I have a question. Currently I am running 3 LED's in series with 8 strings in parallel ( 24 LED's ). I'm using 22ga twisted pair wire. My LED voltage is at 10.8v at the LED with a 12.5v battery power. The LED's are rated at 3.5-4V FV max so I'm good here. My question is what would change if I only built 3 in series with 4 strings in parallel ( 12 LED's )? Either way I'm at 10.8v at the LED right? Would there be any problem running 12 compared to 24 this way?
12 and 24 LED hookup question
- Thread starter mnm99
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Steve K
Flashlight Enthusiast
less current will be flowing through the wires, so less voltage will be dropped across the wire's resistance, therefore you'll have greater voltage at the LEDs, causing them to draw more current.
You'll have to change some aspect of your design to keep the LED current within their ratings, I suspect.
You'll have to change some aspect of your design to keep the LED current within their ratings, I suspect.
Just did a few tests. I have to correct my first post Voltage. Here is what I got.
Battery at 12.7v
24-LED / 25ft wire
9.65v at LED
3.9A draw
24-LED / 50ft wire
9.24v
2.6A draw
12-LED / 25ft wire
10.4v
3.1A draw
12-LED / 50ft wire
9.76v
2.06A draw
Does this make sense? I need to get the 12LED down to around 9.6v.
Last edited:
Steve K
Flashlight Enthusiast
Looks about right to me. Based on my quick calculations, shown below, you are seeing around 0.75 ohms with the 25 feet of wire, and around 1.4 ohms for the 50 feet of wire.
case 1
voltage drop across wires = 12.7-9.65=3.05
current: 3.9A
wire resistance: 3.05/3.9 = 0.78 for 25 ft wire
case 2
voltage drop across wires = 12.7-9.24 = 3.46
current: 2.6
wire resistance: 3.46/2.6 = 1.33
case 3
voltage drop across wires = 12.7-10.4 = 2.3
current: 3.1
wire resistance: 2.3/3.1 = 0.74
case 4
voltage drop across wires = 12.7-9.76 = 2.94
current: 2.06
wire resistance: 2.94/2.06 = 1.43
case 1
voltage drop across wires = 12.7-9.65=3.05
current: 3.9A
wire resistance: 3.05/3.9 = 0.78 for 25 ft wire
case 2
voltage drop across wires = 12.7-9.24 = 3.46
current: 2.6
wire resistance: 3.46/2.6 = 1.33
case 3
voltage drop across wires = 12.7-10.4 = 2.3
current: 3.1
wire resistance: 2.3/3.1 = 0.74
case 4
voltage drop across wires = 12.7-9.76 = 2.94
current: 2.06
wire resistance: 2.94/2.06 = 1.43
Thanks..I have a question about the current. What way is the current calculated/ added up? Say each led is rated 700ma. Is the load calculated across the parallel string or series? For instance with the 25ft of wire I was at 3.9A. Why?
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Steve K
Flashlight Enthusiast
If you have 8 parallel strings of 3 LEDs (i.e. each string is composed of 3 LEDs wired in series), then the total current is being shared between the 8 strings. For convenience, we'll assume that the current is shared equally, but this is some difference from string to string.
Each string gets 1/8 of the 3.9A, or 0.49A. Each LED is seeing 1/3 of the voltage (approximately) that is measured across the paralleled strings, which is 9.65/3, or 3.2V.
Did we discuss the fact that the voltage and current of the LED have a unique relationship that is usually defined in a graph? Actually, all diodes have similar relationship between voltage and current, although the scaling will vary.
Here's the graph from the wiki page for diodes....
https://en.wikipedia.org/wiki/Diode#mediaviewer/File:Diode_current_wiki.png
In the part labeled "forward", the diode is conducting. When the current is small, the voltage increases rather quickly. ...or to put it another way, if you are controlling the voltage across the diode, the current increases very slowly, but then increases a lot once you hit 0.6V or so. With white LEDs, that voltage is around 3V.
Once the diode is conducting in the forward direction, the voltage does change as the current changes, but not a lot.
In your case, as you change the wire resistance, you change the current going to the LEDs, which then causes the voltage across the LEDs to change a small amount, which also changes the current.
There must be a video somewhere on YouTube where Ohm's law is described, where circuit analysis and Kirchoff's current law is described, etc. You really should have some understanding of electronics at a fundamental level to be working with this stuff.
Each string gets 1/8 of the 3.9A, or 0.49A. Each LED is seeing 1/3 of the voltage (approximately) that is measured across the paralleled strings, which is 9.65/3, or 3.2V.
Did we discuss the fact that the voltage and current of the LED have a unique relationship that is usually defined in a graph? Actually, all diodes have similar relationship between voltage and current, although the scaling will vary.
Here's the graph from the wiki page for diodes....
https://en.wikipedia.org/wiki/Diode#mediaviewer/File:Diode_current_wiki.png
In the part labeled "forward", the diode is conducting. When the current is small, the voltage increases rather quickly. ...or to put it another way, if you are controlling the voltage across the diode, the current increases very slowly, but then increases a lot once you hit 0.6V or so. With white LEDs, that voltage is around 3V.
Once the diode is conducting in the forward direction, the voltage does change as the current changes, but not a lot.
In your case, as you change the wire resistance, you change the current going to the LEDs, which then causes the voltage across the LEDs to change a small amount, which also changes the current.
There must be a video somewhere on YouTube where Ohm's law is described, where circuit analysis and Kirchoff's current law is described, etc. You really should have some understanding of electronics at a fundamental level to be working with this stuff.
