Sorry for my ignorance .....If a single phase of a three phase alternator puts out 3.5 VAC at max rpm....
3.5VAC x 1.414(rms)
= 4.9VDC,....4.9VDC - 1.4VDC (diode drop) =3.5VDC x1.73=6.14VDC???
Why is it when I test the alt that max rpm says 3.5VAC..........after going through the diode bridge the meter only reads 2.9VDC
I think I understand rms (an average of the sine wave?) .....the diode drop is obvious....where does this 1
I think this will be pointless but I will try to explain a bit of it.
If one phase of a 3 phase alternator puts out 3.5vac, then that will be the voltage developed across one third of the winding. What I expect you to have measured is the voltage between a pair of the output terminals. This would be equal to phase voltage for a delta connected alternator, but if it is star connected the phase volts will be referenced to the star point and will in fact only be 3.5/1.73 volts.
Assuming what you have in fact measured to be line volts and you are only dealing with one phase then ( assuming the waveform is a sine wave and that is not likely to be exactly correct). The mean dc of 3.5v rms will be 3.5/1.11 or 3.15v. You now loose two diode drops from the bridge. Unloaded the diode drop is not very well defined and even loaded the claimed 1.4v drop is much of a compromise figure.
If you manage to read 2.9v then that seems to be well within the realms of reality although probably a bit on the high side.
All the maths work if you take things in context.
The equation you are thinking of here is for a 3 phase star connected machine feeding a 3 phase rectifier into a battery.
The phase voltage here will be from one line to star point. The line voltage between a pair of output leads will be phase volts x 1.73 ( root 3).
When rectified in a 3 phase bridge the mean dc will be close to 1.4 times the line voltage. Subtracting diode drops is not mathematically very realistic but the intrinsic drop of a silicon diode is near 0.6v depends on temperature and diode construction. Adding the forward slope resistance for a reasonable load current brings you up to something near 0.8 to 1v depending on the diode.
In your example I assumed that you were rectifying the single phase that you mentioned, with a single phase bridge. If you were actually using all 3 phases with a 3 phase bridge then your figure would be 3.5 x 1.4 then less any diode drops. From your 2.9v figure I suspect you were dealing with the single phase case.
Flux