split charge diode and battery charging query

joss

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I, or rather the boat, has two battery banks. To ensure that the batteries get maximum charge (now have an old Eberspacher fitted) I am thinking of getting an alterantor regulator and a split charge diode The query I have though is with respect to the alernator regulator. What happens if one battery bank is discharged and one is fully charged? I know that the diodes will prevent charge from running 'downhill' from one bank to the other, but assuming the voltage sense for the regulator is taken from the common positive, will the fully charged battery get fried as the alternator produces a higher voltage than would otherwise be preferred? Also, would the flat bank get a lower voltage than needed to give a fast charge. I appreciate that the current would flow into the flat bank, but is there a problem with both banks getting a voltage from the alteranor that is a comprimise between the two extremes? Or should I just fit it, forget it and think about more important things?....

thanks,
Jo
 
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I've just looked into this a bit - this http://www.reading-college.ac.uk/marine/MarineE09.html gives a useful fairly simple summary.

As I understand if the 2 batteries are at different states of charge (as will often be the case) the "resistance" (not technically correct I believe - but helpful) in the 2 batteries dictates how much charge is taken by each one and so helps prevent cooking of the more charged battery.

I believe you would normally have the sensing wire connected to the house battery assumingthis was normally the one most in need of charge to prevent the alternator dropping the charge too soon.

I'm sure others will have better / fuller explanations - but thats my simplistic undertsanding

Best wishes
James
 

DaveS

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I've thought about this too, done a little reading up, and concluded that a cheaper, more efficient, and arguably better in other ways, approach is to link the battery banks using a voltage sensitive relay. Putting my money (c. £50) where my conclusion is, I took delivery of the requisite wee beasty today and plan to wire it up this weekend. I'll post my experience of the initial success (or otherwise) of this next week.
 

boatmike

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I always get arguments on this subject but in my opinion you are confusing voltage with amperage. Without a "smart" regulator your voltage will be between 13.8 and 14.1 volts which is the norm for automotive charging. The smart regulator can be set at various levels but the norm here is about 14.4 volts. All batteries will see this voltage therefore when the alternator is running. This is fine as long as you recognise that it will mean you have to check your electrolyte levels more often. The amperage should flow to the most discharged battery however and you are right to ask where to connect the battery sensor wire. You have three choices assuming 2 batteries. Lets call them batt1, batt2 and both. If you only have a cranking battery on the engine start (batt1) and a service battery to everything else (batt2) the service battery is likely to get most discharged as it is supplying current when the alternator is not running. In this case I would connect the battery sense wire to this so the smart regulator won't shut down charge to it until it is charged. The diode splitter will allow more amps to batt2 than bat1 which will just be on "float" while this is happening and no harm should come to batt 1. Connecting to batt1 will reduce charging rate before batt2 gets maximum charge. Connecting to both will cause a compromise situation and provide a path for current to flow between the batteries so this is not a good idea. If you have a cranking battery and 2 service batteries with say instruments etc on 2 and eberspacher, fridge etc on 3 I would connect the sensor wire on 3 as this is more likely to get deeply discharged.
If you then have a changeover switch for distribution between batt 2 and 3 and suspect batt 2 is more discharged than 3 then you can put the switch on "both" to sense the aggregate charge of both which will help to maintain charge until 2 is up to speed.... Hope this is clear and I don't start another lengthy forum argument over it.......... If in doubt talk to Stirling or whoever you buy the smart charger from. You will I think find that they confirm what I have said in principle and you will find it in their installation instructions anyway.
 

SteveA

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I have exactly this arrangement and connected the sensing wire to the service batteries. In the 5 years since doing it I've had no problems and topped up batteries. I know this isn't very technical but I thought it might help!
 

pvb

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Fit and forget...

It's not a problem; you won't fry the starter battery! If you fit a smart regulator, the sense lead should be connected to the domestic bank. Be sure to follow installation instructions to the letter.
 

halcyon

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Are both batteries used for service ? or is one a dedicated engine battery ?

What type of boat are they on ? and what loads are used with the engine running ?

As the above answers can have a major impact on the systems correct operation

Brian
 
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Common and other possible way ...

The sense wire is to 'tell' the alternator the true voltage level at the battery .... instead of many usual alternators that have engine sensed - and therefore do not like having a split diode in the system. So the following has been devised to overcome it :

Most boats seem to have the sense wire go to the domestic battery - as this suffers greater discharge than the starter in normal circumstances. Many report that this works fine.

There is another way .... based on the split diode dropping the charge voltage approximately 0.7V. This is enough to cause a) charge indicator light to fail to go out, b) insufficient charge voltage to maintain batterys. So how to get the voltage back up again ? Simple .... a diode is placed across the alternator itself .... a cheap diode suitable rated is placed across Aux terminla and Regulator .... this being all in the back of the alternator. So what happens here ? The alternator pumps out 14V, but the diode we have jumped the alternator with drops it by approx. 0.7V ... the regulator apparently sees 13.3V and boosts it back to 14V .... making the actual output 14.7V. This arrives at the split diodes and they drop it by 0.7V .... so there you are 14V at battery - regardless of which battery etc. etc. Cheap and effective.

The only drawback with the diode jumper in the second system - is that if the split diode system is removed at any time - the jumper diode should be removed also as now too much is delivered to the batterys.
 

pampas

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Sterling recommend that the "Sensing"connection be made at the common terminal if using a change over switch. Personally I prefer this system as one does not have to contend with another voltdrop being added to the circuit. My refrig batterys is fed only by a split charge relay again minimal vd`s to worry about and is completely idiot proof.Hope that this is of use to you.
 

William_H

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Re: Common and other possible way ...

Hello Nigel Your description of the jump diode connection is a bit brief and to my mind a bit misleading. From theory (I have no experience) the regulator on the alternator senses the voltage between the alternator output terminal and ground. The diode to increase output regulated voltage would need to be between the ground connection of the regulator and actual ground. (negative or alternator frame) this would lift the whole regulatro circuit .7 volt above ground due to the regulator sensing current and so give the output voltage another .7 volts. This whole idea is predicated on using the original auto type regulator. it does seem to me that using a voltage sensing relay to paralell the batteries when charging would be an easier solution unless of course you wanted to increase charge voltage even further.
have I got the wrong idea or missed something. PS nice to see you back on the forum am I right in thinkinng you have been absent for a while sailing? regards olewill
 

boatmike

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Re: Common and other possible way ...

Thats rather an elegant and simple dodge that I had not heard of before Nigel! Would work without a "smart" regulator too..... Well done, just shows, never to old to learn. Thank you! With reference to William_H's comment though I think for the sake of anyone who want's to try it we should be specific regarding the location of the diode and ensure people get it the right way around (!) or the battery might be destined for a very short life.... Would we not achieve the same result with just a resistor?
 

joss

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One battery is for starting. Ther are two other batteries (as a single bank) for services. All are identical lead acid. Boat is beneteau with 18HP volvo and 65A alternator.

Thanks to everyone for taking time to reply to my question.

Jo
 
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jump diode ....

quoted from Alistair Garrods book "Electrics afloat" produced in conjunction with PBO .....

page 64 .... headed "Fooling the senses"

Me - away for a while ... sort of - took a few days rest and went on boat but mainly haven't seen any post to get my teeth into !!! Work also cuts into my forum time !! Nice to be missed though !!

The diode trick is going to be done on my boat - as I was given a split diode block a while back so seems a shame to waste it. Must say though that the old 1-2-off system that I have on my old tub worked well when I was out on her ...

/forums/images/graemlins/wink.gif
 

roger

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If you use a smart regulator you should set it to a lowest voltage required by any of the batteries. If I remember correctly an ordinary lead acid ventet battery can take the highest voltage, next comes the sealed and the deep cycle and lowest is the AGM type. You cannot afford to overcharge the sealed battery in particular as you cant put any more water into it.
One consequence is that you are likely to under charge the higher voltage batteries.
The Sterling regulator instructions set this out in detail and recommend using cheap ordinary vented batteries. Other authorities disagree with Sterling.
 

crms

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Also been thinking about this with two different battery types, eg sealed for engine and lead acid for domestic. Would want a higher voltage for the lead acid than the sealed - approx 0.7 volt. Sense off the domestic and use a schottky diode to get min voltage loss - and avoid running the alt a too high a voltage. Put another silicon diode in the sealed batt line, which will knock the voltage down a little and avoid frying it.
Thoughts??
 

bedouin

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I've spent some time considering the same issue. I thought of using an extra diode to drop the voltage to the sealed unit as you suggest. My concern is that the voltage drop across a diode is not constant, but increases with increasing current. Therefore this approach would drop the voltage when the battery is charging, but once fully charged might allow the voltage to creep up too high.
 

William_H

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To Bedoin and Boat mike "diode votl drop". Bedoin a diode in series in conduction direction would be a very elegant solution to the problem of reducing voltage to a sensitive ie SLA battery.
Just to recap a resistor reduces the voltage only when a current is flowing in a linear progression. ie twice as much current = twice as much voltager drop. If you fitted a resistor in series with a battery it will reduce the current into the flat battery by reducing the voltage but as the battery voltage rises the current falls so the volt drop across the resistor falls so that at low current into the battery there will be vertually no vollt drop so the battery will still try to charge to the higher voltage. In practice the current being limited will limit damage to the SLA battery but may still damage it. However the effect is somewhat opposite to you need for a fast charge initially followed by a slow charge.
Now the characteristics of a diode in forward direction are essentially a .7 volt drop regardless of current. In fact at low currents the drop is .7 volt increasing with higher currents to typically 1.5 volts at 10 amps and more at 20 amps (a power diode of course) thhe little ones would be gone at 10 amps with an infinite and permanent volt drop.
A close analogy is a non return valve in a water pipe. it has a spring nolding down a seal which must be overcome to get any water through say 2 psi. Once the spring is overcome the water will flow with a subtraction of 2 psi from the final pressure.(that might be arguable) If there is a large flow then the pressure will drop even more depending on the size of the valve. Obviously in both cases you can't get a flow backwards.
Diodes come in a variety of types/ mterials. so a schotky diode can hav a volt drop about 1/3 of a silicon diode and a germanium diode about half. however silicon is the type available very cheaply in high power rating. the forward volt drop actually varies from diode to diode and also with temperature. Silicon is generaloly quoted as .7 volt.
Of course with all battery charging the time that the current is charging and the temperature are matters that can not be taken into account by the regulator designer. The car system ie standard alternator regulator is designed for long term no damage to the battery which results in less than maximum charge current in the short term. The smart chargers can address this problem better with larger initial current but low current after full charge and long time. They also tend to be able to get th battery to 100% full caharge.
Here is another anology. Imagine you have a water tank constantly filled to a high level. imagine you couple this to a tank adjacent with a pipe from the bottom of one tank to the bottom of the other. The first tank is your standard voltage regulated alternator. The second tank is your battery. the level of water is voltage, the flow through the pipe from one to another is current. So you connect your batery (second tank) to the first ( start the engine) water flows from the first tank at a rate depedant on how low the receiving tank is. As the receiving tank gets full the current (flow) falls until the levels get close but theoretically never get to be equal. Now a bigger pipe will help. (larger alternator capacity)
Unfortunately lead acid batteries don't show a level (voltage) proportional to actual fullness but rather appear to be more full in level than they really are so it takes even longer to get one filled from the other. Yes more voltage indicates more charge but not in a proportional way.
So the answer is to increase the level of the first tank (alternator regulated voltage) this will push more water into the second tank even though the level appears to be higher than the fullness indicates. The trick is to reduce the level of the first tank (alternator voltage) before the second one overflows. (boils and is destroyed) This is what a smart charger does. Bear in mind that the time the pipe is connected relates to the time your engine is running. In a yacht you want to switch off ASAP while a MOBO would be more concerned with overcharge boiling. incidentally by analogy a solar panel or unregulated wind generator produce a quite high volltage about 50% more than a charged battery so even though the current is quite low it will coontinue to charge right up to full battery. the current is for a small panel not enough to do any damage.
Nigel I have often thought that a semi manual system where a relay would provide a high voltage for charging initially on start up then using a timer for perhaps 15 or 30 minutes minutes (with an optional button to reset for another period of high current) after which the system reverts to standard voltage regulated so is safe for an all day motoring. this would mean that a short engine run would give the max amount of charge into batteries.
Ihope the discussion and analogy is usefull regards olewill
 

BrendanS

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Lots of good stuff there, but gave up after first two paragraphs. too much text

If you hit the return key a few times after every paragraph would be much easier to read
 

boatmike

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William, thanks for that. I think Brendan's rather abrupt criticism of your literary style somewhat unfair as I read what you said very carefully. It was I who asked the question after all not him! I am a mechanical engineer with a reasonable understanding of electrics but I have to think hard when playing with "electricery" that is non standard. Your explanation of why a resistor would not do the same job is very clear and I thank you for it.
 
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