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I have a light that needs to be powered by battery. It is a 12V light that draws 4.5 A of current. If I purchase a SLA battery that is 12V rated 5.0 Ah would that battery be sufficient? I am confused on the whole Amp/hours rating. Is a 5.0 Ah battery able to deliver 4.5 A of current? And for how long? Will be using deep cycles probably lighting the light for 10 seconds at a time. Any help appreciated.

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  • \$\begingroup\$ It depends what kind of light it is and the V vs I curve. \$\endgroup\$ Sep 13, 2017 at 22:03
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    \$\begingroup\$ Unless you find the spec sheet that tell you the discharge rating, any attempt to infer it from capacity is only as good as a educated guess based on the type of battery. \$\endgroup\$ Sep 13, 2017 at 22:22

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Your selected battery will be good to power your light for about 1 hour of continuous work. If you fire your light for 10 seconds, your battery will last for about 360 cycles.

The total capacity rating ("5 Ah") has little relationship with how much current the battery can deliver. I would guess that your SLA battery will be able to deliver much more than 4.5A.

CORRECTION: Tony is correct, the 5AH is usually based on 20-hours discharge time. This means that to get 5AH capacity, the discharge current shouldn't exceed 250 mA. Here is an example of typical SLA 12V/5AH battery. The datasheet says that the battery will last only 30 minutes under 5.7 A load, so the battery capacity gets down to 2.85 Ah. So at 4.5 A it might last about 40 minutes of continous discharge. Impulse-type discharge, however, is more favorable, so my initial estimate of 360 bursts might be correct.

The "initial current" of 1.5A is a requirement for charging mode. And, BTW, the battery under my reference can deliver 75 A, but for a really short time of 5 seconds.

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  • \$\begingroup\$ It says on the battery "initial current less than 1.5 A". That's where I am confused. Would it deliver the 4.5 A needed when the switch is turned on? \$\endgroup\$
    – NJG
    Sep 13, 2017 at 22:49
  • \$\begingroup\$ no @NJG It is under needed capacity for your load. \$\endgroup\$ Sep 13, 2017 at 22:57
  • \$\begingroup\$ @NJG, yes, if this is a regular commercial-grade SLA battery, it will deliver 4.5 A, easily. See above. \$\endgroup\$ Sep 13, 2017 at 23:28
  • \$\begingroup\$ Ali, I see your battery and data sheet. But this makes it confusing. You say yes, others say no. Here is the battery I was going to purchase: homedepot.com/p/… \$\endgroup\$
    – NJG
    Sep 13, 2017 at 23:46
  • \$\begingroup\$ But others say I should get a higher Ah battery such as 15-20 Ah range instead of 5. I'll only use it for maybe 15-30 times at 10 seconds of use before putting it on a charger. \$\endgroup\$
    – NJG
    Sep 13, 2017 at 23:48
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AH rating = Amps*20 for 20 hour load the standard test time.

   Capacity = "5Ah"   >>>>>>>>  4Ah  >>> 3Ah 

(total hours * Amps = Ah)   
 20h* =    10h*  =     5h*  =      1h*  =     ?56 minutes 
 Amps Ah    Amps Ah    Amps Ah     Amps Ah    Amps  Ah
0.25  5.0   0.43 4.3   0.8  4.0    3.0  3.0   4.5   2.5?

So 10 second bursts of 4.5A may yield on a fresh full charged "PowerSonic 5Ah SLA" about 150 amp-seconds or ~ 33 bursts.

Ref data

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  • \$\begingroup\$ Thanks Tony, however it says on this battery "initial current less than 1.5 A". That's where I am confused. Would it deliver the 4.5 A needed when the switch is turned on? \$\endgroup\$
    – NJG
    Sep 13, 2017 at 22:58
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    \$\begingroup\$ @NJG, why don't you post a link to actual datasheet for your battery? \$\endgroup\$ Sep 13, 2017 at 23:03
  • \$\begingroup\$ There are reasons why it states initial current must not exceed 0.3*C or 1.5A in your case so you need to do this or get a 15Ah rating besides at this rate you only get <50% capacity from SLA's \$\endgroup\$ Sep 13, 2017 at 23:08
  • \$\begingroup\$ read your datasheet or mine if you don't understand. These are not LiPo's \$\endgroup\$ Sep 13, 2017 at 23:11
  • \$\begingroup\$ Ok, so I would need a battery where on the specs it says it can deliver more than 4.5 A on initial current like this one? \$\endgroup\$
    – NJG
    Sep 13, 2017 at 23:24

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