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ADP1111 Datasheet(PDF) 7 Page - Analog Devices |
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ADP1111 Datasheet(HTML) 7 Page - Analog Devices |
7 / 16 page ADP1111 –7– REV. 0 INDUCTOR SELECTION–STEP-UP CONVERTER In a step-up or boost converter (Figure 18), 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 () (Equation 1) where VD is the diode forward voltage (0.5 V for a 1N5818 Schottky). Because energy is only stored in the inductor while the ADP1111 switch is ON, the energy stored in the inductor on each switching cycle must be equal to or greater than: P f L OSC (Equation 2) in order for the ADP1111 to regulate the output voltage. When the internal power switch turns ON, current flow in the inductor increases at the rate of: IL t ()=VIN R' 1 − e −R't L (Equation 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. In most applications, the voltage drop across the switch is small compared to VIN so a simpler equation can be used: IL t ()=VIN L t (Equation 4) Replacing ‘t’ in the above equation with the ON time of the ADP1111 (7 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 • I2 PEAK (Equation 5) As previously mentioned, EL must be greater than PL/fOSC so that the ADP1111 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 saturation 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 40 mA from a 9 V battery, assuming a 6 V end-of-life voltage. The inductor power required is, from Equation 1: PL = 12V + 0.5V − 6V ()• 40mA ()=260mW On each switching cycle, the inductor must supply: PL f OSC = 260 mW 72 kHz = 3.6 µJ Since the required inductor power is fairly low in this example, 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: L = V IN IL(MAX ) t = 6V 500 mA 7 µs = 84 µH Substituting a standard inductor value of 68 µH with 0.2 Ω dc resistance will produce a peak switch current of: IPEAK = 6V 1.0 Ω 1 − e −1.0 Ω• 7 µs 68 µH = 587 mA Once the peak current is known, the inductor energy can be calculated from Equation 5: EL = 1 2 68 µH ()• 587mA ()2 =11.7µJ Since the inductor energy of 11.7 µJ is greater than the P L/fOSC requirement of 3.6 µJ, the 68 µH inductor will work in this application. By substituting other inductor values into the same 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 equations shown above result in peak currents > 1.5 A, the ADP1110 should be considered. Since this device has a 70% duty cycle, more energy is stored in the inductor on each cycle. This results is greater output power. 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 ADP1111’s current limit feature can be used to limit switch current. Simply select a resistor (using Figure 6) that will limit the maximum switch 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 circuit if the output is shorted to ground. In this case, current is only limited 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 19. Unlike the step-up mode, the ADP1111’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 21. The first step in selecting the step-down inductor is to calculate the peak switch current as follows: IPEAK = 2 IOUT DC VOUT + VD V IN −VSW +VD (Equation 6) where DC = duty cycle (0.5 for the ADP1111) VSW = voltage drop across the switch VD = diode drop (0.5 V for a 1N5818) IOUT = output current VOUT = the output voltage VIN = the minimum input voltage |
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