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Alternator Paralleler Circuit


chris w

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Yes. Or at least seriously clipped.

 

Oh look.............

 

http://www.aetsa.com/Application%20Note.pdf

 

Look at figure 5

 

Gibbo

Ha ha ha .......... nice try. Now look at fig. 6 which is the CIRCUIT which generates that particular waveform. It is NOT the output of the stator as in a normal alternator. The example you show is a PERMANENT magnet alternator with a special circuit to generate the W signal. At least try to defeat my argument fairly rather than just playing to the crowd who aren't electronic enginers.

 

Chris

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Nonsense.... if you were correct with your "2 diode drops above the battery voltage", he would be reading 8.2v not 7.2v. Anyway the peak to peak voltage isn't 14.4v anyway - it's closer to 30v to 60v peak to peak (see below).

 

That's so wrong.... firstly Smelly had been charging his batteries for 2 1/2 hours before taking the measurement so the charge current was probably very low and hence the W voltage will be low, as I said before.

 

Secondly, I HAVE measured this many times on a scope when doing the alternator experiments of a few months ago. The results were always around 12 to 20vrms depending on rpm and charge current on my alternator. the calculations stack up perfectly with my analysis.

 

 

 

The peak to peak voltage on W on my alternator was more like 35v - 60v peak to peak not 14.4v. You've clearly never measured it! Go stick a scope on the W connection and ground with discharged batteries and you will be in for a surprise.

 

Chris, tomorrow, why not ask Sir Nibble/Snibble as well about all this?

 

cheers,

Pete.

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I'd like Smelly to try the circuit exactly as described using tacho/W, but if that fails to work as intended, I have a good workaround :lol:

 

cheers,

Pete.

 

I do plan to but with the usual smattering of daft mistakes I've still to power it up. I think I've found the last daft one... lamp 2 in parallel, so tomorrow eve will start testing again!

 

OTH I feel like a undergrad stuck between 2 PHd's here. Constructively I'll ask the mods to edit this down to the useful bits once it;s blown over; Theo you were right and it's a good wyt to save a couple of hundred quid but the ongoig debate overshadows the useful info in this topic...

 

Smiles

 

Dan

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Yes. Or at least seriously clipped.

 

Oh look.............

 

http://www.aetsa.com/Application%20Note.pdf

 

Look at figure 5

 

Gibbo

Further to my comment above about the fact that the W signal on this particular alternator is artificially generated by a special circuit, it also states, under fig. 6, that the alternator terminals can rise to over 100V, as I too said, which is true for all alternators but which you guys were disputing earlier.

 

Chris

 

OTH I feel like a undergrad stuck between 2 PHd's here. Constructively I'll ask the mods to edit this down to the useful bits once it;s blown over....

Smiles

 

Dan

........and how will you and the mods determine which are the "useful bits" ????????????? When your circuit doesn't work, maybe you'll see that all of it is useful. :lol:

 

Chris

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(sigh)

 

There was a point about three pages back when agreement, understanding, sweetness and light broke forth and I thought a useable circuit might be forthcoming.

 

That'll teach me to be optimistic.

 

There will be!

 

Dan is doing an excellent job with the current one, should it not work exactly as intended, I have a good alternative with a minimum of modification.

 

cheers,

Pete.

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It's physically isolated not magnetically isolated, as you well know.

 

Yep. And the magnet field cannot be differentiated from that produced by an alternator on the stator.

 

I guess you have no formal qualifications in electronics then if you don't know anything about circuit analysis.

 

I seriously doubt you have any.

 

You did in post #188

 

So your comprehension is so poor that you can't see the extra words in there?

 

Measuring between W and ground is NOT measuring across the stator as no part of the stator is grounded.

 

No one ever said it was. Are you making things up now to try to take the heat off your incorrect statement?

 

Er.......... have you not noticed that the W terminal gives out an AC voltage so it's not likely that a 12v DC relay will work on it, n'est pas!!!

 

So. Are you going to educate us all on what is the difference between an AC coil relay and a DC coil relay?

 

And are you then going to retract your statement and accept that the output from the W terminal is in actual fact DC in that it does not change direction?

 

Nonsense.... if you were correct with your "2 diode drops above the battery voltage", he would be reading 8.2v not 7.2v. Anyway the peak to peak voltage isn't 14.4v anyway - it's closer to 30v to 60v peak to peak (see below).

 

You are this:-

 

Wrong

 

That's so wrong.... firstly Smelly had been charging his batteries for 2 1/2 hours before taking the measurement so the charge current was probably very low and hence the W voltage will be low, as I said before.

 

Secondly, I HAVE measured this many times on a scope when doing the alternator experiments of a few months ago. The results were always around 12 to 20vrms depending on rpm and charge current on my alternator. the calculations stack up perfectly with my analysis.

 

The peak to peak voltage on W on my alternator was more like 35v - 60v peak to peak not 14.4v. You've clearly never measured it! Go stick a scope on the W connection and ground with discharged batteries and you will be in for a surprise.

 

So. Let's get this absolutely clear so you can't do your usual backpedalling trick at a later date........

 

You maintain that the voltage at the outside terminals of the stator can be measured at up to 60 volts peak to peak?

 

For clarification that is one stator terminal connected to the B+ trio and the other connected to the B- trio. They must be those terminals because there aren't any others.

 

Just answer that one question. I know you won't.

 

Or let's look at it another way......

 

You said.......

 

Go stick a scope on the W connection and ground

 

And you say you have measured up to 60 volts at those points.

 

The only thing between those points and the battery terminals is a diode at each end.

 

Where is the other 45 volts going?

 

You are wrong.

 

Gibbo

 

............. playing to the crowd who aren't electronic enginers.

 

That would very clearly include you.

 

Gibbo

 

Further to my comment above about the fact that the W signal on this particular alternator is artificially generated by a special circuit, it also states, under fig. 6, that the alternator terminals can rise to over 100V, as I too said, which is true for all alternators but which you guys were disputing earlier.

 

Why are you dragging up completely irrelevant sh*t?

 

No one has disputed that. Of course it can rise to over 100 volts. That is under open circuit conditions. It can't do it with the batteries connected.

 

That is nothing to do with your incorrect statement that the W terminal has a voltage on it of "60 volts". It is compeltely incorrect.

 

The W terminal will never have a peak voltage of it of more than the battery voltage plus two diode drops (unless the alternator is broken).

 

Gibbo

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So. Let's get this absolutely clear so you can't do your usual backpedalling trick at a later date........

 

You maintain that the voltage at the outside terminals of the stator can be measured at up to 60 volts peak to peak?

 

For clarification that is one stator terminal connected to the B+ trio and the other connected to the B- trio. They must be those terminals because there aren't any others.

 

Just answer that one question. I know you won't.

 

Or let's look at it another way......

 

You said.......

 

 

 

And you say you have measured up to 60 volts at those points.

 

The only thing between those points and the battery terminals is a diode at each end.

 

Where is the other 45 volts going?

Gibbo

Between W and ground I have seen as much as 60v peak to peak (ie: ~20vrms) sine wave output on a scope. There... definitive enough.

The W terminal will never have a peak voltage of it of more than the battery voltage plus two diode drops (unless the alternator is broken).

 

If that were true why would we need a regulator?

 

 

Chris

Edited by chris w
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Between W and ground I have seen as much as 60v peak to peak (ie: ~20vrms) sine wave output on a scope. There... definitive enough.

 

Between W and ground you will only ever see positive voltage. The most negative it will ever go is one diode drop below ground.

 

So that means a signal going from ground up to 60 volts peak.

 

I think you're making this up based on what you think you know as opposed to what you really measured with a scope.

 

You have never seen 0-60 volts peak on the " terminal of a working alternator connected to the batteries.

 

If that were true why would we need a regulator?

 

That is completely irrelevant. If you don't realise that then there isn't much hope for you.

 

Draw us all a diagram of an alternator. A normal 3 phase with full wave rectification. Write the voltages down and draw diagrams of the waveform at the following points.

 

B-

B+

D+

The ends of each stator winding

 

All with respect the battery negative.

 

(I can't wait for this)

 

Gibbo

 

Edit: I've saved you the trouble. Here's a diagram of a typical alternator.............

 

alternator.gif

 

Draw waveforms and show voltages at the end of each stator winding (ignore the fact that it's star unless you think (incorrectly) that it makes a difference in which case draw a delta one) and at the B+ and B- terminals.

Edited by Gibbo
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Lordy!

The ac output on an alternator is either W (one phase) or N (neutral point).

W will read a little (a diodes worth) above output voltage either W to B+ or W to -ve. N will measure about 7V.

Measured thousands, cannot really understand how theoretical predictions can come up with any other figure.

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Of course it can....... it can ultimately go up to about 120v !!

 

Between W and ground I have seen as much as 60v peak to peak (ie: ~20vrms) sine wave output on a scope. There... definitive enough.

 

It can't rise above the level at which the main output diodes start conducting.

 

The peak AC voltage inside the alternator will be 2 diode drops above the DC output voltage.

 

The AC RMS voltage will therefore be somewhat lower than the DC output voltage.

 

I'm reading 7.2 v between W and ground with the engine having run for 2 1/2 hours, does that help?

 

The ac output on an alternator is either W (one phase) or N (neutral point).

W will read a little (a diodes worth) above output voltage either W to B+ or W to -ve. N will measure about 7V.

Measured thousands, cannot really understand how theoretical predictions can come up with any other figure.

 

I know science shouldn't be done by concensus but come on!

 

Gibbo

Edited by Gibbo
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Lordy!

The ac output on an alternator is either W (one phase) or N (neutral point).

W will read a little (a diodes worth) above output voltage either W to B+ or W to -ve. N will measure about 7V.

Measured thousands, cannot really understand how theoretical predictions can come up with any other figure.

 

(Tin hat and running shoes on)

 

Would you get different results with an oscilloscope at W if you measure loaded and disconnected.

 

Richard

 

(faces exit and prepares for flight)

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http://www.amplepower.com/primer/tach/index.html

 

"Most alternators provide a signal that can be used to indicate how fast the alternator is turning. The signal is a half-wave rectified output which has an amplitude about one-half the DC output voltage."

 

Gibbo

Let me deal with this first and then I'll come back to the diagram above.

 

Amplepower must be bad at maths because, if the tacho signal is a half-wave rectified version of the output, it can NEVER be half the output voltage.

 

From simple analysis of a half-wave rectified signal it is easy to show that the average waveform (as shown on a DC meter) will be Vmax/(pi) = 0.32 x the output. This is less than 1/3 of the output not one half, even assuming they are right about the half rectified bit, which you didn't allude to at all. You alluded to a sine wave output in deriving the 7.2v above. So Amplepower neither agree with you or me!!!

 

[for those still interested (!!!) the maths of a half-wave rectified signal is: v = 1/(2pi) (Vmax) {ʃ pi0 (SinΘ.dΘ) + ʃ 2pipi (0)} ]

 

Further, they go on to say that, if the alternator regulator has a float mode then the W output, at float, will probably be too low to drive the tacho and erratic reading will take place. That's strange because float will be around 13.4v say......... yet, when the battery is discharged, the output will be far less than that at maybe only 12.2v, yet the tacho connection works fine.

 

The only explanation for the tacho's still working with a very low battery output initially, yet not working with the much higher float voltage is if the W connection is actually the "internal generator" voltage. This will be high when the battery voltage is low (to drive high charge current), but low when the battery is in float.

 

Draw waveforms and show voltages at the end of each stator winding (ignore the fact that it's star unless you think (incorrectly) that it makes a difference in which case draw a delta one) and at the B+ and B- terminals.

The output voltage from a star-wound alternator will be ROOT(3) higher than if it is delta wound as you well know. Now you are saying that the output will be the same regardless. Errr.......

 

When the battery is discharged, the regulator will be full-fielded so is not limiting the output voltage. The output, in this case, is clamped by the battery to say 12.2v if 50% discharged. Yet my measurements of a few months ago showed quite clearly the different star and delta wound output voltages on the W connection with the same battery output voltage (12.5v in that case). I even discussed my findings at some length with you and Snibble. The relationship was exactly ROOT(3) between the two.

 

Chris

 

I know science shouldn't be done by concensus but come on!

 

Gibbo

I fear you are mistaking consensus for wisdom

Edited by chris w
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Amplepower must be bad at maths because, if the tacho signal is a half-wave rectified version of the output, it can NEVER be half the output voltage.

 

From simple analysis of a half-wave rectified signal it is easy to show that the average waveform (as shown on a DC meter) will be Vmax/(pi) = 0.32 x the output. This is less than 1/3 of the output not one half,

 

The peaks of the stator AC sinewave are quite severely clipped due to appreciable current only being drawn at the peaks (ie when the output diodes are conducting) and the resistance of the stator.

 

This therefore raises the average voltage. It is therefore somewhat higher than 0.32 x the output voltage. And more than 0.32 x the output voltage (say 0.4 x) is "about one-half the DC output voltage" which is what the Amplepower website says.

 

Funnily enough, it's also what the rest of us all said.

 

even assuming they are right about the half rectified bit, which you didn't allude to at all. You alluded to a sine wave output in deriving the 7.2v above. So Amplepower neither agree with you or me!!!

 

I sense some backpedalling starting (as expected).

 

Show me where I said there would be a sinewave on the W terminal.

 

Gibbo

 

The output voltage from a star-wound alternator will be ROOT(3) higher than if it is delta wound as you well know. Now you are saying that the output will be the same regardless. Errr.......

 

You're forgetting that the output is clipped by the batteries.

 

Actually do the excercise I posed. The waveformas and voltage will be the same whether it is star or delta.

 

When the battery is discharged, the regulator will be full-fielded so is not limiting the output voltage. The output, in this case, is clamped by the battery to say 12.2v if 50% discharged. Yet my measurements of a few months ago showed quite clearly the different star and delta wound output voltages on the W connection with the same battery output voltage (12.5v in that case). I even discussed my findings at some length with you and Snibble. The relationship was exactly ROOT(3) between the two.

 

I fear you are mistaking consensus for wisdom

 

And I fear you have now realised you are wrong but still refuse to admit so.

 

Gibbo

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The peaks of the stator AC sinewave are quite severely clipped due to appreciable current only being drawn at the peaks (ie when the output diodes are conducting) and the resistance of the stator.

 

This therefore raises the average voltage. It is therefore somewhat higher than 0.32 x the output voltage. And more than 0.32 x the output voltage (say 0.4 x) is "about one-half the DC output voltage" which is what the Amplepower website says.

 

Nice try.............. now 0.4 = 0.5 :lol: and I note that you do now recognise that there is a volts drop across the stator. :lol:

 

You also glossed over the fact of the tacho's not working on float but working with a low battery voltage. Explain please!!!

 

Show me where I said there would be a sinewave on the W terminal.

 

You told Smelly that he should see 7.2v on the W connection with an output of 14.4v as measured on a DC meter. ie: you were taking the average amplitude of a FULL sine wave.

 

The waveformas and voltage will be the same whether it is star or delta.

 

So you're saying that star and delta output voltages have no difference - I await your white paper on this. :lol:

 

 

Chris

 

I can't argue with you calcs Chris, mainly 'cos I can't be arsed to sing along with them, why should I? I've measured this thousands of times, what's to calculate?

But we had several phone calls when I discussed my results with you from the star/delta thingy a few months back and you never queried the W voltage results then. I discussed the fact that I did indeed see the ROOT(3) output difference. What's changed now?

 

Chris

Edited by chris w
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But we had several phone calls when I discussed my results with you from the star/delta thingy a few months back and you never queried the W voltage results then. I discussed the fact that I did indeed see the ROOT(3) output difference. What's changed now?

 

Chris

To be honest Chris, I simply cannot recall it. And I still cannot understand why when you did the star/delta mod you did not see the same results I have measured on a number of occasions. I cannot support a theoretical effect that flies in the face of what I have observed and measured over a number of years.

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To be honest Chris, I simply cannot recall it. And I still cannot understand why when you did the star/delta mod you did not see the same results I have measured on a number of occasions. I cannot support a theoretical effect that flies in the face of what I have observed and measured over a number of years.

But it's not just theoretical... I still have all my tabulated data from the star/delta experiment.

 

Chris

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But it's not just theoretical... I still have all my tabulated data from the star/delta experiment.

 

Chris

Sorry, cross purposes. I meant the W voltage. I could show you a boat where the star/delta mod has made the difference from flat every morning and new batteries after a month to never flat and batteries going strong. Frankly, on both matters I am getting splinters under my fingernails from scratching my head over why our live measurements differ so much.

Edited to say,

sorry, not feeling well this morning, temperature, sore throat sniffle sniffle (oink oink) and brain is not at optimum. Uxcuse me if I don't make much sense, FISH!

Edited by Sir Nibble
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Nice try.............. now 0.4 = 0.5 :lol: and I note that you do now recognise that there is a volts drop across the stator. :lol:

 

Is that the best you have? 0.4 is "about" 0.5

 

You also glossed over the fact of the tacho's not working on float but working with a low battery voltage. Explain please!!!

 

I didn't gloss over it. I ignored it as I saw it as an attempt to take the heat away from the issue that you say the W terminal can have 60 volts on it yet theory shows otherwise, Amplepower say otherwise and everyone who has measured one says otherwise.

 

You told Smelly that he should see 7.2v on the W connection with an output of 14.4v as measured on a DC meter. ie: you were taking the average amplitude of a FULL sine wave.

 

You're just making things up now.

 

So you're saying that star and delta output voltages have no difference - I await your white paper on this. :lol:

 

Doing the same again.

 

A 14 volt star alternator has the same output voltage as a 14 volt delta alternator.

 

So.......

 

Back to the issue.......

 

1. How can you get 60 volts on the W terminal when it is clamped to the battery voltage plus 2 diode drops?

2. How come no one else has measured this apart from you.

3. Why have you not put the voltages and waveforms on the diagram as requested?

 

Gibbo

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I didn't gloss over it. I ignored it as I saw it as an attempt to take the heat away from the issue that you say the W terminal can have 60 volts on it yet theory shows otherwise, Amplepower say otherwise and everyone who has measured one says otherwise.

 

 

Gibbo

Not everyone......

 

So explain why a float voltage won't drive the tacho but a very low battery voltage will?

 

It's not glossing over the issue, it's a very importaant and germane point to understanding what's going on. My theory at least addresses this issue satisfactorily. Your's doesn't - unless you believe there's a god of course.

 

So how do you explain this apparent anomaly?

 

Chris

Edited by chris w
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Not everyone......

 

So explain why a float voltage won't drive the tacho but a very low battery voltage will?

 

It's not glossing over the issue, it's a very importaant and germane point to understanding what's going on. My theory at least addresses this issue satisfactorily. Your's doesn't - unless you believe there's a god of course.

 

So how do you explain this apparent anomaly?

 

Chris

 

I'll explain it after you answer these:-

 

1. How can you get 60 volts on the W terminal when it is clamped to the battery voltage plus 2 diode drops?

2. How come no one else has measured this apart from you.

3. Why have you not put the voltages and waveforms on the diagram as requested? (when you do so you will see why you are mistaken)

 

Gibbo

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I'll explain it after you answer these:-

 

 

Gibbo

ie: you can't explain it and/or you need more time because even you recognise it IS an interesting anomaly.

 

I have answered the questions in my earlier posts. The AC side of the alternator can be represented as a perfect AC generator in series with the resistance of the stator (which will vary depending on whether it is delta or star wound).

 

Chris

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ie: you can't explain it and/or you need more time because even you recognise it IS an interesting anomaly.

 

There is absolutely nothing mysterious or interesting about it whatsoever. Put a small load on the batteries and it will start to work again. It's because the battery holds the voltage above the regulation voltage so the altertnator spends large amounts of time completely shut down. Very old and common problem. Heart Interface added a feature to their external regulator (the Incharge, also sold by Mastervolt in the early days as the old Alpha Pro) to always keep the alternator producing a bare minimum in order to keep the tacho working. The proviso was that the operator had to be sure the voltage didn't continue to rise during float. Usually by switching things on. Not an ideal solution, but that's what happened.

 

The problem, understanding and solution goes back at least 25 years that I know of.

 

So..............

 

1. How can you get 60 volts on the W terminal when it is clamped to the battery voltage plus 2 diode drops?

2. How come no one else has measured this apart from you.

3. Why have you not put the voltages and waveforms on the diagram as requested?

 

Gibbo

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There is absolutely nothing mysterious or interesting about it whatsoever. Put a small load on the batteries and it will start to work again.

 

It's because the battery holds the voltage above the regulation voltage so the altertnator spends large amounts of time completely shut down.

 

Very old and common problem. Heart Interface added a feature to their external regulator (the Incharge, also sold by Mastervolt in the early days as the old Alpha Pro) to always keep the alternator producing a bare minimum in order to keep the tacho working. The proviso was that the operator had to be sure the voltage didn't continue to rise during float. Usually by switching things on. Not an ideal solution, but that's what happened.

 

 

Gibbo

How can the float voltage (say 13.4v) shut the W terminal down? If you are correct about the relationship between the output voltage and the W terminal, then the W terminal should be at around at least 13.4v + 2 diode drops = 14.6v minimum if the regulator goes into float. But when the battery is discharged to say 12.2v, your W voltage would only be 12.2 + 1.2v = 13.4v. But that works!!!!!

 

When charging at say 14.4v, your output voltage would be 14.4 + (at least) 1.2v = 15.6v and that works too!!! At a 14.8v charge, the W terminal would be nearly 17v according to you (assuming 1v drop per diode at high currents).

 

There is an anomaly somewhere.

 

Chris

Edited by chris w
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