Posted: 3/10/2013 1:28:07 PM EDT
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I've searched and not found an answer that satisfies me. GD don't let me down... Let's say I've got a coffee pot that draws 1200W @ 120V AC - 10 amps. I've got a DC inverter that supports up to 120V/2000W constant (4000W peak) from an automotive 12V power source. How many amps @ 12V DC will the inverter draw if the coffee pot (and only the coffee pot) is plugged into it? |
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I do know that a device's power draw spec is the wattage that it pulls from the power source. Your inverter's spec seems very specific about it being able to supply 2000W. The inverters power draw from the DC supply will be more than that 2000W though. It may be up to 87% efficient. (not a joke, the numbers do go up there for switching power supplies)
ETA: dont even *try( to run it off anything but directly connected to the batt terms. |
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Quoted:
I found a formula shortly after posting. DC amps = AC Watts / 12 x 1.1 The 1200W coffee pot draws ~110 DC amps. At least in theory. That's a pretty good number to use. The 1.1 factor assumes that the inverter is 90 percent efficient at converting 12 VDC to 120 VAC. Note that when drawing that much current from a moderately-sized battery, the battery's voltage will start dropping pretty quickly. As it drops, the inverter will draw more and more amps from it, in order to continue supplying a full 1200 watts to the coffee pot. Example: If the battery voltage drops to 11.0 volts, the inverter will need to draw around 120 amps to supply the full 1200 watts (again, assuming 90 percent inverter efficiency). |
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Quoted:
IF the inverter was 100% efficient, (which it is not) power out would equal power in. Thus 1200W out would be 1200W in. 12V x 100A Better to get a normal pot and put it on the intake manifold This is correct. A good inverter design should be capable of achieving efficiencies in the mid 90% range, but I have no idea how good the inverters are that are commonly available or what their efficiencies are like. |
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Quoted:
Quoted:
I found a formula shortly after posting. DC amps = AC Watts / 12 x 1.1 The 1200W coffee pot draws ~110 DC amps. At least in theory. That's a pretty good number to use. The 1.1 factor assumes that the inverter is 90 percent efficient at converting 12 VDC to 120 VAC. Note that when drawing that much current from a moderately-sized battery, the battery's voltage will start dropping pretty quickly. As it drops, the inverter will draw more and more amps from it, in order to continue supplying a full 1200 watts to the coffee pot. Example: If the battery voltage drops to 11.0 volts, the inverter will need to draw around 120 amps to supply the full 1200 watts (again, assuming 90 percent inverter efficiency). You are correct about the voltage drop/amp rise, but most inverters now have a low voltage shut-down feature to protect such a thing. Seems like it may be around 11 volts, can't remember? |
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most inverters now have a low voltage shut-down feature to protect such a thing. Seems like it may be around 11 volts, can't remember? Some of them sound an alarm at around 11.0 volts, but 10.5 volts is pretty typical for the shut-down. With big loads and small batteries, the battery voltage can sag quite a bit - even though the battery still has some charge left in it. So, inverter manufacturers tend to set the shut-down voltage pretty low in order to prevent nuisance shut-downs. |