I want to power this device, which is a tplink wr703n. It is a small travel router. When I looked at the specs, it says that it uses 100mA for wifi, and 85mA to run it. This is incredibly low. Now, what I want to do, to run this device along with a webcam and a flash drive (a flashed openwrt, and flash drive is acting as the hard drive storage).

I currently have it running on a 5V / 1A usb port. The battery pack is 10,000mA (10A)
at the time of this post it has been 11 hours (edit; it just died past the mark when I start it by the minute) since I started this. It is down to <25% battery.

I am looking at solar panel / battery(ies) to power this for 24/7. If I want to know what I will need would I take 10,000 / total hours (running device with webcam streaming) to get the mA it draws per hour?

So, lets say it ran for 11 hours, so 10,000/11 = 909mA?

Seems about right. Now to get watts.

I would do w = v * a, so 5 * 909 = 4.5 watts.

Now to get how many watts for 24 hours would I do 4.5 * 24 = 108 watts? That seems high. Well, then again the sun is only out (in my area) would be 11 hours average.
http://aa.usno.navy.mil/cgi-bin/aa_durtablew.pl .

But maybe only 4-6 peak sun hours (if that is the correct term)

So we will say around 9 hours (just for the day not just peak) 4.5 * 9 = 40.5 watts .

If that is true, that were a pretty big solar panel for such small device.

Now this part is what I definitely help with. Battery. I was on a website looking at their 3.7V 6600mA Li-ion batteries, and made me question would I really want to run my stuff on 3.7V?

If I want to run smaller solar panel would I need a bigger battery to even it out? How would I determine adequate battery for the project?

  • \$\begingroup\$ For a load of 4.5 watts continuously, given 9 hours to recharge, one needs (4.5*3600*24)/(9*3600) or 12 watts to recharge the battery and power the load. If 6 hours, 18 watts are needed. If 4 hours, 27 watts. \$\endgroup\$
    – Dan D.
    Commented Oct 1, 2014 at 6:44
  • \$\begingroup\$ Thank you for the comment, how would I determine battery size for this setup, too? edit; I came across this solar panel that is 6v 5.6watts, just curious is that 5.6 watts per hour or the day? here is the link adafruit.com/products/1525 \$\endgroup\$ Commented Oct 1, 2014 at 7:40
  • \$\begingroup\$ Your units are confused: you multiply watts by hours and declare an answer in watts, which should be in watt-hours. Similarly amp-hours for the battery. This is why your 40.5 number is meaningless. \$\endgroup\$
    – pjc50
    Commented Oct 1, 2014 at 8:45
  • \$\begingroup\$ Insolation levels by month and area worldwide www.gaisma.com \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 10:53
  • 1
    \$\begingroup\$ Assume your 4.5W load is correct and steady. Call it 5W. In a day thats 24 x 5 = 120 Watt hours. See www.gaisma.com for sun hours. Not knowing where you are allows not a good figure. Assume mid winter sunshine hours = 2 hours full sun equivalent a day. That's typical US from Houston to well up the coast. Half that in NYNY. SO 120 Wh/2 SSH = 60 W. That's at 100% with optimum pointing and perfect battery charging and perfect energy recovery from battery. You'll get maybe 50% actual. So panel is 60W/50% = 120 W panel [!!!!!!!!!!!!!!!!!]. \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 11:00

2 Answers 2


Your units are confused. For reference: $$Power(W)=Voltage(V)*Current(A)$$ $$Battery~Capacity(Ah)=Current(A)*Time(h)$$ $$Energy(Wh)=Power(W)*Time(h)=Battery~Capacity(Ah)*Voltage(V)$$

For solar panel output,

  1. calculate total energy of one day
  2. divide 1 by minimum peak sun hours
  3. divide 2 by about 0.85

For battery capacity, you have to consider the day when there is no sun. Multiply 1 by longest no-sun period length.

  • \$\begingroup\$ Insolation levels by month and area worldwide www.gaisma.com \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 10:52

Load: 5V x 900 mA = 4.5W.
Call it 5W. That's power.

In 24 hours that's 24 x 5 = 120 Watt.hours.
WH daily energy = 120 ...... 1
That's energy

Despite the sky being bright for many hours the equivalent full sunshine hours a day are typically 4 to 5 annual average, 2h/day in Winter, 5-6h/day in summer.
Kabul Afghanistan is about 6.5/d in summer and 4 or 5 in winter. This is great for solar but there are drawbacks with the location.

To find daily SSH for your location web search: gaisma city_name
or look at www.gaisma.com

eg Houston, Texas, USA.
2.34 SSH/day in December. That is with optimum orientation all day - ie tracked panels.

THEN you must store the energy in a battery. Then you must recover it.

Overall figures on about 50% of panel rating as available energy out of battery - and that can be optimistic if care is not taken.

Battery energy availability = 50% ...... 2

From above we need 120 Watt.hours of energy.
Assume 2 SSH a day of equivalent full power sun at your site mid-winter.
SSH/d minimum = 2 ...... 3

Win = Wh per day /SSH = 120/ 2 = (1)/(3) = 60 Watts ...... 4

At 50% available 60/50% = (4)/(2) = 120 Watts. ...... 5

A 120 W panel is needed !!!!!!!!

And that will operated "hand to mouth" in winter. Get a bad day and you will run out of power next evening.

The truly serious would probably use a 200-250W panel. Or larger !!!

Reducing panel size:
If the web cam does not have to operate 24/7 then a motion detector or timer etc will greatly help the energy needed. The router uses about 10% of the total.


Battery Watt hours for one day of operation:

Battery Wh = Vbat x Ah of bat at time period in specs.

For 5W x 24 h you need 120 Wh/day.

If you have a 5V battery (you don't) you need
~= 120/5 = 24 Ah of capacity (!)
When the PV panel is on-sun battery is charging and not discharging but in winter that is not most of the time.

Also battery must tolerate 2h charge time AND be capable of it.
LiIon and LiPo will NOT 100% charge in 2 hours.
They allow about 70%-=80% in one hour but take about 3h to 100%, so you need > 24 Ah if you charge in 2h in winter.

If you used a 12V battery and a buck converter the 120 Wh/day is equal to 12V at 10 AH.
Then add factor of 2x so battery is not deep discharged each day.
4x is better.

Then add buck converter efficiency of say 75% to 5V.
Partially this has been allowed for in my prior figurings as has panel to battery losses.
Minimum = 10 Ah x 2 = 20 Ah to sto deep discharge daily.
Better = 10 Ah x 4 = 40 Ah.
This will discharge 25% on an average winters day at 2 SSH/dsy, give you several days reserve usually and be vast overkill in summer.
40 Ah is a large car battery.

So 120 W PV panel
12V 40Ah battery. Deep cycle rated if possible.
12V-5V buck converter. Charge controller - battery will be overcharged in summer without a proper controller.

OR reduce web-cam mean power substantially by whatever means.


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