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ADP1073AN-5 데이터시트(PDF) 7 Page - Analog Devices |
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ADP1073AN-5 데이터시트(HTML) 7 Page - Analog Devices |
7 / 16 page ADP1073 –7– REV. 0 limit feature can be used to limit switch current. Simply select a resistor (using Figure 4) 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 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 ADP1073’s power switch does not satu- rate when operating in the step-down mode. Switch current should therefore be limited to 600 mA for best performance 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. The first step in selecting the step-down inductor is to calculate the peak switch current as follows: IPEAK = 2 × I OUT DC VOUT +V D V IN –VSW +V D (6) where DC = duty cycle (0.72 for the ADP1073) 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 As previously mentioned, the switch voltage is higher in step- down mode than in step-up mode. VSW is a function of switch current and is therefore a function of VIN, L, time and VOUT. For most applications, a VSW value of 1.5 V is recommended. The inductor value can now be calculated: L = V IN(MIN) –VSW –VOUT IPEAK × t ON (7) where tON = switch ON time (38 µs) If the input voltage will vary (such as an application which must operate from a battery), an RLIM resistor should be selected from Figure 4. The RLIM resistor will keep switch current con- stant as the input voltage rises. Note that there are separate RLIM values for step-up and step-down modes of operation. For example, assume that +3.3 V at 150 mA is required from a 9 V battery with a 6 V end-of-life voltage. Deriving the peak current from Equation 6 yields: IPEAK = 2 ×150 mA 0.72 3.3 + 0.5 6– 1.5 + 0.5 = 317 mA The peak current can than be inserted into Equation 7 to calcu- late the inductor value: L = 6–1.5 – 3.3 317 mA ×38 µs =144 µH Since 144 µH is not a standard value, the next lower standard value of 100 µH would be specified. To avoid exceeding the maximum switch current when the input voltage is at +9 V, an RLIM resistor should be specified. Inductor Selection—Positive-to-Negative Converter The configuration for a positive-to-negative converter using the ADP1073 is shown in Figure 17. As with the step-up converter, all of the output power for the inverting circuit must be supplied by the inductor. The required inductor power is derived from the formula: PL = |VOUT|+V D ()× I OUT () (8) The ADP1073 power switch does not saturate in positive-to- negative mode. The voltage drop across the switch can be modeled as a 0.75 V base-emitter diode in series with a 0.65 Ω resistor. When the switch turns on, inductor current will rise at a rate determined by: IL (t) = V L R' 1– e –R't L (9) where R' = 0.65 Ω + R L(DC) VL = VIN – 0.75 V For example, assume that a –5 V output at 75 mA is to be gen- erated from a +4.5 V to +5.5 V source. The power in the induc- tor is calculated from Equation 8: PL = |−5V|+ 0.5V ()×(75mA)=413mW During each switching cycle, the inductor must supply the fol- lowing energy: PL f OSC = 413 mW 19 kHz = 21.7 µJ Using a standard inductor value of 330 µH, with 1 Ω dc resis- tance, will produce a peak switch current of: IPEAK = 4.5V –0.75V 0.65 Ω+1Ω 1– e –1.65 Ω× 38 µs 330 µH = 393 mA Once the peak current is known, the inductor energy can be calculated from Equation 9: EL = 1 2 (330 µH) × (393 mA)2 = 25.5 µJ The inductor energy of 25.5 µJ is greater than the P L/fOSC requirement of 21.7 µJ, so the 330 µH inductor will work in this application. The input voltage varies between only 4.5 V and 5.5 V in this example. Therefore, the peak current will not change enough to require an RLIM resistor and the ILIM pin can be connected di- rectly to VIN. Care should be taken, of course, to ensure that the peak current does not exceed 800 mA. |
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