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Alternator Paralleler Circuit


chris w
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Remove the one from the small relay. You might have to play around a bit to stop the diode across the rotor from acting as a quencher. This will massively speed up the opening time of the smaller relay.

 

But keep the diode on the paralleling relay. This will slow that relay opening time long enough so the other reg doesn't kick back in while the smaller relay is still opening.

 

If that doesn't work and there's no other suggestions from Gibbo, I've got something that is likely to.

 

cheers,

Pete.

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Why not? The PDAR does, does it not complete the rotor circuit to earth, and the warning light is connected to rotor +ve so the lamp should light as normal until the D+ voltage from the field diodes is established, again, as normal.

 

You could be right but there's something telling me that the PDAR doesn't bark up until it sees 12 volts on the D+ terminal, which it will never see if the internal reg isn't working. I'm not certain on this, I could well be wrong which is why I asked it as a question.

 

Gibbo

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Do you two ever sleep?

Yes......but not together!!!!!

 

I sleep like a log every night but sometimes burn the candle at both ends. I've luckily never suffered from insomnia; I don't think it's anything to lose any sleep over. :lol:

 

Now....I gotta read these new posts.

 

Chris

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You could be right but there's something telling me that the PDAR doesn't bark up until it sees 12 volts on the D+ terminal, which it will never see if the internal reg isn't working. I'm not certain on this, I could well be wrong which is why I asked it as a question.

 

Gibbo

I have seen many many alternators modified for sterling regs where the internal reg has been isolated conpletely or even removed altogether.

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I have seen many many alternators modified for sterling regs where the internal reg has been isolated conpletely or even removed altogether.

 

Then I stand corrected. I know the older one was ok, as is the Adverc, but I had a feeling the PDAR was different and needed the internal reg leaving intact.

 

Gibbo

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I did a really long reply to this addressing all your points but I really fail to see the point.

 

I'll just say this:-

 

Get two identical batteries at the same state of charge, (let's go for 50%). The internal resistance will be identical.....

 

Put a 10 amp charger on one, and a 100 amp charger on the other. The one with the 10 amp charger will start to increase the terminal voltage while throwing in 10 amps. It might take 4 hours to reach acceptance. The bulk stage lasted 4 hours. It has put in 40 amp hours. The battery is at 90% state of charge. Acceptance starts. The "clever" adaptive charger calculates a 6 hour acceptance cycle.

 

The 100 amp charger will immediately raise the voltage to acceptance. The bulk stage lasted 2 seconds. The battery is at 50.00000001% state of charge. The "clever" adaptive charger calculates an acceptance cycle of a few minutes then goes into float. The battery is still half flat.

 

Using a charger somewhere in between these two extremes doesn't make things suddenly jump into a new arena. It's a straight line graph from one to the other.

 

Gibbo

But that doesn't happen in the real world..... no-one puts 100A through a single battery (ie: charging at C/1). Yesterday, when my batteries (3 x 135AH) were well down, the paralleler put a maximum current of 93A through them. The terminal voltage still took well over an hour to get anywhere near 14.2v. I appreciate that's around 30A per battery (C/4.3) not 100A per battery but this is real life.

 

The terminal voltage rose quickly to the early 13.x v region but further quick rise was negated by the now decreasing internal resistance and (therefore) the decreasing voltage dropped across that.. The bulk stage time didn't halve because the charging current had doubled over what it was before I fitted the paralleler.

 

Chris

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But that doesn't happen in the real world..... no-one puts 100A through a single battery (ie: charging at C/1). Yesterday, when my batteries (3 x 135AH) were well down, the paralleler put a maximum current of 93A through them. The terminal voltage still took well over an hour to get anywhere near 14.2v. I appreciate that's around 30A per battery (C/4.3) not 100A per battery but this is real life.

 

Which is why I wrote...........

 

Using a charger somewhere in between these two extremes doesn't make things suddenly jump into a new arena. It's a straight line graph from one to the other.

 

Edit: To clarify that bit, if on a certain installation, from a certain state of charge, a 10 amp charger reaches acceptance in 5 hours, and a 30 amp charger reaches acceptance in 3 hours, then a 20 amp charger will (not might), reach acceptance in 4 hours. And in each case, the higher the charger size, the lower will be the state of charge when acceptance is reached.

 

The terminal voltage rose quickly to the early 13.x v region but further quick rise was negated by the now decreasing internal resistance and (therefore) the decreasing voltage dropped across that.. The bulk stage time didn't halve because the charging current had doubled over what it was before I fitted the paralleler.

 

Once you get rid of the preconceived (apparently fixed) idea that adaptive charging works (because you've been told it does) and actually work it out properly or do some measurements and tests you will come to the only honest conclusion which is that it doesn't work. Trust me, you will. Even the bloke that designed it knows it doesn't work!

 

Chris, instead of fighting against it because you don't want it to be the case, just look upon it as something that you haven't yet discovered. Then, after you've played with your parelleling system in anger for a few weeks, and you realise it is indeed correct, you won't have to do a load of backpedalling :lol:

 

Gibbo

Edited by Gibbo
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Chris, have you got some sort of voltage/current data logging equipment you could run against your setup?So we could get some fancy graphs over time ?

Edited by stuart
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I have given this some thought and it seems to me that the crux of the problem is that the standard reg on the engine alternator is machine sensed and therefore will not be maintained in an open circuit condition by the higher voltage of the domestic alternator. Now horses for courses, bloke like Chris is likely to be most happy with a pocketful of semiconductors juggling milliamps but this is what I'd do.

Remove the domestic alternator regulator, cut the link between -ve brush and regulator, I would leave enough metal on the brush to curl it under the lip on the plastic brush box to secure it in place the same as the opposite side of the brush. Then connect a wire to the regulator side of the break and earth the relay through it. Now when the reg opens at 14.2 volts the voltage will not fall in response as the PDAR will continue to conduct, therefore the reg should remain open, not just long enough to drop the relay but until the engine is stopped. That's what I'd do, but then I'm used to precisely that kind of work.

I understand what you're suggesting. The domestic reg simply becomes a brush holder for the alternator, the rotor current is controlled only by the PDAR and the original domestic reg simply acts as a 14.2v comparator to switch the relay.

 

The relay -ve is connected to one side of the break (the output transistor collector) whilst the PDAR is still connected to the other side of the break, viz: the field connection, ie the brush. Neat - have you considered a career in alternators?

 

I haven't got a reg around me at the moment to look at the mechanicals of cutting the regulator -ve connection whilst still leaving the brush retention and support in place.

 

Chris

PS: I agree with you that the PDAR will still function without the original reg (so long as the brushes function of course)

Edited by chris w
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Chris, instead of fighting against it because you don't want it to be the case, just look upon it as something that you haven't yet discovered. Then, after you've played with your parelleling system in anger for a few weeks, and you realise it is indeed correct, you won't have to do a load of backpedalling :lol:

 

Gibbo

I am not a paid-up member of the "Save Adaptive Charging Society". It either works or it doesn't...... I'm happy either way. But I need to understand fully the electronic process that is taking place and cannot just accept it's true or not because you (or anyone else) says it is true or not. By stating that adaptive charging doesn't work you have definitely piqued my curiosity. Now I need to tread the path to enlightenment on my own.

 

Where I have difficulty at the moment is accepting that a higher current leaves the batteries in a lower state of charge once the acceptance stage is reached. That is what your argument turns on. With a much higher charge current, I am not seeing the battery's voltage suddenly jumping up to a higher value quicker. As I said earlier, the decreasing internal resistance counters that to a large degree. If the internal resistance were constant, I would agree with you immediately. But it's not constant.

 

Even using your example of a 10A charger taking 5 hours, a 20A charger taking 4 hours and a 30A charger taking 3 hours to reach acceptance, then the relative AH into the battery (ignoring charging inefficiencies) will be 50AH, 80AH and 90AH. So the batteries will be in a HIGHER state of charge with the 30A charger not a lower state of charge as you suggested. Further, because of the longer time taken to reach acceptance, the acceptance cycle time for the 10A charger will be longer than the 20A charger which will be longer than the 30A charger acceptance time which is the correct way you want it.

 

Chris

 

Chris, have you got some sort of voltage/current data logging equipment you could run against your setup?So we could get some fancy graphs over time ?

I don't have any data logging equipment but do have voltmeters and ammeters on all channels. I would have to log it all manually (at constant revs) which would clearly need me to get a life afterwards as I would have to do it twice, once with and once without the paralleler for several hours.

 

The problem with logging it whilst cruising is that the revs are obviously not constant which screws up the validity of the data.

 

Chris

Edited by chris w
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I am not a paid-up member of the "Save Adaptive Charging Society". It either works or it doesn't...... I'm happy either way. But I need to understand fully the electronic process that is taking place and cannot just accept it's true or not because you (or anyone else) says it is true or not. By stating that adaptive charging doesn't work you have definitely piqued my curiosity. Now I need to tread the path to enlightenment on my own.

 

And when you do, believe me, you will be converted to the school of "Oh you're right, it can't (and doesn't) work".

 

Even using your example of a 10A charger taking 5 hours, a 20A charger taking 4 hours and a 30A charger taking 3 hours to reach acceptance, then the relative AH into the battery (ignoring charging inefficiencies) will be 50AH, 80AH and 90AH. So the batteries will be in a HIGHER state of charge with the 30A charger not a lower state of charge as you suggested. Further, because of the longer time taken to reach acceptance, the acceptance cycle time for the 10A charger will be longer than the 20A charger which will be longer than the 30A charger acceptance time which is the correct way you want it.

 

They were uncalculated examples to show that the graph is a straight line not to show what the times would be!

 

If a 10 amp charger took 4 hours to reach acceptance (thus returning 40 amp hours), then a 20 amp charger, into the same battery, from the same state of charge would take less than 2 hours to reach acceptance, thus returning less than 40 amp hours. Conversely a 5 amp charger would take longer than 8 hours to do the same thing, thus returning more than 40 amp hours.

 

I find it a bit odd that you can't see that when you accept that a 100 amp charger would instantly wack the voltage right up to acceptance before it had even returned a single amp hour.

 

I'm not guessing at this. Having looked at so many charge graphs (literally thousands) I can still see them in my head!

 

Gibbo

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They were uncalculated examples to show that the graph is a straight line not to show what the times would be!

 

Gibbo

The graph cannot be a straight line as can be seen by plotting the examples you gave. (It doesn't matter if the actual figures for the 10A charger etc are accurate, it's the principle).

 

The equation for your examples works out to be t = 6 - (i/10) where "t" is the charge time and "i" is the current ie: 10A needs 5 hours, 20A needs 4 hours; 30A needs 3 hours and so on.

 

Plotting this equation gives:

chargetimelinear.jpg

 

It can be seen straight away that whatever the real numbers, there will be a current at which the charging is instantaneous (zero time) and at zero current the battery would still charge in 6 hours !!!! Clearly both of these are a nonsense.... but be patient.

 

Further if one plots "t*i" to obtain ampere hours (AH) the graph looks like this:

 

amperehourslinear.jpg

 

This shows a peak ampere hours into the battery at a current of 30A. (multiplying the above equation by "i" to obtain AH's ("i*t"), then differentiating (dAH/di) with respect to "i", setting the result equal to zero and solving for "i" will give the same answer mathematically).

 

The interesting thing here is that, irrespective of whether your original figures are correct, the "shape" of the ampere hour curve wil always have a peak in it. This shows that it's not just a case of too much current causing an incorrect acceptance cycle calculation, but too low a current too.

 

However, it is obvious to me that the charging curve must be exponential as at zero current the battery will never charge and at a huge current the charging time will never quite reach zero but will always remain finite. So the charging curve will have the form t = Ce-si where t = charging time, i = charging current and C and s are constants based on the actual data.

 

Using your original data again, this equation can be derived and works out to be t = 6.45e-0.026i

 

The plot of this is below:

 

chargetimeexponential.jpg

 

If, once again, one plots ampere hours, then once more a peak can be seen whose actual value (from the curve or derived mathematically by calculating d(AH)/di again ) turns out to be at a charging current of 39A in this example.

 

amperehoursexponential.jpg

 

This graph is the real proof that neither a larger current nor a smaller current give the best acceptance cycle time. Above and below 39A (in this example) the acceptance time will be incorrect, viz: too short at low currents and too short at high currents too. In actual fact, owing to the shape of the curve, the acceptance cycle calculation will be a little more tolerant of higher currents (ie: >39A in this example) than of lower currents (<39A in this example) because the curve has a flatter slope above 39A than below 39A.

 

So, what I have shown mathematically is that adaptive charging will always be a compromise and that it will only be truly accurate for a narrow band of charging current. However, it also shows that charging at too low a current has just as detrimental an effect on acceptance cycle times than too high a current because of the sharp drop off of the ampere hour curve on the low current side of the ampere hour curve.

 

Furthermore, we don't know (and how can the manufacturer of chargers know either) how to massage their acceptance cycle to give optimum cycle times as each battery bank will give a slightly different curve?

 

The best we can say, taking this example as a kind of typical figure, is that the optimum ampere hours into the battery in the bulk stage is 90AH (in this example)and that if we agree a band of currents that puts at least 80AH into the battery, either side of the maximum, then any current in the range from 20A to 65A (in this example) should give acceptable results for the acceptance cycle.

 

So, bottom line is that I agree with Gibbo that acceptance cycle times will never be truly accurate. However, it's not a case of low currents giving too long an acceptance cycle or high currents giving too short an acceptance cycle but that currents that are too high OR too low will both give acceptance cycles that are too short. ie: the expected AH into the batteries at the end of the acceptance cycle has a peak at a certain current and, either side of that optimum current, the battery will be less charged at the end of the bulk stage than might be thought intuitively.

 

Chris

Edited by chris w
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The graph cannot be a straight line as can be seen by plotting the examples you gave. (It doesn't matter if the actual figures for the 10A charger etc are accurate, it's the principle).

 

The equation for your examples works out to be t = 6 - (i/10) where "t" is the charge time and "i" is the current ie: 10A needs 5 hours, 20A needs 4 hours; 30A needs 3 hours and so on.

 

Are you being mischievous again?

 

Plotting this equation gives:

chargetimelinear.jpg

 

It can be seen straight away that whatever the real numbers, there will be a current at which the charging is instantaneous (zero time) and at zero current the battery would still charge in 6 hours !!!! Clearly both of these are a nonsense.... but be patient.

 

Well not quite....

 

We are not talking time to full charge, we are talking time to acceptance. With a high enough current this will approach zero (I say "approach" simply because nothing can technically take zero time, it is impossible) but for all practical purposes the time to acceptance, at a high enough charge current is zero.

 

Now, if you take my other figures (the ones actually meant to show time to acceptance not to show that it goes up and down as an inverse of the charge current) you will see I gave:-

 

5A = (more than) 8 hours

10A = 4 hours

20A = (less than) 2 hours

 

Ignore the "(more than/less than)" for the time being as they are due to internal resistance and (obviosuly) the product of the charge current and time has has to be equal in each case if giving no regard to internal resistance.

 

Plot those on log paper and it is indeed a straight bl**dy line. It is obvious that it could not be straight on a linear graph. And I suspect you know that!

 

And now you completely lose me...........

 

Further if one plots "t*i" to obtain ampere hours (AH) the graph looks like this:

 

If you plot t*i you will get a perfectly straight line if you use the correct figures (ie the ones I have just given above).

 

So I've snipped the rest of the post. Do the graphs again using the proper figures and you will get a straight line for t*i. This is obvious. Otherwise where is all that power going?

 

Now add in a parameter for the internal resistance (even if you allow for it decreasing with increasing state of charge) and you will indeed find that the state of charge is lower when acceptance is reached with a higher charge current.

 

The rest of the graphs are innapropriate because they are based on incorrect figures plucked out of the sky to show a trend.

 

Gibbo

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Are you being mischievous again?

 

We are not talking time to full charge, we are talking time to acceptance. With a high enough current this will approach zero (I say "approach" simply because nothing can technically take zero time, it is impossible) but for all practical purposes the time to acceptance, at a high enough charge current is zero.

 

Plot those on log paper and it is indeed a straight bl**dy line. It is obvious that it could not be straight on a linear graph. And I suspect you know that!

 

And now you completely lose me...........

 

I think you're the one being mischevious. :lol: I did the maths to show that what you stated is TRUE, viz: that acceptance cycle times will be (mainly) inaccurate. What the maths threw out (surprisingly) is that currents which are too low are just as bad as currents which are too high so far as acceptance cycle calculations are concerned.

 

If you plot t*i you will get a perfectly straight line if you use the correct figures (ie the ones I have just given above).

 

Since the graph of your data is t = 6 -i/10 (a straight line) then how can t * i be a straight line too?

 

ie: t * i = [6 - i/10] * i = 6i - i2/10

 

Since there is a "squared" term in the equation, it can never be a straight line.

 

Ampere hours = t * i so AH = 6i - i2/10

 

Differentiating and equating the result to zero to solve for the maximum gives:

 

d(AH)/di = 6 - 2i/10 = 0 at maximum

 

therefore imax = 30A

 

That equation is based on your original figures BUT even if one disregards those figures, clearly charge time to acceptance is a function of current, viz: t = f(i).

 

Ergo t * i can never be a straight line, whatever the figures, since there will always be an i2 term since t*i = i * f(i)

 

The rest of the graphs are innapropriate because they are based on incorrect figures plucked out of the sky to show a trend.

 

 

That's just not so.... the graphs show a trend that will obtain regardless of the actual figures you chose. It is clear that there cannot be a direct linear relationship between charging time and charging current, because of the anomalies which this throws up. So the relationship is exponential which then fits both your data and real life.

 

However, whether one accepts this or not as a mathematical fact, the fact also remains that plotted as a supposed straight line equation (your data) or as an exponential curve equation (also your data), the product of t * i gives a curve with a peak in ampere hours. ie: there is an optimum current which will deliver the maximum AH's into the battery in a given time.

 

This can be envisaged intuitively as well in that high currents (large i) will reach acceptance quicker (small t) and so the product of i * t will be less than some middle value of current multiplied by some middle value of time as the graph shows. At low currents, although t is large, i is now small and so the product of the two terms is again less than optimum.

 

Chris

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We need charge graphs.

 

Excuse me for a lack of the finer points but...

1A charge <more than> 10 hours

5A 5 hours

10A <less than> 1 hour

 

doesn't look linear to me unless it's plotted logarithmically which is inherently not linear isn't it?

 

Admittedly it's been a long time since I've studied sums but...

 

edit for formatting

 

edit again, wouldn't Chris's hypothesis give someone a chance at another million by coming up with teh circuitry that could identify that ideal charging cycle?

Edited by Smelly
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Since the graph of your data is t = 6 -i/10 (a straight line) then how can t * i be a straight line too?

 

ie: t * i = [6 - i/10] * i = 6i - i2/10

 

USE THE CORRECT DATA!!!!!!!!!!!!!!!

 

5A = 8 HOURS

10A = 4 HOURS

20A = 2 HOURS

 

!!!!!!!!!!!!!!!!!!!!!!!!!

 

t*i is a straight line

 

If you maintain it is not then please explain where the several hundred watts of power is going.

 

The product of time and current has to be the same (disregarding losses)!

 

Gibbo

 

We need charge graphs.

 

Excuse me for a lack of the finer points but...

1A charge <more than> 10 hours

5A 5 hours

10A <less than> 1 hour

 

doesn't look linear to me unless it's plotted logarithmically which is inherently not linear isn't it?

 

Correct

 

Admittedly it's been a long time since I've studied sums but...

 

edit for formatting

 

edit again, wouldn't Chris's hypothesis give someone a chance at another million by coming up with teh circuitry that could identify that ideal charging cycle?

 

I already know it. There's just no pracitcal way to achieve it.

 

Gibbo

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USE THE CORRECT DATA!!!!!!!!!!!!!!!

 

5A = 8 HOURS

10A = 4 HOURS

20A = 2 HOURS

 

!!!!!!!!!!!!!!!!!!!!!!!!!

 

t*i is a straight line

 

If you maintain it is not then please explain where the several hundred watts of power is going.

 

The product of time and current has to be the same (disregarding losses)!

 

Gibbo

t * i is a straight line only because you have massaged the figures to make it a straight line, ie: chosen figures such that t * i = 40AH in this example.

 

But that doesn't make it so.... here's a plot of your current vs time figures above.

 

chargetimeGibbo.jpg

 

Clearly this is not a straight line and neither will it be a straight line on log paper either because the figures do not fit an exponential curve. For that to happen, the 10A figure would have to be 5 hours instead of 4 hours to satisfy the equation t = 12.7e-0.0924i which is the exponential equation that fits your end points on an exponential curve.

 

If you maintain it is a straight line on log paper then your figures have to be:

 

5A = 8 HOURS

10A = 5 HOURS

20A = 2 HOURS

 

and so the AH curve will be:

 

5A = 40AH

10A = 50AH

20A = 40AH

 

which (again) has a peak in it as predicted mathematically. Differentiating and equating to zero, gives that optimum value of current to be 10.8 amps whereby t drops out at 4.7 hours and the maximum AH as the product of these two which is 50.6AH.

 

Here's the plot of it:

amperehoursgibbo.jpg

 

There is no earthly reason why charging at some particular current should charge the battery the same in AH as at some other current. The two scenarios are not connected.

 

Chris

Edited by chris w
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................For that to happen, the 10A figure would have to be 5 hours instead of 4 hours to satisfy the equation t = 12.7e-0.0924i which is the exponential equation that fits your end points on an exponential curve.

 

Hang on, it was you that said it was exponential.

 

I'll try a differenbt approach and then I give in:-

 

(Let's assume for all of this an average voltage of 13.8 volts during bulk)

 

In your 2nd graph in post #62 you show a 30A charge (for 3 hours) delivering 90Ahrs to the battery. A total of 1240 watt.hours sent to the battery and actually charging the battery.

 

Your same graph shows a 5A charge (for 10 hours) delivering about 25 amp hours. A total of 690 watt.hours but only delivering 414 watt.hours to the battery.

 

Question:-

 

Where did the other 276 watt.hours go if there is nothing wrong with the maths and considering there is nothing in the maths to allow for inefficiencies or losses etc?

 

I'll tell you where it went. It went into broken maths!

 

If you take the proper figures I gave you, you get:-

 

5A * 8 hours * 13.8 volts = 552 watt.hours

10A * 4 hours * 13.8 volts = 552 watt hours

20A * 2 hours * 13.8 volts = 552 watt.hours.

 

Oh look! A straight line!

 

Gibbo

 

which (again) has a peak in it as predicted mathematically.

 

But your maths is broken!

 

Explain where the other power is going with the different charge rates.

 

Gibbo

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I have pinned this topic. I think that it is worth ensuring that this idea does not get lost in the mists of future ones!

 

Nick

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Hang on, it was you that said it was exponential.

 

I'll try a differenbt approach and then I give in:-

 

(Let's assume for all of this an average voltage of 13.8 volts during bulk)

 

In your 2nd graph in post #62 you show a 30A charge (for 3 hours) delivering 90Ahrs to the battery. A total of 1240 watt.hours sent to the battery and actually charging the battery.

 

Your same graph shows a 5A charge (for 10 hours) delivering about 25 amp hours. A total of 690 watt.hours but only delivering 414 watt.hours to the battery.

 

Question:-

 

Where did the other 276 watt.hours go if there is nothing wrong with the maths and considering there is nothing in the maths to allow for inefficiencies or losses etc?

 

I'll tell you where it went. It went into broken maths!

 

If you take the proper figures I gave you, you get:-

 

5A * 8 hours * 13.8 volts = 552 watt.hours

10A * 4 hours * 13.8 volts = 552 watt hours

20A * 2 hours * 13.8 volts = 552 watt.hours.

 

Oh look! A straight line!

 

Gibbo

 

 

 

But your maths is broken!

 

Explain where the other power is going with the different charge rates.

 

Gibbo

It was YOU that said that it was "a straight line when plotted on log paper" not me.

 

For it to be a straight line on log paper the numbers must lie on an exponential curve - that's a mathematical fact.

 

So if I take your 2 end points, viz: 5A for 8 hours and 20A for 2 hours, I can derive the equation they lie on because all exponential curves have the form y = Aesx where in this case y is time (t) and x is the current (i).

 

That curve has one unique solution and its t = 12.7e-0.0924i (I am happy to show the maths steps to this but I want to avoid suicides on the Forum :lol: )

 

So, having derived the equation, the other point at 10A can ONLY lie on this curve if the time is 5 hours NOT 4 hours because the point 10,4 doesn't lie on the curve.

 

You have chosen 10,4 simply because 10 x 4 = 40AH = 552 watt.hours and gives you your right answer.

 

Just because I choose to charge at 5A or 10A or 20A doesn't necessarily mean that the same amount of charge will go into the battery at the end of the bulk stage. That's what you were saying too; the problem with charging at different rates is that at the start of the acceptance cycle the battery can be in a different state of charge depending on the charging current.

 

I AGREE WITH YOU.

 

However, the difference between us is that I have shown mathematically that there exists an optimum charge rate which will deliver the maximum AH's to the battery in a particular time. Either side of this optimum charge rate, the charge delivered is less than the optimum. So I AGREE with you that acceptance cycle calculations are a compromise.

 

If you now say that the points don't all have to lie on a straight line on log paper then all bets are off as my mathematical analysis is entirely predicated on what you said about straight lines and log graphing.

 

If all the points must lie on a straight line on log paper (ie: an exponential curve) then my maths is correct. If you're now saying that they don't have to lie on a straight line then the analysis will be very difficult because the points could be anywhere.

 

Chris

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I AGREE WITH YOU.

 

But for the wong reason.

 

There is no optimal charge current. Well, actually there is, because there is a certain charge current below which it cannot keep up with the battery internal self drain. But just above this charge level, the required acceptance time gets continually longer whereas an adaptive charger will calculate a prgrssively shorter and shorter one all the way through the graph.

 

Adaptive charging isn't a compromise. It simply does not work!

 

Honest.

 

Gibbo

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I think that what we see here is two people arguing slightly different points so as an involved and not quite innocent bystander I will elaborate...

 

Gibbo, you mentioned "efficient" charging. At that I assume you meant the best energy into the battery assuming the amount of energy generated to put into it.

 

Chrisw, were you in fact talking about "quicker" charging, whereby higher charge currents will more speedily charge things, although not so efficiently in terms of energy consumed?

 

Us boaters with our diesel engines are somewhat fettered as to "efficiency" in energy terms, unless, Gibbo, you can design a system to do what you've already said is not practical. Accepting that isn't "fast" nearly as good as "efficient"?

 

If I'm right then so are you both and sorry for not intervening earlier but I was learning stuff. If I'm wrong then I'll happily stand corrected.

 

The relay lands tomorrow; I think it's a really good idea and will be following your circuit closely Chris as I like the deliberate act of switching it on and not being able to re-instate that system without further deliberate action from me come engine stopping time.

 

Depending on whether Bagpuss is game for putting a coat of paint on (and we've got extra red in the strive for a purple boat so it's entirely possible :lol: ) them it'll either be this weekend or next that the job gets jobbed. The PV panel will just have to keep the batt's happy in the meantime. :lol:

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