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ADP5050ACPZ-R7 数据表(PDF) 34 Page - Analog Devices |
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ADP5050ACPZ-R7 数据表(HTML) 34 Page - Analog Devices |
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34 / 60 page ![]() ADP5050 Data Sheet Rev. 0 | Page 34 of 60 DESIGN EXAMPLE This section provides an example of the step-by-step design procedures and the external components required for Channel 1. Table 15 lists the design requirements for this example. Table 15. Example Design Requirements for Channel 1 Parameter Specification Input Voltage VPVIN1 = 12 V ± 5% Output Voltage VOUT1 = 1.2 V Output Current IOUT1 = 4 A Output Ripple ΔVOUT1_RIPPLE = 12 mV in CCM mode Load Transient ±5% at 20% to 80% load transient, 1 A/µs Although this example shows step-by-step design procedures for Channel 1, the procedures apply to all other buck regulator channels (Channel 2 to Channel 4). SETTING THE SWITCHING FREQUENCY The first step is to determine the switching frequency for the ADP5050 design. In general, higher switching frequencies produce a smaller solution size due to the lower component values required, whereas lower switching frequencies result in higher conversion efficiency due to lower switching losses. The switching frequency of the ADP5050 can be set to a value from 250 kHz to 1.4 MHz by connecting a resistor from the RT pin to ground. The selected resistor allows the user to make decisions based on the trade-off between efficiency and solution size. (For more information, see the Oscillator section.) However, the highest supported switching frequency must be assessed by checking the voltage conversion limitations enforced by the minimum on time and the minimum off time (see the Voltage Conversion Limitations section). In this design example, a switching frequency of 600 kHz is used to achieve a good combination of small solution size and high conversion efficiency. To set the switching frequency to 600 kHz, use the following equation to calculate the resistor value, RRT: RRT (kΩ) = [14,822/fSW (kHz)]1.081 Therefore, select standard resistor RRT = 31.6 kΩ. SETTING THE OUTPUT VOLTAGE Select a 10 kΩ bottom resistor (RBOT) and then calculate the top feedback resistor using the following equation: RBOT = RTOP × (VREF/(VOUT − VREF)) where: VREF is 0.8 V for Channel 1. VOUT is the output voltage. To set the output voltage to 1.2 V, choose the following resistor values: RTOP = 4.99 kΩ, RBOT = 10 kΩ. SETTING THE CURRENT LIMIT For 4 A output current operation, the typical peak current limit is 6.44 A. For this example, choose RILIM1 = 22 kΩ (see Table 11). For more information, see the Current-Limit Protection section. SELECTING THE INDUCTOR The peak-to-peak inductor ripple current, ΔIL, is set to 35% of the maximum output current. Use the following equation to estimate the value of the inductor: L = [(VIN − VOUT) × D]/(ΔIL × fSW) where: VIN = 12 V. VOUT = 1.2 V. D is the duty cycle (D = VOUT/VIN = 0.1). ΔIL = 35% × 4 A = 1.4 A. fSW = 600 kHz. The resulting value for L is 1.28 µH. The closest standard inductor value is 1.5 µH; therefore, the inductor ripple current, ΔIL, is 1.2 A. The peak inductor current is calculated using the following equation: IPEAK = IOUT + (ΔIL/2) The calculated peak current for the inductor is 4.6 A. The rms current of the inductor can be calculated using the following equation: 12 2 2 L OUT RMS I I I ∆ + = The rms current of the inductor is approximately 4.02 A. Therefore, an inductor with a minimum rms current rating of 4.02 A and a minimum saturation current rating of 4.6 A is required. However, to prevent the inductor from reaching its saturation point in current-limit conditions, it is recommended that the inductor saturation current be higher than the maximum peak current limit, typically 7.48 A, for reliable operation. Based on these requirements and recommendations, the TOKO FDV0530-1R5, with a DCR of 13.5 mΩ, is selected for this design. |
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