Which is the best charger?

[DHutch]

I think using a multistage charger on a generator may be a fundamentally pointless thing to do unless you run the thing for a full 'cycle' each time.

I too have often thought this.

- It may even be linked to ernies problems. Although i have only skim-read the thread.

 

We occasional use our honda ex1000 to run out 25amp (24v, 17yo) sterling charger, and although it works well, if the batteries are only a bit down it quickly throttles back to 10amps or less for an hour or so, then down again to maybe 3-4amps for a while, at which point it usually get switched off.

- I know it would properly shorten the batteries life a bit but i would have thought in terms of (generator) fuel efficiently to be able to hold it onto 25amps for maybe half an hour and then knob it off and let the batteries sort themselves out.

 

 

 

 

Daniel

[chris w]

Well the higher the current you charge each battery in a smaller bank with, the less % it's charged when it reaches 14.8v, OK?

 

Once it reaches 14.8v, the less % each battery is charged, the longer it needs in the absorption stage to reach full charge.

 

So Gibbo is right, the absorption stage for the smaller battery bank should be longer than for the larger bank.

 

cheers,

Pete.

 

I fully understand what you're trying to say, but you're getting confused. The smaller bank will reach 14.8v faster than the larger bank.... on this we agree. But you're not seriously suggesting that the half-sized bank needs more than the 6 hours for which the controller was going to charge the full-sized bank.

 

As I say, run the maths and you'll see how nuts that it. Let me do it for you.........

 

My batteries (400AH), have a true (Peukert) capacity of 982AH so 50% discharge is 491AH. From 50% discharge, they reach 14.8v in about 30 minutes at a current of approx 50A.

 

50A for 0.5 hours equates to a (Peukert) charge of 80AH, of which, only about 70% will be absorbed by the batteries due to battery charging physics. So the batteries will receive about 56AH in that half an hour. So they are now charged to 491AH (the original 50% charge left in them) + the 56AH from the bulk stage. This gives a total charge of 547AH which means they are now 56% charged.

 

For the smaller bank (200AH nominal) the Peukert capacity is actually 400AH so 50% charged is 200AH. If the bulk stage is only 15 minutes @ 50A then the charge absorbed by the batteries (using the same maths as for the large bank) will be 28AH. Thus the total charge in the small bank is now 200AH + 28AH = 228AH. This means the smaller bank is now charged to 57% of its total capacity.

 

ie: the large bank AND the small bank are both charged to the same percentage level after the bulk stage. If the large bank need an additional 6 hours from the controller to take them to 100% charge (as is the case with my controller and batteries) you're not seriously suggesting that the smaller bank will need even longer to take them to 100% charge. The amount that each bank needs to get to 100% is 44% which, for the large bank is 432AH and for the small bank is 176AH. Taking into account battery charging physics (70% efficiency), the actual amount needed to be injected into the batteries to actually fill the remaining 44% is:

 

432/0.7 = 617AH for the large bank and 176/0.7 = 251AH for the small bank. Ergo, the smaller bank to larger bank time ratio will be 251/617 = 41%

ie: the smaller bank will charge in around 41% of the time needed to charge the larger bank. So, rather than needing MORE time, the half size bank actually needs LESS than HALF the time to charge.

 

This follows logically and intuitively anyway as a battery bank which consists of 4 equal sized batteries compared to a battery bank which consists of just 2 equal sized batteries will NOT have twice the capacity of the smaller bank but a capacity ratio which is:

 

[(C_L/20)^1.3 x 20] / [(C_S/20)^1.3 x 20] where C_L and C_S are the capacities of the large and small battery bank respectively.

 

Since, in this case, C_L = 2C_S then the expression above may be written as:

 

[(2C_S/20)^1.3 x 20] / [(C_S/20)^1.3 x 20] which simplifies to 2^1.3 = 2.5

 

So the full size bank is actually 2.5 times as large as the smaller bank. Thus, given that they are at exactly the same point after the bulk stage (as illustrated by the maths above) the small bank will take 1/2.5 the amount of time to charge to the full 100% as compared to the large bank. And, if you calculate that reciprocal, 1/2.5 magically comes to 40% which is exactly the time reduction we calculated above using actual figures. So theory and practice are hand-in-hand.

 

You can argue the actual times and substitute your own figures above eg: say a bulk stage of 50A for a whole hour for the large bank and 30 minutes for the small bank but you will still come up with the same answer of close to 40%. In fact if the bulk stage is proportionately longer for each battery bank the time difference to charge the smaller bank gets even better. (ie: around 37% instead of 40% if the bulk stage doubles to 1 hour for the large bank and 30 minutes for the small bank). That's why adaptive charging DOES work.

 

Chris

[blackrose]

I too have often thought this.

- It may even be linked to ernies problems. Although i have only skim-read the thread.

 

We occasional use our honda ex1000 to run out 25amp (24v, 17yo) sterling charger, and although it works well, if the batteries are only a bit down it quickly throttles back to 10amps or less for an hour or so, then down again to maybe 3-4amps for a while, at which point it usually get switched off.

- I know it would properly shorten the batteries life a bit but i would have thought in terms of (generator) fuel efficiently to be able to hold it onto 25amps for maybe half an hour and then knob it off and let the batteries sort themselves out.

Daniel

 

I still don't see what the problem of running a multi-stage charger from a generator is exactly? Ok, so if you don't run a full charge cycle it's isn't good for the batteries. So, run a full charge cycle!

 

In addition, nobody's said what the alternative is? Run a single stage charger from a generator - is that really going to be any better for your batteries? Well no, not if you don't charge them fully... surely it amounts to the same thing?

 

Also, since my knowledge of how battery chargers work is quite limited, I've no idea whether adaptive charging works or not (except that it sems to work on my boat). However, it might help the discussion if all those who say it doesn't work could please tell us which make and model of battery charger they have on their boat? Gibbo's probably made his own massive output charger from the windings of old alternators, various bits & pieces he found in his shed and a couple of old shoe boxes, so that doesn't count. I'm only interested in commercially available chargers.

 

Remember, we're not arguing about multi-stage chargers vs. single stage chargers, we are comparing multi-stage chargers featuring adaptive charging against multi-stage chargers with timed charge cycles (I think?)

 

Apart from the Victron Centaur which was already mentioned, can we have some models in the latter group that you are using please?

Edited March 13, 2008 by blackrose

[jayjayranger]

does anyone change the acid in there batterys

[smileypete]

the large bank AND the small bank are both charged to the same percentage level after the bulk stage

 

That simply cannot be true, as internal resistance means the batteries in a smaller bank which are each taking a higher charge current will start charging at a higher voltage. Simple Ohms law.

 

If you put an extremely high charge current in, they will go to 14.8V instantaneously and be 0% charged when they reach 14.8V

 

If you put a very low charge current in, they will be 100% charged before they reach 14.8V

 

Now which of the above two extreme cases needs a longer absorption charge? The first one, with the higher bulk charge current.

 

So batteries in a smaller bank that each take a higher bulk charge current will need an longer absorption charge.

 

cheers,

Pete.

Edited March 13, 2008 by smileypete

[chris w]

That simply cannot be true, as internal resistance means the batteries in a smaller bank which are each taking a higher charge current will start charging at a higher voltage.

 

cheers,

Pete.

 

Of course it's true. I just ran the maths. Your possibly not understanding the maths, doesn't make it wrong. The lower internal resistance is indeed the reason the smaller bank gets to 14.8v first; so I'm not disagreeing that the smaller bank will get to 14.8v first - that's obvious. What you're ignoring is the mathematics of precisely what happens and putting more weight in some overall gut-feel. As you can see above the theory AND the practice were exactly in accord.

 

But, if you take the time to follow the maths above (and, by all means, substitute your own numbers) you will see that the smaller bank will always have approx the same (or greater) charge as the larger bank at the end of the bulk stage.

 

Even if you took a massive charger delivering say 200A, the maths still works. Try it. Of course, if you think my maths is incorrect, please point out where and feel free to proffer your own mathematical analysis.

 

Chris

[smileypete]

Even if you took a massive charger delivering say 200A, the maths still works. Try it. Of course, if you think my maths is incorrect, please point out where and feel free to proffer your own mathematical analysis.

 

Okaaay.

 

If you start charging a battery at 100A instead of 50A, the terminal voltage will rise twice as fast, agreed?

 

But to reach 14.8V in exactly half the time, the terminal voltage would have to start at exactly the same level, agreed?

 

Now in reality, the terminal voltage will start higher at 100A charge due to internal resistance, agreed?

 

So with 100A charge the battery will reach 14.8V in less than half the time, agreed?

 

cheers,

Pete.

Edited March 13, 2008 by smileypete

[Gibbo]

OK, suppose I have 4 batteries discharged to 50%, and the alternator controller gives them a 6 hour absorption cycle. I now discharge them again to 50% and take away 2 of the batteries. I agree the bulk stage will be shorter, but you are saying that the controller should be giving them a longer absorption charge than before (ie: longer than 6 hours), That surely can't be correct as there are only 2 batteries to charge, instead of 4, so one would expect a shorter absorption cycle and that's what the controller would do.Chris

 

This is the common mistake made by the very people who came up with the idea of adaptive charging and is the root of the problem. You make the same mistake.

 

You naturally assume that the 2 batteries need a shorter acceptance cycle than the 4. It is completely wrong.

 

Ok, you agree up the the point that you discharge to 50% then take away 2 of the batteries. You agree that the bulk stage will be shorter.

 

Now here is what happens. The same size charger is now running into a battery bank of half the size. If you ignore state of charge for the moment, those 2 batteries, connected together, have twice the internal resistance that the four together had. This means that the terminal voltage (with the charger on) will be higher for any given state of charge than it would have been for all 4 batteries together. This terminates the bulk stage earlier. With the 4 batteries connected (this depends on the size of the charger) they might be at (say) 85% when the bulk stage terminates. With the 2 batteries the bulk stage will terminate earlier at (say) 75%. So with the 2 batteries instead of 4, they are at a lower state of charge when acceptage begins. Which means they need a LONGER acceptance cycle, NOT a shorter one which is what an adaptive charger will give them.

 

Further. Even if this didn't happen (and the 2 were at the same state of charge as the 4 when acceptance began) this would still mean that the acceptance cycle should be the same NOT shorter. Acceptance cycle is voltage limited not current limited. Each battery will draw the same current whether it is connected on its own or in parallel with 3 others.

 

Gibbo

 

Well the higher the current you charge each battery in a smaller bank with, the less % it's charged when it reaches 14.8v, OK?Once it reaches 14.8v, the less % each battery is charged, the longer it needs in the absorption stage to reach full charge.So Gibbo is right, the absorption stage for the smaller battery bank should be longer than for the larger bank.cheers,Pete.

 

Absolutely spot on. That is exactly what is required.

 

And it is exactly the opposite of what adaptive charging does.

 

Adaptive charging does NOT work. The idea is fundamentally flawed.

 

Gibbo

 

I fully understand what you're trying to say, but you're getting confused. The smaller bank will reach 14.8v faster than the larger bank.... on this we agree. But you're not seriously suggesting that the half-sized bank needs more than the 6 hours for which the controller was going to charge the full-sized bank.As I say, run the maths and you'll see how nuts that it. Let me do it for you.........

 

If you now add in the internal resistance of the battery and re-do your maths you will see that the bulk stage terminates at a lower state of charge in the smaller battery bank because twice as much voltage is dropped across the internal resistance.

 

That then completely messes up your maths and you'll find that the smaller battery bank will be at a lower state of charge when acceptance begins.

 

This isn't speculation. I've been doing this sort of stuff for longer than most people have even known what a battery is!

 

Gibbo

Edited March 13, 2008 by Gibbo

[RLWP]

If you ignore state of charge for the moment, those 2 batteries, connected together, have twice the internal resistance that the four together had.

 

Just to check my understanding, looking at the internal resistance of the battery only, this is because you have two resistors connected in parallel, not four - is that right?

 

 

Richard

[chris w]

If you now add in the internal resistance of the battery and re-do your maths you will see that the bulk stage terminates at a lower state of charge in the smaller battery bank because twice as much voltage is dropped across the internal resistance.

 

That then completely messes up your maths and you'll find that the smaller battery bank will be at a lower state of charge when acceptance begins.

 

This isn't speculation. I've been doing this sort of stuff for longer than most people have even known what a battery is!

 

Gibbo

 

The battery charge state is entirely dependent on the integral of current with respect to time. (OK, only about 70% gets into the battery due to charging physics, but even if we ignore this for simplicity, it doesn't change the fundamantal result).

 

If I charge a nominal 400AH battery bank (ie: an actual 982AH battery bank) for say 1 hour @ 50A from a 50% discharged starting state (491AH) the charge put in will be 50^1.3AH = 162AH (or actually 70% of this to be pedantic) ie: 113AH.

 

If I charge a nominal 200AH battery bank (ie: an actual 400AH battery bank) for say 30 minutes @ 50A from a 50% discharged starting state (200AH) the charge put in will be 50^1.3 x 0.5AH = 81AH (or actually 70% of this to be pedantic) ie: 56AH.

 

So, after the bulk stage, the larger bank is at 491 + 113 = 604AH (62% charged) and the smaller bank is at 200 + 56 = 256AH (64% charged)

 

ie: both batteries banks are at the same relative state of charge.

 

The internal resistance of the battery is irrelevant in the calculation because it has already been taken into consideration by the fact that the charging current is 50A, or whatever for your particular system. The charging current will be determined by:

 

(Av - Bv) / (Ar + Br) where Av & Bv are respectively the alternator voltage and the battery terminal voltage; and Ar & Br are respectively the internal resistances of the alternator and battery. (As Smileypete said above, it's just Ohm's Law!)

 

Since Ar will always be significantly higher than Br, thereby swamping it, the main determinant of alternator charging current will be Ar and so whether or not you have 2 batteries (higher net internal resistance) or 4 batteries (lower net internal resistance) the charging current during the bulk stage will be virtually identical for both set-ups.

 

Further, even if this were not so, the charging current in the bulk stage is usually engine rev limited anyway, so this is another reason the currents will be the same during the bulk stage and my calculations are sound.

 

Chris

Edited March 13, 2008 by chris w

[magnetman]

What does ' 50 1.3 AH = 162AH ' mean ??

[blackrose]

I've been doing this sort of stuff for longer than most people have even known what a battery is!

 

What's a battery?

[chris w]

What does ' 50 1.3 AH = 162AH ' mean ??

 

 

It's not "50 1.3 AH = 162AH". It's 50^1.3AH = 162AH; ie: 50 raised to the power of 1.3 = 162AH, where 1.3 is the typical Peukert factor for a wet lead acid battery.

 

The charging and discharging characteristic of a battery is NOT a straight line so the quoting of a single value somewhere on that curve (eg: 400AH @ the 20 hr rate) means that you cannot use this figure in your calculations of battery life versus current drawn unless you first convert it to the true curved characteristic first postulated by a Mr Peukert in the mid 1800's. Hence the "Peukert factor" or "Peukert calculations". These give the true mathematics from which time versus current may be established.

 

So a typical set up of 330AH (3 x 110AH batteries @ the 20 hr rate) which most people think would deliver 50A for 330/50 = 6.6 hours till the batteries were absolutely flat is not correct.

 

The true calculation would be:

 

[(330/20)^1.3 x 20]/ 50^1.3 = 4.7 hours

 

This leads to the result that there is a significant difference of 2 hours or a 30% "loss" of expected capacity between "simple" linear maths and "Peukert" maths. But it's true and the correct method of yielding the result you will actually get in practice.

 

In practice, you shouldn't take the batteries below 50% or you will damage them longterm. Thus you could draw 50A for 2.3 hours to half discharge (calculated using Peukert maths). Simple linear maths would give the wrong answer of 3.3 hours (again a 30% negative variance).

 

Chris

 

PS: someone came up with a great analogy to this the other day on here. If you get say 25mpg from your car at 60mph, you would not expect to get 50mpg at 30mph and so on. This is because the fuel used against speed characteristic is a CURVE not a straight line and so one has to use exponential curve mathematics to do the calculations not simple arithmetic.

Edited March 13, 2008 by chris w

[Chris-B]

I charge em up

We use them

They go flat

 

I charge em up

We use them

They go flat

 

I charge em up

We use them

They go flat

 

I charge em up

We use them

They go flat

 

I charge em up

We use them

They go flat

 

etc etc etc

 

 

Ok coat !

[magnetman]

Like a cattery only bats not cats.. Wouldn' it be better to use ^ or something?

[Gibbo]

The battery charge state is entirely dependent on the integral of current with respect to time. (OK, only about 70% gets into the battery due to charging physics, but even if we ignore this for simplicity, it doesn't change the fundamantal result).

 

Agreed.

 

If I charge a nominal 400AH battery bank (ie: an actual 982AH battery bank) for say 1 hour @ 50A from a 50% discharged starting state (491AH) the charge put in will be 50^1.3AH = 162AH (or actually 70% of this to be pedantic) ie: 113AH.

 

No it won't. It will be much less than this. Putting 50 Amps into a battery is REDUCED by Peukert NOT increased.

 

If I charge a nominal 200AH battery bank (ie: an actual 400AH battery bank) for say 30 minutes @ 50A from a 50% discharged starting state (200AH) the charge put in will be 50^1.3 x 0.5AH = 81AH (or actually 70% of this to be pedantic) ie: 56AH.

 

Same mistake as above.

 

Re-do your maths with Peukert the right way round and you will still end up with the smaller bank reaching the same state of charge after the end of bulk. But that's because you're doing it wrong.

 

So, after the bulk stage, the larger bank is at 491 + 113 = 604AH (62% charged) and the smaller bank is at 200 + 56 = 256AH (64% charged)

 

ie: both batteries banks are at the same relative state of charge.

 

Incorrect. See below.

 

The internal resistance of the battery is irrelevant in the calculation because it has already been taken into consideration by the fact that the charging current is 50A, or whatever for your particular system. The charging current will be determined by:

 

And here is the root or your error.

 

To take your figures. The charging current is 50 amps. So it is either 50 amps into 4 batteries, or 50 amps into 2 batteries. Assume the internal resistance of each battery is 0.01 Ohms.

 

Further assume that the "on charge" battery voltage at 80% state of charge at 50 Amps is 14.2 volts. This is the actual internal battery voltage, not taking the internal resistance into consideration.

 

With 4 batteries, there is 12.5 Amps flowing through the internal resistance of 0.01 Ohms. The internal resistance therefore drops 0.125 volts.

 

With 2 batteries, there is 25 Amps flowing through the internal resistance of 0.01 Ohms. The internal resistance therefore drops 0.25 volts.

 

The 4 batteries therefore have a terminal voltage of 14.2 plus 0.125 = 14.325 volts when charged with 50 Amps at 80% SoC.

 

The 2 batteries have a terminal voltage of 14.2 plus 0.25 volts = 14.45 volts when charged at 50 Amps at 80% SoC.

 

So clearly (you must see it by now!) the smaller battery bank will have a higher terminal voltage at the same SoC at the same charge current. This, obviously, puts the charger into acceptance earlier and therefore when the batteries are in a lower state of charge. The smaller bank therefore requires a LONGER acceptance cycle NOT a shorter one.

 

You can carry on arguing this one for a month if you like. It doesn't alter the fact that you are wrong.

 

Gibbo

[magnetman]

sorry Chris W.my phone browser doesn't show anything between 50 and 1.3, i am guessing that you did use an operator i can't see it.

[chris w]

Wouldn' it be better to use ^ or something?

 

Why? The message formatting allows one to use an actual superscript so 50^1.3 can be written exactly as it should be written. 50^1.3 is a modern shorthand for people who don't know where to find the "superscript" button

 

Chris

Edited March 13, 2008 by chris w

[magnetman]

Or people with displays which don't show superscript. I can't see it.

[chris w]

Agreed.No it won't. It will be much less than this. Putting 50 Amps into a battery is REDUCED by Peukert NOT increased.Same mistake as above.Re-do your maths with Peukert the right way round and you will still end up with the smaller bank reaching the same state of charge after the end of bulk. But that's because you're doing it wrong.Incorrect. See below.And here is the root or your error.To take your figures. The charging current is 50 amps. So it is either 50 amps into 4 batteries, or 50 amps into 2 batteries. Assume the internal resistance of each battery is 0.01 Ohms.Further assume that the "on charge" battery voltage at 80% state of charge at 50 Amps is 14.2 volts. This is the actual internal battery voltage, not taking the internal resistance into consideration.With 4 batteries, there is 12.5 Amps flowing through the internal resistance of 0.01 Ohms. The internal resistance therefore drops 0.125 volts.With 2 batteries, there is 25 Amps flowing through the internal resistance of 0.01 Ohms. The internal resistance therefore drops 0.25 volts.The 4 batteries therefore have a terminal voltage of 14.2 plus 0.125 = 14.325 volts when charged with 50 Amps at 80% SoC.The 2 batteries have a terminal voltage of 14.2 plus 0.25 volts = 14.45 volts when charged at 50 Amps at 80% SoC.So clearly (you must see it by now!) the smaller battery bank will have a higher terminal voltage at the same SoC at the same charge current. This, obviously, puts the charger into acceptance earlier and therefore when the batteries are in a lower state of charge. The smaller bank therefore requires a LONGER acceptance cycle NOT a shorter one.You can carry on arguing this one for a month if you like. It doesn't alter the fact that you are wrong.Gibbo

 

1. The current is NOT decreased by Peukert because I converted the battery capacity to its Peukert capacity first. If I had not done this and instead had used the nominal (label) capacity in my calculations then I would agree with you. So you are incorrect on the first two points. So my calculations are correct and the two batteries are pretty much at the same percentage level of charge after the bulk stage.

 

2. You ignored the alternator internal resistance which will swamp the battery resistance. You also ignored the point I made about the bulk current's being rev limited which will ensure the bulk currents are the same, regardless of any internal resistances.

 

Chris

 

Or people with displays which don't show superscript. I can't see it.

 

My post crossed with yours, about the fact you're using a phone browswer.

 

Chris

Edited March 13, 2008 by chris w

[Gibbo]

1. The current is NOT decreased by Peukert because I converted the battery capacity to its Peukert capacity first. If I had not done this and instead had used the nominal (label) capacity in my calculations then I would agree with you. So you are incorrect on the first two points. So my calculations are correct and the two batteries are pretty much at the same percentage level of charge after the bulk stage.

 

So by your reckoning and maths, the only conclusion is as follows:-

 

You convert the battery capacity to its Peukert Capacity. So your 400 Ahr (20 hour rate) battery is somewhere round about 850 Ahrs.

 

This means you can discharge it at 1 amp and remove 425 Amp hours to get to 50% state of charge.

 

You can then recharge, from a 50 amp alternator but only actually put in say 300 amp hours (real amp hours) to get back to full?

 

Interesting idea.

 

2. You ignored the alternator internal resistance which will swamp the battery resistance. You also ignored the point I made about the bulk current's being rev limited which will ensure the bulk currents are the same, regardless of any internal resistances.

 

The alternator internal resistance is completely and utterly irrelevant.

 

Get your self 2 batteies and discharge them to the same state of charge.

 

Put them both on a 25 amp charger and measure the voltage. Now remove one of the batteries. With a single battery at 25 amps charge the voltage WILL be higher than with both the batteries connected.

 

This WILL put the charger into acceptance with the single battery in a lower state of charge.

 

The maths is easy. But you get it wrong. The practical test is easy, but you won't do it.

 

Gibbo

[chris w]

So by your reckoning and maths, the only conclusion is as follows:-

 

You convert the battery capacity to its Peukert Capacity. So your 400 Ahr (20 hour rate) battery is somewhere round about 850 Ahrs.

 

This means you can discharge it at 1 amp and remove 425 Amp hours to get to 50% state of charge.

 

You can then recharge, from a 50 amp alternator but only actually put in say 300 amp hours (real amp hours) to get back to full?

 

Interesting idea.

The alternator internal resistance is completely and utterly irrelevant.

 

Get your self 2 batteies and discharge them to the same state of charge.

 

Put them both on a 25 amp charger and measure the voltage. Now remove one of the batteries. With a single battery at 25 amps charge the voltage WILL be higher than with both the batteries connected.

 

This WILL put the charger into acceptance with the single battery in a lower state of charge.

 

The maths is easy. But you get it wrong. The practical test is easy, but you won't do it.

 

Gibbo

 

A 400AH nominal capacity has a Peukert capacity of 980AH. If it were totally discharged (ie: zero capacity) and one could charge it with a constant 50A, the effective peukert current would be 160A (50^1.3) so it would take 980/160 = 6hours. Actually 9 hours if we take the 70% charging efficiency into account.

 

For your second point, you are still ignoring the rev limited current which actually obtains in most cases in practice.

 

Chris

Edited March 13, 2008 by chris w

[Gibbo]

A 400AH nominal capacity has a Peukert capacity of 980AH. If it were totally discharged (ie: zero capacity) and one could charge it with a constant 50A, the effective peukert current would be 160A (50^1.3) so it would take 980/160 = 6hours. Actually 9 hours if we take the 70% charging efficiency into account.

 

For your second point, you are still ignoring the rev limited current which actually obtains in most cases in practice.

 

Chris

 

You're back pedalling now. It is irrelevant what causes the current limit. Whether it is a rev limit on an alternator, a simple ("it's as much as I've got") limit on a cheap charger, or the controlled current limit in a smart charger.

 

The fact is a smaller battery will reach the acceptance voltage before a bigger battery on the same charger because voltage is dropped across its internal resistance. This puts the charger into acceptance with the battery in a lower state of charge.

 

It therefore requires a longer acceptance cycle not a shorter one which is the opposite of what adaptive charging does.

 

Gibbo

[chris w]

You're back pedalling now. It is irrelevant what causes the current limit. Whether it is a rev limit on an alternator, a simple ("it's as much as I've got") limit on a cheap charger, or the controlled current limit in a smart charger.

 

The fact is a smaller battery will reach the acceptance voltage before a bigger battery on the same charger because voltage is dropped across its internal resistance. This puts the charger into acceptance with the battery in a lower state of charge.

 

It therefore requires a longer acceptance cycle not a shorter one which is the opposite of what adaptive charging does.

 

Gibbo

 

 

Well I know mathematics (and physics) and I don't agree. I do agree that the smaller battery will "bulk out" first, I have never argued with that.

 

The maths though shows that, despite the shorter time, the smaller battery will have virtually the same percentage charge as the larger battery, at the end of the bulk stage, for the very reason that, although the time is shorter, so is the capacity and the two effects more or less cancel out.

 

The maths works whether you use Peukert maths (which is the correct way, but non-mathematicians may believe it's some kind of fiddle) or whether you use simple maths. Simple maths will not reveal the absolute answers but will reveal the relative values which will still be virtually identical for both banks.

 

Chris

[Gibbo]

Well I know mathematics (and physics) and I don't agree. I do agree that the smaller battery will "bulk out" first, I have never argued with that. The maths though shows that, despite the shorter time, the smaller battery will have virtually the same percentage charge as the larger battery, at the end of the bulk stage, for the very reason that, although the time is shorter, so is the capacity and the two effects more or less cancel out.The maths works whether you use Peukert maths (which is the correct way, but non-mathematicians may believe it's some kind of fiddle) or whether you use simple maths. Simple maths will not reveal the absolute answers but will reveal the relative values which will still be virtually identical for both banks.Chris

OK. Now we're starting to make some headway.....You state that the smaller battery bank will be be at the same state of charge as the larger battery bank at the end of bulk?So at the very least you HAVE to accept that both batteries require the same acceptance time? Certainly not a shorter time for the smaller bank.Therefore when (not if) I prove to you that the smaller bank will actually be in a lower state of charge at the end of bulk then you will have to also accept that the smaller bank requires a longer acceptance cycle not a shorter one.Gibbo

Just to check my understanding, looking at the internal resistance of the battery only, this is because you have two resistors connected in parallel, not four - is that right? Richard

Yes.Gibbo