I'm not sure how to search this so hopefully someone will answer. I'm rigging a floundering boat for my son and want to use LED light bars on the front and sides. I want to power then with an inverter generator because they are so much lighter and quieter. The LED light bars are 12 VDC, the invertor generator is 2000 watts and has 2 receptacles for 110 VAC and one for 12 VDC. The LED light bars will total around 1200 watts at 12 VDC. Will the single, 12 VDC receptacle handle that many watts or is there another way I should power the LED light bars?
Electrical – Inverter Generator Watt Question
generatorinverterledwatts
Related Solutions
For Problem 1:
A string of LEDs hooked up in series with a current regulator like SuperTex CL220 would do the trick: The component is a simple a 2-terminal device (like a diode) and needs no additional components or configuration. It allows 20 mA (+/-10%) current to pass through as long as there is sufficient voltage headroom: 5 volts above the total forward voltage of the LEDs is sufficient. This current regulation is stable up to 160 volts, enough for your purposes.
Note that LEDs are current-dependent rather than voltage dependent devices. They glow at essentially the same brightness as long as the current is constant.
For your application, if the LEDs need to start glowing from around 60 Volts, half the 120 Volts your students generate, then a string of 25-30 standard 5-mm red LEDs would be optimal. They would glow without intensity change till your maximum voltage.
Too many LEDs = they won't glow till a higher voltage.
Too few LEDs = the CL220 device would overheat in dissipating the surplus voltage.
For Problem 2: The extent of energy storage required to keep a string of around 25 LEDs (from above section) glowing for even half a minute, is pretty high. Capacitors would not be the way to go, unless you have access to big power-line capacitors through surplus channels.
- Your capacitor bank would need to provide 20 mA at a minimum of 60 volts (again from above section), for "a few" seconds.
- Capacitor needs to be rated for a voltage higher than the highest the generator could conceivably generate.
- Though "supercapacitor" is a popular term these days, typical supercaps are rated for 5.5 Volts or 12 Volts, not hundreds of Volts.
- Adding in buck/boost generator trickery to make this work would result in complexity far beyond using a battery and off-the-shelf charger.
I hope this helped.
First, you are assuming that just because the engine is rated for 4 kW it is actually run at 4 kW.
Second, 3 kW electrical power out, converted to your AC standard and probably regulated, from 4 kW in doesn't sound so bad at all. That's 75% effiecient from the shaft to the outlet, with several conversions in between. There is the mechanical to electrical conversion of the generator or alternator itself, probably some sort of DC conversion, then a inverter stage that makes presumably nice AC like you get in your house. Even if there are only 2 conversions from shaft to outlet and the losses are equal per conversion, each would have to be 87% to acheive 75% overall. With three equal conversion each would need to be 91%. Sounds like you're getting quite reasonable performance for a portable generator.
Now compare that to the losses in converting gasoline to shaft rotation and you'll see none of them matter. The gasoline to shaft conversion is probably a order of magnitude less efficient than the 75% you are quoting from shaft to outlet. You are looking at the wrong end of the generator unit to find gross inefficiency.
Best Answer
You will have to look into more detail into your invertor. If the generator is rated at 2000w, then it can handle 1200; however, that rating might be for the AC output and not DC.
Check to see if it contains any rating specifically for the DC output. If you cannot find this, the invertor will probably be fused; if you can determine the value of the DC fuse then you can establish what wattage the invertor DC output was intended for.