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FreeTrack Forum > FreeTrack : English Forum > Support : Tracking System > Looking for a review of this setup
Sabooo | #1 15/05/2008 - 02h58 |
Class : Apprenti Off line |
First, thanks in advance to anyone taking time to read this, and especially anyone who can help:
Looking to do a 3 point cap. I have purchased the following: Switchable voltage PS from Radioshack that I plan to run at 4.5v - 300mAh output. 3 x SFH-485P LED's (1.5v, 100mah forward) According to the calculator, this seems to be the resistor I need: 15 Ohm 1/8W 5% Carbon Film Resistor as seen in this link: http://www.opamp-electronics.com/catalog/ohm-18w-carbon-film-resistor-pack-100-p-451.html Is that right, or has my utter lack of electronics knowledge led me astray? Cheers, Sabooo
Edited by Sabooo on 15/05/2008 at 02h59.
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Deimos | #2 15/05/2008 - 03h35 |
Class : Beta Tester Off line |
Well, the setup seems to be ok, but personally i wouldn't power the LEDs with so much current - about 75mA should be more than enough, even for a unmodified webcam. Of course battery drain won't be an issue since you want to use a power supply, but at 100mA LEDs will probably considerably heat up, and may eventually damage after prolonged use.
Besides, such bright LEDs might be cause lens flare in the camera, making tracking unreliable (it happened to me). So i'd say you need about 18Ohms resistor. [EDIT] By the way, output of power supplies (and forward current of a LED, and current flow in a circuit) is measured in mA (miliamps), not mAh (miliAmpere-hour). Battery and rechargable battery capacity is measured in mAh [/EDIT]
Edited by Deimos on 15/05/2008 at 03h41.
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fatboy | #3 15/05/2008 - 07h17 |
Class : Apprenti Off line |
Ahh good to knew deimos, I'm thinkin of using a plug in transformer as well.
Almost thought you said 180 Ohm resistor a second there, threw me for a loop without the space. I'm on my way to do all my wiring tomarrow perhaps I'll grab a resistor, I didn't know the LED's would be too bright, my original plan was to use 4.5 V (3 AA's) and just hope that each LED would drop 1.5 volts each. But given the difficulty to find IR LED's in my town, the last thing I want to do is burn one out Thanks. |
Sabooo | #4 15/05/2008 - 12h30 |
Class : Apprenti Off line |
Thank you Deimos - I will try to source a lower current power supply, or maybe even reconsider entirely and go with a battery powered affair.
The tiny bit of experience I have with this sort of thing was with rechargable batteries, and of course you are right that the circuit would not be measured in relation to time. Thanks for the correction! I am surprised someone has not created a kit for this yet. I would have paid the markup for a set of electronics (LED's, power source, appropriate resistor(s).) The assurance that I had the right kit would be worth it to me. Of course, half the fun of such a project is the chance to learn something in the middle of it. Cheers, Sabooo |
Deimos | #5 15/05/2008 - 13h28 |
Class : Beta Tester Off line |
Whoa, wait a second - i think you misunderstood me a bit. 300mA rating of the power supply is just the maximum current that the supply can source - even if you chose a very small resistor, the current will be no more than 300mA. But a power supply doesn't "force" the current into the circuit, so if a circuit's resistance allows only 80mA to flow trough it, only 80mA will be sourced from the supply. Theoretically, you could as well power the LEDs from 10A (that's 10000mA) power supply, provided that you have the right resistor (about 18 Ohms in this case). Of course i wouldn't recommend that, since it would very be inefficient setup If you already have the power supply, there shouldn't be need to change it. If you didn't buy it yet, you don't have to get anything stronger than 300mA. You could just take something with a bit higher voltage if you get the chance - about 5V. But even 4.5V should work. And while we're at it - you can as well an use a power supply you already have - and i'm sure you have more than one lying around the house: maybe a charger from a cell phone you don't use, or something like that. The voltage doesn't need to be as low as 4.5-5V. even 18-20V will do, if the supply can source 300mA - you'll just need to use the resistor calculator again, inputting the correct voltage.
Well there's one user on the forum selling (or planning to) kits - just the parts, or pre-built clips. http://forum.free-track.net/index.php?showtopic=788 |
Sabooo | #6 15/05/2008 - 14h09 |
Class : Apprenti Off line |
Head...exploding...must...think...
I already have the powersupply - it is switchable voltage. Here is a link if it helps to the RS page - ugly URL. http://www.radioshack.com/product/index.jsp?parentPage=search&summary=summary&cp=2032056.2818119.2818335&productId=2552559&accessories=accessories&kw=switch+adapter&techSpecs=techSpecs¤tTab=summary&custRatings=custRatings&sr=1&features=features&origkw=switch+adapter&support=support&tab=techSpecs I believe I entered unregulated PS in the wizard, and in re-reading the specs this is claiming to be regulated. I will put the meter to it to be certain of what it's putting out. Is there an advantage to using higer voltage and lower current, then? I admit I'm having a hard time getting my head around this. Thanks for your patience in trying to help. Cheers, Sabooo |
Deimos | #7 15/05/2008 - 14h25 |
Class : Beta Tester Off line |
Well, if you need something to thing about, read:
http://forum.free-track.net/index.php?showtopic=700&message=5339 :D But in short - i thing that setting your power supply to 6V should work fine. It's not that you use higher voltage and lower current - you'll use higher voltage and the same current (so smaller resistor will be needed). The problem with the 4.5V is that the sum of 3 LEDs forward voltage is 4.5V. If you read above, you'll notice that this might mean that it is in fact the voltage drop across all 3 LEDs, so if the voltage drops 4.5V on 4.5V supply, then there'll be no voltage across the circuit, and the current will not flow. Of course, that's just theoretical - in fact the forward voltage in the specification is a maximum value and in reality will be lower than 1.5V - it should work even on 4.5V supply, but better safe than sorry As for the regulated/nonregulated supply - the regulated ones always output the nominal voltage (the one you set - 6V if you follow my suggestion in your case), while the nonregulated ones output the nominal voltage only at maximum load, at lower loads they tend to output voltages higher than nominal. Selecting the nonregulated power supply in the led assembly wizard simply assumes voltage 30% higher than nominal. If you measure the voltage with no load (just attaching the voltmeter to its plug, and you see the nominal voltage (the one you set), it's indeed a regulated supply.
Edited by Deimos on 15/05/2008 at 14h31.
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Sabooo | #8 15/05/2008 - 16h41 |
Class : Apprenti Off line |
I read your post and...I THINK I GET IT! (I can almost hear the sigh of relief.) I needed the formulas for it to make sense. Just to confirm, the value I put into the wizard for the LED's forward current should not be it's maximum rated current, but rather the current I want for the circuit.
So to go with your suggestions, 6v regulated PS, 3 - 1.5v LED's - if I wanted 50ma current, that would require a 30 Ohm resistor. 75mA would need 20 Ohm resistor, etc. The way I had originally entered the values, I was making the circuit 4.5v, 300mA, which would either be hard on the LED's due to high current, or possibly result in a non-functioning circuit due to insufficient voltage. I only had to repeat the phrase "LED's are not light bulbs" a few dozen times to break my way of thinking. cheers, sabooo
Edited by Sabooo on 15/05/2008 at 16h42.
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Deimos | #9 15/05/2008 - 18h22 |
Class : Beta Tester Off line |
Yep, now you DO get it right
Enjoy using freetrack, and don't forget to post pictures of your rig in the gallery tread |
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