Battery loglivity calculations for a small radio project. - Page 1
Forum section: electronics / beginners

author: danshtr
author_rank: Contributor
timestamp: April 27, 2017, 08:36:38 pm
content: Hi All.I have a radio project (LORA) which consumes in average 17.5ma and with 200ms bursts of 150ma every 5 seconds (accounted for in the 16ma average) . All at 3.3v.I want to keep it compact but lasting 24 hours.What battery/regulator combo do you suggest?I would love to use a single AAA battery with step up, but I don't know how to calculate the waste. Do I just look at the graph athttps://www.pololu.com/product/2563and see that with 1.5v battery the efficiency is between 70% to 80%. So 800mah (of AAA battery) * 0.7 (efficiency of the stepup) / 17.5 = 36 hours?Am I missing something?Thanks !

author: james_s
author_rank: Super Contributor
timestamp: April 27, 2017, 08:41:24 pm
content: That should get you pretty close. Given the low draw, the battery should deliver very close to if not a bit over the rated capacity. I'd say build a prototype and run it from a battery until it stops working. I don't think you'll have much trouble getting over 24 hours.

author: ebastler
author_rank: Super Contributor
timestamp: April 28, 2017, 12:58:54 am
content: I think you overlooked the impact of the actual step-up conversion. Let's assume you had a perfect, 100% efficient step-up converter. In order to provide 1 mA @ 3.3V, that converter would draw 1 mA * 3.3/1.5 = 2.2 mA @ 1.5V. I don't see that represented in your calculations. The losses due to real-world efficiency below 100% then come on top of this factor.

author: james_s
author_rank: Super Contributor
timestamp: April 28, 2017, 03:10:26 am
content: Oops, that explains that feeling I had in the back of my mind. You'll need to measure (or calculate) the current drawn by the converter, it's looking like 800mAh AAA won't get you there but a 2,000mAh AA ought to. There are small lithium batteries that might be worth looking into as well, the higher voltage would eliminate the need for a boost converter.

author: digsys
author_rank: Unknown
timestamp: April 28, 2017, 05:15:49 am
content: I agree with the lithium cell path. Even if you used a LDO linear regulator, you're still running at 90% efficiency, with a simple circuit. They have much higher energy capacity as well.For the low energy required, you could use a switched capacitor buck cct, usually run at 98%+, to reclaim most the LDO loss. You'd need a reasonable 3.3V storage cap though. Perfect.

author: sleemanj
author_rank: Super Contributor
timestamp: April 28, 2017, 06:17:03 am
content: Work in Watts.Your radio wants 3.3v * 0.0175A = 0.05775 WYour cell has 1.5v * 0.8Ah = 1.2Wh (ignoring discharge curve!)A boost converter will lose probably 30% from the cell (which includes current to run the converter, and losses inherent in conversion)So we can simply reduce the cell's capacity to account for those losses and call it 1.2Wh * 0.7 = 0.84Wh availableTherefore, after adjustment for losses of 0.84Wh available, and your load needs 0.05775W, you get to run for... 14 and a half hours by the numbers. Real world may vary.You can of course work backwards too, if you want to run for 24 hours, you know you need 0.05775W * 24h = 1.386Wh of power, increase that for those 30% losses, 1.386Wh / 0.7 = 1.98Wh your cell has to provided to the boost converter. So if your nominal voltage is 1.5v then, you need 1.98Wh / 1.5v = 1.32Ah worth of cells (which is probably easily in the vicinity of 2 decent AA cells in parallel) to run for 24 hours.