Electrical question.

[bloodhound31]

I have some LED's that ran off three AAA batteries in an old torch.

I've pulled them out of the torch, de-soldered them from the round circuit-board (Pictured below) and am now running individual wires to separate them for another application.

Do I run a positive and negative from the battery to EACH LED, or do I run a positive to the first one, then join them all up in a row, like one circuit in one end of the LED, out the other end, into the next etc?

I forget all the terms.... was it parallel or series?

Which option in the picture is it?

Baz.

[bojan]

It is probably the combination of those :
2 in series, then two of those sets in parallel.

This is because Red LED has ~2V forward voltage, so 2 of them in series will have 4V
3 batteries are 4.5V.. just enough for two LED in series. I would recommend resistor in series of both parallel sets, to limit the current (when batteries are fresh)

For more details see here:
http://www.kpsec.freeuk.com/components/led.htm

[bloodhound31]

Thanks mate.

How come the torch doesn't have a resistor in it?

[bojan]

Lower cost..
current limitation relies on internal resistance of batteries, and voltage close to threshold voltage of LED's.. but it is not good for LED's working life, because higher currents at switching on will (slowly) cause deterioration of LED's

[bloodhound31]

Right, so what's a resistor, what does it look like, where do I get one, how much does it cost, and how do I connect it in? Oh, and I am crap at formula so calculating the resistance needed is beyond me....

Working my way towards 20 questions... ;)

EDIT: I found a 1K resistor online, brown, black red and gold bands. Now to look around in my shed for something old and electronic that I can pull one out of.....

[bloodhound31]

So, the anode (a) is the little element inside the LED and the cathode (k) is the larger one?

Positive to resistor then to anode, negative to cathode?

I found an old Nokia phone charger and busted it open thinking there might be a 1k resistor inside. Is this one? The printing on the circuit board reads R1.

[PCH]

Hi Baz,

that resistor is 20 ohms ...

see here http://www.csgnetwork.com/resistcolcalc.html for the colour coding.

So you need brown black red for a resistance of 1k.

Just to point out how it works. In your pic, you start at the red end, or actually the end away from the gold band in this case, and you'll see three colours, then gold or silver, then the last colour means something else but you don't need to worry about that.

So you have Red, Black, Black.

Just write down the digits, instead of the colours, from the chart in the link I supplied. So counting down, red = 2, black = 0, next black also = 0.

You write the number (resistance value) down left to right as follows. The first two bands represent a part of the numberical value of the resistor, then the third band is the number of '0's to add to the first two numbers you just wrote down.

Therefore, in this case, write down a '2' and a '0', and the third band which is also black (number zero) means add zero '0's to the 20. So the value is just 20 ohms. If the third band were a red, it would mean add two zeros to the 20, - therefore 20+00 = 2000, or 2k ohms.

Hope that helps ;)

Paul

[bloodhound31]

Oh ok, thanks mate. I'll look around for something else electronic in the shed to bust open.... :P

[PCH]

They're only a few cents from the shop if you have to purchase :thumbsup:

[bloodhound31]

Which end does the positive go in?

[bloodhound31]

So positive through the resistor to all the anodes of the LED's, running all the cathodes to the negative end of the battery?

Does that sound like it would work?

[bartman]

Barry ....you crack me up!!!!!!

The man who set to task
a shed and mask

oh my, a LED to shine
.......that would be devine

To power the shed and mask
A mystical power source needs to be had

Resistance is the answer
not the power

Listen to others
for I have no idea

:lol:

[PCH]

Correct Baz :thumbsup:

[bloodhound31]

Right, well, I did all that - only to find that adding the resistor dims all the LED's down to absolutely useless.

So, I removed the resistor and hooked it up to four LED's. Perfect and as bright as they were in the torch. Then I hooked it up to the remaining four that were in the torch also. Perfect again. Looks like the torch didn't need a resistor, so just because they are joined now via wires instead of a board, it seems it doesn't necessarily mean they need a resistor all of a sudden.

[Shark Bait]

Electronics is good fun Baz. I have just started down this path with some other projects and have had to learn about resistors etc.

My book shows resistors have been used in projects that are similar to the circuit you have put together. As mentioned before, they are cheap and protect the LED's from burning out. Jaycar sell a packet of 8 for approx 40 cents. The simple circuit that I have in front of me at the moment shows a 6V power supply with positive running through a 470 ohm (yellow, violet, brown, gold) resistor to a flashing red LED, which then runs back to negative.

I don't have an answer for your question at the moment and I am not up to speed with the calculations either. I am curious now. Tonight I will try to copy what you have done and experiment with the resistors that I have here. I will let you know what I come up with.

[Scorpius51]

Looks like you are using 'white' (actually more blue) LEDs that have a forward voltage of ~ 3.5V. So you can probably run them in parallel without a current limit resistor. But, as Bojan stated earlier, you may limit their life. You will also drain you batteries more quickly. Perhaps you should use at least a 50 Ohm resistor (or 51 or 47 Ohm preferred values). If you're using the batteries in series (4.5V total), then you want to drop nearly 1 volt before dumping the rest across the LEDs.

Cheers
John

[bloodhound31]

Thanks everyone for the education. I'll be needing it a lot more in future for the rest of the Iron Man suit (http://www.iceinspace.com.au/forum/showthread.php?t=94432&page=8).