I am usually in need of powering my microcontroller projects with 5V. Typically this is not a problem, as I use any power supply (converted PC power supply, wall adapter, USB, etc.).

However, when I need my projects to be portable and lightweight, I am not quite sure what the cheapest/easiest way to accomplish this would be. The additional requirement is that I would like to use 3.7V LiPo batteries. This is simply because they are much smaller than any other solution, and can pack a lot of power. Also because I have many lying around.

If I was using something higher than 5V, then I would just use the L7805 to drop it down, or some other solution. So how would I step up the voltage from a 3.7V battery to 5V in the cheapest and smallest way possible? Is there a widely used IC which accomplishes this task?

  • \$\begingroup\$ If 4.5 V would suffice, then 3x AA batteries in series might do the job. \$\endgroup\$ – m.Alin Mar 3 '13 at 14:01
  • \$\begingroup\$ @m.Alin Thanks, but I specifically want to use 3.7 V LiPos because 3x AA batteries are too large and heavy for my application. \$\endgroup\$ – capcom Mar 3 '13 at 14:08
  • \$\begingroup\$ 2-3.7V cells + L7805. Don't use unequal or dissimilar cells. \$\endgroup\$ – Optionparty Mar 3 '13 at 14:10
  • \$\begingroup\$ Then you could use a DC-DC boost converter to step up the 3.7 V to 5 V. Examples from TI. \$\endgroup\$ – m.Alin Mar 3 '13 at 14:16
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    \$\begingroup\$ May I suggest taking a step back and asking whether you really need 5V? Lots of uCs are capable of going to 3.3V and less, and doing so generally decreases consumption and will help with battery life. Often, it doesn't require a complete redesign unless you're using some especially intransigent parts. \$\endgroup\$ – Chintalagiri Shashank Mar 4 '13 at 9:09

Two options for you:

  • One is to use a boost converter for ~3.7V to 5V (i.e. if the supply range is always below the required rail voltage) There are hundreds out there to choose from, you can get e.g. very simple fixed 5V boost regulators which only require a few external components (see TI, LT, ON Semi, Analog Devices, Microchip, Maxim, etc - they all make switching regulators).
    Or if your supply is above and below 5V over the battery discharge curve, then as Madmanguruman suggests, use a buck-boost/SEPIC/Cuk converter. Make sure the input range of the IC fits with your battery voltage range. There are plenty of questions on here that discuss the use of switching regulators in this type of situation, so I won't cover that again here.

  • The other is to use a microcontroller and surrounding ICs that run from +3.3V - then a simple LDO can be used, which at this drop (say for a Li-Ion from ~4.2V down to 3.5V or so to +3.3V - you can get LDOs with less than 100mV dropout voltage nowadays) will be efficient and simple. Nowadays +3.3V is very common, and many microcontrollers with a "nominal" 5V can run fine from 3.3V - a random example is the PIC16F1828 which boasts a 1.8V - 5.5V operation range. So check to see whether this might be an otpion with the chips you are using, or whether swapping a couple would be easy enough.

  • \$\begingroup\$ Awesome, thanks a lot! I am using the ATmega8, and it isn't really rated for less than 4.5 V. So I am not sure if something around 4 V would be under the rating. Is there any risk in trying to use something under-volt, or will it simply not work if it doesn't like it. \$\endgroup\$ – capcom Mar 3 '13 at 15:14
  • \$\begingroup\$ If the specs don't include +3.3V, then don't use it (always keep within the specs, otherwise the manufacturer does not guarantee what will happen - it could appear to work okay, but then break with a rise in temperature of 5 degrees, firmware may intermittently do odd things, RAM may lose it's contents, etc) Undervoltage wll almost certainly not damage the micro, so you can try it for curiosities sake. However, I notice there is an ATmega8L which is the low voltage version of the same micro with a range from 2.7V to 5.5V, so you may want to consider using this. \$\endgroup\$ – Oli Glaser Mar 3 '13 at 15:43
  • \$\begingroup\$ @OliGlaser I suggested a buck-boost, not a boost as you stated. It will buck or boost depending on the battery voltage. \$\endgroup\$ – Adam Lawrence Mar 4 '13 at 1:34
  • \$\begingroup\$ @Madmanguruman - Yes, I did actually notice that, but generalised with just the boost scenario as well as the buck-boost or other over/under options - I mentioned both cases in case the OP is using a supply always lower than 5V. Sorry for misquoting you, I'll fix it. \$\endgroup\$ – Oli Glaser Mar 4 '13 at 8:58

What you need is an efficient buck-boost converter that will reduce the voltage when the cells are above 5V, and increase the voltage when the cells are below 5V.

This is important to maximize the useful energy that the cells can provide. A fixed linear regulator can never fully utilize the stored charge in the cells since the input to the regulator must always be higher than the output.

As an example, Texas Instruments has a series of high-efficiency buck-boost converters that are intended for use with rechargeable cells.


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