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Li-Poly battery voltage changes a lot while discharging — from 4.2V to 2.7V. Most modern digital ICs (I'm currently thinking of Cortex-M3 microcontroller and an RF module) require input voltage levels of 2.7V-3.6V.

So, I think that connecting them to 4.2V power supply will most probably result in a dead IC. How this problem is solved in actual devices?

I can only think of two solutions: the first is a buck-boost converter, which should certainly work, but is really hard to build (at least for me; I never had success with self-made DC-DC converters, not to say the ICs for them are hard to obtain in local Russian stores); and a linear LDO regulator, which will still incur some dropout, maybe as low as 0.7V, but it means that the smallest battery voltage with which the circuit is usable is around 3.4V, and quite a big part of charge is then left unused.

Catherine
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  • As a small note, LDO can be used in other contexts as well. You are specifically talking about a linear LDO regulator. – Kellenjb Jul 18 '11 at 19:33
  • @Kellenjb thanks for clarifying, I've edited the question – Catherine Jul 18 '11 at 22:18
  • I understand that the "Maximum Voltages" section of the datasheet says that the performance is spec'ed from 2.7 to 3.6V. However, these are 3.3V parts, and you need to seriously consider how much you want to cut corners. A little deviation from the nominal value is acceptable, none is not enough, and too much will sometimes result in a dead IC, but sometimes that's OK. You'll have to do some analysis to find out where those numbers are. Have you tried running your design at 2.7V? 3.6V? How about 4.2V? Have you tested a statistically meaningful quantity of them at those voltages? – Kevin Vermeer Jul 18 '11 at 22:32
  • LDOs can have less than 0.7 V dropout. So a linear LDO might be the best choice for low currents. – markrages Jul 18 '11 at 22:38
  • @Kevin, your comment about "statistically meaningful quantities" would make sense if I'd want to commercially manufacture such devices. But this is just an one-time hobby project. Also, this is the reason I don't feel comfortable with a design which can suddenly die because of a slight deviation. – Catherine Jul 19 '11 at 20:11

1 Answers1

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Linear has literally a few dozen buck-boost switchers which take 2.5V-4.2V in with 3.3V out, for example the LTC3534. This uses only common "chicken-feed" like Rs and Cs, a small coil and offers high efficiency. You may find the controller at the usual suspects, but it's not cheap. This is probably the best solution (apart from price).

The LDO is another option. NXP CortexM3 controllers like the LPC1343 work on voltages down to 2V, I don't know about the RF modules. LDOs with dropout voltages less than 100mV are not uncommon, so even if the battery's voltage gets as low as 2.7V you still have 2.6V left at the LDO's output.

A third solution is a switched-capacitor voltage doubler, followed by a buck switched regulator. This may look stupid at first sight, but you avoid the expensive buck-boost regulator, and will have much more choice for the buck (in both meanings of the word). Having the voltage doubler followed by an LDO is also an option, but then your battery will drain much quicker.

PeterJ
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stevenvh
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  • Linear devices are great, I know, but there are really few their parts available from local vendors, so that's not an option.

    Also, I've rechecked the datasheets, and my parts can actually work with voltages under 2.7V, so the LDO becomes the best possibility. Thanks!

     – Catherine Jul 18 '11 at 09:43
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    If cents count then an MC34063 IC will allow a constant voltage of your choice solution at an all up price of around $US1.

    People should say where they are located. In the US your nearest "local supplier" is only an internet session away.

     – Russell McMahon Jul 18 '11 at 10:34
  • @Russell: I did ("in local Russian stores"). Interestingly, MC34063 is available, and is ~$1, too. (btw, buying from DigiKey/Mouser is not a price problem, but rather delivery one; it can take literally months to get here.) – Catherine Jul 18 '11 at 12:55
  • @Russell - I checked the 34063's datasheet and from the block schematic you can see that it doesn't do buck-boost. But maybe you would use it with the voltage doubler? – stevenvh Jul 18 '11 at 13:01
  • MC34063 can be used as a building block to do a wide range of things with some thought.
    1. Buck-boost is achieved with either an external switch at either end the end of the inductor (as in ref below) or you can probably remove Q702, R708, R709 and ground pin 2 (sw emitter) for lower current loads 9< 1A?). "Glue parts" are not very major and one external low cost transistor. More on this if of interest.
    2. Or you can implement a SEPIC converter to provide buck-boost.

    http://www.datasheetdir.com/circuits/2/Schematic-For-The-Mc34063-Buck-boost-Leaded-Demonstration-Board.jpg

     – Russell McMahon Jul 18 '11 at 16:08