LED Lighting Project ~ Advice & Guidance

Thread Starter


Joined Oct 7, 2014
Evening All!

I'm undertaking a project in a field I've never embarked before and it's totally alien!

Past few days I've been reading up on the Direct Current Textbook who have available on the site.

Fantastic by the way! Written in a manor I can understand especially someone with learning difficulties.

Learning off my on back is difficult and trying to decipher what's being said I find difficult to comprehend and understand. However when taught by another is where I learn quick and able to grasp what's being taught when shown. Also, doesn't help when I have dyscalculia attempting to workout what resistors I need :D.

I'm familiar with wiring and soldering having done numerous guitar rewiring's but this is a different ball.

On to the project!

What I'm attempting to make is two 12vdc LED circuits with two LED On/Off switch's and one dimmer. The first circuit is going to be a strip of UV LEDs to illuminate my computer and powered by the PSU that's wired to an LED off on switch. My first incline was to use purple LEDs however after research purple LEDs appear more white then purple and UV LEDs have a stronger purple/violet light with the bonus of UV glow.

The second circuit is comprised of 30 or more warm white LED's as a monitor back light with the same setup as above with the addition of a dimmer. The source I don't know yet if its viable to run another 12vdc from the PSU.

I would like both circuits dimmer and switches to be on one box/house with the 3 switches. Something I'll make myself I have a ton of plexiglass and tools to make something fancy to house the switches/dimmer.

For now I'll just concentrate on the first LED circuit and when you fine people make it all clear to me I'll attempt to make the second one with little assistance to see if I've learnt anything.

I would like to make a 24 UltraVoilet LED 12vdc circuit with a LED switch.

The LED's I've purchased are:

Size: 5mm Round
Brightness: 3000mcd
Forward Voltage: 3.2v - 3.8v
Forward Current 20mA (typical) 30mA max.

Wavelength: 395nm

If my maths is right, 24 LEDs will need to be wired in series of 3 (3 x 3.8v = 11.4v) which makes 8 series's that will be wired in parallel to the source. The source is a 12v rail from my computers power supply. I have looked at buying premade LED Strip/Tape but where's the fun in that?

This is the schematic I have come up with.

24 LED Schematic ~ EDITED.png

Now I've read on Omhs Law how to work out what resistor I need but it still doesn't make sense to me.

I'm not sure if this is the correct method of working out the required resistor but I'll give it a shot.

The LED consumes 3.8v of the 12v supply and forwards 8.2v to the next LED which consumes another 3.8v forwarding the remaining 4.4v which leaves 0.6v to return to the positive?

The 0.6v left over is what I need to resist or soak up before the current enters the LED series?

So if each LED is rated between 20 & 30 mA so lets call it 25 mA for headroom, do I divide 25/1000 by the source volt which is 12?

(3.8 - 12) / (25/1000) = 328 Ω

As I understand it, that resistor is for one LED? So to calculate the resistor needed for three LED's I would just divide the remaining voltage by the current which is 25mA.

So 0.025 / 0.6 = 24 Ω

By that maths, I would need 8x 24 Ω resistors?

I feel I'm about to have the chalk thrown at me haha!

I'm not sure how to include the switch which is a 3A/250VAC push button switch with a 12v LED. As for the junctions I'm not sure if that's the correct way of connecting 8 wires to one wire I thought it would be neater method of wiring up the circuit. Wouldn't 8 wires directly connected to one terminal cause heating issue?

Major thanks in advance, looking forward to getting this project under way and learning about basic circuitry.


Joined Apr 5, 2008

The value for the resistor will be (Vsupply - (n X Vled)) / Iled.
You can use 3 leds in series.
Use the least voltage for the leds, this will give the maximum current.
The current limiting resistor will be ( 12 - ( 3 X 3.2 ) ) / 0.025 = ( 12 - 9.6 ) / 0.025 = 2.4 / 0.025 = 96 Ohms.
The closest E series value for the resistor is 100 Ohms.
The current will be 2.4 / 100 = 0.024 A = 24 mA.
If the leds would have a voltage drop of 3.8 Volts, the current would be ( 12 - ( 3 X 3.8 ) ) / 100 = ( 12 - 11.4 ) / 100 = 0.6 / 100 = 0.006 A = 6 mA.


Thread Starter


Joined Oct 7, 2014
Hi and thanks :)

Made it allot clearer and I was able to follow most of it however the current part of the equation.

Why wouldn't the current be 25 mA and instead 24 mA? Does the remaining voltage (2.4v) have to match the current?

This is the updated schematic.

24 LED Schematic ~ EDITED.png

I've tried to work out what resistor wattage I would require. From research you can work it out by using this formula.


I=0.024 mA

(2.4 x 0.024) = 0.0576w ~ 56.6mW
= 1/8 Resistor ~ 0.125w

Is this correct?

My only concern now is how do I add the LED switch in the loop? I've read I need not worry about the Amp from the switch however there isn't much information about the LED in the switch other then just connect a 12v wire to it.

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