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Maker Pro

Need help with some basic electronics, please.

I'm working on a project that involves running a wireless video camera by using a solar panel and rechargable batteries as back up.
The camera runs on 5V, the solar panel is rated for 9V and I'm going to use 6AA rechargable batteries.
I need help with how to connect these so the solar panel can also charge the batteries while running the camera at the same time due to excess power output during sunny days.
I know that I will need a couple of diodes, voltage regulator etc.
Can someone explain/show me how to put this together in the most efficient way?
I do have a breadboard I can use to test this circuit out with.
You probably need some more information from me, but this is all I can think of right now.
Thanks!
 
We Need amps and or watts for the device load, and the panel.

In/out diodes on the battery is one way. Use the battery as a regulator. Probably more efficient ways though. You might even be able to put the whole lot in series with no other components? I wouldn't advise it though.

Once we know amps and volts we can help you make the circuit you need. Should be pretty simple.
 
Solar panel:
Power output: 1W
Voltage output: 9V -5%
Current output: 112mA -5%

Camera:
Power output: 5V , 2A
Power consumption: 5W


The reason I'm going to use 6AA rechargable is that I was told that rechargable AA only puts out about 1.2 to 1.3V . 6x1.2= 7.2 V.
That way I know it has enough to power the camera as long as I put a voltage regulator between them, or maybe the small difference in Voltage will make it ok without one?
It also means that the solar panel will be able to charge the batteries when it is putting out more than 7.2V. Should I put the voltage regulator between the panel and the batteries instead?

Do I need to go bigger on the W on the solar panel, or does that just mean how long it will take to charge the batteries? Will a 2W panel shorten the charge time?

I also believe that the batteries I will be using are like 2000mAh, or more.
Thanks!
 
Solar panel:
Power output: 1W
.
.
.
.
Camera:
Power consumption: 5W

Do you see the issue here? In FULL light at OPTIMUM performance the solar panel will only produce 1/5 of the power needed to run the camera... Realistically the panel is only producing maybe about 1/20 or so of the power to run the camera when you factor in daylight hours and other loses... Maybe a little better, probably worse since the camera numbers you posted really don't match up, 5v@2A is 10 Watts not 5 Watts...

That means there will never be any surplus charging of the batteries at all, the camera will be draining the batteries at a rate of give or take 20 times the charge rate...

Your 2000mAh batteries might last about 20-30 minutes in this application...

Hook up 25-30 of those solar panels and you might be in the ballpark to a sustainable power source if the battery capacity will last for the duration of the 'dark' times... The batteries will need to be rated at upwards of 60+ aH to coast for 16 hours or so of 'darkness'

BTW those are just numbers off the top of my head real quick, but the point remains you are on opposite sides of a field in getting this to work...
 
Not sure about the 2A. I was just pulling numbers that was on the box. For the input, it says: 100-240v~ 50/60Hz, 0.5A. ?
Output: 5V. 2.0A ?
It is a wireless mini IR camera, so it shouldn't take that much to run it.
I have run a similar mini camera with the solar panel itself with no problem. Even in cloudy conditions.
 
After some more research I found that the camera is pulling only about 350-500mA, but I guess I have to find a solar panel that puts out 5W.
I found a12V 5W solar panel. Will that do the trick?
Would I burn up a 5V voltage regulator with that type of panel?
What should I do to make sure the 6 AA battery pack doesn't get damaged?
Thanks!
 
So the camera is using 5V x .5A = 2.5W and the solar panel is producing 5W. This is enough to run the camera in full sunlight, but only if you use a switching regulator, a linear regulator would eat up 58% of the power which does not leave enough to run the camera.

So with a 90% efficient regulator you would need 2.8W to run the camera. This leaves you 2.2W to charge the batteries.

With a regulator and the typical charging effiiciency of batteries this would give you maybe 1.5W worth of power to charge the batteries. So for each hour of full sunlight you would be able to capture enough energy to run the camera for an additioanal 36 minutes. And to achieve this you would need an efficient design.

How much time do you expect to run the camera each day? You have enough to run it about 8 + 4.8 = 12.8 hours on each day that you have full sunlight. For 24/7 operation, I still think you are off by at least a factor of 2, and probably more like a factor or 4.

Bob
 
Thanks everyone so far. I think I'm getting closer to figure this out.
I was planning on running it 24/7, but it looks like that is not going to be possible.
Guess I can install a simple switch to turn it off when I don't need it running, or if there is a switch that can sense when the panel isn't putting out enough (like at night) that would automatically turn it off.?
Also, could you guys put together a schematic on how these things should be connected and email it to me? I think I have a pretty good idea, but I'd still appreciate it.( PM me for email adress).
So, I'll have the camera, solar panel, battery pack, couple of 12volt zener diodes, 5volt regulator and some sort of switch.
Thanks!
 
I was just pulling numbers that was on the box. For the input, it says: 100-240v~ 50/60Hz, 0.5A. ?
Output: 5V. 2.0A ?

Box or power pack? Bob just because it uses .5a @ 250v.... tells us nothing... That's mains load of a transformer- dc stepdown.... Output of witch seems to be 5v2a. Or 10w.

It puts out 2A probably to charge a Li battery in the device... I.d guess you'd want to supply about the same to keep it running. It may run on lower amps, once the battery is fully charged. Wonder what kind of charging circuit the camera has inside to control the Li?

Sounds like you need 5v constant with variable amps, if you want to try and run it on less when charged/no battery fitted. Until you can verify if it will draw less than 2a (which does seem like a lot) I would assume you need 10w panel. More if you want the batteries to charge as well. Unless you are prepared to charge a lot more than film..

Got a Multimeter? Can you measure what the device actually uses? Ask for help if your not sure you can do it right.. I'd test in a number of modes/functions. Zooming the lens for eg may need more power than other operations. I know my camera chews the battery with GPS tagging on...
 
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