Hello,
I’m hoping to use the Buck/Boost Converter to convert changing voltages from a solar panel into 12V power for a DC fan. My 10W solar panel outputs between about 12 to 20 volts, with a current range of about 0.25 to 0.62 Amps. It looks like the Buck/Boost Converter might require a minimum of 1 Amp of input current. Will the Buck/Boost converter not work if the input current from the solar panel is below 1 Amp?
Thank you.
According to the data sheet, the regulator “no load loss” (usually called the quiescent current) is typically 25 mA, so that would be the minimum input current.
The data sheet has a specification for “input current” at various input voltages, but it is not clear what the output current is for that test. I suspect those values are the input currents, measured for 12V/3A (36 Watts) output. So the 40V/1A input figure leads to efficiency = 36W/40W or 90%.
How much current you need on the input side depends on how much power your fan needs, plus a little bit for inefficiencies in the buck/boost.
If your fan needs 12 volts at 1 amp to run, it needs 12 watts. (watts are volts x amps, 12V x 1A = 12W)
So that means you need at least 12 watts on your input side. If your panel is outputting 12 volts at 0.25A, that’s only 3 watts of power and won’t be enough to run your fan.
On the other hand, if it’s outputting 20 volts at 0.62A, that’s 12.4 watts which theoretically is more power than you need but remember that some power is lost in the conversion process. (About 15% is lost at 20 volts input)
It depends on your fan, if it needs 200mA to run, you might be OK with your current panel and the buck/boost but any more than that and you’re probably going to need a bigger (or more) solar panels.
A good rule of thumb is that you’re going to need 20% more power (watts) available on your input side than what your fan needs to operate. If your fan needs 36 watts to run, you need a minimum of a 44 watt solar panel to run it in the worst case conditions.