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Twospoons
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| Quote: | Originally posted by chemoleo
It' s a shame that LEDs don't do less than 360 nm.
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You can bet on Nichia working towards ever shorter wavelengths. They seem to be the leaders in this field - they built the first blue diode laser
IIRC.
Maybe their violet laser could be frequency doubled to 200nm? Though at this point mercury discharge starts to look simpler!
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BromicAcid
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Well, I just bought some resistors, 470 ohm, 1/4 watt, 5% tolerance. They also had 1/2 watt but they didn't have enough of them, will these 1/4
watt resistors work?
Anyway, I realize regular glass absorbs a lot of UV but I'm not too worried about it. I plan to use my setup to make some carbon tet from
tetrachloroethane in a closed system, no HCl made so I can keep everything nice and closed up. The only difficulty is in breaking that C-C bond that
is left, it happens readily with strong enough UV radiation but this might not be good enough now that I've had these opinions on the lack of
intensity of these UV lights. Anyway, going to go for three LED's in series in parralel with one another with these 470 ohm resistors on each
set of three.
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IrC
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Well you got me to playing so I ordered a hundred from the Hong Kong guy. 470 won't work for three in series, just one, and with a 12 volt supply
they will not really be running quite bright enough. For three you need about 68 ohms. Just figure that 3 in series is 9 or 10 volts, say 10, minus 12
= 2 volts. I=E/R so I= 2/68 = 29.4 mA, about right. P=E*I, so in the resistor P=2*0.0294 = .0588 watt. Since the resistor is rated 0.25 watt, you are
safe. You should always derate 50% if you don't like failures. For your 470 ohm on a single LED you have 12-3=9, so you need to drop 9 volts in
the resistor. I=E/R = 9/470= 19.1 mA, not quite enough. This also depends upon your supply voltage which you still have not stated, but 15 volts would
be better for the 470 with a single LED. You need to calculate these things out before you buy resistors, and before you can do this you must choose a
supply voltage.
Compromising let's say you are going to go with what you have, 470, and guess that 3.3 volts at 29 mA is closer to the proper turned on LED
voltage and current. To get 29 mA through 470 ohms: E=IR = .029*470= 13.63 volts, plus 3.3 volts of the LED = 16.93 volts. Rounding to 17 volts gets
you there. The power in the resistor is E squared/R = .6 watts so you would need 1 watt resistors here, kinda wasteful. I would go back to the three
in series, guess 10 volts, and pick the resistor for a supply of 12 volts, meaning the resistor needs to drop 2 volts at 29 mA, giving a 68.9 ohm
resistor with a power dissipation of .058 watt. A quarter watt resistor with a standard value of 68 ohms is fine. If you would decide upon a supply
voltage we can give you exact calculations. 470 ohms usually works great for the visible 4,000 MCD LED's but to get the UV you want you really
need to get them close to 25 to 29 milliamps.
Building a large array like you want is much work, so do not begin until you have calculated everything out and are firmly set in all parameters.
PS: U2U me a mailing address and I will mail you 40 of the half watt 68 ohm resistors for free. I have more than I will ever use in my life. I do not
think it would cost me more than a buck to get them out to you and they likely cost me 2 cents each to buy, so call it my donation to making your UV
stuff work. 40 would do 120 LEDS in strings of 3 and your address will not get out to anyone.
[Edited on 13-7-2005 by IrC]
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BromicAcid
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Thanks for the offer but I have quite a bit of store credit at radio shack that I have to burn up in a somewhat short period of time.
My problem so far with this thread is that I've been telling everyone what kind of LED lights I have and how I want to set them up and I want
everyone to tell me what kind of power source and resistors I need. The problem with that (that I'm just realizing is) is that I am giving an
equation with two variables and it's hard to solve for both of them with only one equation. So, let me revise this, what would be the minimum
requirements for a power supply to connect these bulbs in sets of three, fifty of them.
I will try to get something close to the supply specified here and once obtined I will post what I got and it should be a simple matter of myself or
someone else using the formulas so graciously supplied in order to determine the resistance necessary in the system.
My problem as of now is that I know next to nothing as to how to determine the specifications for the powersupply and such. I mean, do I figure out
the number of amps that I need by multiplying the amps required for each bulb by 50 to get the total amps? Does this chance since I want to do series
of three in parrallel with one another? The voltage should be what? These are 3.4 - 3.6 V bulbs so should it just fall in this range, could it be
more, less, does the voltage I get affect the amps I should have? I know none of this, I can try to learn it but so far whenever I learn an aspect of
it what I know in that one aspect throws everything else off....
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12AX7
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| Quote: | | Originally posted by BromicAcidI mean, do I figure out the number of amps that I need by multiplying the amps required for each bulb by 50 to
get the total amps? Does this chance since I want to do series of three in parrallel with one another? The voltage should be what? These are 3.4 -
3.6 V bulbs so should it just fall in this range, could it be more, less, does the voltage I get affect the amps I should have? I know none of this,
I can try to learn it but so far whenever I learn an aspect of it what I know in that one aspect throws everything else off....
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Bromic:
LEDs are constant-voltage devices. Apply a current to them and a certain voltage (within a relatively small range, hence "constant
voltage" will be developed on the terminals. As voltage rises, current rises
exponentially (on the current vs. voltage curve) until it explodes, hence a constant voltage power supply is a big no-no.
The reason is stabiity. Say voltage rises by one volt, from 3V to 4V. Current rises from say, 20mA, to 20mA * 10^(4/3) = 430mA! (That sounds
possibly a bit high, I don't know what the actual base is. e gives too low a result. Oh well, I'm no semiconductor physicist.) This
raises the junction dissipation from a comfortable 3 * 0.02 = 60mW to a whopping 4 * 0.43 = 1.72W, easily leading to smoke buildup inside (then
bursting and going outside) the little epoxy case. Remember, the magic smoke has to stay inside for it to work.
Since by far the easiest power supply to build is constant-voltage, we have to use something to interface them safely. A constant current source (or
sink) is perfect, but a simple resistor works as well.
Since an LED is essentially a CV element, and a power supply is also CV, a resistor placed between the two passes a current proportional to the
difference in voltages, no matter the current level: i.e., I = (Vcc - Vf)/R (derived from I = V/R, ohm's law). If you connect LEDs in series,
you can subtract more of the PS voltage. Too many in series and no current will flow at all because Vf(tot) > Vcc! (Yes, in reality some will,
but not much because you're on the short side of the exponential curve.)
So let's say you have a 12V supply and a bunch of 3.6V LEDs. Three can be connected in series for a total 10.8V, but four cannot (for a total
14.4V). Two would be even more comfortable, as with three, it takes a change of merely +10%/-5% to change the current by a factor of two! The AC
line is not controlled quite this tightly, though YMMV.
Anyways, the resistor is R = V/I = [Vcc - Vf(tot)] / If, where If is the rated forward current of the LEDs.
Do NOT parallel LEDs, as CV sources do not parallel well. You want series so they all have the same current (the same current going into a wire has
to come out, right?).
Else what will happen is, inevitably, one LED will have a lower forward voltage, thus pulling down the voltage on the other LEDs paralleled to it. If
the resistor is sized so that all LEDs in parallel get the rated current, most of it will go to just one or two, putting those well over limits and
making a micro-cheronobyl. The other reason is, LEDs, indeed most - if not all (I don't know) semiconductors - have a negative temperature
coefficient. Forward voltage DROPS as temperature increases. The lower voltage causes one to hog current, causing it to dissipate more power, which
raises its temperature, which ultimately explodes it!
So to connect a whole whack of LEDs, you have to use high voltage (something greater than Vf, that is) and many resistors (one resistor per series
string). If you want to run 100 x 3.6V LEDs in series, you'll need more than 360V, however twenty strings of five each will run something more
than 18V (25VDC or so would be good, with one 220 ohm resistor per string). Current adds in parallel, so if each string takes 30mA, you'll
consume 0.03 * 20 = 0.6 ampere.
I don't know how much of this is patronizing but I'm too lazy to read the entire thread and see what I'm repeating. It's
summarized in one at any rate.
Tim
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IrC
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I think he just wants to know what to buy rather than the theory behind it all. So here goes. Yes, multiply the number of strings by 29 mA: (50/3)
(and round to 17) which gives 51 LEDS, with 17 strings. 17 times 29 mA is about a half amp, so derate this and buy a one amp 12 volt wall wart. Wire 3
leds in series with a 68 ohm, half watt resistor. Do this 17 times. Put all 17 strings in parallel with the wall warts 12 volts. Mount on a nifty
twospoons board and sink with lead length as he mentioned. Plug it in and glow away. You need 17 resistors, each 68 ohm half watt, 51 of your UV LEDS,
a one amp 12 volt wall wart, and whatever circuit board or mounting of some kind. I emailed the MFG for a graph of MCD versus nanometers and MCD
versus current but have not heard back yet.
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Archimede
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The best way to do it should be with one resistor per led. If you group them in mixed serie/paralled setups, and one burn the others get a higher
current.
The leds dont have much internal resistance , thats why they burn if you dont limit the current with a resistor.
In theory you could put em all in parallel with one big resistor but I would not recommend that. And if you put em all in series and one burn, the all
thing stop working.
[Edited on 17-7-2005 by Archimede]
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IrC
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Ya think? Like the very first answer to the question back on page 1 of this thread. Does anyone actually read these threads?
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Twospoons
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One thing about the 12V 1 Amp wallwart - don't buy a cheap one! They're nasty, unregulated things, subject to line voltage variation etc.
I know they're more expensive, but get a properly regulated one - the guys at rat shack should be able to help you (if they're not just
clueless store clerks). The output voltage accuracy from the regulated supplies is usually better than +/- 5%, they're thermally protected, and
will usually survive all kinds of abuse without the smoke leaking out.
[Edited on 18-7-2005 by Twospoons]
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IrC
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I think the little regulated 4 amp supply rat shack carries would be better, then he could also use it for other stuff down the road. As to your
comment "(if they're not just clueless store clerks)", this one he needs to worry about. I lived near one that had a Pearl Optical next
door. Going in there and looking around, the guy came up trying to be overly helpful like they usually are there. He asked what I was looking for and
I told him "contact cleaner". He stated "I'm sorry, that is the store next door". Getting pissed that I started ignoring him
while I walked on around looking for the can of lubricating contact cleaner he came up and started arguing with me that they did not have contact
cleaner. He must have had some other thing going on that day as he was getting real angry with me until I found it, picked it up and showed him the
words clearly marked on the can "contact cleaner". So when in radio shack it is helpful to already know what you are doing and totally
ignore the help..
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Lambda
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Newnes - Power Supply Cookbook (2001)
I just had to throw this one in here, but IrC may turn out to be right afterall. Will anyone actually read this book ?.
Newnes - Power Supply Cookbook (2001) has 278 pages of crystal clear PDF-scans and is 3.76 mb in size.
For those who will read this book:
http://rapidshare.de/files/3123636/Newnes_-_Power_Supply_Coo...
[Edited on 18-7-2005 by Lambda]
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Archimede
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| Quote: | Originally posted by IrC
Ya think? Like the very first answer to the question back on page 1 of this thread. Does anyone actually read these threads? |
Yes I was trying to point out that this is the best way to do it and why. After reading 2 more pages that shifted the project (IMO) in the wrong
direction.
Sorry, next time I will quote the original poster of this idea.
[Edited on 18-7-2005 by Archimede]
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BromicAcid
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Hard to say this but I ended up ignoring everyone....  I wired 10 LED's in
parrallel and hooked them to a variable power supply a friend of mine let me borrow and cranked it up into the 3 - 4 V range and viola the little
LED's lit up powerfully, very nicely. It disconnected one and the amps put out by the power supply dropped accordingly. So, I scaled up. I
took a wood salad bowl and put 75 LED lights into it. Good thing I noticed they have a positive and negative lead before hand.
Anyway, I wired them all in parralel and it looked really neat, now I hate soldering though, I ended up doing about 160 solder spots and messed up a
lot. Then I hooked it up to the power supply and ::bink:: 
One of the lights blew out and none of the rest were lighting, strange since they were in parrallel... So I looked at a voltage carefully and the
needle was jumping all over the place, sometimes over 20 V. So, I started cutting wires and sectioning it off into smaller sections, and they would
light up readily when current was applied until I got to one section. One of the lights would glow orange and none of the other ones would light.
Normally when a LED dies it does so quitely, I don't know if this was a factory defective LED or it just chose some strange path to die on, but
somehow it was taking all the current I put into the setup and canceling everything out.
So, a few days and frustrations later I had most of it re-wired along with a few blulbs that I burned out by holding the soldering iron on them too
long. I kept the leeds long to act as heat sinks like was suggested earlier in the thread. Out of the 75 origional lights 6 died along the way so I
have 69 left in the setup. I left it running for 30 minutes without it warming up or any more lights dying. The test run will be tommorow, weather
and time permitting.
Attached are some pictures, 1) Inside of the bowl where all the lights point 2) Outside of the bowl with the wiring all in place 3) Picture of the
thing turned on, everything else appears dark but it was taken in a lit room.
BTW, sunglasses are a must, UV radiation is of course bad for your eyes but even looking at them for a few seconds makes things fuzzy for a while
afterwards.
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12AX7
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Anyone wanna put down bets on when the whole set is going to go, completely?
Ten dollars says two hours run time.
Tim
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BromicAcid
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Oh, wasn't that a major vote of confidence. I'll have you know I've distilled nitric acid from drain cleaner and fertilizer in glass
pop bottles with a propane torch as a heat source with no problem until I found out soda-lime shouldn't be heated, whereupon my next attempt
ended in disaster, it's amazing how far ignorance will take you. And now you've ruined it, now I have a nay sayer and my experiment is more
likely to fail 
Honestly, if I take a light out of the setup the amp output of the supply drops as a result, the voltage remains constant throughout and nothing heats
up. I'm not going to be running these things at 100% of the recomended voltage and I have purposely destroyed LED's with this new toy of
mine and it takes me purposly bringing the voltage into the dangerous region for these LED's. I have this one experiment to do that should take
a few hours and I see no problem in completing it with the contraption that I have made.
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12AX7
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Well yeah, current drops because you're reducing the number in parallel, Kirchoff says so. That doesn't mean it's good, or for that
matter the least bit stable! 
Tim
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BromicAcid
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And 6 hours later not a single light had gone out
Ran my apparatus for photochlorinatoin today, it went well, except that towards the end the sulfuric acid was falling into the same area of the
hypochlorite and it wasn't reacting so too much pooled there and the stopcock of the addition funnel was opened so it all started reacting at
once, blew the stopper out and caused a rain of concentrated sulfuric acid. The cloud of chlorine gassed me more then I had ever been gassed before
by that element....
As for how it worked, when I lifted up the chlorination cover the air and lights were the same temperature as the surroundings but the flask was hot,
the chlorinated constituents were condensing on the inside of the flask it was so hot. Definately a good reaction pace (being that the heat must have
been coming from the exothermic chlorination). Going to fraction the product later considering most of the products have markedly different boiling
points.
Attached is a picture of the setup. I made one neat improvement though, the bubbler through the sodium hydroxide solution at the end was actually
inserted into a sepretory funnel, into the stem of it. In the end a glass wool plug was inserted. In this way I could close the stop cock to stop
the suck back when necessary and in addition there was a lot of water to suck back to fill the inside of the sepretory funnel.
Oh well, picture attached.
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Tacho
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I agree with 12xy. The leds are likely to fail soon if you have too much current flowing through them. Measure it! It should remain bellow 20mA.
Besides if you put too much current through a common green led, they go yellowish and stay that way. Maybe the same will happen to your UV. You may
have a different wavelength and not notice it. Long UV?
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Twospoons
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Well, they're your LEDs, so do with them as you please. Just be extra careful not to knock the power supply, or heat the LEDs too much. You
have no room for error.
Helicopter: "helico" -> spiral, "pter" -> with wings
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IrC
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He has some room for error, he isn't using them all so he has some left. Besides, that Hong Kong guy is cheap! I cannot believe what I have paid
for those things in the past. Anyway, his statements tell me all I need to know. His experiment seems to have worked ok so he must have enough UV to
do the job. Also he said they were not getting hot so the current cannot be too high even if he does not have a meter. Reaction working and leds not
hot, it ain't broke don't fix it. I do disagree with the under 20 mA statement though, for these leds they just would not work very well
down there.
As a side note, Bromic Acid and Twospoons got me going playing with them and I found out one of the aluminate phosphors I thought had failed actually
did not glow at all until I hit it with the UV LEDS, then it glows great in the yellow. I can shine a 3 watt white LED on it and nada, hit it with a
dozen UV LEDS which are way dimmer and great yellows come out after I kill the light. Very bright, and interesting but the powder does not seem to
glow in the UV other than a slight white similar to a brightner. I think the MgO quenching I was playing with somehow made this batch only respond to
UV, and then only glow in the yellow (the patent formula was only blue here until I added a quencher). So next I will stir some up in some glass and
see if I can make a cool UV to yellow frequency shifting material. I don't think I'm off topic here as it did involve the UV LEDS.
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Organikum
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An interesting approach, but.......
Actually looking at your setup Bromic I would say that removing the UV-diode mantel and exposing the flask to the sun which shines so nicely in the
picture will produce much better results. A simple mirror would boost efficiency.
The actual power of the diodes compared to plain sunlight is neglectable, compared to a simple 150W halogen lamp with the glass-UV-shielding removed
it is still small.
Like this:
Remove glass, be careful the lamp is HOT. Worked perfect for me every time.
A dash of red phosphorus makes the UV obsolete - the shit scraped from matchboxes works.
/ORG
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Mr. Wizard
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I was wondering why nobody mentioned the bare halogen light bulb. Those tubes are supposed to be fused silica, so they will transmit UV.
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IrC
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A good idea but they do put out a large amount of heat as well. Robbing a copy machine of the exposure light and matching power supply is one idea,
you also need a cooling fan. There also is the thought about duty cycle, 100 percent for a long time might not be a good idea. In any case handle the
bulb with gloves and clean it well with some kind of solvent that has no denaturing or other additive that would leave a film. Any finger oils on the
quartz envelope will cause dark spots to appear. The light also needs to be run a proper duration so that the halogen cleaning cycle will be
completed, to ensure maximum life and light output. Like I mentioned, there will be the issue of a great deal of heat. That was why I mentioned the
mercury bulb used in UV erasers earlier, lots of UV with little heat, but another mentioned they do not put out enough overall candlepower. I'm
still not so sure about that though, if the LEDS work in the experiment I see no reason the eraser light would not also work.
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Organikum
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Sunlight is the first choice, the heat of the halogen lamp can be hancled with a 12V computer fan. The halogen is to handle with care when addition of
compounds to the reaction flask is necessary, but me at least survived this too.
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vulture
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Organikum, do you have any indication as to which are the shortest wavelengths a halogen bulb will emit? Because 400nm for those LEDs isn't
really UV, IMHO.
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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