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ADP1073 Datasheet(PDF) 6 Page  Analog Devices 

ADP1073 Datasheet(HTML) 6 Page  Analog Devices 
6 / 16 page ADP1073 –6– REV. 0 COMPONENT SELECTION General Notes on Inductor Selection When the ADP1073 internal power switch turns on, current begins to flow in the inductor. Energy is stored in the inductor core while the switch is on, and this stored energy is then trans ferred to the load when the switch turns off. Both the collector and the emitter of the switch transistor are accessible on the ADP1073, so the output voltage can be higher, lower or of opposite polarity than the input voltage. To specify an inductor for the ADP1073, the proper values of inductance, saturation current and dc resistance must be deter mined. This process is not difficult, and specific equations for each circuit configuration are provided in this data sheet. In general terms, however, the inductance value must be low enough to store the required amount of energy (when both input voltage and switch ON time are at a minimum) but high enough that the inductor will not saturate when both VIN and switch ON time are at their maximum values. The inductor must also store enough energy to supply the load without satu rating. Finally, the dc resistance of the inductor should be low so that excessive power will not be wasted by heating the windings. For most ADP1073 applications, an 82 µH to 1000 µH inductor with a saturation current rating of 300 mA to 1 A is suitable. Ferrite core inductors that meet these specifica tions are available in small, surfacemount packages. To minimize ElectroMagnetic Interference (EMI), a toroid or pot core type inductor is recommended. Rod core inductors are a lower cost alternative if EMI is not a problem. Calculating the Inductor Value Selecting the proper inductor value is a simple threestep process: 1. Define the operating parameters: minimum input voltage, maximum input voltage, output voltage and output current. 2. Select the appropriate conversion topology (stepup, step down or inverting). 3. Calculate the inductor value, using the equations in the following sections. Inductor Selection—StepUp Converter In a stepup, 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 power that must be stored is calculated from the equation: PL = VOUT +V D –V IN(MIN ) ()× I OUT () (1) where VD is the diode forward voltage (≈ 0.5 V for a 1N5818 Schottky). Energy is only stored in the inductor while the ADP1073 switch is ON, so the energy stored in the inductor on each switching cycle must be must be equal to or greater than: PL f OSC (2) in order for the ADP1073 to regulate the output voltage. When the internal power switch turns ON, current flow in the inductor increases at the rate of: IL (t) = V IN R ′ 1– e –R ′t L (3) where L is in henrys and R' 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 com pared to VIN, a simpler equation can be used: IL (t) = V IN L t (4) Replacing t in the above equation with the ON time of the ADP1073 (38 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: EL = 1 2 L × I2PEAK (5) As previously mentioned, EL must be greater than PL/fOSC so the ADP1073 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 satu ration 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 equa tions above. For example, consider a supply that will generate 5 V at 25 mA from two alkaline batteries with a 2 V endoflife voltage. The inductor power required is, from Equation 1: PL = (5V + 0.5V –2V )×(25 mA) = 87.5 mW On each switching cycle, the inductor must supply: PL f OSC = 87.5 mW 19 kHz = 4.6 µJ Since the inductor power is low, the peak current can also be low. Assuming a peak current of 100 mA as a starting point, Equation 4 can be rearranged to recommend an inductor value: L = V IN IL(MAX ) t = 2V 100 mA 38 µs = 760 µH Substituting a standard inductor value of 470 µH, with 1.2 Ω dc resistance, will produce a peak switch current of: IPEAK = 2 V 2.0 Ω 1– e –2.0 Ω× 38 µs 470 µH =149 mA Once the peak current is known, the inductor energy can be calculated from Equation 5: EL = 1 2 (470 µH)× (149 mA)2 = 5.2 µJ The inductor energy of 5.2 µJ is greater than the P L/f OSC re quirement of 4.6 µJ, so the 470 µH inductor will work in this application. The optimum inductor value can be determined by substituting other inductor values into the same equations. When selecting an inductor, the peak current must not exceed the maximum switch current of 1.5 A. 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, then the 1.5 A limit may be exceeded at the maximum value of VIN. In this case, the ADP1073’s current 
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