I'm working on a network setup for the middle of no where. I want to try and use solar to run the network hop completely. I have never built anything like this before so I wanted to run it by some people who will know more than me. Please let me know where I'm wrong here and help me understand if I'm making a mistake with the calcs. Thanks.

So the setup has 3 network devices.

I'm using an UBNT Rocket AC PTP radio on the backhaul. 24V, 0.5A Gigabit PoE Adapter Max. Power Consumption 8.5W

and a Rocket AC PTMP Radio on the AP Also 24V, 0.5A Gigabit PoE Adapter Max. Power Consumption 8.5W

To power both the units and provide some basic routing and monitoring I'm using an UBNT Tough Switch 5 POE 24VDC, 2.5A Power Adapter Max. Power Consumption 60 W PoE Out Voltage Range 22-24VDC

At most the switch will draw 60W. Each POE can do up to 8.5W max so 5x8.5 = 42.5W. The switch probably draws 17.5W (Probably less but I'd rather overestimate.)

So at most I should be drawing 8.5W + 8.5W + 17.5W = 34.5W.

34.5W / 24V = 1.4375amps. I want this to run 24 hours a day so that's 34.5ah. Because I don't want my batteries to drain more than half and because we will have cloudy days I will use 50ah batteries (2 of them for 24 v). That will give me 50ah at peak usage.

I will mount the switch and the batteries and charging unit in a vented box at the bottom of the pole. The pole will be 40ft high and have both the radios at the top with cable ran down to the box. On the pole I'm mounting 2 50w 12v Solar Panels.

50w at 12v = 4.16amps of output. 4.16 x 2 panels = 8.32 amps per hour. It won't be 100% efficient so let's say 70% so figure 5.824amps per hour.

I'll only get that, at best 7 hours a day at the lowest point so that generates 40.768amps. Which is more than my useage for the entire day of 34.5ah.

Even at 6 hours of sunlight I'm at 34.9ah so I should still be ok?

This should work shouldn't it?

Thanks for your time and thoughts.

  • \$\begingroup\$ Go through your calculations again, paying close attention to the units. One clue: "amps per hour" is a nonsense unit. Also, make sure you're adding up amps at the same voltage. You should probably connect your panels in series, which means that they'll supply a peak of 4.16 A @ 24V. \$\endgroup\$
    – Dave Tweed
    Commented Feb 4, 2017 at 20:29
  • \$\begingroup\$ So if a 12v battery is rated at 30ah does that not mean it provides 30amps of power at 12v? And if a device draws 1.5 amps at 12v would that battery not be good for 20hours of service if drained to 100% and assuming everything else was perfect? I'm not understanding that part I guess. \$\endgroup\$
    – ahackney
    Commented Feb 4, 2017 at 20:55

1 Answer 1


You're confusing yourself with amps and volts, so it would probably be a good idea to just stick with watts, and convert at the end.

You say your total load is 35 W, so you need 35 W × 24 h = 840 W-h per day of operation. I would recommend having enough battery capacity for at least 3 days of operation, or about 2500 W-h. (This would be 100 A-h at 24 V.)

You'll want the batteries to recharge within a day, even if they're substantially depleted. Battery efficiency is on the order of 70%, so you'll need 2500 / .7 = 3600 W-h of energy for the recharge. If you get a minimum of 6 h of sunlight in a day, you'll need 3600 W-h / 6 h = 600 W of panels.

So I would say that your estimates are very much on the low side. You can either expand the power system, or find lower-power devices to run the application.

  • \$\begingroup\$ I could definitely reduce the power consumption by not using the switch to power the radios but it looks like Id still need more charging power. \$\endgroup\$
    – ahackney
    Commented Feb 4, 2017 at 20:56

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