24 volt to 12 volt converter

tdevery wrote:
I like the idea of having at least 200watts for power if needed. I pick my campsites on google earth to try to get some sun. But with older state parks in PA, this is hard to find.

You will like it better having 200W portable and 500W permanent :)
Permanent will work on the road, and on open camps, and (a little) under the trees. It still works in leafy shade on a sunny day - poorly - but it works all the time.

You already have 40A controller for 500W permanent. Being MPPT, it can accept 24V panels, they are cheap per watt. Your 2 portable arrays will need 15-20A PWM controller, one for both, it can be found for ~$50. It is not likely that you will have to use both PWM portable and MPPT permanent at the same time, but don't see anything that would make such a pairing impossible.

Why the obsession with cell resistance
It doesn't matter in this case
It is the current thru the wire that creates the voltage drop between the batteries and the inverter
The battery resistance to the current is. A moot point
50 amps per 12v battery *2 for 100 amps ,1200 Watts
Or
50 amps thru 24 volt battery for 1200w
Each and every cell is still passing 50 amps
There are no internal circuit differences

There is no difference

I like the idea of having at least 200watts for power if needed. I pick my campsites on google earth to try to get some sun. But with older state parks in PA, this is hard to find.

I need to play...

"Honey I shrank the panels"

When crossing back into Mexico, or, maybe mount them lengthwise and vertical on the roof and plaster cardboard over them

"Eat At Juanita's Restaurant"

tdevery wrote:
... 12.5 amps with two 100 watt Renogy panels wired up for 24 volts to the 40 amp charge controller. Think I will permanently mount two more 100 watt panels on the front of my 5er roof.

Looks like you are trying to use the same 40A controller for either small portable or small permanent (or maybe both at the same time). This is going to be some pain.

Mount 2*250W on the roof of your 5-er. They will cost about the same as 2*100W. Better yet, 3*250W. With a proper 60A MPPT. You will rarely need to mess with portable panels then, and (hopefully) your generator won't get used much either. Build it right from the beginning.

Common situation, people are buying portable or a couple of small permanent panels because they don't know or not sure about this whole solar thing. Then they are trying to use these old parts for expanding the system...

If money is really a problem, use that 40A MPPT for 400-500W permanent install, and get cheap 15-20A PWM for a portable.

brulaz wrote:

OldSmokey wrote:

BFL13 wrote:
"- battery V drop under load is halved on a % basis at 24v"

Is that true? It might be halved or doubled as such, but I would think on a Percentage

Basis it would be the same. Please elaborate--my brain quit :)

assume T105RE with internal resistance of 0.001 Ohms per cell and 1200 Watt load

for 12V we have 6 cells so 0.001*6*100 A = 0.6 V
0.6/12*100% = 5 percent drop

for 24V we have 12 cells so 0.001*12*50 A = 0.6 V
0.6/24*100 = 2.5 percent drop

OK, East coast is back online with a cup of coffee, and even after sleeping on it I'm still confused by OldSmokey's example.

The 24V example uses four T105REs in series to get 12 cells and 24V.

But it appears that the 12V example only uses two T105REs to get 6 cells and 12V. What about the other two T105REs in parallel?

Wouldn't they effectively reduce the internal resistance by half and the V drop by half?

Making the %V drop identical to the 24V example.

too much coffee... LOL

what about the other two RE's in parallel.. ?? I dunno, no one mentioned parallel batteries ?

so lets do it now and before we do, let's have an Ohms law refresher

Ohms law.. in plain language states that the current through a conductor between two points is directly proportional to the voltage across the two points.

using the standard symbols we have:

V = Volts
I = Current in Amperes
R = resistance in Ohms

If we know two of the three variables then we can derive the third so we get

V = I * R
I = V / R
R = V / I

also relevant to this topic is resistors in parallel and Power in Watts. watts can be

derived from any of the other variables we have here so we have

W = V * I
W = I*I * R
W = ( V * V ) / R

the resistance of any number of resistors in parallel is defined as
the reciprocal of the sum of the reciprocals of all resistors

so we have R = 1 / ( 1/R + 1/R + 1/R ...)

example: we have two resistors of 10 Ohms each
the total is 1 / (1/10 + 1/10) and 1/10 = 0.1 so we get 1.0 / 0.2
and that's now 5 Ohms

for resistors in series, well you just add the values together..

now, batteries also behave like a resistor. In my post I used the figure from Trojan for the T105RE. It doesn't matter really what battery you use just as long as we know the internal resistance. for this example we are going to IGNORE any cable or connection post
resistance and focus on just the battery.


for the T105 it's 0.001 Ohms per 2V cell, so a 6V battery has R = 0.003 Ohms

so we get 1 / ( 1/0.003 + 1/0.003 ) = 0.0015 Ohms for the 2P pack
and 0.0015 + 0.0015 = 0.003 for the series pack

using my earlier load of 1200 Watts and 12 Volts
we get 1200 / 12 = 100 Amps. 0.003 * 100 = 0.3 Volts

as a percentage of the battery voltage we now get 0.3/12*100 = 2.5 percent. same as the 24 Volt string.

so your answer is yes, the voltage drop will be half.

any advantage ? well, not really.. you need more heavy cables to interconnect the batteries and you still have the same total energy ( 450Ah*12 ) vs ( 24V*225Ah ) you also need a bigger CC to charge the 12V pack..

I also have 4- 190AH 6volt batteries, so I believe it is a 380 AH at 12 volts.
But you are correct. 50% use is 190 AH

Like I posted earlier, on a 45 degree stand at the Outerbanks this summer. I was showing 12.5 amps with two 100 watt Renogy panels wired up for 24 volts to the 40 amp charge controller. Think I will permanently mount two more 100 watt panels on the front of my 5er roof. Keep the other two mobile in case I park in shade.

2oldman wrote:

tdevery wrote:
I have 4-6volt AGM at 380 amps. .. Keeping two (panels) on a 45 degree portable setup in case I have shade when camping.

I'm thinking you have effectively 190ah.

Shading will drastically harm your output on series-connected panels unless they are well-equipped with bypass diodes. I guess you mean you'll just be using those 2 panels by themselves.

Usable charge 190AH, yes. (It's AH, not Amps).

Don't know where and how 2 portable panels came from, in this topic, (and what power). Portable panel off the shelve is 100W, split in 2*50 squares that fold onto each other like a suitcase. Buying 2 of them will give you mere 200W total and will cost not much less than 600W permanent install. Will be a bit of pain to set up and break up every time, not sure what to do with it when you're away from the camp, and this is financially inefficient, compared to permanent install. You could buy your own panels as big as you like, even 250W, and make your own portable array, but panels bigger than 120-140W are heavy and bulky, a nightmare to carry around and store inside.

If you camp in the woods most of the time, portable could be the only choice, otherwise - forget about it. As an addition to permanent 600W, maybe. One 100W suitcase, for those occasions. Just one - you might not be able to use even one, depending on distance to the closest sunny spot. I would make that permanent more than 600W if possible.

voltage drop R*I
resistance times amps

technically 1200w /12v 100amps, 1200w/24v 50amps

R*100 versus R*50

now lets say the voltage drop at 12v 100amps is 1v

now 24 same resistance less amps voltage drop is 0.5v

so 12v-1=11v alarm! 24v -0.5 = 23.5v No alarm

so that should work, engineering math wise
even if there is consistent 1v drop, it becomes 11v verus 23v still above alarm stage

but it has to be a system wide change, 24v batteries, 24v inverter

all advantage is lost and more 'LOSS' is introduced when you install a converter
from 12v to 24v just to power the inverter, the DC converter has its OWN losses