n00b needs help with solar panel kit – inverter shutting off

strawmann

Hello all – I'm starting to learn about the electrical aspects of this and could use some insight from a veteran.

I picked up a starter solar panel kit with a 100 watt, 12 volt panel and a controller.  The controller is connected to a 12 volt, deep cycle marine battery that reads 12.09 volts on the meter when it's fully charged and in direct sunlight.  I believe it's 65 amp hours, but it could be 75.  The battery is connected to a 1000w inverter (2000w peak).

I get a solid 6-8 hours of sun a day, depending on the season.  During the day, there's no problem powering my TV, cable box, and surround sound (about a 600w load) through a 16-gauge extension cord that's 10' long.  But, at sunset the inverter shuts off and starts beeping, even though the battery still has 11.86 volts.

I thought maybe the inverter was too small, so I upgraded the original 1000w to a 2500w (5000w peak), but it still beeps and shuts off soon after losing direct sunlight.  Do I need to add another battery?  Is it something else?

Also, someone told me that at 12.09 volts, the battery is only about 30% charged. At 11.86 volts, they said it's got a 0% charge, but neither of those make sense to me.  Would someone mind walking me through how that math works, please?

mcgivor

There are a series of issues which are obvious 

The 100W panel is too small for the battery and load demand, running a load whilst charging will deprive the battery of charge current leading to cronic undercharging. When the battery is near to fully charged the voltage should be ~14.4 V,  at 12.09V resting without any load or charging would mean the battery is at ~50%, if charging it would be significantly less.

The battery capacity is too small for the load and inverter, in fact the 2500W inverter alone without any loads used would be more than the system can provide, just being switched on, it's self consumption.

Before building a system there are a series of steps which need to be followed, firstly an inventory of loads and the amount of time used, this is probably the most important step, which must include the inverter self consumption. Once this figure is established the battery capacity can be determined, remembering that a 100Ah battery’s useful capacity  ( lead acid  ) is 50Ah as you don't want to go less than 50% state of charge , SOC. Then you need to figure how many days of no sun you want the system to support the loads for and multiply by that figure.

Once the battery capacity is established the means to charge can be calculated, remembering the array capacity must be able to replace more than the what is taken out of the battery, by at least 125%, this figure will be determined by location and how many hours of useful sunlight is available. It's always better to oversize to some degree to,account for low product days.

Attempting to provide all figures and calculations without information  would be difficult, as well as being information overload, the best approach would be a step by step calculation starting with load demands. But what I can say is what you have will never work, no offense.

dennis461

For battery state of charge, go here and scroll down until you see the chart.
https://www.solar-electric.com/learning-center/batteries-and-charging/deep-cycle-battery-faq.html

and, you need to measure battery voltage right before the invert cut's out.
I suspect it will be around 10.5, then creep up to your 11.86.
The inverter may be cutting out too soon.

Can you measure the inverter input voltage and the battery voltage at the same time or are they too far apart?

BB.

Yea--You are "deficit charging" the battery. Basically, your loads (Amp*Hours @ 12 volts or Watt*Hours) you are taking out is more than is being replaced by your solar panel(s). You are running your battery bank down to near zero state of charge--And simply running the loads from your solar panel+"slowly" discharging your battery.

A larger inverter just makes this work (larger inverters "waste" more energy running vs a very small inverter sized to your loads). A larger battery bank will not fix anything (long term either)...

I would suggest you get a Kill-a-Watt type power meter, a DC current Clamp DMM (really AC+DC current clamp digital multi meter), and do some measurements of your loads and some basic math to figure out what you really want/need. Some links to devices (the first is our host, the others you pick and choose as you see fit, or ask more questions):

https://www.solar-electric.com/kiacpomome.html kill-a-watt meter (120 VAC energy measurements)
http://www.sears.com/craftsman-digital-clamp-on-ammeter/p-03482369000P ("good enough" DC current clamp meter)
https://www.amazon.com/gp/product/B019CY4FB4 (better current clamp meter)
http://rc-electronics-usa.com/ammeters/dc-amp-meter.html (DC voltage AH/WH meter)
Brady Instruments #1-77-8 Battery Hydrometer
MidNite Solar Hydrovolt Battery Hydrometer

In general, solar power systems do not generate near as much energy as people think. You need to be very conservative in your loads (small and very efficient loads) and a relatively large solar array to keep things happy. Also, you design your battery bank size to support your loads.

Just a quick example of how this could work. Say you want to power your loads over night (battery charging during day, run loads in evening). Say you watch TV on a reasonable 33 Watt LED based set for 5 hours per night (very quick example of math using "nominal" conservative off grid assumptions):

• 33 Watts * 5 hours = 165 Watt*hours per night
• 165 WH * 1/0.85 AC inverter eff * 2 nights of usage (no sun/bad weather) * 1/0.50 max battery discharge * 1/12 volt battery bank = 64.7 AH ~ 65 AH @ 12 volt battery

Notice we are talking about a 33 Watt load on a 120 VAC inverter... A large inverter may draw 20-40+ watts just "turned on". A small inverter may draw 6 watts just "turned on" (Tare Load).

To recharge the battery, need 5% to 13% rate of charge--10% is good starting point for full time off grid usage:

• 65 AH * 14.5 volts charging * 1/0.77 panel+charge controller derating * 0.10 rate of charge = 122 Watt solar array nominal

And then there is sizing the array to where you live... Say a fixed array in Los Angeles California:

http://www.solarelectricityhandbook.com/solar-irradiance.html

Los Angeles
Average Solar Insolation figures

Measured in kWh/m2/day onto a solar panel set at a 56° angle from vertical:
(For best year-round performance)

Jan	Feb	Mar	Apr	May	Jun
4.50	4.82	6.05	6.78	6.83	6.80
Jul	Aug	Sep	Oct	Nov	Dec
6.69	6.67	6.40	5.85	5.07	4.41

On (long term solar) average, the system would see (worst case) sun of 4.41 hours... The minimum solar array needed to support the 165 WH per day load in December would be:

• 165 WH per day * 1/0.52 end to end solar system AC efficiency * 1/4.41 hours of December average sun = 72 Watt array minimum
  

For a Los Angeles based system, running a small laptop or LED TV with a SMALL AC inverter would work pretty nicely with a 122 Watt array.

Anyway, a way at looking at your system (note, I did the calculations based on what I thought your present system could support--I do not know anything about your true loads).

-Bill

BB.

Forgot to add.. A 65 AH @ 12 volt battery bank would support a maximum AC inverter load (reliably) in the range of ~80 Watts to 163 Watts maximum (basically 250 Watts per 100 AH @ 12 volts for a flooded cell lead acid battery).

A 1,000 Watt inverter would require a battery bank about 10x larger to "reliably" run that inverter in the long term.

-Bill