chris,
If by "blocking diode" you mean a diode after your rectifier prior to your battery/inverter, then no, you should not need this. If by "blocking diode" you mean you have a DC motor configured as a DC generator, then I would absolutely use a "blocking diode", I would also add either a filter capacitor or, as suggested above, perhaps a battery, to clamp the voltage spikes coming from the DC generator. I personally would use a capacitor rather than a battery. The inverter will suck your battery to zero during periods of no wind while a capacitor will merely filter the DC output. Selection of the capacitor voltage and value should be done with care. I would suggest at least a 25% safety margin on the voltage (ie if the "peak voltage" you see is 75V, select AT LEAST a 100V capacitor). The value of the capacitor needs to be selected based on the "ripple" current, which in your case is likely to be rather high. The capacitor should be connected across the leads (in parallel, not series).
You might look at a capacitor like this:
http://search.digikey.com/scripts/DkSea ... =P10659-ND,though with no knowledge of your system, this is just a stab in the dark.
You might also consider a capacitor like this:
http://search.digikey.com/scripts/DkSea ... 99-3507-NDconnected directly across the motor/generator terminals. This should help limit the voltage spikes associate with the slip rings. Again, selection of optimum value would require intimate knowledge of your system.
Factors that should ultimately determine capacitor selection include 1) Average DC Output Voltage 2) Peak Spike Voltages (dv/dt) 3) Average Current 4) Peak current spikes (di/dt) 5) Spike Frequency (Poles * RPM)
Experimentation with the capacitor across the motor terminals will likely be the best method of determining its size. I would suggest using a 500V to 1kV ceramic capacitor for this purpose. The value I would simply test in "order of magnitude" steps. You might start with 10pF, .001uF and 1uF values. Install the 10pF and compare system performance to no capacitor at all. If system performance improves, add the .001uF capacitor (no reason to remove the 10pF). If system performance again improves, add the 1uF capacitor (again no reason to remove the others).
The filter capacitor after the blocking diode can be as large as you can afford, 4700uF 200V is a good starting place. The general formula is:
Q = CV
where
Q = the charge in coulombs
C = Capacitance in Farads
V = Voltage in Volts
If you will remember that an Amp is 1 coulomb/second and you have an average of 10 amps @ 30V and you have a 5 pole motor turning at 600RPM, here's what you should see:
5 pole motor @ 600RPM = 3000 Pulses/Min = 50hz => 1 pulse every 20mS.
10 Amps suggests 14.1A peaks
30V suggests 42V peaks
10 Amps = 10 Coulombs/Second * 20mS = .2 Coulombs/Pulse
So.....
.2C = 4700uF * V => 42V/Pulse as the charge on our 4700uF capacitor.
With no load other than our capacitor the voltage would quickly ramp up to no load voltage which could be quite high, hence the suggestion of a 200V capacitor. If you wanted to include over-voltage protection you could reduce the Voltage rating of the capacitor and increase the capacity (it turns out the size/cost of capacitors are a function of the voltage rating and the value) to say 22,000uF @ 100V......
.2C = 22,000uF * V => 9V/Pulse as the charge on our capacitor. This would ramp up to working voltage in ~ 3 pulses (~60mS).
Assuming your inverter is either 50hz or 60hz output, it is fairly easy to see that the larger value capacitor would produce significantly less ripple voltage than the smaller one; however, it must be ASS-U-MEd that the inverter has a fairly large capacitive input filter built-in. Your purpose is to filter the DC output so that it does not damage the inverter input. Is larger safer/better? Yes, to a point. There is a certain amount of energy lost in the storage process, among other pitfalls of over-filtering, but for this application, to keep things fairly simple, you can pretty much assume that bigger is better. You just need to be aware that as the RPM decreases, Voltage AND Current drops. You would want to ensure that it doesn't take more than a couple hundred mS to reach cut-in voltage across your filter capacitor at low wind speeds. (Re-Do the above math at 200RPM/20V/4Amps, and you will see what I mean).
Fish
Sat Jan 02, 2010 11:20 am 