LED Poi (Simple LED circuit advice required)

G33k · 11-19-2003, 07:30 AM

Okay, I realise this isn't case related, but it seemed like the closest to what I wanted to ask. Basically I want to make some LED Poi a little like these, but longer and brighter; I want ~20+ cm rods of light.

I was thinking of powering 4 x Super Bright LEDs per POI, but I'm not much of an electronics wiz. From what I've read, I can power these if I run them in parallel off 3xAAA NimH Batteries. The LEDs I've ordered can be seen here. Now I gather I need a current-limiting resistor, but I'm not sure how to calculate the value... can anyone shed some light on this for me / suggest a better way of doing this?

Many thanks in advance

bigben2k · 11-19-2003, 09:09 AM

I've done the "LED thing" a few times...

First, you have to identify the forward voltage. In this case, it's 3.4, with a max of 4.0.

Then you have to size your supply, and this bit is going to be a bit tricky here: Ni-Cads usually drop off from 1.5 to 1.2 pretty quickly, then drop gradually to 0. I cna't remember how Ni-mHs behave, so you might have to do a little digging.

They are however, AAA batteries, so I'll work with the assumption that they provide 1.5 volts; you can figure out the details on your own.

The LEDs are also rated for 20 mA typical (30 max) current.

So the first thing is to figure out what voltage is left for the resistor. Assuming that you'll be shooting for a typical setup (you can run them a llittle higher, for a bit more luminosity, as long as you don't exceed the specs), you'd have 3 * 1.5, or 4.5 volts as a supply. Subtract 3.4 for the LED, and you have 1.1 volts going through the resistor.

Current stays the same, at 20 mA (0.02 amps). Using V=R * I, where V is voltage, I is current, you can then calculate R: 55 Ohms. Of course you won't find a 55 Ohm resistor, so you pick the next highest value available (56 Ohm, I believe), so that you don't let more current through, than you intend.

That's it. Good luck!

G33k · 11-19-2003, 10:26 AM

Just had a quick check and the batteries say 1.2v on the side. Some fully charged ones read at 1.3v on a multimeter.

so, 1.3x3 = 3.9v
and 3.9-3.4v = 0.5v over the resistor right?

then R = V/I = 0.5v / 0.2A = 2 Ohms?

Seems simple enough, but don't you have to do something when you use 4 instead of 1 LEDs? Or is that 2 Ohms per LED?

bigben2k · 11-20-2003, 09:20 AM

You got it.

If you have multiple LEDs, you have to determine if they're in series or in parallel.

If in parallel, the voltage stays the same, but you'll want to add 20 mA per LED, so if you have 4 LEDs for example, you'd recalculate for 80 mA.

If in series, the current stays the same, but you add your voltages. If you had 4, that would be 4 * 3.4 = 13.6, which you can't supply with three AAAs.

With such a low difference, you could run the LEDs without any resistor (as long as the max voltage isn't exceeded).

G33k · 11-20-2003, 09:22 AM

Okay, that sounds good to me - I'll have a try when the LEDs arrive with 4 in parallel. Thanks very much for the help

I'll let you know how this all turns out - should be pretty good if all goes to plan!

G33k · 11-25-2003, 09:42 AM

Right, the LEDs arrived last night and a quick test with some AA batteries (still waiting for the AAAs). The batteries (3 in series) gave me a little over 4v when fully charged and I was getting 0.5A through the four LEDs. So that's ~125mA per LED which is WAY too much (they were DAMN bright though!

).

The thing is, 0.5v / 80mA is only 0.625 Ohms if I'm not mistaken. That's quite a small value for a resisitor isn't it? Maybe I'm missing something here - wish I'd listened more in my electronics lectures now :/

How can I bring the current down?

Groth · 11-25-2003, 10:09 AM

0.5V / .080A = 6.25 Ohms. 5 Ohms will work, and will be easier to find.

G33k · 11-25-2003, 12:33 PM

Thanks for that - I'll give it a go...

G33k · 05-23-2004, 11:37 AM

Resurrection time

Here's how they turned out :

Pretty sweet I think you'll agree, and they seem to work fine with 3xAAA batteries and no resistors! I was going to add the resisitor, but got bored waiting for my order to arrive so tried it without.. It works fine so I left it as it is

I'm moving onto the next stage in my project now thoough - just ordered a PIC microcontroller + programmer

11-19-2003, 07:30 AM	#1
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	LED Poi (Simple LED circuit advice required) Okay, I realise this isn't case related, but it seemed like the closest to what I wanted to ask. Basically I want to make some LED Poi a little like these, but longer and brighter; I want ~20+ cm rods of light. I was thinking of powering 4 x Super Bright LEDs per POI, but I'm not much of an electronics wiz. From what I've read, I can power these if I run them in parallel off 3xAAA NimH Batteries. The LEDs I've ordered can be seen here. Now I gather I need a current-limiting resistor, but I'm not sure how to calculate the value... can anyone shed some light on this for me / suggest a better way of doing this? Many thanks in advance __________________ "Success is 99% Failure"

11-19-2003, 09:09 AM	#2
bigben2k Responsible for 2% of all the posts here. Join Date: May 2002 Location: Texas, U.S.A. Posts: 8,302	I've done the "LED thing" a few times... First, you have to identify the forward voltage. In this case, it's 3.4, with a max of 4.0. Then you have to size your supply, and this bit is going to be a bit tricky here: Ni-Cads usually drop off from 1.5 to 1.2 pretty quickly, then drop gradually to 0. I cna't remember how Ni-mHs behave, so you might have to do a little digging. They are however, AAA batteries, so I'll work with the assumption that they provide 1.5 volts; you can figure out the details on your own. The LEDs are also rated for 20 mA typical (30 max) current. So the first thing is to figure out what voltage is left for the resistor. Assuming that you'll be shooting for a typical setup (you can run them a llittle higher, for a bit more luminosity, as long as you don't exceed the specs), you'd have 3 * 1.5, or 4.5 volts as a supply. Subtract 3.4 for the LED, and you have 1.1 volts going through the resistor. Current stays the same, at 20 mA (0.02 amps). Using V=R * I, where V is voltage, I is current, you can then calculate R: 55 Ohms. Of course you won't find a 55 Ohm resistor, so you pick the next highest value available (56 Ohm, I believe), so that you don't let more current through, than you intend. That's it. Good luck! __________________ WBTA (Water Block Testing Alliance) Nordic Hardware WC 101

11-19-2003, 10:26 AM	#3
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	Just had a quick check and the batteries say 1.2v on the side. Some fully charged ones read at 1.3v on a multimeter. so, 1.3x3 = 3.9v and 3.9-3.4v = 0.5v over the resistor right? then R = V/I = 0.5v / 0.2A = 2 Ohms? Seems simple enough, but don't you have to do something when you use 4 instead of 1 LEDs? Or is that 2 Ohms per LED? __________________ "Success is 99% Failure"

11-20-2003, 09:20 AM	#4
bigben2k Responsible for 2% of all the posts here. Join Date: May 2002 Location: Texas, U.S.A. Posts: 8,302	You got it. If you have multiple LEDs, you have to determine if they're in series or in parallel. If in parallel, the voltage stays the same, but you'll want to add 20 mA per LED, so if you have 4 LEDs for example, you'd recalculate for 80 mA. If in series, the current stays the same, but you add your voltages. If you had 4, that would be 4 * 3.4 = 13.6, which you can't supply with three AAAs. With such a low difference, you could run the LEDs without any resistor (as long as the max voltage isn't exceeded). __________________ WBTA (Water Block Testing Alliance) Nordic Hardware WC 101

11-20-2003, 09:22 AM	#5
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	Okay, that sounds good to me - I'll have a try when the LEDs arrive with 4 in parallel. Thanks very much for the help I'll let you know how this all turns out - should be pretty good if all goes to plan! __________________ "Success is 99% Failure"

11-25-2003, 09:42 AM	#6
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	Right, the LEDs arrived last night and a quick test with some AA batteries (still waiting for the AAAs). The batteries (3 in series) gave me a little over 4v when fully charged and I was getting 0.5A through the four LEDs. So that's ~125mA per LED which is WAY too much (they were DAMN bright though! ). The thing is, 0.5v / 80mA is only 0.625 Ohms if I'm not mistaken. That's quite a small value for a resisitor isn't it? Maybe I'm missing something here - wish I'd listened more in my electronics lectures now :/ How can I bring the current down? __________________ "Success is 99% Failure"

11-25-2003, 10:09 AM	#7
Groth Cooling Savant Join Date: Mar 2003 Location: MO Posts: 781	0.5V / .080A = 6.25 Ohms. 5 Ohms will work, and will be easier to find.

11-25-2003, 12:33 PM	#8
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	Thanks for that - I'll give it a go... __________________ "Success is 99% Failure"

05-23-2004, 11:37 AM	#9
G33k Cooling Neophyte Join Date: Sep 2002 Location: Bristol [UK] Posts: 73	Resurrection time Here's how they turned out : Pretty sweet I think you'll agree, and they seem to work fine with 3xAAA batteries and no resistors! I was going to add the resisitor, but got bored waiting for my order to arrive so tried it without.. It works fine so I left it as it is I'm moving onto the next stage in my project now thoough - just ordered a PIC microcontroller + programmer __________________ "Success is 99% Failure"

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)