# Solving problem of long DC wires

DC current through a very long wire losses its first current, because the wire work as a resistor.

This is the biggest problem when using DC appliances in solar home systems, because there is a very long distance between the battery bank and the load.

So, I consider distributing the batteries through the wire, for example if the wire is 60 meters long and we have 3 batteries, I will put one battery in each 20 meters.

1. Is the diagram below correct? Will it have effect on battery charging process?

2. Are there other solutions for this problem?

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Good morning! what´s the load current? and what is the section of the wire? – Bruno Y Mar 16 at 12:08
Larger diameter wire = lower resistance. For instance 16 AWG is 13.2 ohms / per 1000 meters while 10 AWG is 3.28 ohms / 1000 meters. From here: hyperphysics.phy-astr.gsu.edu/hbase/tables/wirega.html – Tyler Mar 16 at 12:14
Your “solution” will have the same total losses (all the current produced by the solar panel still has to travel the same distance). As Phil pointed out it will also affect charging/discharging. Get better wires or use a higher voltage. – Michael Mar 16 at 17:15
If the wire resistance is a significant factor then you need larger wires. And when choosing where to place the battery you need to decide whether the peak current you're worried about is during charging or discharging. – Hot Licks Mar 16 at 17:17

If your wire is long or small enough that there is a significant voltage drop across it, that means in your 2nd configuration each battery will see a significantly different load.

The battery farthest from the load sees the long wires as a resistance in series with the load. Thus, less current will be drawn from this battery.

The battery closest to the load sees very little resistance from the wires in series with the load. Thus, more current will be drawn from this battery.

Thus, the battery closest to the load will discharge faster than the others. This is probably not good, as it puts more wear on that one battery.

There is also a similar problem with dissimilar resistance between the batteries from the perspective of the charge controller. When the batteries are not full, the battery closest to the charge controller will take more of the current. Only if you are charging at a low enough current to make the voltage drop across the wire significant will all three batteries reach an equal charge.

Perhaps if the loads are also distributed across the wire, then there could be some merit to this system. However in that case, I think it might make more sense to run each load from an independent battery.

I would say the best solution is to use a thicker wire, which will reduce the resistance of that wire, and consequently the voltage drop and energy loss (and fire hazard, assuming a properly selected circuit breaker).

Another solution is to step-up the voltage on one end, then step-down the voltage at the other end. Or just design the entire system to work at a higher voltage. Holding power constant, increasing the voltage decreases current, and consequently resistive losses in the wiring. This is precisely the technique used by electric utilities, where wires are even longer.

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Sixty meters upgraded to heavier wire... could be a major rebuild. It takes bigger boxes, bigger conduits, different connectors- it isn't an easy alternative. – Whit3rd Mar 18 at 2:27

Sometimes, in this sort of system, it makes sense to operate the power storage and distribution system at a higher voltage — say, 36V or 48V — and use point-of-load DC-DC converters to provide the specific voltages needed by each load.

The currents flowing in the long wires will be reduced by the same factor (i.e., 1/3 or 1/4), and the I2R losses will be 1/9 or 1/16. This can more than make up for the losses in the DC-DC converters.

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Your resistive losses will be significant unless your wire is very thick. 120m of 16 awg wire has a resistance of 1.6 Ohms, or 1.6 volts per amp. even 10 awg will loose 0.2V per amp. I suggest taking Dave's advice and moving to 48V. Use a 48V solar charger to charge 4 12V batteries in series and connect to a 48V to 12V converter near the load.

Your solution will make things worse. The first battery will be charged the fastest and discharged the slowest. The last battery will be charged the slowest and discharged the fastest. After many cycles, the first battery will remain mostly charged and the last battery will remain mostly flat leaving the middle battery to do most of the work. When connecting batteries in parallel they should be connected like this

simulate this circuit – Schematic created using CircuitLab

So that the path through each battery is the same length.

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+1 how is this not addressing the question?? Pence has explained why the proposed design is incorrect, and presented another solution, exactly as the OP asked. – sebf Mar 16 at 22:55

Distributing the batteries as shown won't really help as Phil mentioned above.

What you need to do, is determine where the highest current flow is, and minimize the wire distance there. If you have a high current load, but a low current charger, put the batteries by the load. That will minimize the wire losses.

The best solution is just to use heavier wire, or even buss-bar, depending on what your actual charging capacity, battery capacity, and load are.

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Your problem is that you're not using a "solar home system". You're using a system designed for caravans/RVs/boats, and you're making it do something it fundamentally isn't designed to.

If you want to do this for real, you need something that'll step the voltage up, so that you can get power to your house with minimal losses. Typically the easiest way to do that would be to buy an inverter off the shelf, and fit that on the battery bank. You then have mains-level AC driving your house, so transmission losses are much lower. And then you can then use all your normal AC appliances, light fixtures, etc., instead of using 12V DC fixtures.

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The bus-connect diagram is workable, but not optimal. A variant electric connection would be to wire a LOOP around your house, of +12V and ground, and position multiple batteries and loads around the loop. This means that electricity to a load comes from clockwise-located batteries as well as counterclockwise-located batteries (and all points in a 60m loop are no more than 30m from any given battery).

The common solution to power distribution is to use a high distribution voltage (120VAC for compatibility; for simplicity or shock safety, 48VDC might be a good choice). Insulation for higher voltage is inexpensive compared to copper for higher current. If your appliances use 12V only, there are step-down converter modules at modest prices with good (not great) efficiency.

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Your picture for the load looks like an outdoor light. Can you locate your load batteries and solar panel all close together for minimal resistive losses and only run a tiny signalling current the full length to connect a light switch (which may be 12V 2mA) to a transistor and relay at the load and panels and battery. Telecoms multistrand wire would be sufficient for that, expecting to need N+2 strands to work switches for N loads.

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Voltage drop over the wires between the charger and batteries will hit your charge rate hard, as has been said.

In your specific application, a further reason to put the batteries on the left of the line is that the LED driver is almost certainly able to run off <12V with minimal effect on performance.

I'd hazard a guess that to get an appreciable drop in light intensity you'd have to feed the lamp with below 10V. These sorts of light generally use a constant current LED driver IC. The more basic driver ICs (e.g. MAX16803) need a little headroom, and as white LEDs drop up to about 3.8V, they'd struggle to run 3 in series off 12V nominal, so are more likely to run 2 in series, meaning you'd be OK down to ~8.5V without any loss of brightness. If the lamp uses something like a MAX16840 it can be designed run off a much wider voltage range (but this would be more expensive.

You should probably check the specifications on the lamp.

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