Illuminated switch question

**AZNmastermind** · 03-13-2012, 01:56 AM

Hello,

I am just slightly confused.

I want to use this switch:
http://www.thecustomsabershop.com/16...itch-P702.aspx

With a Seoul P4 Green LED

All powered by a 4AA Battery Holder

Now from my understanding I would need a 150ohm 1/4w resistor

But is that only for the switch? Meaning I need 2 resistors? One for the switch, One for the LED? If so then how would I go about wiring that?

**TuxedoCartman** · 03-13-2012, 03:14 AM

The LED in the switch in on a completely different circuit than the rest of the switch itself. Resistors inhibit the flow of electricity along the path that they're on, so yes you'll need two resistors: one for the main LED, and one for the switch LED.

Hopefully more experienced electricians will correct me if I'm wrong, but I believe that if you want the switch to light up all the time, splice into the positive line somewhere before the switch. If you only want it on when the saber itself is on, splice into the positive line somewhere between the switch and the main LED. Splice the negative for the switch LED anywhere along the negative line.

**AZNmastermind** · 03-13-2012, 09:34 AM

Ok cool, so if I want the switch off until I turn the blade on, my wire setup would be.....

Battery with positive line to switch, then positive line to resistor, positive line to another resistor, then to LED?

And the negative would just go from Battery to LED?

**TuxedoCartman** · 03-13-2012, 10:06 AM

Like this...

SimpleDiagram.jpg

The blue squiggly rectangles are resistors. I have no idea what size you'll need; you're going to have to figure that one out on your own. There's resistor calculators elsewhere on this site as well as other places on the internet.

**Zzan** · 03-13-2012, 10:07 AM

No.

The leads on the switch for its LED are NOT the leads on the switch that actually operate the switch. The switch will operate without the LED wired or even if the LED is blown, provided the switch leads are wired correctly.

You need to think of them as two seperate systems.

Switch LED+ to resistor to Battery +
Switch LED- to Switch lead 1
Switch lead 2 to Battery -

Activating the switch closes the connection between switch lead 1 and switch lead 2 causing the LED to light. De-activating the switch causes the connection between switch lead 1 switch lead 2 to be broken and the LED goes off.

For your main LED you do the same thing,

LED + to Resistor to Battery +
LED - to switch lead 1
Switch lead 2 to Battery -

**TuxedoCartman** · 03-13-2012, 10:27 AM

Yeah, listen to Zzan. His sounds more correct.

**AZNmastermind** · 03-13-2012, 11:35 AM

Ok cool so im starting to get it then.

Then the battery will have 4 wires connected to it correct?

Will the 6v be enough to power both LEDs?

**Zzan** · 03-13-2012, 11:57 AM

Your switch LED needs 3.3v @ 20ma so yes, 3.3v < 6v.
Your Main LED needs 4.1v @ 1000ma so yes, 4.1v < 6v.

Remember you are sending 6v to each, they are seperate circuits. The resistors you are using in each path filter down that 6v to the appropriate level for that path.

The switch LED isn't going to be a significant drain on your system. Your main LED is what will drain your batteries. Now as the batteries lose power the voltage will drop and your main LED will go out before your switch LED.

If you find that you aren't getting acceptable runtime with 4x1.5v batteries you can always upgrade to a 7.4v rechargable battery pack, though you will have to replace your resistors with appropriate replacements if you go that path (6v < 7.4v so you will need more resistance to keep your LED's from blowing.) A buckpuck is also more efficient than a resistor and just as simple to wire, they cost more but if you are using replaceable batteries instead of a rechargable battery pack it will pay for itself in a short amount of time.