[solved] Can someone help me pick the right battery/voltage regulator combo?

Hello,

I'm working on a project using an arduino Yun and there no regulator on it. So I started looking for battery, regulators, step-down/step-up regulator, etc... and now I just don't know anymore.

Here's my two founds after excluding the L7805 from my choices because of the 2V step-down thing.
StepUp/Down 5V regulator: looks pretty cool
http://www.coolcomponents.co.uk/5v-step-up-step-down-voltage-regulator-s7v8f5-1135.html
Battery 3.7V at 2000mAh
http://www.coolcomponents.co.uk/lithium-polymer-battery-2000mah.html

What do you guys think of that combo?

Also, a technical question. I always though that the higher the mAh the longer the battery will last before charge (obviously depending on what's sucking on it) but that feels a bit naive. I think I read during my search that higher voltage regulated to lower voltage will last longer than low voltage regulated to higher voltage. Even if the latter has a higher mah.

So I'm confused....and the analogy of the water only took me so far in understanding electricity.

I hope someone can help me pick the right one.

Cheers

ps: The arduino Yun seems to need around 250mah, a photodiode and IR led, I might also have a 2 or 3 leds, and maybe four IV-9 numitron, which need something like 3.5V at 20mah apparently. So a total around 350mah maybe, hope that helps.

The step up/step down regulator you linked is a switching regulator and can make good use of a battery of any voltage. The other type of regulator, a linear regulator, needs to have the battery voltage matched more closely to the regulated voltage -- preferrably ~2V above. The limitations of linear regulators is likely the source of your confusion.

So, assuming you're using a switching regulator, the best way to measure the capacity of a battery of a given voltage and Ah rating is by converting to watt-hours. As an example, your 3.7V battery with 2000mah (2Ah) capacity contains (3.7 * 2 =) 7.4 watt-hours of power. The s7v8f5 regulator suggests 90% efficiency so you'll get (.9 * 7.4 =) 6.66 watt hours of power. Converting that to 5V, you'll get (6.66 / 5 =) 1.3Ah of current, or running at .35A it would run for (1.3 / .35 =) 3.71 hours.

Hi Chagrin,

Thanks a lot for your very clear explanation and taking the time to write down the whole process. I'm going to have to bookmark your answer for future projects!

I guess I'm going to need a bigger battery for this project as 4h of autonomy won't be enough. I was hoping to find a way to get 5 days of autonomy, that was a bit naive of me. I'm not even sure that will be possible.

I've learned something today and I will be going to bed less stupid :slight_smile:

Cheers!