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AP64060 数据表(PDF) 13 Page - Diodes Incorporated |
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AP64060 数据表(HTML) 13 Page - Diodes Incorporated |
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13 / 18 page ![]() AP64060 Document number: DS44123 Rev. 2 - 2 13 of 18 www.diodes.com August 2022 © Diodes Incorporated AP64060 Application Information (continued) 9 Setting the Output Voltage The AP64060 has adjustable output voltages starting from 0.8V using an external resistive divider. An optional external capacitor, C4 in Figure 1, of 10pF to 220pF improves the transient response. The resistor values of the feedback network are selected based on a design trade-off between efficiency and output voltage accuracy. There is less current consumption in the feedback network for high resistor values, which improves efficiency at light loads. However, values too high cause the device to be more susceptible to noise affecting its output voltage accuracy. R1 can be determined by the following equation: ������������ = ������������ ∙ ( ������������������������ ������. ������������ − ������) Eq. 4 Table 1 shows a list of recommended component selections for common AP64060 output voltages referencing Figure 1. Table 1. Recommended Component Selections AP64060 Output Voltage (V) R1 (k Ω) R2 (k Ω) L (µH) CIN (µF) COUT (µF) C3 (nF) C4 (pF) 1.8 27.4 22.1 4.7 2.2 10x2 100 OPEN 2.5 47.5 22.1 6.8 2.2 10x2 100 OPEN 3.3 69.8 22.1 8.2 2.2 10x2 100 OPEN 5.0 115 22.1 10 2.2 10x2 100 OPEN 12.0 309 22.1 22 2.2 10x3 100 OPEN 10 Inductor Calculating the inductor value is a critical factor in designing a buck converter. For most designs, the following equation can be used to calculate the inductor value: ������ = ������������������������ ∙ (������������������ − ������������������������) ������������������ ∙ ∆������������ ∙ ������������������ Eq. 5 Where: ∆IL is the inductor current ripple fSW is the buck converter switching frequency For the AP64060 , choose ∆IL to be 20% to 30% of the maximum load current of 1A. The inductor peak current is calculated by: ������������ ������������������������ = ������������������������������ + ∆������������ ������ Eq. 6 Peak current determines the required saturation current rating, which influences the size of the inductor. Saturating the inductor decreases the converter efficiency while increasing the temperatures of the inductor and the internal power MOSFETs. Therefore, choosing an inductor with the appropriate saturation current rating is important. For most applications, it is recommended to select an inductor of approximately 4.7µH to 22µH with a DC current rating of at least 35% higher than the maximum load current. For highest efficiency, the inductor’s DC resistance should be less than 7 0mΩ. Use a larger inductance for improved efficiency under light load conditions. |
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