long wire use in my led light near 200 meters and it connect to an inverter battery

Question

We are use white high brightness 1 watt led which voltages 3.3-3.6V and current 350mA. Now We are connected 24 led which in 4 led in series of 6 groups and wire one sq. mm are use. Wire is use nearly 200 meters(two color wire for main supply) and at some distance led of each group is connect to two wire of these and these wire is supply from inverter battery(which is continues charge and it output voltage is 13.10V). now all leds are working but frist group of led give full light and second group also give full brightness or light but then third group is low light then first and second group. then continue decreses in light of led and last group of leds are give very very low light.How to solve these?

And also suppose we are not use these inverter battery then how many power are need and what is use for it?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

If I read correctly, the poster is doing this, and is experiencing voltage drop due to the length and resistance of the wire.

The reason for the voltage drop is because the wire has resistance, and you are drawing high current (350mA) for each series string.

The \$1\space\text{mm}^2\$ wire has a resistance of approx. 20 ohms/kilometer, so it is 0.020 ohms / meter. Assuming that you have the LED strings equally spaced across 200 meters, then your circuit actually looks like this, electrically:

^{simulate this circuit}

At each point along the wire, you are increasing the wire resistance, and requiring that wire to take more and more current. As an example, take the LED string in the middle, represented by I3.

At that node, I3, in my picture above, there is 6 x 0.666 = 4 ohms back to the source voltage, and that 4 ohms of resistance is carrying the current of I3, I4, I5 and I6 = 350mA x 4 = 1.4 Amps ( and that's a simplification, as I neglect the current and voltage drops contributed by the LEDs before it, at I1 and I2 -- forget about them for now) .. so that 1.4A across 4 Ohms causes a voltage drop of 1.4 x 4 = 5.6 Volts, according to ohms law.

So you see, there is a voltage drop at that mid point of 5.6V or more, and this is too low to light the remaining LEDs (I4, I5 and I6).

The simple solution is to raise the input voltage, if your LED can handle it, or increase the thickness of the wire to maybe \$4\space\text{mm}^2\$ wire. This will have much less resistance, dropping only a few volts along the 200m length.

The proper and correct solution is to raise the voltage to about 18-20V, use thicker wire, maybe \$2.5 - 4\space\text{mm}^2\$, and use a constant current regulation for 350mA at each string. This is the only way to get each string to be equal brightness.

Answer 2

The first answer given probably covers most of your problem. however, I think there is one other major factor to consider.

From your description, I can't quite decide how your LEDs are exactly running, whether they are in series or in a series/parallel configuration.

In a series configuration, each Led that your voltage passes through, your voltage will drop by the 3.3-3.6 volts. So if you start at 13.10 volts, and lose 3.3 volts with every LED in series that you pass. Assuming that each string of 4 is in parallel and then 6 strings are in series, then each string you pass by drops you voltage by 3.3-3.6 volts for the next string.

This is addition to what line losses you will get as explained by the other answer, and your voltage will be very low by the time you hit the last few LED strings.

So how much voltage do you need?

This depends on what your actual setup is. I will assume it is 4 LED in parallel, with six strings of them in series. Then you need 3.6 volts X 6 and add several volts for safety margin and use a thicker wire so you don't get line losses.

Answer 3

Brian Onn has given you details of why your problem is occurring.

The way that I'd approach this problem requires a bit of electronics. This would be something that is relatively easy for me to build but I'm not sure of your capability. Nonetheless, I'll describe my approach.

Simply said: I'd put all 24 LEDs in series. This requires a constant-current switch-mode boost converter that runs from your battery and delivers both the required voltage and regulated current.

You have 24 LEDs, each needing about 3.2V. 24 * 3.2 =~ 77Vdc at a regulated current of 350 mA. Your wire will also add voltage drop - but less than the voltage drop that you are currently having because the current is so low compared to the way that you are doing it now. I'd design the boost converter to have a maximum output voltage of about 100V - this allows you to use thinner wire if you want and / or go for a longer distance.

There is a plethora of pre-built switch-mode LED power supplies available with exactly this specification - but the ones that I've seen all run from AC Mains supply. But you want to run from a battery.

No problem. Linear Technology and TI and Maxim and a great many other chip manufacturers make ICs that turn this into a fairly easy project.

Answer 4

It sounds to me like you've got too much power going through your LEDs and are burning them out. I've had similar problems with this. I would suggest using resistors if you are not already doing so.