I experimented with resistor values untill i got to 17.7mA as measured with my multimeter, at that current if i looked directly into the LED then looked around the room, i had a red dot in my eyesight so i didn't go any brighter , that was with the red LED.

Well, that’s not right. If you stared at a red LED, you should be seeing the complementary colour, bluish-green or cyan, as the after image.

 

Well, that’s not right. If you stared at a red LED, you should be seeing the complementary colour, bluish-green or cyan, as the after image.

Maybe it was a bluish colour, i can't remember, i do know i couldn't see properly for a full minute.

 

LEDs come in diffused types (colored body) and non-diffused types (crystal clear body). As you've found, non-diffused types can be blindingly bright. For indicators, diffused types are best (easiest to look at).

 

What are you talking about?

Sorry. This comment belongs on a different thread. I'm deleting my mistake.

 

Shouldn't stare into a non-defused LED. The brighter the more harm they can cause to your eyesight. It probably won't be instantaneous but the harm can be cumulative.

 

Shouldn't stare into a non-defused LED. The brighter the more harm they can cause to your eyesight. It probably won't be instantaneous but the harm can be cumulative.

Yeah i know, i was just curious as to how bright the thing was...it's BRIGHT!!!

 

it's BRIGHT!!!

That's why there's no automatic rule of thumb to always supply "15 mA" or use "150 ohms".....

In the olden days of LEDs, 20mA resulted in a dim glow. Today's LEDs are much more efficient with 5mA often being very bright. I have some LEDs where half a mA is plenty for a very visible indicator.

 

i was just curious as to how bright the thing was...it's BRIGHT!!!

The 4 LEDs with brightness information in your original post indicate that they're all what's considered ultrabright or high output. Anything in the 1000+ mcd range is in that category. LEDs from the 1970's were typically in the less than 20mcd range.

 

I have several reels of early blue LEDs:

 

As has been said, I think the diffused variety make better indicators, I bought a box of ultra bright clear ones with the intention of building a colour organ at some point.

 

As has been said, I think the diffused variety make better indicators, I bought a box of ultra bright clear ones with the intention of building a colour organ at some point.

You can sand the lenses with fine grit sandpaper to make them more like diffused. That'll be a pain-in-the-ass, but at least you can use what you already have.

 

You can sand the lenses with fine grit sandpaper to make them more like diffused. That'll be a pain-in-the-ass, but at least you can use what you already have.

No I want the bright ones for my project, at the $10 or so for a box, I can buy more.

 

Hello,

Read the following page about leds:
https://micro.magnet.fsu.edu/primer/lightandcolor/ledsintro.html

There is a picture that shows the difference in viewangle between a clear and a diffuse led:


Bertus

 

You probably won't find this to be an issue using only 5V and less than 20mA, but it is worth checking how much heat will be dissipated in the resistor (watts of heat = V * V / R). In your example, 5V supply minus maybe 3V for the LED leaves 2V across the resistor. To get 20mA, your resistor would be 2V/20mA = 100 ohms. So the heat dissipated by the resistor is 2V * 2V / 100Ω = 40mW. That isn't very much; any through-hole (leaded) resistor and even a tiny 0402 SMT resistor would probably be fine, especially if you connect it with sizeable PCB traces (they serve as a heat sink for the resistor).

Now consider running the same LED from 12V. Now the resistor needs to drop 9V (12-3). To get 20mA, you need a 450Ω resistor. It will dissipate 9V*9V/450Ω = 180mW. A through-hole 1/8W resistor isn't going to be happy about that; you need a 1/4W part. Similarly, some 0603 SMT parts are rated for only 100mW, so you need to be careful about the part selection or move up to an 0805 package.

 

You probably won't find this to be an issue using only 5V and less than 20mA, but it is worth checking how much heat will be dissipated in the resistor (watts of heat = V * V / R). In your example, 5V supply minus maybe 3V for the LED leaves 2V across the resistor. To get 20mA, your resistor would be 2V/20mA = 100 ohms. So the heat dissipated by the resistor is 2V * 2V / 100Ω = 40mW. That isn't very much; any through-hole (leaded) resistor and even a tiny 0402 SMT resistor would probably be fine, especially if you connect it with sizeable PCB traces (they serve as a heat sink for the resistor).

Now consider running the same LED from 12V. Now the resistor needs to drop 9V (12-3). To get 20mA, you need a 450Ω resistor. It will dissipate 9V*9V/450Ω = 180mW. A through-hole 1/8W resistor isn't going to be happy about that; you need a 1/4W part. Similarly, some 0603 SMT parts are rated for only 100mW, so you need to be careful about the part selection or move up to an 0805 package.

You don’t need to calculate the value of the resistor to find the wattage.

P = I x V

20 mA x 2 V = 40 mW

20 mA x 9 V = 180 mW