TibFreak

Don't know if this would be the proper forum or not, so feel free to move it.

I am in the process of making my little boy a "cell phone tower" for his little town/scene I am making him to play with(Example pic provided below). It will have the flashing red light and all the red lights as a real one does. I am just unsure of how I should be wiring it. I know it needs to be wired parallel via the Chat support guy from PartsExpress.com, but when trying to explain it, he talked circles around me and I didn't soak in anything really.



i also will provide links of the LED's. This is all something that is probably cake to do, but I just don't understand any of it really.

8 of these LED's -------> http://www.parts-express.com/pe/showdetl.c...tnumber=070-015


Not sure which of these 2 yet ----------> http://www.parts-express.com/pe/pshowdetl....tnumber=070-026
for the flashing light on top. ----------> http://www.parts-express.com/pe/pshowdetl....564&scqty=1

I also have all the proper resistors needed for all of these according to PartsExpress guy.

So how I do this? Any diagram on how I should be doing it would merit a Chocolate chip cookie, maybe.

01tibby

what is your power source, voltage wise?

TibFreak

Sorry, forgot to mention that. It is 9volt, unless someone nixes that one, but I have been told it will be sufficient.

01tibby

yeah, it will probably work.

with the eight main ones, you can wire 4 of them in series, and then connect them to the battery. so,

battery + to long lead of one led ->short lead of another led. do this so that you have 4 leds wired together, and then the short lead of the last led needs to go to the - of the battery.

do the same for the other set of 4 normal ones, and then connect them to the battery in the same way together(in parallel) with the other set.

the blinking led depends on which led you go with. looks like you can connect the blinking one directly to the 9v battery with the other ones. the red tinted one will need a resistor.

this setup will drain a 9v battery semi-quickly. i'd guess like 20 minutes or so(without doing any math) of decent brightness.

TibFreak

20 minutes? That suxors.

Any suggestion of what power source to use? I want something that would last at least a week or so if possible.

01tibby

well, let's do math...

an alkaline 9v battery capacity is 565mAh according to wikipedia. the 8 red leds draw 30mA max, the clear blinking one is 70mA.

so that's 310 mA total draw. so maybe more like an hour or two(very rough approximation). brightness will decrease the longer it is left on as voltage decreases.

you can get a simple 9v power supply, just make sure it's capable of 310mA draw. you can also wire up a couple 9v batteries in parallel (like leads connected together). every additional battery will just add to the overall capacity.

TibFreak

Any body remember what those batteries was(don't know if they still make them) that looked like a huge 9 volt battery sort of, and a flashlight top hooked to the top of it. My dad used to have one, can't remember how many volt it was, but I was thinking that might be a little better for my needs.

I also thought about powering it from a wall outlet too, by wiring some kind of plug-in to it.........would that be possible?

01tibby

yeha man, just like a cell phone charger, but make sure it's rated for 9v at around 300 something mA. those big batteries would work provided they are 9v.

Stocker

Most LEDs will be plenty bright enough with less than max. current. If you run them at 10 or fewer milliamps each, you'll get way more run time and also not blind yourselves. There are a sh*t-ton of novice-level LED tutorials out there, and you can wire them up about a jillion ways if you can figure out how to do it. Time to go a'readin'.

DTN

<div class='quotetop'>QUOTE (tibby01 @ May 8 2010, 01:12 PM) <{POST_SNAPBACK}></div><div class='quotemain'>well, let's do math...

an alkaline 9v battery capacity is 565mAh according to wikipedia. the 8 red leds draw 30mA max, the clear blinking one is 70mA.

so that's 310 mA total draw. so maybe more like an hour or two(very rough approximation). brightness will decrease the longer it is left on as voltage decreases.

you can get a simple 9v power supply, just make sure it's capable of 310mA draw. you can also wire up a couple 9v batteries in parallel (like leads connected together). every additional battery will just add to the overall capacity.</div>

The average If=20ma on that LED. Depending on how you bias them via resistors, they should all be 20mA. You should be looking at 180mA. After you get it working on battery power, find a 9V charger at Radio Shack. Rember E=IR(Volts = Current * Resistance
Back in high school algebra you remember that those expressions are interchangable. I= E/R R=E/I.
You can find LED calculators online which will provide you the information as far as resistors go. With 9V you should be fine using two LEDs in paralell, which will save you about double on current usage. Try to make 2, or if on wall power, use 3 in a single circuit. They will be just as bright and work just fine if you decrease resistance. Resistance = wasted power. Foreward bias = held back power. Use as much foreward bias as possible to decrease current usage.
Foreward bias (Fv=) is cumulative. This will reduce voltage.
The three values you are looking at are Fv=, If= and your resistance.

Fv=3.2 iF=20mA (or.020A) This is an average LED.
You take total voltage which is 9V and subtract Fv, then divide by iF to get the resistance you need.
9V-3.2 = 5.8 then 5.8/.020 = 290 which is the resistor you need for the LED. Resistors sap power by turning it into heat.
So, if we take 2 LEDs in paralell, then you can get Fv=6.4.
9v-6.4 =3.2 then 3.2/.020 = 160 ohms of resistance required

This means you would effectively reduce your current usage from .18A to .1A. or 100ma, It should last 4 hours assuming flat current, but because current decreases you should get 8 hours off of a 9V battery using the series-paralell circuit I described with 2 LEDs inline.