Author Topic: Battery loglivity calculations for a small radio project.  (Read 1046 times)

0 Members and 1 Guest are viewing this topic.

Offline danshtr

  • Contributor
  • Posts: 17
  • Country: us
Battery loglivity calculations for a small radio project.
« on: April 27, 2017, 08:36:38 pm »
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 at https://www.pololu.com/product/2563 and 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 !

 

Offline james_s

  • Super Contributor
  • ***
  • Posts: 9800
  • Country: us
Re: Battery loglivity calculations for a small radio project.
« Reply #1 on: April 27, 2017, 08:41:24 pm »
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.
 
The following users thanked this post: danshtr

Offline ebastler

  • Super Contributor
  • ***
  • Posts: 3040
  • Country: de
Re: Battery loglivity calculations for a small radio project.
« Reply #2 on: April 28, 2017, 12:58:54 am »
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.
 

Offline james_s

  • Super Contributor
  • ***
  • Posts: 9800
  • Country: us
Re: Battery loglivity calculations for a small radio project.
« Reply #3 on: April 28, 2017, 03:10:26 am »
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.
 

Online digsys

  • Supporter
  • ****
  • Posts: 2062
  • Country: au
    • DIGSYS
Re: Battery loglivity calculations for a small radio project.
« Reply #4 on: April 28, 2017, 05:15:49 am »
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.
Hello <tap> <tap> .. is this thing on?
 

Offline sleemanj

  • Super Contributor
  • ***
  • Posts: 2420
  • Country: nz
  • Professional tightwad.
    • The electronics hobby components I sell.
Re: Battery loglivity calculations for a small radio project.
« Reply #5 on: April 28, 2017, 06:17:03 am »
Work in Watts.

Your radio wants 3.3v * 0.0175A = 0.05775 W

Your 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 available

Therefore, 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.

« Last Edit: April 28, 2017, 06:18:38 am by sleemanj »
~~~
EEVBlog Members - get yourself 10% discount off all my electronic components for sale just use the Buy Direct links and use Coupon Code "eevblog" during checkout.  Shipping from New Zealand, international orders welcome :-)
 
The following users thanked this post: danshtr, sokoloff


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf