Hello Sparky01, s4w2099.
I almost hoped you would not ask this question, not a simple answer.
My example below is worked out with shorted diodes, with Schottky the benefits is less.
For a given wind speed the blades produce a given max Power.
The blades will supply this power at various RPM's with the efficiency changing based on the blade design. The lower RPM occur with a low impedance load on the generator, the blades go into stall, the current goes up, with RPM going down. Stall = Lower RPMs than based on wind speed.
Think of it as a constant power which changes with wind speed. (works for me:-)
What does this have to do with this question? Everything or I would not have spent couple hours writing this up.
Assumptions: 10' system at 36% eff. Vb = voltage of battery. Vd is diode voltage. Ra is generators + line resistance. Pb is power into battery.
Vb = 13v, Vd = 2.4V (two diode drops as in a bridge) and Ra = 1 ohm.
At low currents the Vd drops well be less, fixed to ease simulation.
Cut in is 5mph, available wind power is 18watts, want 1 amp charge current. Setting this up to establish the voltage needed at cut in is 16.4v = 13v + 2.4v + Vra (1 * 1).
Now you know what determines cut in speed.
For a PMA alternator we have an open circuit voltage Voc, based on RPM.
Voc is calculated here based on wind speed as a ratio to cut in RPM.
For a given wind speed there is a limited power output, if loaded greater then the available power out the RPM will drop lowering the forcing voltage to match the load to available power output. The power out will remain based on the wind speed not the RPM, the blades are being somewhat stalled, so the output power will be based on the profile of the blade. Here we will assume is within range of the blade and is the same at different RPM, when in reality it will be less.
Example just pass cut in speed:
6mph we have 31 watts, 19.68Voc, 1.58a
So, V = 13 + 2.4 + 1*1.58 = 16.98v or 2.7v mis-match causing Voc to exceed power limit.
RPM will lower until 17.2v is reached and I = 1.8 a for 31 watts.
Now lets short out the diodes. Vd = 0. For 5.1v mis-match. New I will be 2.05amps. 0.25a different.
Had to use a quadratic solver to get the numbers above.
Equation is for power: I^2 + 15.4I -31 = 0 for the first, and I^2 + 13I - 31 = 0 for second.
R*I^2 + VI -31 = 0, here R = 1 so looks like it's gone, when it's still here.
Example: 15mph, 485 watts, 49.2Voc, 9.87a
With diodes: 15.6amps, 31.0 volts & Pb = 202watts. (solution)
Short diodes, I = 16.46amps, 29.46 volts, 213 Pb.
3.25watts extra for your trouble for the first, 11 watts extra for the second.
When doing the math, remember the Power is Maxed for the different wind speeds, so the blades RPM will be lowered to limit the power to stay within the power available, the efficiency changes based on the blades. Above 12mph, more power is wasted as heat in the generator than is deposited into the battery.
You will see a few watts improvement, that's it. For the cost many other things can be done, like larger blades, star-delta switching (60-70% improvement), taller tower, etc.
A Few have reported adding in series resistors. The math shows the power in the battery will drop a few percent, but they recover the other power lost as heat in the generator into a resister bank for heating. Thus making it worth while. A controller is required, no resistance at cut in.
The IR math for the diodes does not tell the true story. It's just the heat of the diode and not worth the cost to make a cheap heat sink smaller. (For me:-)
Have fun,
Scott.
