How would i wire 3 LEDs together

I am trying to make something, and i want to wire 3 leds together so that i have 3x brightness of course. I would be using 6v of energy...

can anyone help?

Sure, but we'll need a bit more information. You are using white LEDs (or what color)? Normally, you need to put them in series (so the same current goes through each in turn, causing 3 times the total light), but this will raise the total operating voltage too high.

So what are the details? It may well be that the best idea is to drive each of the three (or any other number for that matter) as an individual (at the expense of more battery power, of course).

Anyway, can you give us a few more clues?

Doug Owen

well... lets see how well i can explain this...

Say i have 3 leds... and for the matter of this discussion, 1 6v battery (or a power source of 6v)

How would i put the leds together? Would i make each positive (+) and each negative (-) go to that proper pole on the battery? or could i put one positive of the led to the battery, then the next positive to the negative of the current led, and the last positive to the negative of the 2nd led, and then the negative of the last led, to the negative on the battery... (i hope you understand what i mean)...

I would want to use different colors of lights, mainly blue, green, and possibly yellow/orange and white..

my source of power would either be... 4 AA batteries. --OR-- two 3v lithium watch batteries (like those found in the photon light)

Thanks for the additonal information. In this funny world of ours, six Volts from AA cells is not the same as from two 'coin cells' to us. This is because the coin cells are more self limiting in terms of their ability to deliver true power (current). It's quite possible to run white LEDs directly from two such cells, red ones from a single one, which is what most 'keychain' lights are. Just hook the plus LED lead (the long one) to the plus of the battery and the minus to the minus. I'll be very bright at first, but it should work OK.

The more general case needs a resistor to limit current. With AA cells, I'd suggest 47 or 51 ohms. This should give you a modest overdrive (30 mA) with fresh cells, but should be fine. Understand that it will dim out after a farily long time (days). And it won't be easy to get the last little bit out of the battery as alkalines aren't dead until .8 Volts or so each.

But I suggest 47 ohms for each LED. Battery plus to resistor. Other end of the resistor to LED plus. LED minus to battery minus. Put the next LED and it's resistor in parallel with the first, and the next....

You can change the resistor value up or down to change the current of course.

Doug Owen

It depends on the LED, each color LED has a different operating voltage - usually higher for shorter wavelengths, ie 2.2v for RED and 4.1v for UV

If I assume you are using WHITE, then they will need approx 3.2v, so you can't wire more than one per group

Good old ohms law to the rescue:

Assume

LED Voltage 3.2v
LED Current 20ma

R=E/I
R=(6-3.2)/.020
R=2.8/.020
R=140 Ohms

If you were to use RED LEDs @ 2.2v Each, then you could wire up two in series

R=E/I
R=(6-4.4)/.020
R=1.6/.020
R=80 Ohms

Hmmmmm... I have not a clue about what you said...

Here is why i ask...

Recently i bought a light, that has 3 LEDs in the head. At the time of purchase, there were 4 different colors available. Red, Green, Blue, and White. Each light, only required 4 AA batteries. This i guess, is where I am confused. How does each color run on 4 AA batteries.

by using coin cells, i was going to try and imitate the light, only condensing it, not having to have a long tube to use the 4 AA batteries... you know what i mean?

http://www.eknifeworks.com/webapp/eCommerce//product.jsp?A=&range=21&Mode=Cat&Cat=15&SKU=CTT7934

this is the light i bought, but you cant really tell what it is.

what it is is a clear rod in which the 3 leds shine down to make the rod appear as a "solid rod of 'clear' light"

what i want to do is make one of my own, just only smaller, like i said above, getting rid of the space the 4 AAs take up...

and by the way, what do you mean by wiring it in parallel

and for the resistors, could i use the calculator on ledsupply.com to help me out???


Thanks for all of your help guys, i hope i can figure this out

LEDs are current driven. The current passing through a LED is directly related to the voltage across the LED. eg,for white LEDs, at about 3.6V, you will have 20mA passing through the LED and at 4V, the current will be about 30mA. The 4V, 30mA combo is the maximum recommended values for white Nichia 5mm(size) LEDs.

Resistors in your application will be used to reduce the voltage going to the LED. Lets say that you want to power three white LEDs in parallel from a 6V source. 1)We know that the max V that should go through the LED is 4V.The resistor is therefore required to reduce the voltage by 2V. 2) since you will be powering 3 LEDs in parallel, the total current draw will about(30mA *3) 90mA. To calculate, you divide the difference in voltages (V in -V required by LED) by the total current in Amps. ie, 2V/.09A= 22ohms .

If the 2.2V/ 20mA numbers Inretech quoted is correct, then, as he said, you can put two red LEDs in series to get a total V requirement of 4.4v. You will need the resistor to drop the battery voltage by only 1.6V. Since the LEDs are in series, the current passing through one must pass through the other and therefore only 20mA will be required. To calculate, Vin- V required/ current. ie 1.6V/.02A= 80ohms.

Series/ Parallel. If you hook up 4 AA cells in series, you will have a total of 6 V comming off of the negative and posative ends of the line of cells. Here, the +ve of cell 1 touches the -ve of cell 2, which touches the +ve of cell 3 which (as you can see the trend) touches the -ve of cell 4. In series you add the individual voltages to get the total V. The 6V is measured between the -ve of cell 1 and the +ve of cell 4. If you hook up the same 4 AA cells in parallel, you will have all the negatives connected together and all the posatives connected together. The total output voltage will be 1.5V. The V remains the same, however, the 4 cells in parallel will last approximately 4 times as long as a single AA cell.

With LEDs, series/ parallel connections are the same. In series, you connect the -ve of the first LED with the +ve of the second which connects to the -ve of the third etc.In series, you add the individual V requirements of the LEDs to get the total V requirement. eg, 3LEDs in series will require about 4V *3 = 12V to power them.They will however only need 30 mA from the V supply. In parallel, you connect all the posatives together and all the negatives together. Here, the V requirement for the 3LEDs will be the same 4V however, the current requirement will be (30mA*3) 90mA. Makes sense?

When "experimenting" with LEDs and resistors, always MEASURE the current to make sure - you can purchase "cheap" DMV's from Harbor Freight for under $5

The LED companies design the LEDs for a specified current, overdriving them will make them brighter - but at a cost of operating time

From "Blade Runner"

Tyrell: The light that burns twice as bright burns half as long - and you have burned so very, very brightly, Roy