the wattage of the inverter is its max capability..
this means if you connect an input that can give the inverter enough power it will be able to produce 250 watts output.. MAXIMUM
IF you try and draw by hooking up say a 1500 watt heater to the inverter YOU WILL DAMAGE IT.. or it wont work..
now that said your question was the SOURCE.. obviously if the source cant supply 250 watts then the inverter will only be able to supply what the source can provide.. and thus will NOT BURN ANYTHING OUT it will just not be able to supply 250 watts out of the inverter only what you put in as the source..
so if you put a 100 watt source you can only get MAXIMUM 100 WATTS OUT OF THE INVERTER ...
if you try and draw more than 100 watts the source will degrade and maybe blow a fuse or something..
but the inverter will be fine..
First off you need to understand that the wattage ratings for both the source and the inverter are just maximum output ratings. Unless you connect a load to the inverter then you are not drawing any power ( maybe a little for bias power of the inverter ).
You can do as you say but be sure that the load you connect to the inverter does not draw more than 180W or else you will exceed the limit of the source.
Yes, it's OK to do what you propose - connect a 180 watt source to a 250 watt inverter. Bear in mind that the load must be rated at less than 180 watts. The inverter will convert the DC source to AC.
The source is DC driving the inverter?
that is fine. It just means the inverter is not working at it's full capacity.
If it's 90% efficient, then you will get 0.9x180 = 160 watts output before the source gets overloaded.
Will this blow my circuit? I know if I aim for an inverter too low for my source it will overheat. How would I go about inverting this power from DC to AC?
