Using a DC-DC Buck/Boost Converter with an input of 8 - 30 V DC, and an output of 12 V DC and maximum of 10 Amps. The generator that is attached to the input of the converter has an efficiency of 72.5% and the converter has an effiency of 85%. The manufacturer states that the converter can output 100 watts continuous and 120 watts peak.
so, with this DC-DC Buck/Boost converter is Vin*Iin*generator efficency=Vout*Iout*converter efficiency?
or can you input less power than you put out?
or do you always have to put in more power than you put out?
or is the first equation true, basically Pin=Pout
Thanks
so, with this DC-DC Buck/Boost converter is Vin*Iin*generator efficency=Vout*Iout*converter efficiency?
or can you input less power than you put out?
or do you always have to put in more power than you put out?
or is the first equation true, basically Pin=Pout
Thanks