solar/battery powered wifi camera setup advice

so i want to setup a remote wifi camera, battery powered with solar charging.

Proj Plan: choose smallest setup battery/solar panel / camera that will achieve my goal

Goal: Daytime only monitoring , motion detection and occasionally (2 times day connect and stream for few minutes), camera can be off from sunset to dawn.

some research on web shows the cameras i am looking at vary in range of

5-6V @ 250 -- 750 mA current where camera is (On/No WiFi connection) to (On/Wifi connected/ streaming video / panning / zooming)

The average seems to be around 500 mA (On/Wifi connected / motion control enabled)

So, is the problem as simple as some math of amp hours and power output ?

12 hours daylight gives 12 x 500mA avg camera current = 6000 mA for daytime operation.

so if it is powered by a solar panel it must be ?? capable of producing

6000 mA over the 12 hours ??

If have a 5w 12v solar panel that gives me .6 A of 12 v for 12 hours =

7200 mA power ? which should cover the power requirements ?

However , from experimenting with the solar panel it seems to fluctuate its power output and since the camera is not going to be happy with that for on/off/pan/zoom etc i will likely need a battery for buffering camera power draw fluctuations ??

I suppose then i need some sort of proper regulator to connect the solar panel to and manage charging the battery ?

Is that it ? how to size the battery ?? do i compare solar panel output to required camera and then subtract.

solar panel is capable of say 4000 mA hour reliable power so i need 6000 mA - 4000 mA ==> 2000 mA hour battery ??

much advice appreciated

robb

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Reply to
robb
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Wifi doesn't reach very far. Can't you run wire? CAT5 and PoE maybe?

I've run a USB camera 75 feet from a PC, with two active cables.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

Don't forget cloudy days. Which means instead of a 6Ahr system You'll need a 6Ahr times the number of cloudy days, 4 days = 36Ahr. You'll need more solar charging capacity. The shunt type regulators are inexpensive.

Cheers

Reply to
Martin Riddle

That's 6000mA/Hr -- loosely speaking, 6000mA for one hour, 3000mA for 2 hours,

500mA for 12 hours. But, you also need to ensure you have that ampacity at the *voltage* you require.

No. That assumes the panel produces exactly what the camera consumes *when* it consumes it.

E.g., if the sun "goes behind" a cloud, solar panel output will fall. But, camera requirements will *not*. You either figure out a way of storing power to bridge those times when the panel's output is not sufficient to cover the camera's needs (e.g., battery) *or* increase the capacity of the panel so that even in "low light conditions" it meets those needs.

Note that storing energy isn't a lossless proposition. You won't get *out* as much as you put *in*! So, there are inefficiencies that need to be accommodated in your calculations.

Also, the panel's output may not be entirely usable "as is" (with the current load -- battery+camera). Ideally, you would use a MPPT charger to operate the panel at it's "sweet spot" given current conditions. This, of course, adds still more inefficiencies to the calculation...

No way of knowing. Your camera is ~5V @ 500mA -- 2.5W. Theoretically, you'd only need ~200mA from a 12V panel (0.2A * 12V = 2.4W). But, that assumes 100% conversion (from 12V to 5V) efficiency.

Yes. And the camera's "draw" will vary based on what you are commanding it to do (e.g., PZT)

Yes. And efficiency losses there...

Reply to
Don Y

That would be 6000maH.

The solar radiation curves are always a hump. I've never seen them plateau. So I assume there is some fudge factor to determine how to get an average radiation.

A 5W 12V panel must be rated at some solar radiation level. You need to compensate for the actual solar radiation.

There are COTS charge controllers.

For lead acid, the 7AHr cells are about as small as it gets. I would just use one of those.

The battery doesn't generate power, it merely stores it. So you need to size this based on the solar cell. There is loss in the charge controller, loss in charging the battery, and loss in pulling energy out of the batter.

When you make a "one off", it pays to over engineer the project. Otherwise you end up junking components for different components. When you design for a million units, then you have to fine tune.

Check out the DOE solar charts:

Buy the time you price this all out, POE might be the way to go.

Reply to
miso

Yes, the math is simple. The numbers you plug into the equations are NOT simple.

What does the camera do when you dribble the power down until it malfunctions? Does it come back the next day? Is the storage corrupted? How do you plan to turn it on/off daily?

If you have a battery to even out the power availability.

Yes, if you have such a panel with those measured results.

If those are the numbers on the box, it ain't nearly big enough. The numbers on the box are what you get at high noon on the brightest day of the year when there are no clouds, no dust or bird crap on the panel etc.

Depending on your location, google "insolation" to see if there are maps/graphs that describe how much actual sun power you can get on an average day in a given month for your location. You'll find that it's WAY less than your calculations suggest. And there will be overcast days/weeks and dust and bird poop. You're gonna need a LOT bigger solar array than you think and a LOT bigger battery than you'd think.

Solar power makes sense ONLY if there are no other alternatives. Nobody should dissuade you from whatever hobby you choose. Just do the math before you go spending money on stuff that may turn out to be inadequate.

You need to carefully examine your reasons for wanting this. If you're monitoring animal traffic, it's relatively simple. If it's a property security monitor, you may find that the solar camera system is the most attractive thing on the property and won't be there in the morning.

It's a LOT more complicated and expensive than you think.

Reply to
mike

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