Wireless sensor bar (this will work out right?)

ghosstt · Post by **ghosstt** » Sat Nov 03, 2007 4:14 pm

If i get 4 of these

Link

"High output infrared LED 5mm 1.2VDC 29mA 940nm"

1.2 volts a pop, right?

4 * 1.2 = 4.8 volts in total

a AA is 1.5, so

4.8 / 1.5 = 3.2

so would i just use a 4 AA batterholder, wire one end to a resistor and then resistor to an LED? then LED to LED?

http://doctabu.livejournal.com/64758.html

He used 4 AA's, and 4 LEDs. is 6 volts to much for them to handle?

Post by **bicostp** » Sat Nov 03, 2007 4:22 pm

Why not use rechargeable AAs? They're typically 1.2 volts apiece...

4.8 / 1.2 = 4

Rechargeables are cheaper in the long run, anyways.

ghosstt · Post by **ghosstt** » Sat Nov 03, 2007 4:31 pm

well, i have a set of 4 rech AAs, and they are 1.5 a pop, but i have AAAs that are 1.2v a pop, 700 mAh. How bout if i did something like,

Switch-------------------------(*)--(*)---
|<---red wire ^positive led leg |
| |
| |
batts--------------------------------------<---black

would that work out right?

Post by **Skyone** » Sat Nov 03, 2007 4:46 pm

ghosstt wrote:well, i have a set of 4 rech AAs, and they are 1.5 a pop

What type of batteries are those?

Kurt_ · Post by **Kurt_** » Sat Nov 03, 2007 5:14 pm

As for the actual LED, they will most definitely work. A small red 3mm (possibly 2mm) LED light on my Receiver outputs enough light to be picked up by the Wiimote. So does the lava lamp beside the TV, which is green.

Guess what? I got flux that doesn't conduct electricity today. It's a milestone. Up to this point, I have been using liquid acid flux, for use with Stained Glass and Lead tubing. It is conductive, its residue is conductive, and its residue is also black and gunky. The new stuff is rosin and isopropanol. Woot! This will help me a lot in cart slot relocations.

ghosstt · Post by **ghosstt** » Sat Nov 03, 2007 5:39 pm

well, i just went and picked up 2 battery holders, and 2 LEDs. the leds are rated 1.2V, the batteries are 1.2V. PERFECT!

black wire to cathode? or red wire?

Kurt_ · Post by **Kurt_** » Sat Nov 03, 2007 5:43 pm

Cathode = Positive.

On the LED, it's the long wire and the smaller part inside the LED (the negative end hangs over the smaller positive end)

ghosstt · Post by **ghosstt** » Sat Nov 03, 2007 5:49 pm

so black goes to cathode

(black is negative on mine)

right?

vskid · Post by **vskid** » Sat Nov 03, 2007 5:55 pm

ghosstt wrote:so black goes to cathode

(black is negative on mine)

right?

Kurt_ wrote:Cathode = Positive.

On the LED, it's the long wire and the smaller part inside the LED (the negative end hangs over the smaller positive end)

Red to cathode.
Resistor(s) are still required to limit the current the LEDs take.

ghosstt · Post by **ghosstt** » Sat Nov 03, 2007 5:58 pm

if its 1.2v led, and a 1.2 AA battery, isnt it perfect?

vskid · Post by **vskid** » Sat Nov 03, 2007 6:16 pm

ghosstt wrote:if its 1.2v led, and a 1.2 AA battery, isnt it perfect?

No, it can still draw more mA than are healthy, LEDs are just suicidal that way. Just a small, 1ohm resistor should be enough, it just needs something else to restrain it from suicide.

Kurt_ · Post by **Kurt_** » Sun Nov 04, 2007 10:21 am

Some LEDs have built-in resistors nowadays.

I would recommend closer to 5-10 ohms.

ghosstt · Post by **ghosstt** » Sun Nov 04, 2007 11:37 am

well, it works. i think what ill do it put a 5 ohm resistor on it, and put a 2 AA battery pack, and wire the LEDs in a series. will the 5 oh resistor be to much?

timmeh87 · Post by **timmeh87** » Sun Nov 04, 2007 2:53 pm

If the led is rated for 1.2v, then applying 1.2v to it without a resistor should be perfectly acceptable. You can choose to set either the voltage or the current of an LED, and let the other one find the right balance. A red LED might be designed to operate at 1.2v and 20mA, which is one possible balance point.

At voltages above 1.2v, the balance current quickly approaches infinity, as the diode begins to act like an <s>open</s> closed circuit. It is wise to install a small resistor to prevent accidentally burning out your LEDs. For example, when fully charged, 2 rechargeable batteries may have an open-circuit voltage of 2.7v. Although this is only .3v higher than needed, the current could be more than twice as much, and the leds will look a lot brighter.

If you throw in a 5ohm resistor, at 20mA it will only drop .1v, so the LED will have 1.1v and will still probably work reasonably well* (slightly dimmer).

at 1.7v, an LED without a resistor might draw something like 500mA. However, at 80mA, the voltage across the 5ohm resistor will be 0.4v, which means that the LED will have 1.3v*.

Code: Select all

V     Resistor    No resistor
1.2   <20mA*      20mA
1.7   80mA*	    500mA

So you see a resistor is a good idea. 1ohm is ok, but 5 is probably better. 10 might be too much, but it would be even better. Test out some values to see what works best in your circuit.

*The LED can only take on values that are on its IV curve, which looks something like this (The green line):

http://en.wikipedia.org/wiki/Image:Diod ... Image3.jpg

You cant solve circuit equations for the diode without the graph, or an equation to define it (its complicated). I just made up values that I thought would be pretty close, for illustrative purposes.

vskid · Post by **vskid** » Sun Nov 04, 2007 6:17 pm

Ah, yet another informative post by timmeh.
That makes sense, I couldn't understand why LEDs would draw more current than they needed.