# splitting DC and AC

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**2,511**admin
I was wondering if I set up a DC system in my barn for just lighting, of course purchasing only DC lights etc. Would it be possible in the future to split part of that off and attach an inverter and make a portion of what comes out of the battery AC? So, could I have both an AC and DC line coming out of the same battery?

I appreciate this website and all the great questions and answers! I am hoping to get a small PV system in my barn in the spring and I'm sure I will make good use of the forum.

I appreciate this website and all the great questions and answers! I am hoping to get a small PV system in my barn in the spring and I'm sure I will make good use of the forum.

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## Comments

32,804adminThe best way to think about "solar" is to break it into pieces...

The major piece for Off-Grid solar is the Battery (or battery bank). This battery must be able to supply all of the energy needed to run your loads--Review the basic formulas for electricity:

V=I*R

I=V/R

R=V/I

Power (watts)

P=V*I

P=V^2 / R

P=I^2 * R

Work (Watts*Hours)

Power*Time

Watt*Hours=Watts * Hours

kWatt*Hours= Watts * Hours / 1,000 (basic billing unit on your electric bill)

Amp*Hours

Sort of the same as Watt*Hours, except missing the Voltage Component:

Amp*Hours=I*Hours

AH=Watts/Volts

So--long way around, need to figure out how much power (watts) and how much power over time (Watt*Hours) you want the battery to supply. Your best bet is to use the most efficient lights you can (CFL's, LED, not filament bulbs).

This would be, for example a 13 watt CFL bulb running 2 hours per night would need 26 watt*hours...

A 12 volt battery should be sized to give about 6 days of power (3 days of normal operation, and no more than 50% discharge--or the battery will die young)...

26 WH * 6 = 156 Watt*hours

Typically, batteries are rated in Amp*Hours (at a 20 hour discharge rate--i.e., the battery can supply XX amps for 20 hours before dead).

So,

156 WH / 12 volts = 13 amp*hours (3 nights of use, and 50% max discharge) rated battery...

If you want those lights to be 120 VAC CFL's, then you will need an "inverter" which takes DC and converts it to AC... And the inverter will be operated by the battery (with some additional losses--so the battery should be roughly 20% larger).

Next, you need to recharge the battery (ideally every day with solar)--so you need a multi-stage charge controller... That can be from the utility line, a generator, solar panels, etc... Depending on what the charging source is, you will need to pick the appropriate charge controller.

Roughly, for an off-grid system, as a first approximation, the amount of Watt*hours you will collect from your panel will be ~2x the panel's rated power output in Watt*hours (solar panel rating * 2 hours of sun per day)...

So, to run 26 watt*hours of load:

26 WH / 2 hours of sun = 13 watt panel to run the light two hours per day in reasonably sunny weather.

Of course, the more power you need, the larger all of the components (battery, inverter, charge controller, solar panels, wiring, etc.) need to be.

And, of course, you can get DC power from your battery to run some lights, and run a DC to AC inverter to run your AC loads at the same time... But you have to "size" the system to support all of those loads.

-Bill

9,583✭✭✭✭✭Because of "copper loss" it's much easier to distribute AC, than DC.

|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||

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10,300✭✭✭✭to simplify bb's answer is to look at the inverter as just another dc load to the battery and that battery has to supply all of the current needed by all of the loads big and small. inverters tend to be a big load because of the wattage being (12v x 10 amp = 120w and 120v x 1 amp = 120w) the same and inverters need a base amount of current to operate over that and is a determining factor in inverter efficiency. the bottom line is you can do it if you size the system big enough to cover the future needs of the inverter and its loads.

5,436✭✭✭✭I run two parallel systems. My house is wired with 120vac circuits for the lighting and the outlets with it's own breaker panel and wiring.

The 12vdc system runs the radio, fridge ignitor, telephone etc, as well as a couple of "emergency" lights so that if the ac breakers or inverter is off I can have light to get into the house.

All works very seamlessly,

Tony

6,290adminThank your for all of your input!

I do have a question on the copper loss of a DC system. I was under the impression that a DC system would be more efficient for a simple lighting application... can you go into more detail?

10,300✭✭✭✭that would be due to the efficiency of the inverters i mentioned. it will be the wattage you need for ac plus the operational wattage of the inverter. this is sometimes seen as a percentage of efficiency so if the inverter is say 85% efficient and is at 900w there's another 15% of that 900w to run the inverter or roughly 135w more for greater than 1000w drawn. also inverters will draw power if on even when there's no load and the better inverters try to address this to keep it to a minimum.

that means dc is a good viable option, but not for great distances as dc, being low voltage, loses quite a bit of its power in the wire resistance and is addressed often as the voltage drop and is sometimes given in percent form. this is one of the reasons why 24 and 48v pv and battery systems are popular is because it can lessen losses due to wire resistance. circumstances vary and so do the voltages so in most cases it must be worked out in theory first and evaluate the costs of doing it in those ways that will still satisfy one's power needs.

32,804adminIt is never a black vs white answer on system losses...

First, different lights have different efficencies...

- Filament lights--simple, cheap, "self ballasting to a degree", not very efficient, AC or DC just fine
- CFL's--Very efficient, needs high voltage, must be AC, plus ballast (current limiting)
- LED's--Very efficient, needs DC, needs ballast to limit current

Three very different light sources, three very different types of electricity needed.Filament bulbs can work almost anywhere AC/DC (within reason)--however, a hot filament wastes lots of heat (Halogen bulbs are much more efficient--but may need a stable voltage DC source to operate efficiently).

CFL's need AC... Whether you use a "DC" CFL or an AC CFL, there will still need to be a conversion process from DC to AC with current control (CFL's are mercury vapor arc lights--so need to start arc and limit current. Usually need AC to not erode electrodes).

LED's--very efficient, but need some sort of current control. Very non-linear devices. Need current control to keep light levels optimum, and to prevent device failure (if voltage is a bit low, LED is very dim, if voltage is a bit high, over-current will pop LED). Frequently, "cheap" 12 volt LED lamps simply use resistors to limit 12 volt current to a "6-10 volt" LED Bulb... The resistors waste energy as heat. Better LED devices use electronic power supplies which are more efficient and provide a more stable light level (if battery voltage varies).

So--each lighting source has its pluses and minuses. With modern electronics, the conversion losses can be held down to the 5-15% range--so, other than the price impact for electronics--AC/DC and high/low voltage (in and of itself) is typically not a deciding factor.

Where most issues arise is how much power/current/energy do you need to move around your home/cabin/etc.

This gets back into a couple of the math equations I posted above...

P=I*V (power = current * voltage)...

So, from that equation, we see that if we change from 12 volts to 120 volts, the current required will drop to 1/10th the value required at 12 volts (for the same power requirement).

And that gets us into another issue... Wasted energy/more wire required.

Basically, to pass around 10x as much current either requires 10x thicker copper wire (expensive) or to live with higher voltage drops and losses.

P=I^2 * R

So--if we have a length of wire, and want 120 watts, we can either use 12 volts at 10 amps or 120 volts at 1 amp...

The ratio of the increase in heat wasted with 10 amps vs 1 amp is (10^2/1^2) or 100x as much wasted energy...

Or, you need to pay for 10x as much copper to run the same 120 watts of power at 12 volts vs 120 volts.

Also with a "high" voltage distribution system, the voltage drop allowed is much greater... For example a 12 volt system with 2 volt drop gives 10 volts--most "12 volt" devices will not work at 10 volts.

120 volt system with 2 volt drop--118 volts--all devices will work very well.

And--12 volt DC storage battery systems are not really 12 volt DC systems but 10.0-15.5 volt systems... Most 12 volt devices are designed to operate on a car cigarette lighter outlet which is around 12-14 volts. The very low and very high voltages seen on storage battery banks (near discharged+long wire runs with voltage drops, and the 15-15.5 volts batteries need to properly equalize) will cause many/most "car" type accessories (cell charger, computer adapter, etc.) to fail.

So, frequently, it is better to get a good quality true sine wave inverter to supply 120 VAC 60Hz (regulated) devices (including cell chargers and laptop computers) around your home/cabin (even if you have 85% efficiency by adding an inverter) than to try and run everything using "12 Volts DC".

Notice, we really did not make a difference about AC or DC current--for a first approximation, there is no difference (from our point of view) between AC and DC in power wasted/used.

The advantage to AC is that we can use a simple transformer to change voltage and AC is much easier to "switch" on and off than DC (DC arcs are sustained much longer than AC arcs).

In the end, we get down to what power is needed... If you need a little and have good quality DC powered devices--use DC...

If you are making a larger system, that runs more than a few feet, and want to use non-DC appliances--do it right and build out a system with an AC inverter.

-Bill

5,436✭✭✭✭As usual Bill has it exactly right. The only thing I might add, is that even if you figure out all the theoretical efficiencies and losses to the gnat's [email protected]# you will never build the system to match. For most of us a simple matter of adding up all the potential loads, factor in a number for cloudy days, a rough number for efficiency losses etc and build a system that is marginally larger. I don't (ever) mean to diminish the learned thoughts of those that are way sharper than me, but sometimes keeping the calcs simple works just fine in the real world.

For example I let the Tri-metric do my thinking for me. Instead of calculating each load, plus line losses, plus inverter losses plus charging losses etc, I just let the Tri-metric tell me what I am using at any given time, as well as how much I have used (or charged ) cumulatively. I've now lived with my system long enough to know when the Tri-metric says that I am down X amphours, I can look at the sky and the time of day, the forecast for tomorrow, and know I will (or will not) need to charge. My daily use is remarkably consistent over time. A quick look at the meter every morning tells me how many amp/hours used in the previous 24. The cumulative is almost exactly what my eyeball average is over more than a year.

Tony

6,290adminThank You All! I'll consider all these options as I pull my resources together. I especially like the breakdown of pros and cons of each lighting option.

I knew I should have paid better attention in my circuits class! Oh well, this is more fun anyway.