maximum power output while charging battery

Simon, your question is confusing, because you state that the PMG generates 368 watts under no load.  Since wattage is a product of both voltage and current ( P(watts) = E(voltage) x I (current), you must have a current flow through a load to be able to measure wattage.  So, how can you possibly determine wattage with no load?  Generally speaking, a generator produces an open load voltage that is higher than that when it is under load, but the current is determined by the amount of power that can be extracted from the wind, the efficiency of the PMG, and the losses due to resistance of the wiring in the generator and in connective wires to the load, and the voltage would be clamped at or near your nominal battery voltage, since you are using the battery as a load.  In order to get more power from the PMG, you would need to have less resistance in the windings of the PMG, or increase efficiency of the turbine blades, or have greater strength of the permanent magnets or their coupling to the windings.  Increasing the length of the turbine blades is an easy way to capture more power from the wind, since it is dependent on swept area.  Just my suggestion.  :-)  Jon C.

Hello again, Simon.  You simply do not have enough turns of wire on the PMG to generate over 12v at 200rpm.  The only other way to get more voltage would be to DC - DC convert the output to a higher voltage, but then you would suffer losses in the converter and wiring, and realize a smaller output wattage overall.  You could, as I said earlier, put larger, more efficient blades on the turbine, or increase the efficiency of the turbine itself.  If you don't have sufficient wind at your site to spool up the turbine at adequate speed, you might raise the mast height or find a better location for the turbine.  I got the impression that your PMG is homemade?  Might I suggest buying a commercially made wind turbine that may satisfy your charging requirements?  Jon C.

Hi Simon.  Yes, increasing the number of turns of wire in the PMG will increase the voltage, since each turn contributes to the resultant output voltage.  If you must decrease wire size in order to accomplish that, or keep the same size wire, there will be an effect of increased resistance in the windings, and a subsequent voltage drop under load due to power losses in the smaller wire, so it is a tradeoff........more voltage out, but more internal losses.  This may or may not work for you, and I think that you will have to experiment with this to determine the results.  I wish you good fortune in your project.  Jon C.

You are most welcome, Simon.  :-)