Split Charging: Diodes, VSR, or Series Regulator?

nigelmercier

nigelmercier

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My boat has a Balmar Max Charge regulator, which feeds a diode splitter to charge the two banks. Although the Balmar senses the battery voltage (rather than the alternator output) I can't help wondering if the voltage drop in the diodes would be better eliminated.

I'm considering alternatives, either a VSR (Voltage Sensitive Relay) or a Series Regulator. However, the latter is very expensive.

I also wondered about re-using the diodes from the splitter as a GI, with the addition of a couple of capacitors.

Any thoughts?
 
The point about the battery sensing is that it adjusts the alternator output to the "correct" voltage at the battery thus compensating for the drop across the diodes in the splitter.

The diodes splitter will not I think you will find be easily adaptable to make a GI Even if the diodes can be extracted from it and are of a suitable rating you will need four not two in oder to get the necessary level of voltage blocking. A couple of bridge rectifiers can be used to make a GI.

See HERE

and Here for details of a DIY GI

The capacitor may or may not be necessary. There was some discussion on here some weeks ago about adding indicator lights to a GI
 
Fitting a VSR will give you more charge.
A blocking diodes say drops 1 volt, charge is 50 amp, that equals 50 watt going to heat. So at charge voltage you are using 4 amp that could be going into you battery to warm the globe.

Brian
 
Sorry Halcyon, as Vic says the regulator adjusts the voltage to accommodate the drop through the diodes therefore there is no loss of charge. Removing the diodes completely would not increase charge at the batteries simply lower the output from the alternator.
Nige, Leave things well alone as you will only waste money and not increase your charge at the battery.


and the loss within the diodes is immaterial as
 
Sorry PCUK if you drop 1 volt, and you are charging at 50 amp, you are loosing 50 watt to heat.

This comes out of your alternators available output, thus not going into your battery.

Brian
 
Or perhaps both are right, one claims with some justification that if you sense voltage at the battery you will charge to the full potential, and the other claims that if you use diodes and thus have a voltage drop that power will be wasted as heat which also rings true, with the probable result yes the batteries will be fully charge but this will take a little longer because of the energy wasted as heat.

Of course little old me could be completely wrong.
 
If the alternator is charging at max capability, then yes the diode splitter will reduce the current available. This will manifest itself as Increasing the RPM necessary for a particular charge level.
Once the rpm is above this level, the regulation takes over and the splitter ceases to have nay influence.
Older alternators tend to have an almost straight line current vs rpm whereas some modern ones will produce a lot at tickover. There is data on the web for Bosch ones. The data will not be exactly valid as you are regulating at a higher voltage, but it may give an insight.
I improved the low rpm charging a lot by changing from the Yanmar standard Hitachi to a Bosch alternator from a mondeo.
Also helps if pulleys are optimised to spin the alternator as fast as possible, without over-revving it at max engine revs.
It depends on what problems your current set up gives you. I wanted to charge the batteries fast, while leaving harbour with the fridge cooling, so that I could turn off the motor asap and enjoy the rest of the day sailing.
Hope that's of some interest.
 
Can I but in and ask a question, electrikery and charging batteries are beyond my limited brain capacity.

I have a standard 55amp alternator charging 2 banks, engine battery 1 x 110ah domestic bank 3x 113ah through what I believe is a VSR. I have a Sterling 90 amp alternator to battery charger, will there be any significant advantage to fitting the sterling unit over the existing VSR?
 
Yes I would expect the Stirling unit to work well, maybe looking at your boat pic you run the motor more than I do, so the alternator will not be such a limitation.
All this charging stuff depends to some extent on how you intend to use it. If you get a nice float charge in a marina, you're less dependent on the alternator etc etc. How much engine time do you anticipate vs how much current drain?
It's easy to say a smart regulator and high output alternator is best, but it may not be best value for you. But unless your batteries are quite well discharged, they won't take 55amps for very long, the charge rate will drop over time.
BTW my alternator was £20 from a car breaker, it also means the original one could be left unmodified for the external regulator.
Another case where a big alternator pays is if you have an anchor windlass, it will be a lot easier on the batteries in some scenarios. e.g. if you weigh anchor, then motor for only 20minutes and leave the boat with no charging for a week.
 
[ QUOTE ]
Fitting a VSR will give you more charge.
A blocking diodes say drops 1 volt, charge is 50 amp, that equals 50 watt going to heat. So at charge voltage you are using 4 amp that could be going into you battery to warm the globe.

[/ QUOTE ]

Although it is more likely that the diodes drop 0.7V rather than 1V, this is still 5% of the available power being wasted. Ultimately, this costs money: fuel, wear on the alternator, fan belt etc.

I also like the idea of giving priority to the domestic battery, as this powers the instruments. And I can always use the lower half of my 24V bank to crank in an emergency.

With reference to the GI, I don't see why one would need 4 diodes, they normally kick in at 1.4V DC. The capacitors are there to prevent the diodes conducting with stray AC currents, and you need a big one for LF <u>and</u> a little one for HF!
 
A power diode can have a full load volt drop between about 0.5 - 1.4 volt depending on type, I was just trying to make a general point.

As you appear to be split voltage 12 / 24 volt a relay whould not help, but if you have single voltage systems the split charge relay can also double up as a link start relay, as well as allowing bi-directional charge. All not available with diode systems, high or low volt drop.



Brisn
 
[ QUOTE ]
With reference to the GI, I don't see why one would need 4 diodes, they normally kick in at 1.4V DC

[/ QUOTE ] Four diodes because individually they block about 0.7volts, not quite enough to to give protection from stray "galvanic" currents So two are used in series to block about 1.4 volts which is adequate. Then of course you need two with the opposite polarity making four in total.

Regarding the capacitor its purpose and operation is explained HERE but there is no suggestion a low value one is also required for stray high frequency currents. Maybe if you are running radio transmitters, radar, microwave ovens etc for a large proportion of the time that you are plugged into shorepower or located where you are close to high power radio or radar transmissions it would be necessary.
 
[ QUOTE ]
[ QUOTE ]
Then of course you need two with the opposite polarity making four in total...

[/ QUOTE ]

D'OH! Note to self, engage brain before touching keys /forums/images/graemlins/smile.gif

The problem with large electrolytic capacitors is that they are not very good at conducting higher frequencies, so a smaller ceramic takes over.
 
[ QUOTE ]
The problem with large electrolytic capacitors is that they are not very good at conducting higher frequencies,

[/ QUOTE ]

Also not very suited to the application.

Brian
 
[ QUOTE ]
The problem with large electrolytic capacitors is that they are not very good at conducting higher frequencies, so a smaller ceramic takes over

[/ QUOTE ] Not so much to do with them being electrolytic (In fact are they for this application? I would have thought not) its a question of a small capacitor having a low impedance to HF while a large one has a high impedance.
But my knowledge on all this sort off stuff is growing rather fuzzy
 
Regarding the losses when using diode isolation versus VSR. Yes I agree at full current the volt drop is going to approach or even exceed 1 volt. Now the problem is that we are using the alternator as a charger for batteries so needs to generate at the battery over 14 volts to get a large current in. The smart charger will exceed this voltage.
The alternator will likely be running at full power ie full field current.
As the current of an alternator approaches full power the internal resistance means that the voltage will fall. The actual rating is at 12v output and really meant for a load not battery charging.

So at rated current ie 50 amps the alternator is barely producing 12volts(less if the revs are not at max) and you waste one of those volts in diode you do not have enough voltage for large current charging. Now of course the system is self limiting and charge current falls rapidly as charge voltage falls so it will maintain the 14+ volts but only by reducing the current.

So I reckon that if you are approaching the rated power of the alternator (or the max power it can produce at the supplied RPM) when at max charge current then diode isolator will reduce current a lot more than the simple 1 volt drop times current might suggest. An ampmeter would prove the point of course.

good luck olewill
 
There is a French company that make them (can't remember the name) using MOSFETS (IIRC) and they are fitted to Beneteaus......... or at least to mine!!
I have the data sheet on the boat if anyone is interested.

Alan.
 
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