Magdave
New member
I understand what your saying but as I said too much load also causes it.
Adding a resistor lowers the brightness of the LEDs and if you have to add too much resistance they are too dim.
A little LED knowledge.
Single LEDs have a voltage rating. Some as low as 1 volt and some as high as around 3.5V or more. It varies based on internal resistance.
So if you hook a single LED to 12V it's going to pop. Thus you add a resistor and calculate it correctly so the voltage across the LED is correct.
Now LED strips work a little differently. Lets say each LED is rated at 2.5V. If you hook 5 of these in series the voltage across each then becomes correct. Ohms law V = IR. Now you can hook this strip directly to 12V and not need a resistor as the voltage across each LED is about correct.
So how do you get say 20 LEDs in a strip to work?
Well obviously if you hook 20 in series the voltage across each will be so low they may not even light!
So what you do is hook 4 strips of 5 LEDs in parallel. Now each set of 5 LEDs is correct and all LEDs light the same.
BUT WAIT? Now the current (I) goes up! This can also trigger the flasher unit to fast flash. If current is too high it fast flashes. If it is to low it fast flashes (like a bulb is burned out). Adding a resistor "may not" always fix too high a current draw without dimming the LEDs beyond what is useful. Again this is why some make a reverse load device to solve this (it's basically another flasher in the circuit powered by another 12V line).
I mean think about it. You can't continue to keep adding as many lights as you want as power is needed to drive them. Go too much and the flasher freaks out. Yes in general a resistor can help but sometimes depending on what your doing, you cant get there with just a voltage drop across a resistor.
Bob
In your example you would require a load resistor for multiple sets. Simplest way to solve that issue is with a isolation relay to a fuse block that drives your additions.. In parallel ckts the resistance total actually drops as 1/R1+ 1/R2+1/R3 and take the reciprocal of that total to determine the RT. For example 2, 3 ohm resistors in parallel will give you a 1.5 ohm RT. In series that would be 6 ohms.
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