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ADP1108AN-5 데이터 시트보기 (PDF) - Analog Devices

 부품명 상세내역 제조사 ADP1108AN-5 Micropower DC-DC Converter Adjustable and Fixed 3.3 V, 5 V, 12 V Analog Devices ADP1108AN-5 Datasheet PDF : 12 Pages
 1 2 3 4 5 6 7 8 9 10 Next Last ADP1108 Calculating the Inductor Value Selecting the proper inductor value is a simple three-step process: 1. Define the operating parameters: minimum input voltage, maximum input voltage, output voltage and output current. 2. Select the appropriate conversion topology (step-up, step- down or inverting). 3. Calculate the inductor value, using the equations in the fol- lowing sections. On each switching cycle, the inductor must supply: PL = 315 mW =16.6 µ J f OSC 19 kHz The required inductor power is fairly low in this example, so the peak current can also be low. Assuming a peak current of 500 mA as a starting point, Equation 4 can be rearranged to recommend an inductor value: Inductor Selection—Step-Up Converter In a step-up or boost converter (Figure 15), the inductor must store enough power to make up the difference between the input voltage and the output voltage. The inductor power is calculated from the equation: L = V IN t = 2 V 36 µs = 144 µH IL(MAX ) 500 mA Substituting a standard inductor value of 100 µH with 0.2 Ω dc resistance, will produce a peak switch current of: ( ) ( ) PL = VOUT +V D −V IN(MIN ) × IOUT (Equation 1) where VD is the diode forward voltage (≈ 0.5 V for a 1N5818 Schottky). Energy is only stored in the inductor while the I PEAK = 2V 1.0 Ω  1 – e –1.0 Ω×36 100 µH µs   = 605 mA Once the peak current is known, the inductor energy can be ADP1108 switch is ON, so the energy stored in the inductor on calculated from Equation 5: each switching cycle must be equal to or greater than: PL f OSC (Equation 2) ( ) EL = 1 2 100 µH × 605 mA 2  =18.3 µ J in order for the ADP1108 to regulate the output voltage. When the internal power switch turns ON, current flow in the inductor increases at the rate of: The inductor energy of 18.3 µJ is greater than the PL/fOSC requirement of 16.6 µJ, so the 100 µH inductor will work in this application. By substituting other inductor values into the same IL (t) = V IN R'  1 − – R 't eL   (Equation 3) where L is in henrys and RЈ is the sum of the switch equivalent resistance (typically 0.8 Ω at +25°C) and the dc resistance of the inductor. If the voltage drop across the switch is small compared to VIN, a simpler equation can be used: I L (t ) = V IN L t (Equation 4) equations, the optimum inductor value can be selected. When selecting an inductor, the peak current must not exceed the maximum switch current of 1.5 A. If the calculated peak current is greater than 1.5 A, either the ADP3000 should be considered or an external power transistor can be used. The peak current must be evaluated for both minimum and maximum values of input voltage. If the switch current is high when VIN is at its minimum, the 1.5 A limit may be exceeded at the maximum value of VIN. In this case, the current limit feature of the ADP1108 can be used to limit switch current. Simply select a resistor (using Figure 3) that will limit the maximum switch Replacing t in the above equation with the ON time of the ADP1108 (36 µs, typical) will define the peak current for a given inductor value and input voltage. At this point, the inductor energy can be calculated as follows: current to the IPEAK value calculated for the minimum value of VIN. This will improve efficiency by producing a constant IPEAK as VIN increases. See the Limiting the Switch Current section of this data sheet for more information. Note that the switch current limit feature does not protect the EL = 1 L 2 × I2 PEAK (Equation 5) As previously mentioned, EL must be greater than PL/fOSC so the ADP1108 can deliver the necessary power to the load. For best efficiency, peak current should be limited to 1 A or less. Higher switch currents will reduce efficiency because of increased satura- tion voltage in the switch. High peak current also increases output ripple. As a general rule, keep peak current as low as possible to minimize losses in the switch, inductor and diode. In practice, the inductor value is easily selected using the equations above. For example, consider a supply that will generate 12 V at 30 mA from a 3 V battery, assuming a 2 V end-of-life voltage. The inductor power required is from Equation 1: PL = (12 V + 0.5V – 2V ) × (30 mA) = 315 mW circuit if the output is shorted to ground. In this case, current is limited only by the dc resistance of the inductor and the forward voltage of the diode. Inductor Selection—Step-Down Converter The step-down mode of operation is shown in Figure 16. Unlike the step-up mode, the ADP1108’s power switch does not saturate when operating in the step-down mode. Therefore, switch current should be limited to 650 mA in this mode. If the input voltage will vary over a wide range, the ILIM pin can be used to limit the maximum switch current. Higher switch current is possible by adding an external switching transistor, as shown in Figure 18. The first step in selecting the step-down inductor is to calculate the peak switch current as follows: I PEAK = 2 IOUT DC   V VOUT + V D IN – V SW + V D   (Equation 6) –6– REV. 0