Two Circuits, One Switch

Thread Starter

Edymnion

Joined Feb 16, 2012
7
Hello all, got a project I'm building with LEDs that I've hit a bit of a conceptual roadblock on.

Backstory:
I'm having to build a lighted disk to put in a halloween mask, and my plan is to fabricate a housing with a ring of 8 LEDs pointing in, give the back wall a reflective paint job, and take a piece of fine grit sandpaper to a heavy plastic lens to frost it. With a diameter of about a soda can, I'm thinking it should provide a pretty good overall glow.

Plans:
I've found some LEDs that appear suitable that run on 2.2v each, with an upper tolerance of 2.4v. If I run four of those in serial, I should be looking at 8.8v required, up to 9.6v tolerated. Thats dead on for a 9v battery, so I shouldn't need any resistors, just straight up wiring. I'll alternate between circuit A and circuit B so that every other light is on the same battery, so in case a bit of wiring breaks I'll still have the other half evenly spaced around the ring for a bit of fault tolerance without complete failure of the piece.

What I Need Help On:
And this is almost embarrassingly simple (I hope). For ease of use, I would like to have one power switch for these. I can do two if it comes to that, but I'd rather have one switch and go. Is there a way to wire both circuits up to a single simple switch (like a regular slider or rocker) that would otherwise keep the two strings of LEDs isolated from each other?
http://www.radioshack.com/product/index.jsp?productId=2062551
 

mcgyvr

Joined Oct 15, 2009
5,394
First you ALWAYS need current limiting of some sort with LED's.. relying on the batteries internal resistance is not the right way to go (sometimes it just works, but its not "proper"). LED's really care about current not voltage.. as long as your supply voltage is above the total of the forward voltage of the LED's AND you limit the current you are fine. You would be better to have 3 LEDs in each string with your 9V battery and the proper resistor determined by R=(Vin-Vf)/I
So assuming 20mA LED's R=(9-6.6)/.02 then R=120 ohms

then you figure out the resistor wattage which is W=((Vin-Vf)*I)*2 So ((9-6.6)*.02)*2 =.096W.. (so a standard 120ohm 1/8W resistor is just fine) The *2 is a safety factor.. Its always good to size your resistor to be 2 times what you need to keep it from running too hot.

Second what you want is classified as a "DPST" switch (double pole, single throw)..very common.
 

Thread Starter

Edymnion

Joined Feb 16, 2012
7
Okay, I get you.



And lets see, a normal 9v battery has a mAh rating of 565, and I'll be running nine 10mA LEDs, so 565/90 = 6.27 hours of maximum battery life. That sounds quite sufficient for my purposes. Thanks all. =)
 
Last edited:

bertus

Joined Apr 5, 2008
20,535
Hello,

You have calculated the total current the wrong way.
When three leds are in series, the current for that string will be 10 mA.
You have 3 of these strings parallel, so the battery current is 3 X 10 mA = 30 mA.

The battery will last longer as you have calculated.

Bertus
 

Thread Starter

Edymnion

Joined Feb 16, 2012
7
Lol, I'm really showing off how long its been since I last did any actual hands on circuit design, aren't I?

Build you a working CPU out of nand gates? No problem. String together some blinky lights and I'm in over my head. =P
 
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