Dual battery wotsit thingummy

you can connect the second one to charge just using a relay which is activated by the charge light if you like.

piccy half way down this page:

Never had any problems using a Warn 8274 with no extra battery, just leave the engine ticking over whilst winching and the bonnet up in case something lets go to protect the windscreen and an old blanket or coat on the cable.

Martin

O> Never had any problems using a Warn 8274 with no extra battery, just O> leave the engine ticking over whilst winching and the bonnet up in case O> something lets go to protect the windscreen and an old blanket or coat O> on the cable.

I totally agree. Same here with a Warn 9000i .... never needed another battery.

Hmmm, okay then. Thanks for those opinions as I hadnt even considered the idea of *not* putting in another battery. Okay, I'll give it a go 'as is' and see how I get on. In the meantime, I've got some more questions :o) I have a winch plate on a standard bumper, rather than a dedicated winch bumper. The protrusion of the plate is quite large, maybe five inches or so, at the centre of the bumper. And then of course there is the fairlead on top of that - the car is a fair bit longer at the front than it used to be! Since its so exposed, is there any bag, or cover, or something, that I would put over the winch to protect it from the worst of the mud to avoid having to powerwash all the exposed cable every time? And while I'm here, whats the situation with wading? Should I reduce the wading depths I take because I now have this huge electrical lump hanging over the front? Sorry to be so dense...

Mines been on for nearly 9 years and I've never covered it ...... still works fine :-)

We have a Warn M5000 for occasional use running from a single battery. It is as big a battery as we can squeeze into the gap, but a single one all the same.

If you do go for a split charge system, the X-Charge

may be overkill but has an excellent reputation. Cheers

Blippie

-- Ten minutes of this rain will do more good in half an hour than a fortnight of ordinary rain in a month.

If you do go for a split charge syste, use proper battery cables, the same size as the ones to your main battery. Otherwise, when you've flattened the second battery, when you restart the engine, as the system connects both batteries together, your main battery will try and charge the second one. You can get a very high current at this time, though tends to be for a very short time. Even when the alternator kicks in, that second battery will draw current like mad. So don't skimp on the cabling between the batteries.

Cheers! Graham Carter Harare Zimbabwe

P.S, My second battery is for the fridge. Great to be able to always have a really cold castle just when you want one after a hard days gamewatching in the bush!

That's not really the case at all. The following is explanation that is easy to understand, They're not my words, that's because the explanation is a far better one than I would be capable of:

I have heard that relay based split charge systems suffer from a serious problem with a dangerously high initial current surge measured in hundreds of amps.

Yes, we've heard the same rumour, usually from suppliers of split charge diodes. It has no basis in theory or practice.

Rather than just stating our side of the argument, we will go through the technicalities of the argument.

The proponents of this rumour state that the fully charged battery will have a higher terminal voltage than the flat battery. This is correct.

They state that a lead acid battery is capable of producing extremely high discharge currents. This also is correct.

They state that a flat battery will have a relatively low terminal voltage. This, again, is correct.

They then go on to state, purely as a conclusion of the above three statements, that when the two batteries are connected together by the split charge relay, the difference in terminal voltage between the 2 batteries, in conjunction with the capability for extremely high discharge currents, will produce a very high current through the split charge relay and it's associated cabling thus causing a fire risk and damage to the batteries.

Which sounds almost plausible.

We will take here the absolute worst possible scenario. A very large (200Ahr, 1500A CCA), fully charged, engine start battery connected to a very large (1000Ah, deep cycle antimony/antimony), almost totally flat auxiliary deep cycle battery bank.

An explanation is in order as to why these were chosen. I must state right now that any other combination of batteries and charge states will produce a much smaller effect. These batteries and charge states have been chosen because they will show the highest possible initial current surge. Which as we will show, is anything but high or dangerous.

A fully charged engine start battery was chosen because:-

1. An engine start battery has a much lower internal resistance than any other type of battery. It is therefore capable of producing much higher discharge currents.
2. The voltage of an engine start battery is slightly higher than that of a deep cycle battery so this will produce the greatest voltage difference between the two batteries and therefore the greatest initial current surge.

A flat deep cycle antimony/antimony battery was chosen because:-

1. It has a lower terminal voltage than a flat engine start battery. This will produce the greatest voltage difference between the two battery banks.
2. The internal resistance of a flat deep cycle battery (whilst higher than an engine start battery when fully charged) is lower than that of a flat engine start battery. An almost flat engine start battery will produce almost no current. An almost flat deep cycle battery will produce substantially more. This lower internal resistance will allow the greatest so called surge current to flow.

Now for some actual figures.

The engine start battery is rated at 1500 CCA. CCA means cold cranking amps. It is the highest current that the battery can produce, at 0 degrees Farenheit for 15 seconds without the battery voltage falling below 9.5 volts. (you may also see MCA, which is Marine Cranking Amps, this is effectively the same thing, but measured at a higher temperature and the terminal voltage has to remain above a slightly higher voltage)

A word of clarification is in order here. There are actually around 10 different specifications for cold cranking amps. Different organisations, different standards (by definition that must be a joke) bodies and different manufacturers all seem to have their own version. The only thing they all agree on is that the measurement is at zero degrees Farenheit. Some state the terminal voltage must remain above 7.2 volts, others state 8.4 volts, some go for 9.5 volts and yet others prefer 10.5 volts. Some state 10 seconds, others 15 seconds, some state 30 seconds, some state 60 seconds and one states 180 seconds. If we average them all out we end up with 9.5 volts for 15 seconds.

A fully charged engine start battery, after rest, will show a terminal voltage of around 12.7 volts.

So from the CCA rating of this battery, we know that placing a 1500 amp load on it will reduce the terminal voltage from 12.7 volts to 9.5 volts. This is a drop of 3.2 volts at 1500 amps. Ohms law therefore tells us that the internal resistance of this battery (when fully charged - it changes depending upon the state of charge) is 0.002 Ohms. And this is a typical figure that can be lifted from battery data sheets.

From this figure we can get an idea of what the terminal voltage will drop to under various loads. For instance putting a load of 500 amps on this battery will produce a terminal voltage of 12.7V - (500A * 0.002R) = 11.7 volts.

A load of 100 amps would produce a terminal voltage of 12.7 - (100A *

0.002R) = 12.5 volts

Now let's turn our attention to the flat domestic battery.

This battery, as it is flat, will have an off load terminal voltage of around 11.8 volts. We cannot calculate the internal resistance of this type of battery as they are not rated in CCA. That isn't what they are designed for. But I can tell you that the internal resistance of these batteries is higher than that of an engine start battery. A higher internal resistance would produce a lower initial surge current. So let's give them the benefit of the doubt and assume it is the same as the engine start battery. This sways the argument in their favour.

Further, the internal resistance of a deep cycle battery increases as the charge status falls. So a flat battery will have a much higher internal resistance than a fully charged battery (this is why a very flat battery takes such a long time before it starts to accept a charge - an effect we have all seen). Again we will ignore this fact. This sways the argument even further in their favour.

Finally there is one little fact that they either seem to have forgotten or that they simply don't understand. That is the fact that in order to charge a lead acid battery, simply increasng the terminal voltage will do nothing. If our flat battery is at 11.8 volts, then presenting it with 11.9 volts will do nothing. There is a "dead band" of voltage which it is necessary to exceed in order to commence charging. This dead band varies between batteries and the state of charge, but is typically around 0.3 volts.

So, in order to produce any sort of charge current into our flat auxiliary battery bank it is necessary to increase the terminal voltage of that battery to 11.8V plus 0.3V = 12.1 volt. That is before any sort of charge current begins to flow.

So let's now connect our two batteries together.

The engine start battery is at 12.7V, the auxiliary battery is at 11.8V. The internal resistance of the engine start battery is 0.002R, we have given them the benefit of the doubt and stated that the internal resistance of the auxiliary bank is the same (even though it will actually be much higher) at

0.002R

So we have a voltage difference of 12.7V - 11.8V = 0.9V. But wait, the auxiliary bank will not even begin to charge until it's terminal voltage is raised to 11.8V + 0.3V = 12.1V. So we now have a voltage difference of

12.7V - 12.1V = 0.6V

By Ohm's law we can see that this voltage difference presented across the 2 battery internal resistances of 0.002R each (0.004R) gives us 0.6V/0.004 =

150 amps.

This is nothing approaching a so called "huge, initial, dangerous, surge current". 150 amps may seem like a lot. But remember we are dealing here with an engine start battery sized to suit a 10 litre, 200 horse power, diesel engine and a 1000 amp hour battery bank which will probably have alternators and chargers connected to it in the region of 150 to 200 amps. The correctly sized split charge system will therefore easily cope with these sorts of currents.

If we reduce the entire system by a factor of, say, 0.4 we end up with a 400 amp hour auxiliary bank and a 600 CCA engine start battery. These are probably figures more typical of the installations you are used to. The intital surge current is correspondingly reduced to 60 amps.

If we now factor in the fact that the internal resistance of the deep cycle battery bank would actually be closer to 0.004R (because the internal resistance of deep cycle batteries is much higher than that of engine start batteries) we reduce this initial surge to 100 amps (this is back with the example of the 1500 CCA engine battery and the 1000 Ahrs auxiliary bank).

Now the real stinger. Add in the fact that the internal resistance of the deep cycle battery will actually be substantially higher than this because the battery is flat (remember the internal resistance rises as the battery discharges) we actually find that the internal resistance of this battery will be around 0.02R.

Now the "initial, dangerous, surge current" is 0.6V/0.022R = 27 amps!

We're not quite finished yet. The proponents of this false rumour usually state "one full battery and one flat battery" so that is the example I based the above on. They are obviously unaware that this state of charge does not cause the greatest current to flow.

The worst scenario actually occurs with a fully charged engine start battery and the auxiliary battery bank at around 40% charge state. The reason is that, although the terminal voltage falls as the battery approaches totally flat, the internal resistance rises. At 40% state of charge the internal resistance is considerably lower. The result in the above example would be that the initial current surge would be around 40 amps, and even that would only last a minute or so (thus discharging the engine battery by 40A *

1/60th hour = 0.7 amp hours) at the most until the surface charge on the auxiliary bank raised it's terminal voltage to that of the engine start battery. Once this had happened, the current would be negligible.

A very brief current surge massively exceeding these calculated figures is a possibility. When we say brief, we really mean brief, like a few tens of milliseconds at the most. This cannot cause any problems. It is a current spike, not a potentially damaging (to other equipment) voltage spike.

Those "in the know" are fully aware of all of this. Our colleagues in the true technical section of the industry are fully aware of it. Electrical and electronics engineers involved in any field relating to batteries, chargers, split charge systems and flat battery protection systems are fully aware of this.

Perhaps the proponents of this rumour are merely passing on what they have heard or perhaps they have some hidden adgenda for perpetuating the rumour?

The 8274 solenoids are open to the elements apart from a plastic guard (not sealed) and I just place a bin-bag over it with a bungee strap wrapped round it to keep it in place. It's never given any trouble but I think the 9000 has all the electrics encased and totally sealed, depends which winch you've got I suppose.

Martin

Unfortunatley, Julian's post caused a very brief current surge massively exceeding the calculating capacity of my cerebrum. This voltage spike has fried my brain.

Lukily, I have little use for it.

Cheers

Blippie

-- Ten minutes of this rain will do more good in half an hour than a fortnight of ordinary rain in a month.

Surely the batteries are in parallel and therefore the resistance would be 0.001R.

Not as I see it. We are talking about _circulating_ currents between the two batteries when they are 'commoned up' by the split charge relay. (or whatever) Circulating currents see the battery's internal resistances as in series.

Julian.

So if I have 2 identical Bty's installed in my vehicle, except one is discharged, then I have a varying serial resistance until both batteries are fully charged and then they behave as resistors in parallel?

I'm not at all sure what you're getting at here, what don't you understand? I provided a good explanation to, amongst other things, dispel the myths surrounding one battery charging another with a heavy current via a (for instance) split charge relay. I can follow the explanation easily enough with just a basic grasp of DC electrics.

Julian.

understand?

He's asking the same thing that crossed my tiny mind - simply put, how can 2 resistances in parallel behave as if in series?? Your explanation is indeed good, but there's a bit (to my mind anyway) of "blind faith" going on. From my own personal experiences over the years I came to the following conclusions regarding various split-charge system types:-

1. Blocking Diodes (or "split-charge" diodes) are ok, but only if the alternator is rewired internally to be "battery-sensed" as opposed to internally "machine-sensed", to overcome the voltage drop across the diodes. Downside is, the alternator runs hotter as it is now regulating at a higher voltage, by roghly 1.2volt.
2. Basic relay connecting 2 batteries together in parallel - works ok, never seen more than 75A transfer (measured with a clamp-type ammeter that was in calibration), but if you try and use a winch with the relay made you will share the winch current (possibly as high as 500A!) between the 2 batteries and fry the relay's internals and/or start a fire in the relay wiring.
3. Manufacture bracket, fit second alternator to engine, run two totally separate systems! Absolutely foolproof as all wiring can be put in at a suitable current rating, no issues of one battery draining another etc etc. A master switch (with key) can be used to feed the winch, and a second identical switch placed close by can be used as an internal "jump-start" facility from one battery to the other. 2 switches, one key, no possibility of the winch draining the vehicle battery and leaving you stranded! Badger.

OK, I will try to explain this specific case:

Draw a diagram with just two batteries and wire them up in parallel. Let's say that the batteries are both 12 volt, therefore no current will flow in the circuit that you have drawn connecting the batteries.

Now, let's say one battery is almost flat and only has 11.5 volts. There is now 12v - 11.5v = 0.5volts driving a current around the circuit you drew. (one battery is being charged and the other is discharging) The magnitude of the current flow is determined by the internal resistance of battery 1 plus the internal resistance of battery 2, ie battery internal resistances in series.

Julian

I think what he means is that there are two actual resistors in the total circuit whilst current is flowing from one battery to the other, without them there would be no circuit. Funny explanation but I know what he means.

Martin

Yes, correct, there must be. ie the internal resistance of battery one and the internal resistance of battery two. All batteries have an internal resistance [1] - no getting away from that one!

[1] when I did physics at school anyway - prolly all media studies these days!

Julian.

"Julian" wrote .....

Don't know anything about 'media studies', they didn't have those in the '50s. :-)

Martin