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CM3015-18 Datasheet(PDF) 11 Page - California Micro Devices Corp |
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CM3015-18 Datasheet(HTML) 11 Page - California Micro Devices Corp |
11 / 14 page © 2004 California Micro Devices Corp. All rights reserved. 09/14/04 430 N. McCarthy Blvd., Milpitas, CA 95035-5112 ● Tel: 408.263.3214 ● Fax: 408.263.7846 ● www.calmicro.com 11 CM3015 PRELIMINARY Application Information Figure 32. Adjustable Regulator Application Example Output Compensation Low-dropout regulator topologies use a current-carry- ing transistor ("pass device") that adds additional gain and phase shift to the design. It is necessary to com- pensate for these attributes to prevent the design from becoming unstable. The simplest way to do this is to add an output RC filter to roll off the gain. Since all reg- ulators use an output capacitor anyway, the only com- ponent needed is a suitable resistance. Fortunately, for the usual range and types of output capacitors, this resistance can be supplied by the ESR of the caps. For many years, capacitors have been the largest com- ponents in regulator circuits, but newer types are shrinking in size rapidly. Unfortunately, their ESRs have shrunk as well, taking them out of the range where they can perform the R function described in existing LDOs. The CM3015 is a new breed of LDO regulator that can work with these tiny low-ESR capac- itors and thus permit designs that take a minimum of space. An attribute of the LDO topology is that the pass device gain generally increases with increasing output current. This may affect output stability: a design that didn't need a capacitor now does, or a design that used a small value now needs a larger one. The CMD advan- tage is that this component can be selected from any capacitor family without regard to its ESR. If the circuit calls for a tiny, low-ESR capacitor, it will work just fine. VIN GND EN VOUT SENSE BYP 1 2 6 5 3 4 VIN=VOUT+ RDROP(IOUT) VOUT=VSENSE (1+R1/R2) VOUT= 4.992V 31.6k 10.0k 1.20V For the input voltage, Vin minimum will depend on the output current. For example, if Iout is 100mA, Vin must be at least 4.992V = 1ohm x 100mA = 4.992 + 0.1 = 5.092V. R1 R2 |
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