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SpecterM #31 25/08/2010 - 10h04

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you really know your stuff.  I keep going back reading your information.

I cant seem to find a PTC-fuse at radioshack so I think im going to stick with using batteries.

I am looking at what you described and I am comparing your  diagram.

Any chance you could draw me a diagram for using the 4AA battery holder?
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gguyaz #32 25/08/2010 - 11h23

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6V is for 4x1.5V
15 Ohm is for 100mA current thru LED's.
Edited by gguyaz on 26/08/2010 at 05h04.
SpecterM #33 25/08/2010 - 14h12

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I thought you said 20ohm?
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dewey1 #34 25/08/2010 - 14h46

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I said to use 20 ohm for a 75ma current at 6 volts.

You do not need to run at the max rating of the led which is 100ma.
Why do you you non electronic people insist on doing things always at the maximum?

If you insist on running leds at max ratings or greater, then design your circuit for pulse width modulation. Then you can place your camera across the room and have the tracking system work.

The intensity difference of the led from 75ma to 100ma is very minimal. All it does is waste power in the form of heat. Why do you need to heat the leds up, it just shortens their life span.
SpecterM #35 25/08/2010 - 16h08

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Its not that I insist on anything.  Im ignorant and have NO idea what Im doing.  I just want to do what you think would work best.

I was at Radioshack today and bought my AA battery holder with switch.

I also picked up some resistors. Heres what I got....

150 ohm 1/2W 5% Carbon Film Resistor pk/5

and

22K Ohm 1/4-Watt Carbon Film Resistor (5-Pack)

I didnt understand the different Wattages.

Which should I use? and is there a particular way this needs to be saudered in line?
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gguyaz #36 25/08/2010 - 16h24

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there are no specs for these LED's? And your resistors values are too high. We said ten (10) ohms or fifteen (15) ohms or twenty (20) ohms, all with a quarter (1/4) watt, half (1/2) watt or until one watt of power
Edited by gguyaz on 25/08/2010 at 16h32.
dewey1 #37 25/08/2010 - 17h35

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SpecterM @ 24/08/2010 - 11h50 a dit:

I ordered the lights you suggested.



Which leds did you order? The SFH485P's?

Which battery holder did you get? The 4 cell AA holder with switch?
What battery type do you plan to use, as I asked previously?

If you would please give specific answers to my questions, I can give specific answers to solve the problem.
With 45 years experience in Electronics, I can assure you a solution to your problems!

I would get a 5 pack of 10 ohm 1/2W 5% Carbon Film resistors.
Color code: brown-black-black-gold.

Instead of black tape, get some small heat shrink tubing, it will give much better looking results.

Assuming you got the 4 cell holder with switch, then wire the IR leds (SFH485P) in series as per diagram and then two 10 ohm resistors in series to to red lead of the battery holder and the black lead to the cathode (-) of the led.

The two 10 ohm in series is equal to 20 ohm total.

Please reply with specific answers before continuing. All you are doing is wasting money by being to hasty with inaccurate decisions.
dewey1 #38 25/08/2010 - 17h44

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[quote=gguyaz @ 25/08/2010 - 13h23
6V is for 4x1.5V
15 Ohm is for 100mA current thru LED's.[/quote]

For clarity on your diagram, make the text 15Ohm@100mA look like this:
(15 ohm for 100mA) or for (20 ohm for 75mA) for a 6 VDC source.
Edited by dewey1 on 26/08/2010 at 19h06.
gguyaz #39 25/08/2010 - 18h11

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dewey1 @ 25/08/2010 - 19h35 a dit:

....
With 45 years experience in Electronics...



JUST waow! How old are u  :lol:

dewey1 @ 25/08/2010 - 19h35 a dit:

...For clarity on your diagram, make the text 15Ohm@100mA look like this:
(15 ohm for 100mA) or for (20 ohm for 75mA) for a 6 VDC source...



Sure I could, but I think everything is already said on the forum and website.

But SpecterM just want to follow some "UTube vidéos" to do it. He is buying everything (and nothing). The material list is written everywhere! Freetrack setup is on the website.
dewey1 #40 25/08/2010 - 18h49

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You and I both know that!
Some people need to learn to have patience, learn to do research on there own, and most important not to believe everything that is on youtube is actually true or real!

You have such a nice simple pictorial diagram that is why I suggested you edit it for future references. Also put your name on the diagram if you created it.
I have posted a schematic version but your pictorial is simpler for non-electronic people.

By the way, you show a PTC fuse of 250 ma, recently some USB 2.0 ports are only 100ma.

I have been in the industry since I was 18. You do the math. Retired a few years ago.
SpecterM #41 25/08/2010 - 19h10

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For clarity.  This is what I bought.

IR LEDS   http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=475-1470-ND

AA batter holder with switch.   http://www.radioshack.com/product/index.jsp?productId=2062253&kw=040293150938&sr=1&origkw=040293150938

10 ohm 1/2watt resistor  http://www.radioshack.com/product/index.jsp?productId=2062309&kw=040293114947&sr=1&origkw=040293114947
Edited by SpecterM on 25/08/2010 at 20h16.
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dewey1 #42 25/08/2010 - 20h11

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Great! Thanks for the clarification,
You have the correct leds and a 6 volt power source.

These leds have a 1.5 v operating voltage. So three in series is 4.5 volts.

6v-4.5v=1.5v

We need to drop 1.5v with a current of .075 Amp or 75mA.
Ohms law states: R=E/I or 1.5/.075 which equals 20 ohms.

All you need now are the 10 ohm resistors I mentioned and wire it up as the pictorial shows. Two 10 ohm resistors in series, for 20 ohms with the cell holder red lead and the anode (+) of the top led.

The cell holder black lead gets wired to the bottom lead of the of the lowest led which is the cathode (-).

Just so you understand it better: leds have a plus lead (anode) and minus lead (cathode).
Some people use the lead length as an indicator of + and -.

I never use that because what if the leads are trimmed to a shorter length!
The correct method to reference to, is the flat side of the led plastic lens.
That flat side is always the cathode or the - lead. Just remember that and you will be OK.

Make sure you reinstall the floppy disk media as a filter and you should see a very much improved operation.

Does that help?
benmeijer #43 25/08/2010 - 20h13

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Leds are OK

Batterieholder looks OK

now you need a 22 Ohm resistor to power the leds with 70 mA. 1/4 watt is ok.

If you have enough 150 Ohm resistors you could use them parallel, I mean take 5 or 6 of them and bind all left leads and right leads together. Then you have about 25-30 ohm.  Use this option till you have bought the 22 ohm.
SpecterM #44 25/08/2010 - 20h25

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Dewey, I returned the other resistors and got the 10ohm 1/2 watt ones.  thanks for the clarification however.  Im completely new to this.  In series is option 1 or option 2?

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ALSO:

The correct method to reference to, is the flat side of the led plastic lens.
That flat side is always the cathode or the - lead. Just remember that and you will be OK.



Im having a hard time understanding this.
Edited by SpecterM on 25/08/2010 at 20h26.
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dewey1 #45 25/08/2010 - 20h43

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Option 1 is series wired.

Look real close at the led and you will see a ridge all around the base of the plastic portion. One side will have a flat spot while the rest of it is round.
On this particular led it is the longer lead. Now you should see it.

Some leds use the short lead as the cathode.

That is why I say use the flat side as the cathode (-) because that is universal on all leds of this type.

If you look at the pictorial you can also see it. Just need to know what you are looking for!

edit:deleted link
Edited by dewey1 on 25/08/2010 at 22h12.
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