Hi.

I'm currently working on a project involving approximately 80 UV LED. It would be great if someone checked my math before I go any further.

This is what I have:

R = (Vs - Vl) / I

Vs = 12v
vl = 3,3v
I = 20mA

R = (12v-3,3v)/0,02A = 435 Ohm

I need a 435-ish resistor.

And now for the part I'm not quite sure I've got right:

20mA x 80 = 1600mA

1600mA / 1000 = 1,6A

I need a power supply that give me at leas 1,6 A.

Correct or not?

 

You are forgetting an important part. The draw on the power supply is dependent on how the LEDs are wired. If you have 80 of these in series, then yes, you would need 1.6A.

But, 80 of these in series would require 3.3v * 80 = 264v

264v!

So, in order to determine if what you are computing is correct, a schematic or wiring diagram will be required.

 

Thanks.

Good point.

I thought I might put one LED and one resistor, one LED and one resistor... and so on...

Is that a bad idea?

I'm working on the schematic.

 

Correct, and near enough to 14 watt dissipated in the resistors.

You would be better off running the LEDs as stings of 3 LEDs in series with 1 resistor in each string.
That way you could have 27 strings (81 LEDs), each resistor needs to be 105 Ohm (110Ohm would be near enough) and the total current will be approx 540mA with just over 1 watt lost in the resistors.

 

Nice!

How did you find those numbers?

 

Practice with the math is all it takes.

Divide your SourceVoltage by the LED VoltageForward.

The Integer part of the answer is how many LED's you can have in series.

Multiply that by Vl to get Vlseries and then use your formula with Vlseries instead of Vl. You generally need at least half an LEDs share of voltage left for the resistor. If you had Vl=3V it would not be a good idea to put 4 LEDs in series without a resistor across a 12V source because you have no real current limiting. If you had Vl=2.9V it would still be a bad idea to use 4 LEDs with a resistor across only the remaining 0.4V.
You might get away with this in a case where the source has enough series resitance built in like a pile of low amp batteries. But soon the battery voltage will drop and you would notice the LED's dimming.

It is a basic requirement that you want to allow room across your series resistor for source voltage changes, for variances in Vl and thermal effects of the LEDs, and other things.

Watts per resistor are I\(^{2}\)R
Multiply by the number of resistors (one in each series string.)

Total current is just multiplied the 20mA spec times the number of series strings.

To get the number of series string was just dividing the number of LEDs wanted by the 3 LEDs that fit in 12 Volts. 26.\(\overline{6}\) is close to 27, so 27 strings of 3 works.

 

I think a simple check is to look at the voltages.

You have 3.3V across the LED, meaning you have 8.7V across the resistors.
Both carry the same current, so your efficiency is abysmal, with your resistors dissipating 8.7/3.3 = 2.6 more power than the LEDs.

If you double up the LEDs per string, you're at 6.6V across the LEDs and 5.4V across the resistor, so now the resistors dissipate a bit less than the LEDs.

 

Thanks guys, but I'm sorry. I still don't get it.

I get 12v / 3,3 = 3,6 LEDs, and thats about it...

Feel kind of stupid here...

 

Pick a specific point you're stuck on and ask.

 

Multiply that by Vl to get Vlseries and then use your formula with Vlseries instead of Vl. You generally need at least half an LEDs share of voltage left for the resistor. If you had Vl=3V it would not be a good idea to put 4 LEDs in series without a resistor across a 12V source because you have no real current limiting. If you had Vl=2.9V it would still be a bad idea to use 4 LEDs with a resistor across only the remaining 0.4V.
You might get away with this in a case where the source has enough series resitance built in like a pile of low amp batteries. But soon the battery voltage will drop and you would notice the LED's dimming.

This part I don't get...

Thanks for your time.

 

You got as far as 3.6 LEDs fitting in the 12 Volts. So that means 3 Leds. The .6 of an LEDs voltage is going to be the room for your current limiting resistor.

To get Vlseries you take Vl of3.3V x 3LEDs = 9.9V

With Vlseries you can treat your string of three LEDs as a single LED of 9.9volts.

Rseries = (12V-9.9V)/20mA which is the 105Ω and to keep your currents at or below your target you want to pick a 110Ω from the standard resistor values.

 

Now I get it.

Thank you so much, I appreciate it a lot!