# Learning by Starting Small

jhrusky
Registered Users Posts:

**7**✭
Can someone help my put together a small system so I can play a bit with solar and learn more about it?

I'd like to buy 1 panel, charge controller, battery and inverter and the cabling I need to hook them together so I can experiment.

I'm unsure if I should get 2 batteries / panel, or 2 panels / battery ... I'd like to be able to add to it if possible, but not required if

the cost is going to be prohibitive. Yes, I know I won't power my house or anything with it other than, perhaps, my cell phone but

I have to start somewhere

Recommendations on panel size and voltage, charge controller type, inverter size, and battery size.

Thank you.

I'd like to buy 1 panel, charge controller, battery and inverter and the cabling I need to hook them together so I can experiment.

I'm unsure if I should get 2 batteries / panel, or 2 panels / battery ... I'd like to be able to add to it if possible, but not required if

the cost is going to be prohibitive. Yes, I know I won't power my house or anything with it other than, perhaps, my cell phone but

I have to start somewhere

Recommendations on panel size and voltage, charge controller type, inverter size, and battery size.

Thank you.

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## Comments

30,502adminHere are a couple examples of smaller solar projects:

Emergency Power

Basically a very long thread that starts from the beginning with a few vague requirements through design and assembly for a "portable" solar RE off-grid power box.

And here is another example by Mike90045 called the Solar Monolith:

Update pictures/information here.

Have a lot of information the Solar FAQ thread (information, links to other project websites, etc.):

Working Thread for Solar Beginner Post/FAQ

-Bill

7✭OK ... how can I figure out how many panels I need to charge a specific battery?

For example, if I have a 12V 49amp Battery (Concorde AGM Battery PVX-490T), and I want to charge that battery with a 180watt panel @ 36.3volts (Jinshi Mono 180),

is there a calculation that will tell me how many hours of sun it should take to charge the battery to full capacity?

3,123✭✭✭✭1. You determine the number of watt-hours the panel will produce by multiplying the nominal wattage of the panels by the number of sun hours and then throwing in a fudge factor.

2. You look at the Charge Controller to see whether it is PWM, in which case you divide the watt-hours by 36.3 to get the number of amp-hours or MPPT in which case you divide the watt-hours by the battery voltage during charging to get amp-hours.

3. You determine how far down the battery is. A 49 AH battery at 50% will take 25 AH plus a fudge factor for chemical efficiency of charging. A 49 AH battery at 80% will take only 10 AH. A 49 AH battery at 0% is toast, so don't do that.

4. Now you can compare whether there are enough amp-hours available in one day to bring the battery back up. But if that looks good, you then have to figure that some of that charging time will be spent in the Absorb stage in which the controller is unable to deliver all of the panel power to the battery because of limitations of the battery itself. So multiply the hours needed by two and you will be pretty safe.

7✭OK ... so the panel I am looking at with 3 hours of sun will produce approx. 540 watts-hours/day (on good day, anyways).

Typically for a very small system, an MPPT Charge Controller is all one would need, correct?

If that is the case, then a 45 amp-hours AGM 12V Battery would essentially be fully charged in 1 day (assuming 10% charge to begin with, and 1/2 day assuming 50% charge to begin with)?

If the above is the case, would it make more sense to add a 2nd battery or a larger battery, or is such a system 'balanced' if there is such a thing?

3,123✭✭✭✭To get a balanced system, you first start with knowing how much energy will be drawn by your planned loads during one day.

Then you size the battery bank so that this is less than 20% of the battery capacity (if you want to allow for up to two days without sun and do not have a generator or other emergency charging source.) Depending on the exact battery type and specifications, you

NEVER want to go below 50% except in a true emergency.And you may or may not want to go any lower than 20-25% routinely.Finally, knowing the size of your battery bank and the depth of discharge you plan for, you can calculate how much panel wattage it takes to restore the energy to the batteries.

You can also start with the panels (space and cost limitation?), and go from there to decide on the battery size and the amount of energy you will have for your loads. But that is usually unsatisfactory in the long run. You either spend too much initially or you wish later that you had beefed up one or another part of the system.

7✭Thank you.

OK ... sorry for keeping on asking these, but I need to fully understand...

SO, to work backwards, I figure out how many watt-hours of power at 120VAC I would like to be able to use.

Let's say, for a practical example, that I can get by with only 2 kW hours of power a day.

SO, I size my battery bank to be able to store between 4kW hours (200% of what I need) and 8kW hours (400% of what I need)?

If that is correct, what is the calculation I need to convert, say, the 8kW hours into 12V DC Battery Amp Hours?

3,123✭✭✭✭Piece of cake: Since watts = amps x volts, watt-hours = amp-hours x volts. And kWH = WH / 1000.

That tells us that (ignoring inefficiencies, and assuming a constant voltage or 12 instead of using the actual 12-14 volt range) 8kWH at 12 volts = 8000WH at 12 volts = 667 AH.

I would, however, tend toward the conservative side on the Depth of Discharge you design for, and go for 20%, so 10kWH. (A lot depends on the type of batteries you will be using.)

Then the last step will be to make sure that you have roughly one watt of nominal panel power for each AH at 12 volts in your battery bank. (or more, of course).

While at the same time, having enough panel watts to provide that between 1.5 and 2 times that 2kWH in the number of solar hours you will have on your typical worst case midwinter day.

7✭OK ... therefore, 10kWh = 10000watthours / 12volts = 833.3AmpHours is what I would need to store. This would also show I need about 833Watts of panel power for best case, 1666Watts of panel power for worse case. (833 watts of panel power produces 833 watt-hours PER SUN HOUR, correct?)

Is that correct?

3,123✭✭✭✭I think you've got it! Almost... -:(

There are two conditions on size of panel array, and you have only covered one of them.

833 watts of panel will give you enough peak current to keep an FLA battery healthy. They need a minimum current at some point during charging, rather than just a total amount of AH.

The other condition is that if you use 2kWH each day, you will need to get 2kWH times charging efficiency (between 1.5 and 2) during the course of the day.

To get 4kWH of panel output, your 833 watts of panel would need to see 4.8 sun hours. So the second criterion is going to call for a larger panel array than the first one.