# AC/inverted more effecient than DC?

cizzi
Solar Expert Posts:

**109**✭✭✭✭
After almost a year of experimenting with solar stuff today I did some basic math for a 13W DC light bulb vs a 13W AC light bulb (energy savings spiral thingy), of course I use an inverter connected directly in my battery for the AC bulb and came out with these results.. I also tested them in practice and so far they check out.. So I was mislead that DC lighting is more effecient (in this case). I also read there's 20% inverter innefecienty but hey that's not even close to the savings using the inverter.. am I completely off the ball here or does someone else see what I mean?

13W = 12V x Amps

Amps = 1.08A

13 W = 120V x Amps

Amps = 0.10A

Now I use lighting all night long directly from my 106AH battery and if i use 1A per hour vs 0.1A per hour I think there's a big difference..

let me know thanks

13W = 12V x Amps

Amps = 1.08A

13 W = 120V x Amps

Amps = 0.10A

Now I use lighting all night long directly from my 106AH battery and if i use 1A per hour vs 0.1A per hour I think there's a big difference..

let me know thanks

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## Comments

31,077adminBecause of the way you set up the problem, both loads are 13 watts and there are no other losses other than the inverter... The AC load will be 20% less efficient (80% efficiency) vs the 100% efficiency of the DC 13 watt lamp.

13 watts of power at 12 volts is exactly the same as 13 watts at 120 volts.

P=13 watts=V*I=12 volts * 1.08 amps = 120 volts * 0.108 amps

Amperage does not represent power--so, ignoring voltage, one cannot assuming anything about power/efficiency/losses without knowing the voltage too.

For overall efficiency, you need to look at the source, the wiring, inverter (if present), more wiring, and then the load.

For wiring, losses are frequently known as Ploss=I^2 * R losses...

So, for a fixed R (ohms), if you cut the current by 1/10 (as in this example), then the Ploss is cut by 100x... Or basically, you need 100x thicker wire to pass 13 watts at 12 volt vs 13 watts at 120 volts (AC or DC, does not matter for 60 Hz and short distances).

So, now you need to look at the costs of distributing your energy around your home/property. Saving the cost (and the losses of the inverter) vs the extra cost to have much heavier copper wire to route the higher currents to the remote loads.

Also, you need to look at the loads themselves... How efficient is the load when running at 12 volts vs 120 volts (i.e., computer power bricks, CFL ballasts, TV's, etc.)... Many times, there is a conversion required anyway to a higher voltage (TV) or the to AC (brushless DC vs AC induction)...

So, the answer is not always obvious which is (12 VDC or 120 VAC) better/worse...

My two cents--In general, unless you have relatively low power DC devices that are close to the battery bank--many times it is better to use the inverter (some are better than 90% efficient) for a home size solar RE installation. And in many (most?) cases, the 120 VAC appliances are more efficient and cheaper than their DC equivalents.

-Bill

109✭✭✭✭So my house is partially re-wired for DC lighting based on your answer I did not make a horrible mistake I take. I was under the assumption that bringing my voltage to 120V would reduce the Amp withdrawn directly from my total AH from my battery bank, this is not the case if I understood your answer right?

10,300✭✭✭✭that would not reduce your draw from your batteries, but that conversion to ac may lessen resistive losses if the wiring were to become too long or required to carry the loads of too many of the lights for its size. the conversion sets up a loss too as was pointed out so each case would have to be weighed as to what is best.

31,077adminThat is correct.

Power=Amps*Volts (P is in Watts)

So, whatever Amps*Volts works out to--is independent of voltage

1.08 Amps * 12 Volts = 13 Watts

0.108 Amps * 120 Volts = 13 Watts

Your actual 120 VAC load (assuming 90% efficient inverter):

Pbatt=Pdevice*1/efficiency=13 watts / 0.90 = 14.4 watts from battery

So, your battery needs to supply 1/0.9 or 11% more power to run the 13 watt device.

Not the end of civilization--but certainly something to think about where normally we strive for a maximum of 3% wiring loss.

-Bill

980✭✭✭✭What inverter are you using ? If you use a large (sine wave) inverter, like, 2000 to say, 4000 Watt, then its efficiency running a 13 Watt light bulb will be TERRIBLE. Remember that these inverters draw 25 Watts or so just being on.

This is why running a CF light on the DC will ~normally~ be higher efficiency than making AC to run a normal CF light bulb.

A modified sinewave inverter will be much better, like 5 to 10 Watts of idle power draw.

I know a guy that uses a lot of DC loads, but also has several small inverters (Exeltech) spread around for small AC loads.

boB