CTEK 250S Distance Question

Jersey4x4

Adventurer
Hi all

I am currently designing my electrical system for my 2007 Hilux

I am planning on running a 55 ah yellow top up front with the original battery which is all well and good and the CTEK 250S I have chosen to charge the yellow top will have no problems...

However the problems start when I plan to add two 75 ah blue tops in the tub which is about 4-5 meters away

I looked up the wiring gauge guide and it doesn't give a size for the aux battery for more then 2 meters away... This is saying to me that batteries receiving a charge shouldn't be more then 2 meters away?

Is this correct or am I reading into it a bit too much

Has anyone got this set up or something similar

Thanks in advance

Jay
 

Jersey4x4

Adventurer
Just recalculated allowing a 0.5 volt drop rather then 0.05 volt drop and it's come up with 4mm2 which sound more like it..
 

DiploStrat

Expedition Leader
????

What is the output of your alternator? That is the number you want to use. As a rule of thumb: AWG 1/0 (53mm2) will pass about 75A over 25 feet with just about 0.5v drop. With a pair of 75Ah batteries, you want to get a charge rate of as close to 100A (or better) as you can. I would guess that you need between 50 and 100mm2, if only to assure that the voltage drop is as low as possible.

In my case, with a pair of 125A alternators and a pair of 150Ah starter batteries, I use a pair of 1/0 cables. Doing it again, I might go up to 2/0. In any case, I see a minimum charge rate of 150A at the camper batteries.
 

Jersey4x4

Adventurer
The hilux has a low voltage output of around 13.3v or so which is why I went for the dc to dc charger option which has a max output of 20 amps
 

dwh

Tail-End Charlie
Voltage drop doesn't much matter when CHARGING batteries, though it definitely matters when supplying loads from batteries, such as an inverter.

Why does it not matter when charging? Because voltage drop is a function of LOAD. As the battery approaches full charge, the resistance rises and the amps flowing though the battery (the LOAD) goes down. By the time the battery is approaching fully charged, you are only going to have a couple amps flowing, and at that point the LOAD on the wire is NIL, and thus the voltage drop is NIL.

A .5v drop during the short time when the wire is fully loaded (battery resistance at its lowest point and no surface charge), might add 15 minutes to a charge cycle that takes hours (or days), so as I say, it's not going to matter much.

A .5v voltage drop through the wire won't make your battery end up .5v low like a .5v drop through a diode-type isolator can. The battery will still reach full charge because at the end of the charge cycle, the voltage drop has gone away.

Distance is irrelevant. You can max out the wire and shave a few minutes off the time that the full charge cycle takes, but there really isn't much need unless you are trying to eke out every watt of harvest from a PV (solar) that only operates for a limited time each day.

Size the wire for the max amps that will flow through it, which in the case of the CTEK 250s is 20a.

(Well, 20a on the LOAD side of the CTEK. On the supply side TO the CTEK, it'll draw a bit more than it puts out on the load side. So I'd say you could get away with #12 on the load side, but go with #10 on the supply side. In which case...why bother? Just use #10 on both sides. (With 30a fuses to protect the #10 wire.))



Now all of the above is talking about CHARGING. For SUPPLYING LOADS (like inverters), the voltage drop INCREASES as the battery voltage goes down, so you have to size the wire for the max load and worry about voltage drop.

For example:

1200w inverter.
Draws 100a @ 12v (when fully loaded with a 10a load on the 120v side)
But not really...
When the battery is fully charged, it draws 1200w / 12.8v = 93.75a
And if the inverter shuts down at say 10.5v, then it draws 1200w / 10.6v = 113.2a just before it shuts down.
So you want to size the wire to minimize voltage drop with a 114a load at 10.6v.
Otherwise, if you had a .5v drop, the inverter would shut down when the battery was actually still at 11.1v. (Which might not be a bad thing.)


So voltage drop matters a lot when supplying loads, but doesn't really mean jack when charging, since voltage drop is continuously variable and disappears during the last part of the charge cycle anyway.
 

Jersey4x4

Adventurer
I am guessing this all apply to dc to dc chargers as they regulate voltage anyway?

A simple split charge system would suffer from voltage drop wouldn't it?

Brilliant stuff cheers
 

dwh

Tail-End Charlie
Just noticed you said your rig's constant voltage power supply (voltage regulated alternator) puts out 13.3v.

So, if the CTEK is putting out 20a @ 14.4v, that'd be 20a x 14.4v = 288w. On the input side, it's drawing those watts at 13.3v so the amperage load on the input wire is gonna be 288w / 13.3v = 21.65a (plus a few watts to power the CTEK itself).

Now, that's the worst case scenario - which is possible, but unlikely since when it reaches 14.4v, it's gonna drop out of bulk stage anyway and as you can see from the manual page 23, once it's into absorb stage, the amp flow (bottom of chart) starts dropping. By the time it gets to the end of the absorb stage, there are hardly any amps flowing, and thus no load on the wire and thus no voltage drop. Go back and play with your favorite voltage drop calculator, but specify 1a or 2a @ 14.4v to see what the real voltage drop will be at the end of absorb stage. (But don't trust the answer as more than an estimate; the data will be a bit wrong since the CTEK is going to adjust the voltages anyway when it does temp compensation. 0_o )

So yea, #10. Lemme see...that's what it says in the CTEK manual anyway. Don't worry about using #8 for a few more feet. All it would achieve would be to shave a few minutes off a charge cycle that's gonna take all bloody day anyway.
 

dwh

Tail-End Charlie
I am guessing this all apply to dc to dc chargers as they regulate voltage anyway?

The CTEK regulates current and voltage in the bulk stage, and voltage in the absorb and float stages. All proper multi-stage chargers do that - bulk stage in constant current mode, then absorb and float in constant voltage mode.


A simple split charge system would suffer from voltage drop wouldn't it?

Nah. A voltage-regulated alternator is a constant voltage power supply. Doesn't do a constant current bulk stage. So what happens is the battery pumps up fairly quickly to a surface charge of (in the case of your truck) 13.3v and after that, it's just a trickle charge of however much current can overcome the battery's resistance at a pressure of 13.3v. A few amps. It'd take all day to fully recharge a 50% depleted 55ah battery.

But again, as the battery reaches full charge, the amps flowing through it are down to almost nothing, so there is no voltage drop anyway.

A diode type isolator has usually around a .5v drop through the diode. If your voltage regulator senses voltage at the alternator (ahead of the diode) then it'll never notice the .5v drop and the batteries (both if they are both fed by the isolator) will end up .5v low. If the voltage regulator senses the voltage at the battery, then it'll just keep the alternator switched on until the voltage at the battery rises to the proper set point, and that'll end up compensating for the voltage drop through the diode.

But that's a diode and a special case.

Any normal battery charger or charging system is going to taper off the current at the end of the charge cycle as the battery's resistance rises - and so in all cases the voltage drop is gone by the time the charge cycle gets to the end.
 

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