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There are quite a few similar posts on this subject but I'm a little confused about the best way to proceed in my specific project. All my previous regulators have been linear types (eg: 7805, LM317 at best) - nice and simple. For me anyway.

The project I'm working on needs 3 voltage outputs from a Lithium (lipo or li-ion) battery - 3.3V, 5V, 12V.

  • 3.3V @ 500mA (estimated) main rail for the ARM based system on module. ARM can be at full power, standby, or off. This requirement doesn't look too onerous. Stackexchange has some good posts in this area.
  • 12V @ 250mA for sensor system A. Switchable from the ARM (or a PIC or some such).
  • 5V @ 2.5A for sensor system B. Switchable from the ARM.

Plus the battery needs to be recharged. I can use one or two cells although it might not be up to me. Two cells might be easier given that I can boost to 12V and derive 3.3V and 5V from the 8.4V line.

I'm not sure about the best way to proceed as there's a blizzard of chips out there. I was hoping someone has used a nicely behaved one that combined charging and regulated outputs. For example, Microchips MCP7383 can charge LiIon fine and a chip like the TPS63020 is a buck/boost regulator to give you 3.3V but other chips combine the charger and the regulator which seems a better idea.

If I use a single cell LiIon I get 3.4/3.5V to 4.2V. Is that too tight for 3.3V, especially given the power draw when I turn on the 5V boot circuit. Should I first boost to 5V and use that to get 3.3V? And then use another boost to 12V?

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    \$\begingroup\$ My manager has purchased the TPS65217. The Beaglebone uses that and it has battery charging and LDO/DC-DC outputs. Except that's a processor PMIC not a system PMIC. Good grief. What would I do with 1.8V and 1.1V outputs!? Drive an LED? Now I have two problems :-) \$\endgroup\$
    – carveone
    Commented Nov 15, 2012 at 11:48
  • \$\begingroup\$ Also just realised the sum of the output powers. At 3.3V and 85% conversion efficiency - the 12V requires 3.5W and the 5V nearly 15W max. Take just the 5V and the Arm together needs over 5 Amps from the 3.3V rail. I think I need 2 cells! \$\endgroup\$
    – carveone
    Commented Nov 15, 2012 at 14:15
  • \$\begingroup\$ How long are the sensors to be on? \$\endgroup\$
    – pjc50
    Commented Nov 15, 2012 at 16:07
  • \$\begingroup\$ 18 Watts peak is a heft load for a tiny cell. How long do you want this battery to last ( short term & long term ) \$\endgroup\$
    – D.A.S.
    Commented Nov 15, 2012 at 17:58
  • \$\begingroup\$ That's a darn good question. They said "an entire shift" which made me laugh. I hope they have a car handy. \$\endgroup\$
    – carveone
    Commented Dec 20, 2012 at 10:53

2 Answers 2

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The 3.4/3.5 Volt Li-Ion cell voltage is not necessarily too tight for sourcing 3.3 Volts, as some Low Drop-Out regulators do mention drop-out voltages of the order of 100 mV at small currents. It would be marginal / iffy, though.

If I were doing this, I would boost to 5 Volts, run an LDO or a buck regulator off that 5 Volt rail to get 3.3 Volts, and separately boost directly from the battery for the 12 Volt rail. That way, the 12 volt boost regulator, which seems to be serving your most current-hungry rail, would not suffer 2 stages of efficiency loss, nor would the 5 Volt boost regulator need to be hefty enough to provide the high current needed by the boost regulator for the 12 Volt rail.

Also, although there are integrated boost / regulate devices that may serve your purpose, as a parametric search on Digikey indicates, there is advantage in keeping the blocks separate: That way, the heat dissipation volumes of each of the 3 parts, 2 boosts and 1 buck / LDO, could be physically separate and thus more of a distributed problem than a thermal hot-spot.

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  • \$\begingroup\$ Thanks Anindo. I'd consider the 3.3V line the main rail, mostly because it supplies the ARM SoM. It's the 5V line at a possible 2.5 amps, that's the hungriest when it has to be on. \$\endgroup\$
    – carveone
    Commented Nov 15, 2012 at 13:17
  • \$\begingroup\$ However boosting to 5V initially may make a good deal of sense, even if it has to be very hefty, as it gives me the option of an easy redesign for 2 cells instead of 1. The heat dissipation was something that hadn't occurred to me either and bears some thinking about... \$\endgroup\$
    – carveone
    Commented Nov 15, 2012 at 13:20
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First of all, i would like to mention that a common Li-Ion (18650) battery can not deliver the amount of power you want to be able to use. Normal max current for 18650 is at around 5 A but at the voltage they are at.

You would therefore need to make a pack out of these batteries.

For the two smaller currents i would use a circuit based onLM2577 or LM2596 depending if you step it up or down.

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  • \$\begingroup\$ I was thinking those chips might have switching frequencies that are too low for a Processor/DSP and the components get bigger. But then I don't know what is reasonable in frequency terms as regards transient response. I've also just seen the TI simple switcher modules that incorporate the whole smps in one chip. But yes, you are right about the battery - I've persuaded the boss to go for a pack. \$\endgroup\$
    – carveone
    Commented Dec 20, 2012 at 10:50

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