In relation to this question: Solar power wattage requirement

I'm trying to get to the math of understanding my wattage for my plan solar-setup. I would like anyone here who knew can correct me on my computation.

My main goal for my solar-setup is not much about concerning the type of devices I can plug, but more on the capacity and efficiency.

My setup:

1 80W solar panel as seen in the specs here

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and for storage a 12v 40AH deep cycle battery

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I will be using a battery controller for this, but for the sake of this question, let's leave it out of the equation for now (Or not if it's relevant)

From my reading in the internet, a 12V40AH battery will yeild me a wattage of 480Watts. But given the these kinds of batteries to last longer, the consumption should not go below 40%, so my usage capacity should not exceed 240Watts, am I right?

Battery Capacity: (voltage) * (AH) = total capcity * 0.6 = usage capacity

My next math is how long it needs to charge. Let's assume that I get 6 hours of direct sunlight (equator). I get the advertised, 4.44Amp current from the sun. So if I need to fill the battery for the first time, a 4.44Amp * 6 will give me 26.64AH of current on that time. It will take me 12 Hours of direct Sun in order to fully charge my battery.

Charging capacity: Solar Current * Hours of Sunlight = Total Current Produced

Lastly, I also want to compute to the power consumption of a device I would like to connect it to. In this example, I have a router that is drawing 6 watts and it's open for 24 hours a day. If I'm going to plug it on batteries, I would like to know if my current setup can take this.

Total Power Consumed: 6W router * 24 hours = 144W

Since I have an operating capacity of 240 watts, I could power my router in a battery for a shy of 2 days. Am I getting this correct?

  • 1
    \$\begingroup\$ Can't help you with a full answer, but I notice that you are treating Amps and Amp Hours (and Watts and Watt Hours) as interchangable. They are not. For instance, a 12V, 40 Ah hour battery could hold 480 Wh - but the battery may only be able to give you 1 A of current, in which case you'd get 4 hours of 120W. Another mistake is to treat nominal values as facts. How do you know that after you draw 480 Wh from that battery, it will be at 0%? It may actually already be the usable capacity. \$\endgroup\$ Commented Mar 20, 2015 at 6:39

1 Answer 1


On a very basic level you have it correct(or at least a reasonable estimate), but we can get more detailed and in depth.

First of all, 12v * 40 Ah gives 480 Wh NOT watts. Watt-hours, I.E. 480 watts for 1 hour or 240 watts for 2 hours etc. A technicality that doesn't change your calculations, but might make future calculations less confusing.

Your first assumption about lead acid batteries and not discharging more than 50-60% is, in general, a correct assumption. However, you call it a deep cycle battery, and a good quality deep cycle lead acid battery may tolerate up to 80% depth of discharge. Less discharge is always better, but the better deep cycle batteries do allow a lower discharge and still have a reasonable cycle life. The only way to know is to look at the data sheet from the manufacturer.

Your second assumption is that your battery will simply charge in a linear fashion. Although your calculations for how long it will take to charge your battery are an OK estimate, in reality a lead acid battery charges to 70% in 5-7 hours (if your solar panel can provide enough current) and the remaining 30% in the following 7-10 hours. Lead acid batteries follow a 3 stage charge algorithm (maybe 4 stages depending on your charge controller).

So in your battery if you discharged it 75% = 10a remaining. To get up to 70% need 18a (10+18=28a). 18A÷5h=3.6a/h which your panel is capable of and you said you have at least 6 hours of sunlight and it will take 5 hours to 70% so in the last hour the battery will continue to charge past 70%. So if you only have 6 hours of sunlight and your controller strictly follows the 5 hours to charge to 70% algorithm, then you will not fully charge the battery each day IF you discharge to 75%! In all likelihood your charge controller will take advantage of your panel's ability to provide more than the 3.6 amps and charge the battery a little quicker, but not much - there are limits.

Finally your router. Does it use AC power? If so, then you need an inverter which will convert the battery power to 120 volt AC which you will plug in your router. All inverters are somewhat inefficient and you will lose at least 15% of your solar power to the inverter. So add this to your calculation and you will find out how long your router will be powered.

  • \$\begingroup\$ thanks for the insight. Yes the router will be plugged on an inverter. But is my computation of 6W * 24 correct? \$\endgroup\$
    – Mr A
    Commented Mar 20, 2015 at 8:09
  • \$\begingroup\$ @MrA - In general yes, your calculation of 6W * 24 hours is correct, but I doubt that the 6W number given is what the router is using every second of every hour of every day! If so then your calculation is 100% correct. I suspect that when you are not using your router or when the network traffic is low, it may use less than 6 watts. There is a device called a kill-a-watt that you can plug into the wall socket and then plug the router into that. Then you could see how many watts you are using exactly over any period of time. In other words, the 6W * 24 hours is likely the Maximum used. \$\endgroup\$
    – Filek
    Commented Mar 20, 2015 at 23:22
  • \$\begingroup\$ @MrA - my big answer above was only to say that your calculations are more or less correct, but don't be surprised if when you build the project the numbers don't work out perfectly, especially because of he way batteries charge. Your on the right track so good luck! \$\endgroup\$
    – Filek
    Commented Mar 20, 2015 at 23:26

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