Dear Dr F.J. ter Heide, General Manager MASTERVOLT ....

[Tony Brooks]

23 minutes ago, nicknorman said:

Ah yes that I can agree with. But that is not the same as the waveform on the W terminal.

 

But the W terminal is fed from the same pair of coils that produces one of the ripples. One main diode just chops one half (top or bottom) off the waveform and supplies it to the W terminal.

Edited January 16 by Tony Brooks

[nicknorman]

1 hour ago, Tony Brooks said:

 

But the W terminal is fed from the same pair of coils that produces one of the ripples. One main diode just chops one half (top or bottom) off the waveform and supplies it to the W terminal.

 

I'm not entirely sure what we are disagreeing on, but never mind! The W terminal is supplied directly from one coil, not through any diodes, although of course being a star configuration the other 2 coils are in there somewhere!

 

OK so let us put some numbers on it. Diode drop 0.6v, battery voltage 14.4v, SoC close to 100%, minimal field current, minimal charge current.

 

So when the voltage on the phase that is connected to the W terminal starts to rise, no current flows until its voltage reaches 15v. Because of the minimal field current, this doesn't happen until the induced current is close to its peak. As the diode starts to conduct, this clamps the voltage at slightly over 15v because there isn't enough energy to push it up much further. Very little current flows through the diode. So close inspection of the battery voltage with a 'scope would show the voltage at 14.4v with a little bump up to say 14.5v as the induced energy passes through its peak. This very small waveform, if you eliminated the static 14.4v, would resemble the very top of a sine wave. Meanwhile the W terminal waveform would be much closer to a sine wave, just with the peaks lopped off slightly as a consequence of the diodes conducting. Of course when the W terminal voltage gets down to -0.6v the other diode will conduct hence why the bottom of the sine wave is slightly lopped off as well as the top.

 

Or

 

Diode drop 0.7v, battery voltage 13.5v, SoC 50%, maximum field current, maximum charge current.

So when the voltage on the phase that is connected to the W terminal starts to rise it does so very rapidly because of the high field current. It reaches 14.2v quickly and the diode conducts.  There is a lot of energy due to the high field current so the voltage is pushed up a fair bit above 13.5v and a lot of current flows through the diode during most of the half cycle.

So close inspection of the battery voltage with a 'scope would show the voltage at 13.5v with a significant near-half sine wave up to say 13.7v as the induced energy passes through its peak. Just a little bit of the start of the sine wave is lost before the diode conducts.

Meanwhile the W terminal waveform would be much closer to a square wave - it would rapidly rise to 14.2v where the diode then conducts, and then take on a small sinusoidal peak for the remainder of the cycle before plunging back to -0.7v. Because the change from -0.7v to +14.2v occurs almost instantly and is a largish voltage, to the casual observer this makes the waveform look more or less square. Yes there is the sinusoidal peak to it but that is a much smaller magnitude - only 0.2v or so and barely noticable compared to the 14.9v peak-to-peak value of the squarish component.

 

All that said, with Li batteries (very low internal resistance) and a large combi inverter with huge input capacitors, I don't see any ripple on my BMS that measures to 3 decimal places of a volt - albeit with its own smoothing input filtering.

 

Bottom line for me is that at low loads the W terminal resembles a sine wave and at high loads it resembles a square wave. Not exactly, but those are the predominant characteristics.

Edited January 16 by nicknorman

[Tony Brooks]

Oh dear, as you said, deja vue. There is a diode involved, the negative one, the positive one being bypassed by the W terminal. You agreed this was the case last time.

 

If what you say is true then the alternators in power stations would not produce sine waves, as far as I know that is a characteristic of rotating alternators.

 

In this particular case, I think the OP just wants pulses to tell his electronics the alternator is running so as long as he gets a pulse the shape probably does not matter, but will that always be the case for all applications? I don't know, and would prefer accuracy. What I have seen on scopes does not support the output being a square wave or close to square wave.

[nicknorman]

I think last time the issue under discussion was slightly diffrerent. The diagram shows that the W terminal is connected directly to one of the phase windings. The diodes only conduct when the phase voltage is either lower than the battery negative voltage (minus a diode drop's worth) or above the battery postive (by a diode drops worth). So the diodes effectively clamp the phase voltage at close to 0v or Vbatt.

 

At voltages in between these two, the diodes don't conduct and no current flows through the phase winding. It is effecively open circuit. This by the way is discounting the field diodes.

Since copper in a rotating magnetic field has current induced in it, not voltage, when current tries to flow through the winding, the voltage has to jump up to Vbatt or down to 0v before any can flow, so this happens almost instantly, hence the squareish waveform. Once the diodes conduct then the waveform becomes more sinusoidal, but the voltage magnitude is limited by the battery soaking up the current. So a large sinusoidal current flows, but a small sinusoidal voltage waveform superimposed on a large squareish wave. The square wave element is the much larger in magnitude and so it predominates in terms of the overall look of the waveform. The diodes introduce serious non-linearity into the proceedings, which is why the waveform is not sinusoidal execept perhaps at very low field currents when the diodes aren't actually conducting any significant current

 

I might actually take my 'scope down to the boat next time we go, to see if I can capture the W waveform, it would be interesting. Only thing is that although the alternator is a 9 diode machine, the field diodes are no longer used and so it behaves like a 6 diode machine.

 

Edit: Why do it yourself when someone on the internet has already done it! This is a W terminal waveform, looks fairly square to me although there are quite a few transients presumably caused by the diodes in the other phases conducting/not conducting. And no mention of how alternator load might affect the waverform. I plucked this from here: https://electronics.stackexchange.com/questions/395125/is-this-alternator-w-rpm-input-circuit-ok-and-how-can-i-improve-it

 

 

 

Definitely deja vue!

 

Edited January 17 by nicknorman

[Tony Brooks]

What you say is probably true for the  wave form from D+ but:

 

Two points, what does the wave form look like if you alter the time base so the peaks spread out?

 

Secondly, I don't think the W terminal is acting into battery voltage because whatever it is connected to be it rev counter, relay, or electronic gizmo has the other side of THAT circuit connected to negative so no battery is involved, just the resistance or impedance of the circuit.

 

All I will concede is that at the start of the phase generation there will be an extended zero volts as the output builds up enough to open the diode and at that point there will be a vertical rise of around the 0.6 ish volts, the rest will be more or less a sine wave. The same applies once the output has fallen to about 0.6 volts.

[nicknorman]

3 minutes ago, Tony Brooks said:

What you say is probably true for the  wave form from D+ but:

 

Two points, what does the wave form look like if you alter the time base so the peaks spread out?

 

Secondly, I don't think the W terminal is acting into battery voltage because whatever it is connected to be it rev counter, relay, or electronic gizmo has the other side of THAT circuit connected to negative so no battery is involved, just the resistance or impedance of the circuit.

 

All I will concede is that at the start of the phase generation there will be an extended zero volts as the output builds up enough to open the diode and at that point there will be a vertical rise of around the 0.6 ish volts, the rest will be more or less a sine wave. The same applies once the output has fallen to about 0.6 volts.

We weren't really talking about the waveform from D+ but that will be 3 phase full wave rectified (ie a sinusiodal ripple for each phase) but I suspect that will be smoothed into a more level voltage by capacitance within the voltage regulator. In the case of my alternator with nothing connected to the field diodes, the D+ will be a straightforward rectified 3 phase.

 

I can only zoom the internet image, there is a hint of a sinusoid at the peaks but also a lot of noise. Predominantly it is a square wave.

 

The W terminal is connected to a phase and a couple of diodes. It is connected to Bat+ through the diode when the voltage exceeds Bat+ by the diode drop. The load from the diodes is massive when compared to that from a tacho (when significant charging is happening) so it is the diodes, the battery load and the phase winding that are the predominant creators of the waveform, I doubt the tacho load on the W terminal makes a significant difference.

Notice that the waveform goes slightly negative compared to battery negative, because the negative diodes only conduct when the phase voltage is less than 0v by a diode's drop worth. Your "concede" sentence is not born out by the waveform diagram.

 

[GreenHeaven]

2 hours ago, Tony Brooks said:

.

 

In this particular case, I think the OP just wants pulses to tell his electronics the alternator is running so as long as he gets a pulse the shape probably does not matter, but will that always be the case for all applications? I don't know, and would prefer accuracy. What I have seen on scopes does not support the output being a square wave or close to square wave.

Correct, and I suspect it is not the shape of the wave produced on the up cycle, but the gaps between them that are the trigger for rpm counting

As you have both correctly surmized, my next problem is getting a wire to the right place on the alternator

I bypassed the marine stores in my search for a high output alt, and laughed at the Balmar prices. I headed striaght down to a farm equipment breakers and found a quality big body alt from a combine. 

 

 

 

B+ B- D+ terminals only (((

 

The tach on the yanmar is driven by a sensor above the flywheel

 

removing the old regulator (low side reg) and replaced with Alpha Pro (high side reg) but

to get to the diodes and attach a W wire, the case must be split. Not something I want to do

I have to remove it again (very difficult job) and take it to an alt shop

Then figure out where to attach a lead - hopefully it will be obvious. 

 

 

Those circular rings are the rails for the diodes I assume

 

This is a 6 dipole it seems, so I set that setting in the alpha pro

 

just getting this far has been a lot of work, but it will put out 160a @ 1200rpm all day long at 50-60 deg c