# Suggestion for correct battery pack

I am trying to run 2 12V Geared DC motors which has No-load current = 800 mA(Max), Load current = upto 9.5 A(Max). Runtime to be atleast 3-4 hours.
The motor takes about 10-12 V for operation.
I need a proper battery pack for these, but how can I determine the specs I should go for?

• This is like the 5th question from you about a motor and a battery. No problem there, ask as many questions as you want. But none of your questions has an accepted answer. – Bending Unit 22 May 25 '16 at 13:20
• @BendingUnit22 actually I solved most of them on my own. I agree I've been irresponsible and irritating but rookie here, this community is the only help I got so, my apologies for being such a clumsy fool... – YaddyVirus May 25 '16 at 15:42
• Most of the batteries on the linked page are 11.1V Li-Poly. I'm not sure I would recommend those, unless you were experienced with that sort of thing. I would go for a 12V ni-mh pack made for model cars. Or maybe a simple 12V lead-acid battery (of the leakproof gel-cell type) if you don't mind a little extra weight. – Andy May 25 '16 at 16:09
• How are you a fool if you managed to solve your problems? Congratulations. There's nothing wrong with posting an answer to your own question. It's actually encouraged. This way you show others what worked for you and create a valuable resource for others. – Bending Unit 22 May 25 '16 at 16:23
• @BendingUnit22 thanks for that bro, I'll make sure I update all my questions... I'm a bit lazy as well though... :P – YaddyVirus May 25 '16 at 17:11

You haven't given a nominal (average or anticipated "normal") current draw for what you expect during the routine operation, so what I'm going to do will probably give you an over-specified battery pack for what you need.

Typically you would have some "worst case" load (all your sensors powered on, microcontroller powered on, all motors at full load, etc.) - this will provide the worst-case current draw and will establish what kind of a "C" rating the battery pack needs to have (more on this later).

Also, you typically aren't going to be operating at worst-case all the time, so you assume maybe your average duty cycle has all your electronics on and maybe your motors at 60% - again, this would be the average motor power during the entire time your device is powered on. This would establish what total capacity was required to provide the desired run time.

A "C" rate is the current that would drain (or charge) 100% of battery capacity in 1 hour. So, for a 1000mAh battery, 1C is (1000mAh/1h) = 1000mA, or 1A. Different battery styles and chemistries have different charging and discharging C rates, or maximum allowable current limits. The nice thing about using a C rate instead of an actual current limit is that it scales correctly for all batteries of a particular style and chemistry, regardless of capacity.

So, the easiest way to determine capacity is to figure your nominal total load and multiply that by your run time requirement. The best way to work with load is to multiply the current for each component by its operating voltage and get an operating power. Sum all of your component power requirements, then divide that by the battery pack's nominal voltage. Once you get a total current requirement in terms of your battery voltage, then multiply that by your desired run time to get required battery capacity.

In your case, you don't give an average load current, just a maximum current. So, based on that, you need a $(12\mbox{V})*(9.5\mbox{A})*(4\mbox{h}) = 456\mbox{Wh}$ battery. This is a significant amount of capacity.

If you are using a lead-acid battery, with a nominal voltage of 12V, then you need a capacity of 38Ah. If you are using a LiPO battery, with a nominal cell voltage of 3.6V, then you would need 3 (10.8V) or 4 (14.4V) in series (3S or 4S pack), and your required capacity is 42.2Ah or 31.7Ah, respectively.

Depending on how detailed you want to be with your battery pack specifications (or how stringent your requirements are) you may want to dig a little deeper with the pack design.

For instance, quality battery manufacturers should provide a curve that shows battery capacity versus discharge limits. As battery discharge current increases, the pack heats up and the internal resistance of the battery increases. This means that more power is wasted in the battery pack, reducing the effective output capacity of the battery.

More important even than that is checking the battery cycles versus depth of discharge. Some batteries could give you 2000 cycles if you discharge down to 40 percent state of charge but only 500 cycles if you went to 20 percent state of charge. Then it's up to you to determine what the benefit is of driving to a lower state of charge.

For instance, if you need to have 10Ah of capacity, you could either go to a (10/0.6) = 16.6Ah battery to go to a 40% state of charge during one "run" or you could use a (10/0.8) = 12.5Ah battery and push to a 20% state of charge. Both packs get you the capacity you need, but you'll wear out the smaller pack first because you're driving to a deeper depth of discharge. However, the benefits (reduced weight, reduced volume) may outweigh the cost (literally, cost of replacing the packs on a 4:1 basis) of using the larger pack. The smaller pack may also actually be less expensive than the larger pack, which could also help justify that decision.

I'll add too that, while you may commonly "float" a lead-acid battery to near 100% state of charge, it's bad (read:fires happen) when you try to do the same for lithium chemistry batteries. That is, lithium batteries may typically only ever get charged to 90% state of charge.

So, in summary,