Originally posted by Turion
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Interesting data. Is the 300 volt measured on one coil? At what RPM? Unloaded means open circuit or nothing connected to the coil terminals except the voltmeter, right?
And loaded (with a resistance?) to draw 0.6 amps, the reading on each of two coils separately is 85V at 2820 RPM and 90V 2835 RPM, right? 90V / 85V = 1.06 and 2835. RPM / 2820 RPM = 1.005. These two ratios should be the same if the coils were identical.
0.6A at 85 or 90V is 51 or 54VA (W if load was purely resistive). For example purpose let's use 300V open circuit and 90V at 0.6A with purely resistive load, both at same RPM. To find the coil resistance, (300V-90V) / 0.6A = 350 ohms. Also, means load resistance = 90V / 0.6A = 150 ohms.
I have mentioned Jacobi's law before. Look it up. What it says is maximum power transfer occurs when load resistance equals source resistance. Which for this coil and speed, is 350 ohms yielding 64W at 0.428A. And this (using a 350ohm load) will give you 150V across the load or coil output voltage.
Adding turns of the same wire size will not alter the maximum power at that RPM with those magnets. I'd suggest that you take time to do the math before taking time (& cost) to add an inch of core and copper to the existing coil as you mentioned.
Well, just checked wire table. Something is screwy. Room temp resistance of AWG #23 copper is 20.8 ohms/1000ft. That means your coil should be about 7 ohms, right? Ever measure it? Perhaps there is large reactive component at that frequency. Some analytical tests would help. Generated voltage vs frequency sweeps with various loads would tell. Scopes shots also.
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