My first project - Solar power computer....I need your help

Re: My first project - Solar power computer....I need your help

It depends... On the type of charge controller, the ambient temperatures, typical/best case/worse case numbers, etc...

The details. the rating of a solar panel is given by Pmp=Vmp*Imp at STC (standard test conditions---Basically full noon time sun equivalent of 1,000 Watts/m² and ~75F cell temperature). Note that Vmp actually falls as solar cell temperature increases. Upwards of 20% drop in Vmp on a hot day under full sun (~worst case 30C/54F temperature rise under full sun, little wind). So Pmp=Vmp*Imp*~0.80 on a hot day/roof.

Next, based on the charge controller. A PWM charge controller is nothing more than an On/Off switch. On, the panel is charging the battery. Off, the panel is not charging the battery. A solar panel is a (mostly) a current source. It puts out ~Imp (current at rated sun) at typical battery voltages. So the energy going into the battery is:

140W panel at 18V
12V battery

which is the best way to calculate the Amperes?

140W/18V = 7.7A

or

140W/12V = 11.6A

• Pbatt=Vbatt*Imp*some losses~ 14.5 volts charging * 7.7 Amps = 111.65 Watts into battery (near full battery)
• Pbatt=Vbatt*Imp*some losses~ 12.0 volts charging * 7.7 Amps = 92.4 Watts into battery (near dead battery)

Now, for a (more expensive) MPPT (maximum power point tracking) charge controller... These are constant power converters, they can take high voltage/low current from a solar array and efficiently (~95%) and down convert (think sort of like the DC equivalent of a variable transformer) to low voltage/high current needed to charge the battery bank.

• Pmp panel = Vmp*Imp = 18 volts * 7.7 amps = 138.6 watts (round up to 140 watts)
• Pbatt=Vmp*Imp*some losses~ 140 Watts panel * 0.81 hot panel losses * 0.95 controller eff = 108 Watts charging battery
• 108 Watts / 14.5 volts battery charging = 7.44 Amps into battery (near fully charged battery)
• 108 Watts / 12.0 volts battery charging = 9 Amps into battery (near dead battery)

For different reasons, we end up using ~0.77 derating of panel "marketing wattage" as the available charging energy (PWM "matching losses", MPPT conversion losses, Hot panel losses, dust on panel losses, aging losses, battery voltage variations, etc.).

An MPPT charge controller is a "switching power supply" (typically a Buck mode switch mode down converter).

Buck converter

There are lots of variations and "adjustments" we can make--But for a back of the envelople calculation, the way we calculate the overall charging energy is "close enough" for solar worth. Anything within ~+/-10% power calculations is pretty much "dead on" when all of the variables (weather, measurement, equipment selection, how the system is operated) that most people will get approximatly the numbers we use here over the life of the system.

So, if you have a 140 Watt panel and want ~5% to 13% rate of charge into the battery bank:

• 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.05 rate of charge = 149 AH battery minimum charging specification.
• 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.10 rate of charge = 74 AH battery nominal charging current
• 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.13 rate of charge = 57 AH battery maximum "cost effective" charging rate

And based on a typical sunny day, minimum available sun for ~9 months of the year in much of the US:

• 140 Watt panel * 0.52 end to end system efficiency * 4 hours minimum sun per day = 291 Watt*Hours per day of AC power

Based on charging during the day and using the power during the evening/night (panel losses*controller losses*battery charging losses*AC inverter losses = ~0.52 end to end system efficiency).

Yes, about 1/2 of the "marketing number" of the panel wattage actually makes it out to your AC appliance. Get an AGM battery, use DC load--You will get a bit better end to end efficiency.

We use these numbers to give you a conservative output energy. I use 14.5 volts battery voltage--I think Marc/Cariboocoot uses 12.0 volts in the MPPT charging calculations.

Mine is a bit more "pessimistic"--either is pretty close to what you will see in actual operation.

-Bill

Re: My first project - Solar power computer....I need your help

Kellez wrote: »

The inverter i suggested (link below) does not have an attached cigarette lighter plug but bolted connections

http://www.amazon.co.uk/Inverter-inverter-notebook-emergency-MRI3013J2/dp/B008HOECP0/ref=sr_1_6?s=electronics&ie=UTF8&qid=1398783755&sr=1-6&keywords=bestek+inverter

It appears to come with two cables--One clips, the other a cig plug. As long as you hard wire it with heavy gauge cable--you should be fine.

300 Watts * 1/0.85 inverter eff * 1/10.5 battery cutoff * 1.25 derating for wiring/fuses/breakers = 42 amps minimum (40 amp internal inverter fuse supplied)

Can you please better explain the result of this calculation? and what is the purpose.

i see that 42 amps > 40 amps of inverter, so what does this mean?

Many times, we do not have the specifications of the AC inverters (paper design phase, sometimes the manual does not state the operational specifications). The above calculations are "close enough" that we can "ball park" the operational requirements without going into the specifications for a 1/2 dozen AC inverters.

AC inverters have conversion losses (typically 85% efficient). And they are constant power devices. XYZ Watts out to AC load, the Pin=Vin*In -- Looking at the equation, the maximum input current to the inverter is going to happen at minimum input voltage (which causes maximum input current). The 1.25 derating is a United States building/electrical code derating for ciruit breakers and wiring (and how "we" rate breakers and fuses--Other countries do not need to follow these customs, and some specialty fuses/breakers are rated differently even in the US--We have "trip curves" that are used to design for specific applications--slow blow, fast blow, etc. protective devices)... I.e., the breakers and wiring need to be 1.25x larger than the maximum continuous current rating of the load (circuit breakers/fuses are supposed to "blow" at 100% of rated current).

So, my generic calculations show that a 42 amp minimum branch circuit would support a 300 watt AC inverter running at 300 Watts AC loads. Typically, there are no 42 amp fuses, so the mfg. either rounds down to 40 amps or up to 50 amps--In this case, either is fine (close enough). And perhaps their inverter runs at better than 85% efficiency or not quite down to 10.5 volts input--Anyway--It was just a conformation that you need wiring+any fuses+battery connections capable of ~40 amps minimum for reliable operation (you don't want false trips on your fuses/breakers or overheated wiring+connections).

Also, very few people will run a 300 watt AC inverter at exactly 300 watts for hour on end--So 40 amp internal fusing will be just fine. If you choose the run the inverter at >300 watts for minutes-hours--You will probably have the internal fuses trip and/or the inverter overheat.

-Bill