DC to DC chargers, do you really need one?

[T1 Terry]

Tomorrow I'll explain why there needs to be more amps (current) coming into the DC to DC charger than what is coming out, it is not because the DC to DC charger is wasting all that energy, it's all to do with that evil watts word.

See if I can make this a shorter one.
Basics, a watt is 1v x 1 amp eg 12v x 1 amp is 12w, 10v x 1 amp is 10w etc
The DC to DC charger boosts the voltage in to 14.4v or the current to what ever it is rated to, 20 amps or 40 amps are the most common on the output side. If the battery is absorbing the 40 amps and could handle a bit more the voltage will not reach the 14.4v mark, but in reality unless they are lithium batteries or a big lead acid battery bank deeply discharged the 40 amps will be more than the battery can absorb without the voltage climbing above the 14v mark. So let's use 14v and 40 amps as the example, 14v x 40 amps = 560w, now the input voltage while pulling 40 amps out of the start battery and through 15 mtrs of cable (remember the return path thing with DC) is not going to be 40 amps at 14v no matter what size alternator you have, we will be generous and say 12v on the input side. We know the output is 560w, allow 10% loss through the device because nothing is perfect when electronics are involved so some gets wasted as heat, so 616w including the losses is required on the input side, we know the voltage is 12v so 616w / 12v = 51.34 amps in to get 40 amps @ 14v out.
Let's step it up a bit, we have now reached the 14.4v mark but the amps are still flowing in so 14.4v x 40 amps = 576w plus 10 % loses = 634w, but this vehicle has a computer controlled smart alternator and the start battery voltage has been lowered to 13.6v rather than the 14v in the first example. Remember voltage is like pressure in the water analogy, so the pressure (voltage) is lower this time so there is even less to start with and we are asking for more current (amps) through the cable, by the time it reaches the DC to DC charger it is likely to be less than 11v.... sound excessive? Let's do the numbers, 634w / 11v = 58 amps. Over the 15 mtr round trip the voltage drop would be around 9% or 1.2v, yet put a multimeter on the DC to DC input side when it's under full load and you will see 11v or less using the recommended 6 B&S cable (13.5mmsq conductor).... why?

T1 Terry

[El Gringo]

Ohm, Ohm, dunno. Bit of resistance maybe.

Cheers.

[T1 Terry]

Ohm, Ohm, dunno. Bit of resistance maybe.

Cheers.

But where, the cable was already considered in the voltage drop. Something no one ever mentions on the forums, they talk about cable resistance but never about the resistance in each and every connection, the connector crimped on the cable, the joint where the connector mates up the terminal, the terminals in the Anderson plug, 2 connecting faces and 2 crimped joints, as the voltage drops and the amps increase the effect also increases.
What about the voltage drop at the battery itself, modern alternators sense their voltage at the start battery so connecting the DC to DC cable at the alternator is a step in the wrong direction, but again, when the alternator output voltage is reduced the resistance in the cable from the alternator to the start battery has more effect, less pressure pushing the current through yet that cable is the same size or smaller these days than back in the '70's and 80's and often a much longer path as it is bundled up in the wiring harness.
Another thing never mentioned is that the computer only sees the need to seriously ramp up the output at a much lower voltage than the target 13.6v, so a slower reaction to the drop in start battery voltage results in a lower voltage (pressure) being applied to push the current (amps) through the 15 mtrs of cable.

T1 Terry

[El Gringo]

I reckon this is one of those cases where big is good, bigger is better..
Cable and connector size I mean.

Cheers,

[T1 Terry]

There is still another cause for the low voltage, it is resistance but what makes it worse than that rated for the copper cable on it's own? A hint, the water analogy here is a similar effect and commonly seen when first running the shower....

[El Gringo]

In water it's a type of inertia, getting something moving from a standstill, there is always some resistance to initial movement.
Like pushing a car, getting it to start moving is often harder than keeping it moving.
However, I didn't think that applied to electrons, at least to any measurable extent?
Though electricity is like water in that the drops of water entering the pipe are not the same ones leaving it at any given time.

Hmmm, thinks..... maybe i need more learning about this.

Cheers,

[T1 Terry]

Ok, the water analogy with the shower is when you turn the hot water on at first there is plenty of water, but cold, by the time it heats up it is a trickle so you need to turn the tap on more. Part of that is because hot water doesn't flow through a pipe as well as cold water but for a whole different string of reasons that is unimportant to this thread, the other reason is pressure drop, the water is under a greater pressure in the cyl because it has expanded from the heating, after that pressure is bleed off the water flow is reduced.
These are the two things in common with the water analogy, resistance in the copper cable conductor increase as it heats up, the greater the resistance the more heat, the higher the current the more watts of electrical energy are converted to heat. Just figures out of the air here, but cables near 0*C 15 mtrs long might loose 20w to heat with 50 amps of current passing through. That 20w of heating raises the cable temp so it is no longer 0*C and this causes the resistance to increase and more of the current is converted to heat and this increased resistance increases the voltage drop. This means to get the same number of watts at the DC to DC charger the current (amps) must increase.... and so on continues the cycle, the voltage at the end of the cable continues to fall unless the conductor cable is large enough to minimise the resistance in the first place.

All this leads to explaining why a DC to DC charger would allow small cable to be used on the supply side, it must be large enough to carry the expected current and this can be 1.5 times the output current, I.E. 60 amps in for 40 amps out, the DC to DC charger hasn't wasted the other 20 amps and it wasn't necessarily converted to heat in the supply cables, it is the result of reduced supply voltage needing higher amps to equal the same number of watts required for the 1DC to DC charger to operate at its true potential.

Are there any aspects of DC to DC charging I need to cover that I haven't so far or any bits I need to explain further for clarity?

T1 Terry

[Bob & Syrenna]

YES (in answer to your question) Good explanation Terry

[El Gringo]

OK, so here's my understanding of all the technical stuff you have gone to all the trouble to explain - see if I got it right?

To charge a house battery from a start battery, it doesn't work too good straight off the alternator, if at all.
so you use a dc to dc charger.
This needs to be placed as close as possible to the battery being charged, with as few connections in the wiring as possible and using appropriate size wire.
The supply to the charger needs to be in a straight - continuous run, also with as few connections as possible.
And for me, probably double the size wire i think i need.

Now, can you also do this through your smart box to charge the lithiums?

Cheers,

[bagmaker]

Perhaps you could add the preferred position of the charger?
Do they put out any interfering RF to be aware of?