12v Solar travel charger - custom built.. needed help..

[pyother]

Dear trekkers,

I am designing a solar travel charger to charge all my gadgets while in remort places...

i got few doubts... it would be help full for me if i got my doubts clarified..

1) i got 12v 10W solar panel with 0.8a.. and a charge controller reads 12v 10A.. unfortunately i cant carry heavy, big Ah battery, so i selected 12v 2ah sealed lead acid [SLA] battery..

what i afraid now is, will my 12v 10a charge controller is too much for my 12v 2a battery..?


2) whats the Maximum safe Volt & Amp to charge my 12v 2a SLA battery...?


My panel reads 20.1V at full sunlight
The charge controller reads 13.5-14V at full sunlight..
I dont know about the output Amp of the charge controller...


3) as i also have a 12v - 220v inverter with 50w capacity... can i connect it to the "load out" of the charge controller, without using a battery and charge only at full sunshine...?

sorry.. i dont have keen knowledge in electronics so my doubts would be childish..

Regards

 

[Ted_Z]

1. Yes
2. I wouldn't charge faster than C/4 or 0.5A.
3. No
12V 2Ah is a really small battery. I don't think you'll be happy with the performance of a 2ah battery. I would use a 9 ah battery at least (these are only about 4"x2.5"x6"). Questions 1&2 can be fixed with a larger battery.

A couple other points:
* A 10W panel is 10W at peak power which is typically around 16V or 17V, at 13.5V you'll be more like 8W (0.6A).
* With your inverter you'll be pulling close to 4A from the battery at 50W. This translates into about 20 minutes of run time from the battery you've selected.

 

[pyother]

Thanks for the reply Ted,

Can i use resistor to limit the current supply to the battery... if so what resistor should i use..? for the same situation..
as you said that i could get only 0.6A from my panel... was the work of charge regulator increases the current and supply..? or to supply the same current from the panel
:-(

i just need to charge my camera battery from the inverter out, max of 6 - 7 watts.. thats y i selected this battery..

Thanks

 

[dwh]

You won't hurt that battery with 10w of PV - and you'll almost never see 10w into the battery anyway.

Unless the charge controller is an "MPPT" type, it does NOT boost the voltage or current - it merely passes the PV output through to the battery and disconnects when the battery voltage rises to "full charge".

Normally, you do NOT connect inverters to a charge controller "load out". That is for lightweight stuff like a light.

You cannot run an inverter directly from a PV/charge controller - the voltage/current varies too much. You must connect the inverter to a battery which serves as a buffer to stabilize the voltage/current to the inverter.



It sounds as if what you have will work - more or less.

I would expect that perhaps the 10w PV just won't ever get the battery fully charged.
I would also expect the battery to die an early death due to A) not being fully charged, and B) being heavily drained too many times.


[EDIT: Oh...and if that PV module actually puts out 12v - then that's not enough to charge a 12v lead-acid battery. Check the PV - it should have a "Vmp" (Voltage, Max Power) of 15v or more to fully charge a lead-acid battery.]

 

[pyother]

Wow..!

Thats superb... happy to know much about the Solar charger..

My battery is 12v 2A

and the panel outs
Voltage at Pmax (Vmp): 17.2v
Current at Pmax (Imp) : 0.584A
Short Circuit Current: 0.631A

so i hope it will take 5 hours of full sunlight to get charged full... Am I right..?

also i will be using it only to charge my camera batteries input 100v-240v and output 8.4 V and 0.55 A. ie. 4.6 watts for 3 hrs once in 2 days... if 2A is not enough... I could go for 3A..

Thanks-

 

[dwh]

PV modules are rated at a standard temperature - 77 degrees F. Since they are pointed at the sun, they are usually hotter than that and they put out less when they are hot. General rule of thumb is to expect 80% output on average.

Then the battery is not 100% efficient at absorbing power either. So again, figure 80% as a rough estimate of how much will be absorbed out of what you feed it.

So .58a x .8 = .46a x .8 = .37a
2a / .37a (per hour) = 5.4 hours

So yup, you're in the ballpark with your charging guesstimate.


As for taking power back out; The inverter has losses as well, so another rough rule of thumb is that whatever the battery has, you'll get around 80% of that as useful watts after passing through the inverter (and camera charger, which also has losses). So factor the load (0.55a) by 1.2 to figure the run-time...

.55a x 1.2 = .66a
2a / .66a = 3 (hours run-time)


Looks about right.

 

[pyother]

Awesome...

Thanks sir, and thanks to Ted, for your esteemed support, Iam now equipped good for a 8 day trek into himalayan wilderness... Hurry...!

regards-