# DC vs AC Load with the same wattage

lightsky
Registered Users Posts:

**13**✭✭
I am currently using 3watt LED AC bulb, would 5watt DC LED blub save more energy since there's no power conversion (DC to AC) as in the case of using AC bulb.

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## Comments

32,997adminIf you already have the AC running, that would make the 3 watt bulb * 1/0.85 = 3.5 Watts DC side.

However, if you have no other loads, an AC inverter consumes power just "turned on"... Maybe 6 Watts for a smaller inverter (~300 Watts), and 10-40+ Watts for larger to much larger AC inverters (1,000-4,000+ Watts).

So a 3 Watt LED on a small AC inverter (no other loads) could draw ~ 3.5 Watts + 6 Watts = 9.5 Watts from the battery bank...

Aslo, there are different LEDs... New versions are more efficient than the older versions (lots of improvements in the last 10 years for LED efficiency)... Also, there are focused vs flood lamps... (lighting a 12" circle of light requires much less energy than a bulb that is outputting a 180 or 360 degree sphere of light).

There are other issues too... Simple DC LEDs may use a resistor "ballast" to control current flow to the LED (can easily be 50% or less efficient). Others may use an active electronic circuit to control LED current (upwards of 95% efficient).

-Bill

1,477✭✭✭✭Another item is power factor on AC-DC LED. A 3 watt LED light will have just a simple rectifier-filter cap for AC to DC conversion prior to a flyback DC-DC downconverting switcher. It will likely have a power factor in the 0.6 to 0.65 range. For US residential you do not pay for appearent power for utilitiy grid, you pay for real power so the poor power factor is not extra cost for electric.

On an inverter or generator, the losses are based on current through the devices so poor power factor will have higher inverter or generator losses. The extra inverter or generator losses will be about 10% of difference between real power and appearent power. Appearent power is current on AC line times AC line voltage (VA) where the the real power is VA * cos( phase angle between V & I) or VA * power factor. The simple rectifier-filter cap AC to DC conversion gives discontinuous pulses of current on the AC line at two times the AC frequency.

On an inverter/battery system there can be significant idle power consumed by a sinewave inverter electronics that can be 20 to 100 watts depending on size and design of inverter. You would not want to run a 1000 watt sinewave inverter to power one LED light bulb. The inverter idle overhead is likely in the 30 watt range. If you need inverter to power other things then the overhead for inverter is diluted.

So you need to consider the whole system. The hassle of running extra lines for a DC source is a consideration.

The 3 watt AC run LED real power is about 3 watts / 0.8 = 3.75 watts and appearent power is about 3.75w / 0.65PF= 5.8 VA. Extra inverter loss just due to LED power factor is (5.8VA - 3.75W) * 10% inverter efficiency loss = 0.2 watts. Total battery power just due to LED light is 3.75 watts / 0.9 inverter efficiency + 0.2 watt PF extra loss = 4.4 watts from battery.

Again, inverter running overhead power consumption is not included in this number.

The 5 watt DC run LED is about 5 watts / 0.85 = 5.9 watts.