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I've developed a simple circuit using a Photon to send data to my database. It has to be a portable device and because of this I looked into lead batteries. My circuit runs on 5V and draws a maximum of 1A per day. I need it to run for about 2-3 weeks, continuously.

When looking at lead batteries, I see a lot about nominal capacity. For example: with this battery the nominal capacity at 1Ah at around 5.25V gives a C20 value of 20A.

Can this information give me any kind of indication about how long the battery will last if I draw only 1A per day? In an ideal situation, if I only draw 1A per 20 hours, this would mean I could run my circuit for 20 times 20 hours so a little over two weeks. But I doubt this is an ideal situation. What does this C20 tell me about a longer usage of the battery, if anything?

Based on the information of my circuit, what kind of battery can you recommend? I'm aware I need a buck converter as well, which power dissipation I have not taken into account for my calculations yet.

Thanks for any insights you can give me.

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    \$\begingroup\$ You have your units all mixed up. You can't draw "1A per day", and battery capacity is measured in Ah. So look at the current consumption of your circuit (1A) and look at your battery capacity (20Ah) -> 20Ah/1A = 20h. So that's less than a day, not 2 weeks. \$\endgroup\$ – brhans Jul 19 '16 at 12:34
  • \$\begingroup\$ Thanks for your response. But afaik, I don't. My circuit draws about 90mAh and goes into a sleep mode at night. That's why I've calculated it at about 1A per day. But let's assume for sake of argument it draws 100mAh continuously without going into a low power mode. What answers can you give me based on my question? \$\endgroup\$ – Len Jul 19 '16 at 12:42
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    \$\begingroup\$ "A per day" is not a meaningful set of units. If you actually mean 1Ah per day then that would make more sense. \$\endgroup\$ – brhans Jul 19 '16 at 12:44
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    \$\begingroup\$ As I said in my first comment, you're getting your units mixed up. If your circuit runs at 50*mA* then it uses 1200*mAh* per day (since 50mA x 24h = 1200mAh). The C20 capacity of a battery is the capacity in mAh if you are discharging it at 1/20th of its standard rated capacity. \$\endgroup\$ – brhans Jul 19 '16 at 12:57
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    \$\begingroup\$ At 100mA (or 0.1A) you will get 20 / 0.1, ie approximately 200 hours backup. When the current drain drops below about 100mA you have to take into account the self-discharge current of the battery. This is rarely specified in the data sheet, but for a battery of this size could be 20mA to 50mA depending on a lot of factors. You would add the self-discharge current to your current drain, so for a 50mA current drain this would be equivalent to say 70mA. This gives 20 / 0.07 or 286 hours rather than the 400 hours that you may otherwise expect. \$\endgroup\$ – Steve G Jul 19 '16 at 13:15
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The C rating often used by RC LiPos is telling you about the ability to deliver the current mentioned. In your case 20A which will empty the Battery in less than 1/20 of an hour so after 3min you're done. The capacity you mentioned is 1Ah so given you draw 1A you'll get 1h of fun (at best). The math isn't really complicated. You mentioned that you expect the average current might be around 90mA so here you go 1Ah/0.09A=11h then your battery is empty. The C rating isn't at all important in your case.

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  • \$\begingroup\$ C20 is like C/20 which means capacity for 20h discharge, mostly given for Lead-Acid batteries. You are talking about 20C (given for high rating Lithium batteries) which means discharge with current 20 times the capacity rating. The difference is 400 times ;) \$\endgroup\$ – Todor Simeonov Jun 8 '17 at 18:49

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