Jump to content

Alternator Paralleler Circuit


chris w

Featured Posts

Well, that's all the big wires in place and having switched everything back on nothing's exploded :lol:

 

I wonder whether to crack on with the switching... I'll need to drill the holes first!

 

edit... and of course, now I need to use the big drill outside it's started hammering down so that, as they say, was a day! Off to the pub

Edited by Smelly
Link to comment
Share on other sites

.

Well, that's all the big wires in place and having switched everything back on nothing's exploded :lol:

 

I wonder whether to crack on with the switching... I'll need to drill the holes first!

 

I'd start the engine, and using a low watt 12v test lamp, connect it between the tacho/W terminal of one alternator and 0V, and check it lights up. Then do the same with the other alternator.

 

Then manually parallel the alternators, and after the charge voltage has stabilised for 10 minutes, repeating the above test should show that that one alternator no longer lights up the test lamp.

 

That said I understand one of your alternators has an external controller, so it's very likely to be the other alternator that no longer lights up the test lamp.

 

This will give some confidence the circuit will have the right signals to work as intended.

 

cheers,

Pete.

Edited by smileypete
Link to comment
Share on other sites

The output load (batteries under charge) will act like a shunt regulator.

 

Ask Gibbo and Snibble if you like.

 

cheers,

Pete.

I don't need to ask, I understand how alternators work. There's only 14v or so on the output because it's REGULATED and it's DC not AC because it's RECTIFIED. Inside the alternator it's just a simple 3-phase AC generator, whose internal output (eg: the W connection) depends on revs and charging load. The W connection is simply attached to one of the phases. Any 12v bulbs attached to the W connection are likely to blow.

 

If you run a sinple Ohms Law calculation, you will be able to work out the internal AC voltage of the generator.

 

Chris

Edited by chris w
Link to comment
Share on other sites

I don't need to ask, I understand how alternators work. There's only 14v or so on the output because it's REGULATED and it's DC not AC because it's RECTIFIED. Inside the alternator it's just a simple 3-phase AC generator, whose internal output (eg: the W connection) depends on revs and charging load. The W connection is simply attached to one of the phases. Any 12v bulbs attached to the W connection are likely to blow.

 

If you run a sinple Ohms Law calculation, you will be able to work out the internal AC voltage of the generator.

 

Chris

 

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.

 

Gibbo

 

Edit: Because of this, when people run split charge relays from the W terminal they usually have to use a lower voltage col relay. Often 6 or 9 volts. A 12 volt relay very often won't pull in.

Edited by Gibbo
Link to comment
Share on other sites

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.

 

Gibbo

You're forgetting about the voltage drop across the stator. I know what these voltages are because I measured them all (at the W connection) when we were discussing the star/delta wiring thing some time back. On a large alternator this could easily reach over 30VAC RMS.

 

Chris

Edited by chris w
Link to comment
Share on other sites

You're forgetting about the voltage drop across the stator. I know what these voltages are because I measured them all (at the W connection) when we were discussing the star/delta wiring thing some time back. On a large alternator this could easily reach over 30VAC RMS.

 

Chris

 

No I'm not.

 

No it couldn't.

 

You're confusing real life with simulation models.

 

If you get 30 volts RMS on the stator that means the peak voltage on the stator is 45 volts. If the DC output is 14.4 volts that means the diodes are dropping roughly 15 volts each.

 

If the DC output is 100 amps that is 1440 watts. Yet the stator is (according to you) producing 30 volts * 100 amps = 3000 watts so 1560 watts is simpy dissappearing into thin air.

 

Gibbo

Link to comment
Share on other sites

Test number 1... connect a 12v supply to the contactor coil and... the cable melted.

 

i ran a multimeter over the contactor and cabled as I thought appropriate but maybe I was wrong. With the studs facing up there are 4 connectors, do i place the supply across them horizontally or vertically?

Link to comment
Share on other sites

No I'm not.

 

No it couldn't.

 

You're confusing real life with simulation models.

 

If you get 30 volts RMS on the stator that means the peak voltage on the stator is 45 volts. If the DC output is 14.4 volts that means the diodes are dropping roughly 15 volts each.

 

If the DC output is 100 amps that is 1440 watts. Yet the stator is (according to you) producing 30 volts * 100 amps = 3000 watts so 1560 watts is simpy dissappearing into thin air.

 

Gibbo

 

You're doing it again.... you are overlooking the voltage drop across the stator. The alternator (from an AC viewpoint) is equivalent to an AC generator plus a series resistance (the stator windings). There is also some inductance but we'll ignore that for these purposes.

 

So if, say, the alternator is delivering 100A and let's say that the stator windings only have a resistance of 0.15 ohms then 15 volts will be dropped across the stator, plus a volt across each of the 2 diodes plus the actual battery voltage DC output. So the PEAK AC voltage will be 15 + 2 + 14 = 31 volts peak which is 22VAC RMS, more than enough to blow a 12v bulb.

 

The diodes will dissipate about 100W each (100A x 1v drop), the battery will be absorbing (including its own series resistance loss) 1400W and the remainder will be heating up the alternator stator (1400W). There's no missing watts.

 

Alternators are only around 50% efficient which is why they get so bl**dy hot as in this example.

 

Chris

 

Test number 1... connect a 12v supply to the contactor coil and... the cable melted.

 

i ran a multimeter over the contactor and cabled as I thought appropriate but maybe I was wrong. With the studs facing up there are 4 connectors, do i place the supply across them horizontally or vertically?

You shorted out the cable by choosing one of the 4 connectors incorrectly. They are connected together in 2 pairs. Use a multimeter to determine which pairs are NOT connected together and connect your + and - to those (either way round). IIRC choose a pair that are side-by-side and connect to those.

 

Chris

Edited by chris w
Link to comment
Share on other sites

You shorted out the cable by choosing one of the 4 connectors incorrectly. They are connected together in 2 pairs. Use a multimeter to determine which pairs are NOT connected together and connect your + and - to those (either way round).

 

Chris

 

that's where I went wrong... i thought as it was across the coil there'd be a circuit!

 

Thank you

 

Needing to charge my batteries up I've started the engine now so it'll have to wait until a dry evening.

Link to comment
Share on other sites

that's where I went wrong... i thought as it was across the coil there'd be a circuit!

 

Thank you

 

Needing to charge my batteries up I've started the engine now so it'll have to wait until a dry evening.

It's a bit daft that they don't provide any wiring instructions but leave you to work it out.

 

Chris

Link to comment
Share on other sites

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.

Gibbo! :lol:

 

Glad you're here, I was beginning to lose my sanity trying to explain this!

 

 

You're forgetting about the voltage drop across the stator. I know what these voltages are because I measured them all (at the W connection) when we were discussing the star/delta wiring thing some time back. On a large alternator this could easily reach over 30VAC RMS.

You measured them with a DVM?

 

cheers,

Pete.

Link to comment
Share on other sites

Gibbo! :lol:

 

Glad you're here, I was beginning to lose my sanity trying to explain this!

But Gibbo's made an error and overlooked the stator losses (that's why an alternator gets hot!!) see my post above.

You measured them with a DVM?

 

Pete.

I measured them with a scope.

 

Chris

Link to comment
Share on other sites

You're doing it again.... you are overlooking the voltage drop across the stator. The alternator (from an AC viewpoint) is equivalent to an AC generator plus a series resistance (the stator windings). There is also some inductance but we'll ignore that for these purposes.

 

No I'm not. You are getting yourself in a mess confusing reality with models which are designed to allow analysis of the external nodes of a device. I've said this before and I'll say it again:

 

Your maths is great, but if you base it on a fundamental misundertstanding (as you are doing here), you're going get bullsh*t results (as you are doing here).

 

So if, say, the alternator is delivering 100A and let's say that the stator windings only have a resistance of 0.15 ohms then 15 volts will be dropped across the stator, plus a volt across each of the 2 diodes plus the actual battery voltage DC output. So the PEAK AC voltage will be 15 + 2 + 14 = 31 volts peak which is 22VAC RMS, more than enough to blow a 12v bulb.

 

How on earth you reach that statement is a complete mystery to me. Where can we measure this 22 volts?

 

The stator and diodes are in parallel with the battery not in series. The voltage at the battery side of the diodes has to be the same as that across the battery.

 

Therefore the voltage at the ends of the stator can only ever be two diode drops above the battery voltage. It simply cannot ever be anything else.

 

Come on Chris, this is elecronics for 6 year olds.

 

The theoretical model in the above scenario is:-

 

Zero volts dropped across the stator. (It is an EMF not a potential difference and therefore has zero volt drop). 17.5 volts produced at the ends of the modelled stator. The (zero resistance) stator is in series with the modelled stator resistance of (say) 0.015R. The modelled resistance drops a voltage of 1.5 volts. But that is an internal modelling device. The voltage drop doesn' actually exist so that it can be measured (where would you stick your meter probes?). The voltage at the ends of the real world stator is therefore 17.5 - 1.5 = 16 volts. Less two diode drops = 14 volts at the alternator terminals.

 

Even if the stator resistance was 0.15R (much higher than reality) the modelled stator would have to produce roughly 30 volts RMS in order to pump out 100 amps at 15 volts but even then you would not ever be able to measure 30 volts anywhere. It is a theoretaical model to separate the EMF of the stator with its own resistance.

 

The diodes will dissipate about 100W each (100A x 1v drop), the battery will be absorbing (including its own series resistance loss) 1400W and the remainder will be heating up the alternator stator (1400W). There's no missing watts.

 

Cr*p.

 

You have, in your above scenario, 22VAC RMS (Notice it has dropped form the previous figure of 30 VRMS). 22VRM at 100 amps = 2200 Watts. You have accounted for 1600 watts. Where is the other 600 watts? Your maths is wrong, your theory is wrong, your measurements are wrong.

 

The peak voltage across the stator terminals will be the battery voltage plus two diode drops. The RMS voltage will be less than the battery voltage.

 

Alternators are only around 50% efficient which is why they get so bl**dy hot as in this example.

 

Actually a typical automotive alternator is about 80% to 90% efficient.

 

Roughly 1/3rd each of those losses are in the rectifiers, the rotor and the stator.

 

The other 30% losses are in the belt/pulley arrangement which is why very big alternators use chain drive.

 

A simple "Whoops" will suffice.

 

Gibbo

 

Gibbo! :lol:

 

Glad you're here, I was beginning to lose my sanity trying to explain this!

 

The theory is pretty straightforward. It's not even difficult to test: You stick a voltmeter between B- and W and you get about 9 volts on a 12 volt system

 

The non theoretically minded fitters also know that a 12 volt split charge relay will very often not work from the W terminal and they have to use a 9 volt relay, sometimes even a 6 volt one. I know this, because I've seen many boat mechanics do it and they asked me "why does it need a lower voltage relay on the W terminal", then when I try to tell them their eyes glaze over followed by "I'm not really that interested".

 

Gibbo

Link to comment
Share on other sites

How on earth you reach that statement is a complete mystery to me. Where can we measure this 22 volts?

As I said....... at the W connection.

 

The stator and diodes are in parallel with the battery not in series.

NONSENSE!!! Of course they are in series. The diodes form a full wave rectifier and the stator is the generator driving the current. Now who's reading electronics for 6 year olds?

 

The peak voltage across the stator terminals will be the battery voltage plus two diode drops.

Ah so the diodes are now mysteriously no longer in parallel with the battery???

 

Actually a typical automotive alternator is about 80% to 90% efficient.

 

Roughly 1/3rd each of those losses are in the rectifiers, the rotor and the stator.

 

The other 30% losses are in the belt/pulley arrangement which is why very big alternators use chain drive.

 

Gibbo

Firstly only VERY expensive military alternators are 80% efficient - you of all people should know that. Car alternators of the types on boats are around 50% efficient only. You know that's correct.

 

Secondly, if there are no volts dropped across the stator as you say, then why is the alternator getting so bl**dy hot !!!!

 

Re equivalent circuits, have you never heard of a Thevenin equivalent circuit? The alternator is fundamentally a "perfect" AC generator with a series resistor (the stator resistance) feeding the battery and 2 diodes. Simples

 

When I was playing around with the star/delta thing a few months back, I took lots of data which I still have. I was measuring the stator volts at the W connection.

 

For instance, at 1200rpm, and with a battery voltage of 12.5v, I was getting 20.2v peak at the W connection and a charge current of 38A. (The voltage is lower than that used in the example above because the current is lower than that in the example above).

 

So the loss across the stator was therefore 20.2 - 12.5 - 2v (for the diodes) = 5.7 volts. At 38A this is an equivalent resistance of 0.15 ohms which is where I got the figure I used in the example above. I did lots of readings at different revs (and therefore currents) and was even able to predict the current I would get based on the W connection volts I was seeing. Checking the ammeter confirmed exact correlation.

 

 

You're also being a naughty boy by multiplying rms volts by peak currents. Even rms volts x rms current doesn't give rms power - now I know that you know that I know that you already know this!!!!

 

Chris

Link to comment
Share on other sites

The non theoretically minded fitters also know that a 12 volt split charge relay will very often not work from the W terminal and they have to use a 9 volt relay, sometimes even a 6 volt one. I know this, because I've seen many boat mechanics do it and they asked me "why does it need a lower voltage relay on the W terminal", then when I try to tell them their eyes glaze over followed by "I'm not really that interested".

I'm thinking that the latch relay will work better on half wave rectified DC than AC, especially with a quenching diode that will allow current in the coil to 'freewheel' until the next half cycle.

 

The problem with full wave rectification is it may allow the latch relay to be reverse powered through the contactor coil. (Edit thinking about it, full wave rectification wouldn't do much good in this case!)

 

 

NONSENSE!!! Of course they are in series. The diodes form a full wave rectifier and the stator is the generator driving the current. Now who's reading electronics for 6 year olds?

Chris,

 

Think of a basic battery charger with a simple single phase transformer feeding a battery via a bridge rectifier.

 

Do you not agree that the peak voltage across the transformer secondary is the battery charge voltage plus 2 diode drops?

 

cheers,

Pete.

Edited by smileypete
Link to comment
Share on other sites

Chris

 

You are 100% wrong.

 

1. The voltage across the stator cannot be higher than the battery voltage plus two diode drops unless the diodes are broken.

 

2. There is no such thing as RMS power. It is marketing speak from Hi Fi sellers. But the fact remains, there is no such thing as RMS power.

 

Gibbo

 

Think of a basic battery charger with a simple single phase transformer feeding a battery via a bridge rectifier.

 

Do you not agree that the peak voltage across the transformer secondary is the battery charge voltage plus 2 diode drops?

 

I think he confused himself when he had his alternator stripped down and brought the winding tappings out.

 

He played with it in star and delta. I think he measured the voltage across one winding in star and, seeing that they were effectively in series, doubled to get his (incorrect) stator winding voltages forgetting that they are 120 degrees out of phase.

 

If you run the maths it ties in roughly with his (now massively reduced from 45 volts peak) figure of 20.2 volts peak.

 

Gibbo

Link to comment
Share on other sites

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

 

I wonder whether to move the relay connection to d+ as that's more reliable ...

Link to comment
Share on other sites

2. There is no such thing as RMS power. It is marketing speak from Hi Fi sellers. But the fact remains, there is no such thing as RMS power.

 

Gibbo

 

That's what I said too...

 

I think he confused himself when he had his alternator stripped down and brought the winding tappings out.

 

He played with it in star and delta. I think he measured the voltage across one winding in star and, seeing that they were effectively in series, doubled to get his (incorrect) stator winding voltages forgetting that they are 120 degrees out of phase.

 

If you run the maths it ties in roughly with his (now massively reduced from 45 volts peak) figure of 20.2 volts peak.

 

Gibbo

 

No I didn't - I took the root 3 bit into account. The volts are reduced because in your example you were using 100A and I was using the 38A I actually got on my particular alternator.

 

Remember that by measuring at the W connection we are measuring bewteen W and ground not across one of the windings because none of the stator connections is grounded.

 

Chris

 

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

 

I wonder whether to move the relay connection to d+ as that's more reliable ...

Your reading doesn't give any info unless we also know the current that the alternator is delivering. Even I agree that at low currents the W connection will have a small voltage but it increases with current. So, if your batteries are not accepting much current, after 2 1/2 hours, then the voltage at W will be low.

 

When you say the W voltage is 7.2v, by the way, is that peak, peak-peak or rms?

 

Chris

Edited by chris w
Link to comment
Share on other sites

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

 

I wonder whether to move the relay connection to d+ as that's more reliable ...

 

I'd make the circuit exactly as in post #168 (clicky) including the extra diode.

 

Then follow the steps at the end of that post.

 

cheers,

Pete.

 

Edit:

 

What you're measuring on W is very likely to be a waveform that varies from 0v to 14.4v

 

The extra diode will help keep the relay closed when the waveform isn't at 14.4v.

 

When the latch relay and contactor is closed using the push button, and the alternators are paralleled, the engine alternator will produce more power and the waveform at W will be much more favourable to keeping the latch relay closed.

Edited by smileypete
Link to comment
Share on other sites

Chris,

 

Think of a basic battery charger with a simple single phase transformer feeding a battery via a bridge rectifier.

 

Do you not agree that the peak voltage across the transformer secondary is the battery charge voltage plus 2 diode drops?

 

cheers,

Pete.

Yes I do agree but it's not the same analogy. The transformer secondary, in the charger, is not the "generator" as with the stator. The generator, and the equivalent currents and voltages, is the primary side not the secondary. Using basic circuit theory, we can "reflect" the secondary and represent everything as if driven by a generator and series resistance on the primary side.

 

I hope you will agree, contrary to what was written above, that the diodes are not in parallel with the battery but in series, as in the alternator.

 

Chris

 

What you're measuring on W is very likely to be a waveform that varies from 0v to 14.4v

 

 

cheers,

Pete.

I did all this on a scope some months ago. Try it yourself and you wil see that W is much higher than you currently believe, when high charge current is being delivered. take a scope and try it - don't take my word for it.

Edited by chris w
Link to comment
Share on other sites

Yes I do agree but it's not the same analogy.

 

It's actually a very good analogy.

 

The transformer secondary, in the charger, is not the "generator" as with the stator.

 

Yes it is.

 

The generator, and the equivalent currents and voltages, is the primary side not the secondary.

 

No it isn't.

 

The secondary of a transformer is an isolated circuit from the primary.

 

The secondary of a transformer is a winding in a changing magnetic field. This is exactly the same as the stator in an alternator.

 

For analysis we can remove the primary, forget it exists, and deal with a winding in a changing magnetic field. From that point on it makes no difference what causes the changing magnetic field. It could be because of an electromagnet (the primary in a transformer), a moving permanent magnet (as in a wind turbine) or a rotating magnetic field (as in an alternator). It makes no difference. From the point of view of analysis of the currents and voltages in the winding they are all the same. They have the same effect.

 

Using basic circuit theory, we can "reflect" the secondary and represent everything as if driven by a generator and series resistance on the primary side.

 

As we can also reflect the currents into the counter rotational force in an alternator.

 

I hope you will agree, contrary to what was written above, that the diodes are not in parallel with the battery but in series, as in the alternator.

 

And you're arguing with yourself there because no one ever said the diodes were in parallel with the battery.

 

I did all this on a scope some months ago. Try it yourself and you wil see that W is much higher than you currently believe, when high charge current is being delivered. take a scope and try it - don't take my word for it.

 

No it isn't. The peak voltage of the stator is two diode drops above the output voltage of the alternator.

 

I have measured it.

 

Theory predicts it.

 

Boat mechanics know that a 12 volt relay doesn't get enough voltage to pull it in when used on the W terminal.

 

Smelly measured 7.2 volts on the W terminal which is a typical reading when measured on the DC range of a normal multimeter.

 

Just admit you're wrong then we can all move on to something else.

 

We won't laugh.

 

Gibbo

 

What you're measuring on W is very likely to be a waveform that varies from 0v to 14.4v

 

Assuming you mean 14.4 volts peak to peak that is spot on (with due allowance for the diode drops).

 

Gibbo

Link to comment
Share on other sites

Assuming you mean 14.4 volts peak to peak that is spot on (with due allowance for the diode drops).

 

I'm expecting it will start to look more like a squarewave than a sinewave when the alternator is under heavy load.

 

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.

Link to comment
Share on other sites

The secondary of a transformer is an isolated circuit from the primary.

 

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

 

For analysis we can remove the primary, forget it exists,

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

 

And you're arguing with yourself there because no one ever said the diodes were in parallel with the battery.

 

You did in post #188

The stator and diodes are in parallel with the battery not in series.

 

 

The peak voltage of the stator is two diode drops above the output voltage of the alternator.

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

 

I have measured it.

 

Theory predicts it.

 

Boat mechanics know that a 12 volt relay doesn't get enough voltage to pull it in when used on the W terminal.

 

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!!!

 

 

Smelly measured 7.2 volts on the W terminal which is a typical reading when measured on the DC range of a normal multimeter.
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.

 

 

Assuming you mean 14.4 volts peak to peak that is spot on (with due allowance for the diode drops).

 

Gibbo

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

Link to comment
Share on other sites

(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.

 

Richard

 

Anyone want to borrow my bulb with two wires and crocodile clips?

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.