My LED's are dim, how do I get them brighter?

Question

I installed 9 LED's into my car with a wire run from the battery.


Battery---15A fuse---20A switch---LED---LED---6LED cluster---LED---Ground


(Sorry about the text diagram) That's how I did it. There's 20 guage wire in between each LED and the switch (it was all I had). I tested the switch with my multimeter and got nothing. I did a cross check to see if the meter worked with the battery and it read normal. The lights are so dim I almost couldn't see them at 10:00pm. Am I running to large a wire? Am I running too many lights? Am I just screwed?

Answer 1

You don't need heavier wire. You need a different wiring method as you are stretching them too thin. If you have resistors ANYWHERE already in the circuit, you must consider them when you go to the site.

Go here:

http://led.linear1.org/led.wiz

This calc will show you exactly how to wire up what you want. You will need to know some specs about the LED's. Otherwise, you can guess with the voltage being 3, current being 25.

I hope you know that LED's are polarity sensitive meaning they have a (+) and (-) wires. The shorter lead (flat side of the LED) is the (-) pole.

http://www.eleinmec.com/figures/006_02.gif

...and yes, the current draw of most LED's is only 25mA (.025 amperes) or less, so even a 5 amp fuse is almost too big. A 1 amp fuse would be ideal.

This explains a lot:

http://www.theledlight.com/LED101.html

Some of the terms may be confusing if your not familiar with electronics.

Answer 2

well obviously you arent getting voltate to them. Im pretty sure that red leds need about 3V a peice. If you wire them in series you will get voltage drops across each one. Wire them in paraelle not series. Make sure each has there own 300OHM resistor if using a 12V source.

BTW, why are you using a 15A fuse with a 20A switch? At most 20 LEDs would draw at tops an amp.

Answer 3

the present by way of each LED is an analogous, yet that "comparable" cutting-edge is decreased using fact the whole resistance of the circuit provides as Rt=R1+R2+R3...+Rn. So the present by way of one LED is The voltage around the sequence combination divided via the whole resistance Rt V/Rt. in case you connect in parallel the voltage for the time of each LED is alway the source voltage V. and the present by way of each LED is V/R the place R is the resistance of a unmarried LED. for the reason that R < RT... the parallel subject places extra cutting-edge by way of each LED. i think of its extra handy to think of roughly from a voltage source perspective.

Answer 4

try a heavier wire..20 gauge isnt really heavy, im not sure how much juice 6 LED's drain but i guess its worth a shot