# 5v power supply from 18650s - high power, low profile

I am trying to use 2 18650 batteries to power 288 APA102 LEDs. The LEDs require 5V, and have a max current draw of 60mA/LED for a total of 17.28A. I will also have a microcontroller attached (currently arduino micro), powering some other low-draw devices. So let's say I'm aiming to support up to 18A. Most of the time though, the current draw will be much lower.

My understanding is that I can use a buck or boost converter (depending how the batteries are wired), and outboard bypass transistor(s) to shunt additional current around the converter (with proper heat-sinking). I am curious though, are there considerations for the buck vs boost configurations? Is my stated understanding even correct? And lastly, will I need to protect my arduino somehow if I'm drastically changing the power coming out of the batteries (e.g. Strobing the LEDs from fully on to fully off)?

I'm open to alternative battery suggestions as well, but I need everything to fit in a cylinder of ~20mm, and energy density is very relevant for this project.

• Are you sure it is 60uA (microamps) instead of 60mA (milliamps)? 60uA * 288 is well under 1A, in fact it is around 17.3mA. – Ron Beyer Mar 23 '17 at 20:27
• Are you sure that 2 18650 batteries are able to give you 17 amps? – Chupacabras Mar 23 '17 at 20:31
• My batteries say they can each provide 25A "max pulse discharge". – Brian Faires Mar 23 '17 at 20:33
• That's 2.5 Amp at 3.7 volts or 9W . You need 5V * 18A = 90W, plus conversion losses. So more like 110 Watts in. And that's before problems with heat and everything. Your looking at 10 or more cells. Minimum. For any reasonable usage time – Passerby Mar 23 '17 at 20:39
• Pulse discharge is typically like 10 ms on 1 second off or something like that. You likely need it's continuous discharge. – Passerby Mar 23 '17 at 20:40

The LEDs require 5V, for a total of 17.28A.

OK, let's round that to 100W.

One 3000mAh 18650 rated for 20A continuous current, as used in e-cigarettes, contains 11 Wh. With two, that's 22 Wh. With 100W power draw you cannot expect more than 15 minutes battery life. Factoring in converter losses, this will most likely be 10 minutes.

I assume you're okay with that, since you say "I need everything to fit in a cylinder of ~20mm". I assume you did your homework, and if you needed more than 10 minutes battery life, you would have specified an adequate size and weight for batteries. It is your problem.

My understanding is that I can use a buck or boost converter (depending how the batteries are wired)

For this level of current, you do not strictly need a multiphase buck converter, but it would make the job easier, use smaller inductors, reduce I2R losses... Ex-National Semiconductors has several chips which will do what you want.

and outboard bypass transistor(s) to shunt additional current around the converter (with proper heat-sinking).

Uhhh? What?

I am curious though, are there considerations for the buck vs boost configurations?

At high output currents, buck converters tend to have higher efficiencies. They are also easier on the batteries, since drawing power from a higher voltage source will require less current in the batteries.

• Do you really expect the converter to eat up 1/3 of the power in the system? And the bypass transistor turns on before the converter hits its max rating, allowing current to flow around the circuit as necessary. I've never used one but you can see an example here. – Brian Faires Mar 24 '17 at 20:00
• A well done DC-DC will have >95% efficiency. Doesn't change the fact that battery life will be tiny. Since you link a linear regulator, I'm quite pessimistic about your ability to pull it off, though... – peufeu Mar 24 '17 at 20:07
• @peufeu I agree. – BeB00 Jan 2 '18 at 4:15
• BrianFaires, If you do the calculation (22wh battery, 90% efficiency, 100W required) you actually get 11.9 minutes. However, the batteries are probably not 22wh. If you look here (powerstream.com/z/LG-18650HE2.png) you can see how the voltage fluctuates. For this battery, at 0.2C discharge it provides about 9.1Wh of energy. At 20A, it's about 8.2Wh, a 10% loss. Factoring this in gives a battery life of about 10 minutes. Your supply is also going to be big. If you dont want to use a large premade one, youll have to design your own. – BeB00 Jan 2 '18 at 4:34
• I would advise against this project at the moment. You're trying to do something that is already not trivial (high power conversion with li-ion), and then fit it into a small space. If you get it wrong, you could damage your equipment, and maybe have the batteries catch fire. I'm 80% sure you wont be able to find a ready made 100W regulator that will fit in a 22mm cylinder, so you'll have to design one. Designing a boost is again not simple, plus you need to maximise efficiency (otherwise your regulator will melt), so it becomes difficult and requires a fair amount of engineering knowledge. – BeB00 Jan 2 '18 at 4:44

This isn't a direct answer to your question, but pointing out something you should look at.

The LEDs require 5V

No, they don't. LEDs that produce visible wavelengths require roughly in the range of 1.6 to 2.5 V. Some "white" LEDs really emit UV and use phosphors to convert that to visible colors, and can require up to 3.5 V. Maybe you have LED modules that require 5 V, but the LEDs certainly don't natively.

Since power requirement is a serious problem in this project, you should look carefully at how to require less power in the first place. Start by defining what kind of light at what intensity you really need, then work backwards to find the sets of LEDs to emit that directly. If possible, drive them with constant current directly from a switcher. Regulating voltage instead of current forces you to put resistor in series with each LED. That drops voltage and therefore wastes power.

Even if you do try to regulate voltage, use a separate rail for each type of LED. Each rail can then be tuned to be only a few 100 mV above the LED nominal voltage so that the series resistor doesn't waste much power. With a fixed 5 V in, there is either another conversion going on, or a considerable fraction of the power is wasted in series resistors.