – 12 (kHz) T SW R f Ω= Thus, when fSW = 500 kHz, the value of RT = 122 kΩ. Select the standard resistor value of 121 kΩ for RT. INDUCTOR SELECTION The peak-to-peak inductor ripple current, ∆IL, is set to 33% of the maximum output current. Use the following equation to estimate the inductor value: ( ) SW L OUT IN f I D V V L × ∆ × = – where: VIN = 12.0 V VOUT = 1.2 V D = 10% ∆IL = 4 A fSW = 500 kHz This results in L = 0.54 µH. Select the standard inductor value of 0.68 µH. Calculate the peak-to-peak inductor ripple current using the following equation: ( ) SW OUT IN L f L D V V I × × = ∆ – This results in ∆IL = 3.176 A. Calculate the peak inductor current with the following equation: 2 L PEAK OUT I II ∆ = + This results in IPEAK = 13.588 A. Calculate the rms current flowing through the inductor by the following equation: 2 2 12 L RMS OUT I II ∆ = + This results in IRMS = 12.035 A. According to the calculated current value, select an inductor with a minimum rms current rating of 12.035 A and a minimum saturation current rating of 13.588 A. However, to protect the inductor from reaching its saturation point under a current-limit condition, the inductor must be rated for at least a 20 A saturation current for reliable operation. Based on these requirements, select a 0.68 µH inductor, such as the 7443330068 from Würth Elektronik, which has 1.35 mΩ dc resistance (DCR) and a 38 A saturation current. OUTPUT CAPACITOR SELECTION The output capacitor must meet both the output voltage ripple requirement and load transient response. To meet the output voltage ripple requirement, use the following equation to calculate the ESR and capacitance value of the output capacitor: RIPPLE OUT SW L RIPPLE OUT V f I C _ _ 8 ∆ × × ∆ = L RIPPLE OUT ESR I V R ∆ ∆ = _ This results in COUT_RIPPLE = 66 μF and RESR = 3.78 mΩ. To meet the ±5% overshoot and undershoot transient requirements, use the following equations to calculate the capacitance: ( ) OUT OV OUT OUT STEP OV OV OUT V V V L I K C − ∆ − × × ∆ × = _ 2 _ 2 ( ) UV OUT OUT IN STEP UV UV OUT V V V L I K C _ 2 _ 2 ∆ × − × × ∆ × = where: KOV = KUV = 2, and are the coefficients for estimation purpose. ∆ISTEP = 6 A, and is the load transient step. ∆VOUT_OV = 5% × VOUT and, is the overshoot voltage. ∆VOUT_UV = 5% × VOUT, and is the undershoot voltage. This results in COUT_OV = 332 µF and COUT_UV = 38 µF. Rev. 0 | Page 17 of 23 |
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