Interesting stuff! This detail may seem off-topic, but is related in that we are trying to estimate the stator wiring current which of course relates to voltage losses and wiring temperature rise. dariv - Thanks for questioning my initial estimate - I had to shake out the cobwebs.

Don't really know what you are getting at here......

I did not state it very well, but my earlier comment meant that the Magnitude of the Current produced by either a single of multiphase alternator is not dependent on phase angle with regard to voltage - Power is.

I don't believe so. The current entering the R/R should be much less.

Here is the way I see it, so please tell me where I'm going wrong:

Thermodynamics dictates power in must equal power out...... so our energy balance would look like this: Stator Watts input to R/R = output from R/R + heat losses, shunt, R/R power factor, ...

If we assume max output of 24 Amps @ 14VDC then we can eliminate the power losses due to the shunt. In other words, if we are using all the power available none will be sent to ground. Therefore, we are left with power from the stator = power output of R/R. (let's assume the R/R power factor is a non-issue as well other small losses due to wire resistance and such)

So, if the max output is 24A @ 14VDC that's 336 Watts. Now we know the stator puts out around 70 VAC max, 3-phase. (the max R/R output condition)

Hence, 336 Watts/(70VAC*sqrt(3)) = 2.77 Amps.....

Of course, it will be slightly higher due to the R/R power factor....

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Good analysis. My initial estimate of current (24A / 3) was based on simple current balance in the R/R. Your Power balance method is another approach that I hadn't thought of. I agree with your assumptions with the exception of the 70 VAC stator voltage. I am certainly not an expert in this area, but here is my thinking.

Current delivered by the stator causes a back emf in the windings that reduces it's output voltage. This is how the shunt regulator controls voltage - by drawing current that "bypasses" the rectifier stage. I think of the R/R as two boxes, an upstream shunt regulator and a 3-phase diode bridge rectifier. The DC output Voltage of the rectifier is somewhere between the input RMS and Peak AC Voltage input. (I believe this is true - I don't know the conversion calculation. Also the stator output is probably not a pure sine wave form.)

So, I'm thinking that in order to regulate the DC voltage to 14-15 volts the AC side must be somewhere around 16 - 20 VAC. (The 70VAC that you assumed I believe was based on our stator open circuit test which says it should be 50-70VAC. BTW this test is not specified in the Service Manual or Clymers) By using 20 VAC instead of 70 VAC in your power calculation the result is 9.7 A.

As you assumed, at rated conditions there should be little to no shunt current. If the load decreases the shunt regulator draws current to keep VDC between 14-15 volts. So at any given engine speed the stator wires see about the same currrent regardless of DC load. My guess is that the stator output at 4000 rpm is something like 80% of rated which is specified at 8000 rpm. So I'm estimating that the stator wires carry something between 7 and 10 amperes during normal riding. (Assuming that the load does not exceed the charging system capacity and results in low DC voltage.)

The OEM stator wires appear to be about 18 guage? (or 1 sq mm metric wiring) Resistance is about 20 mOHM per meter. Assuming 1 meter? from stator to R/R and 10 A per wire the power loss (discounting connector resistance) would be ~ 2 W. This "feels" about right - the wires do get quite warm.

Anyway, this is my thought process. Let me know if I'm off-base - I've been there before.