Why can't I figure this out?

I have a simple switch

http://rsk.imageg.net/graphics/produ...160402w345.jpg

One connector says power, one says acc, and the third ground
(the switch lights up)

What goes where?? Lol
I tried different combos... I am trying to power an led light :)
(this is a fun project for my son)

So I have the battery
I hook POS to power
NEG to one of the LED wires... then what?

Thanks!

OK, here is a little secret. If you extend the flipper, you energize the contacts that the other end points too.

So, the middle is a common. The ends are connected to common based on the position of the switch.

In your picture, the bottom two contacts would be connected. When the switch is flipped the other way, the top two contacts will be on.

Use the middle and one end. If the position of the swich is backwards, use the other. Now you have 2 wires.

You neglected to tell use what battery your planning to use or if the LED is rated for say 12V operation. Makes a BIG difference.

Let's assume that the LED is not for a specific voltage. e.g. a 12V LED lamp.

Lets' assume that you have an LED.

That LED typically has a voltage drop across it. It is usually about 1.2V; So don't use a 1.5V battery.

First, the LED is a current device, rather than voltage. A typical current is 20 mA. Therefore a resistor must be used to limit the current to somewhere between 10-50 mA.
R (ohms) = (Voltage-1.2)/10E-3; Pick the next lower resistor.

Next, that LED may be polarity sensitive unless it's a bipolar LED or something.

A typical LED has a flat side. That side is negative.

So, you need a switch, DC supply > 1.2 V, a Resistor and a switch.

Without the resistor, you just let out the magic smoke.

Usually if the LED is in backwards, nothing bad happens, but it can.

... lmao.. OK... lost me

It's a bright blue LED that has a volt range of 3.4 to 3.8 max
I am using 2 AA's (3V)

The LED specs:

High visibility
Intensity: 8000mcd (typical), wavelength: 465nm (minimum), 467.5nm (maximum)
Viewing angle: 16°
FW current: 20mA, FW supply: 3.2V (typical), 3.8V (maximum)
http://rsk.imageg.net/graphics/produ...769735w345.jpg


OK, so the flat "wire" is negative?
I'm still confused as to what wires go where?


BATTERY + to SWITCH POWER (far left 'tab')
BATTERY - to LED -?

Or do both wires go to the switch first? LOL

Thank for helping!

Funny, I can wire light switches, ceiling lights, outlets, etc... but I can't figure this one OUT! LOL

won't work as you found out. 3V is less than 3.2 V, so it won't light.

Use a 9V battery, Get a 9V battery and clips.

R=(9-3.2)/20e-3; that's a 290 ohm resistor, so use a 220 ohm. A 1/4 W resistor will be fine.

Connect it to the battery in the correct polarity and it will light.

Actually I already tested, the LED lights fine with 3.0 v (I know.. odd) and its bright too!

But I will try the resistor


THANKS!

Okay, let me see if I can explain this. The part where you are getting confused at, is that the switch is not just a switch, it is also a light. I'm no expert in low voltage wiring, but I believe this is what you are aiming for:

From the positive on the battery, run a wire to the "power" tab on the switch. Now, run a wire from the "acc" (accessory) tab on the switch, to the positive (anode) side of the LED.

http://www.pyrouniverse.com/show/fir...ext/LEDS01.gif

Now, from the negative (cathode) side of the LED, run to the negative on the battery. At this point, the switch will turn the LED on and off. If you want the switch to light up all of the time (since it is a lighted switch) you simply need to add a wire from the "ground" tab on the switch to the negative on the battery (this means that, in the end, there will be TWO wires going to the negative on the battery).

I hope that makes things clear for you.