Magical deep UV LEDs - Cubane Ep 12
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- Опубліковано 22 вер 2024
- A modern light-source for a modern man (who doesn't really know how electronics work), what could possibly go wrong? We need the UV for our cubane ok. Twitter: / explosions_fire
Subreddit: / explosionsandfire
Join the Discord!! / discord
Patreon: / explosionsandfire
Music: from the Aphex Twin soundcloud dump. Sorry once again to people who have sent me cool music, I am too short of time and it sucks. Track name: 22 Pearls
References for the whole cubane synthesis: 1997 Lab scale paper: www.researchga...
2013 Pilot Scale: pubs.acs.org/d...
"Cubane Derivatives for Propellant Applications" 1989
apps.dtic.mil/...
I think this video is a good demonstration of the difference between science and engineering
I'll teach you hobbyist level electricity if you teach me hobbyist level chem
I work IT night shift so time diff don't matter
Depending on heat output thermal glue MIGHT be appropriate, as it also provides binding to the LED backing. However it is much less conductive than a paste or liquid metal.
And you are doing neither
Did you not want to ask William Osman for a second opinion to save the potential $60 of LEDs?
I like how other people give genuine chemistry advice in the comments and I'm just sitting here laughing at the funny science man and his magic cubes
As far as I’m concerned this man is the missing link between occult alchemists and actual modern scientists
Wait this is not an occult alchemy channel?
Yeah not gonna lie I know nothing about Chem, I'm just gear for funny Aussie chemist hijinks
Can't help much tbh. This guy is getting his PHD while I'm still doing A levels
Me too I must admit
Speaking as an electrical engineer, this episode is painful.
But entertaining, in a "monkeys playing with power tools" kind of black comedy way.
Speaking as an electronics tech, yeah.. this hurt to watch. But he's learning. Baby's first diode :'-)
I expected nothing less, and I am sorry
@@ExtractionsAndIre your wallet is the one that will suffer the most. Letting the magic smoke out of IC components is painful.
@@lairdcummings9092 and the reaction will suffer too. The overdriven and overheated LEDs will make a very different light with much higher wavelength.
In addition to the usual explosive chemicals in dirty glassware, my favorite Australian meth-lab has expanded into mixing electricity and ionizing radiation. I'm feeling spoiled today!
I feel like the Dark Age of Technology is just months away now.
and all in milo tin too
@@JainZar1 Could you imagine if he started programming too?
Whoever gave you that advice on twitter regarding the LEDs "drawing as much current as they need" is very misinformed. That is true for *constant voltage* loads, but completely wrong for *constant current* loads like these unregulated LEDs. The current-voltage curve on any diode is exponential. That is why the current should be limited to 40mA and then the voltage will correct itself to whatever it needs to be. After the threshold voltage is passed, tiny increases in voltage will lead to huge increases in current, which often leads to the deaths of LEDs. Great vid, looking forward to new instalments.
Cascade over-current failure is *always* an option...
Yea this is correct. Constant current is also important because the resistence is dependent on the temperature. When it heats up, the resistence drops, it draws more current (assuming a constant voltage), thus heating up more etc.
@@orange422 et voila, regenerative feedback loop, and soon the magic smoke escapes.
@@lairdcummings9092 Creating DEDs.
Dark Emitting Diodes.
@@jimmyb1451 le ded
“I keep getting hung up on this 40 mA. An amp should be fine.” Me an electrical engineer screaming: “No it’s 40 mA!!!”
"This bushing requires a 12mm pin. 17mm should be just fine... More is better, right?"
Me, a physicist who's seen her fair share of components go pop when I've accidentally fried them: *grabs popcorn*
@@lairdcummings9092 me, a beginning design engineer: if it fits, it sits... But make sure you document that this was *your* doing... Also, replace the bushing, if it does fit... Walking out 2.5mm radially isn't normal, shop ape.
Me, not an electrical engineer but just an electrician: "No it's 40mA!!"
@@douglasboyle6544 me as softwar engineer NO ITS 40mA
"You're not gonna illuminate a room full of people with UVC, that would be bad."
-Big Clive has joined the chat
A crossover I need right now
It immediately brought up memories of that art installation that was pretty much made out of flourescent UVC tubes that made a lot of people sick. I _think_ that was the one featured on Big Clive...
@@OOZ662 He did indeed discuss an event that was supposed to have had UVA tubes but somehow got UVC ones instead. Everyone at the event got sunburned eyes.
@@smartyhall the phrase sunburned eyes is distressing.
@@andrewgreenwood9068 I've had it a couple of times from beach trips. Feels a lot like pink eye, both have a feeling of sand in your eye and dryness. Another term for sun burnt eyes is snow blindness
If it says 40mA it's 40mA, man. Blasting amps through those LEDs will kill them very quickly. I'm impressed that they even hold up that long. Also: you're not supposed to see the LED's when they're on. They don't create visible light at normal operation AT ALL !!!!! What you're seeing is the death scream of an LED. Similar to normal LEDs which will always glow red shortly before they die, even if they're green LEDs. I would not be surprised, if the LEDs don't generate any useful (for you) UV radiation at that state.
Also, also: UV-C should not make paper fluoresce that strong. Overdriving the LEDs could cause them to drift into longer wavelengths.
absolutely this!
That was brutal to watch. It would have taken just one google search to learn how to drive LEDs. If you can do advanced organic chemistry you can definitely do basic electronics.
Edit: get a resistor: R = ( (Your laptop PSU 19V) - (led voltage 13V) ) / (LED current 0.04A) = 150 Ohm
Original: To not just be "Mr. You're Doing It Wrong". Do the following: take your power supply, set it 13V, turn the current knobs all the way left (voltage should drop to zero-ish, that's fine). Then short the output (or see note below). Turn the "fine" knob of the current settings so that the current meter shows 40mA (which is crazy hard with that meter, sorry). Remove the short from the output. The voltage should pop back to 13V. Then never ever touch those knobs. To be save(er), connect the LEDs to the already switched on PSU. The cheapish ones tend to overshoot the output voltage at power on.
note: instead of shorting the output, use your multimeter in current mode. That even helps with getting the 40mA right.
Yeah one thing that confused me was the lack of drivers available at such low currents?? Like I can’t seem to buy anything off the shelf at 40 mA? And then when I tried with the bench top supply, it seemed fine at 1amp, but I guess that was wrong oops.
@@ExtractionsAndIre Because at low currents a simple resistor will replace the driver :)
Local Australian mad chemist decides he doesn't get enough UV on a daily basis already.
Being an electrician, I think I finally understand how a chemist must feel watching your videos, lmao.
You playing it fast and loose never gave me anxiety before but holy fuck.
Makes me laugh thinking about previous videos where things went awry.
"I figured it would just take the voltage it needs but apparently not"
I think the difference is that he more or less knows what he's doing in regards to chemistry
when the current spiked to 2 amps i actually yelled at my computer
finally relatable to our crowd
Wow! I'm thoroughly impressed those UVB LEDs can survive running at 22x their rated current without dying within 60 seconds, they've certainly come along way from being super sensitive to the slightest abuse. Either that or they've sold you some 1,000 mA 365 nm UVA LEDs... A lot of UVB & UVC LEDs on Aliexpress are actually UVA 365 nm LEDs, might be worth checking they're genuine with the price of that listing being so low.
But oh god your electronics and like my chemistry, I really doubt those poor LEDs are going to survive the several hours for the reaction. I wouldn't be surprised if you're driving them so hard the internal protection zeners are clamping the voltage which is just wasting power as heat. Pretty sure even if you run them at their 40 mA you'll still get more optical power out than those previous lamps. Also *DONT* put the timer on the LED driver output, you should always have the LED connected constantly otherwise the unclamped output capacitance can pop those LEDs at turn on like you so very well demonstrated on the first victim. Also you can ignore the earth, the LED driver is an ungrounded isolated design.
I recommend you abandon the on-off relay timer contraption (the switch on characteristic might also lead to sudden LED death syndrome) and instead just use the bench power supply and to limit the current you can use some resistors, super easy and makes it easier to adjust. You can probably safely shove 50 or 60 mA though them for a 100 hours without them dying.
Happy to provide further guidance or calculations, you can reach me here in a reply, on Twitter or Patreon.
Yes, properly driven LEDs do not need time offs.
well you gotta give those LEDs credits for surviving that kind of abuse
I recall losing an LED to my lab PSU during some low current tests, even with 10mA current limiting and the supply voltage matching the LED!
Was probably an inductive event, because when the leads shorted there was a spark as the output capacitor dumped it's load - likely the inductive spike that followed punched through the LED; halving it's brightness...
@GsaUce Rug Don't know - when flying blind I assume 20mA max current, as it's most typical...
Have seen filament style LEDs lose their shiz on 10mA - but those are their own story...
So half of the "rated" current is my standard procedure...
AAAAAAA NO DON'T GIVE LEDS WHATEVER THEY'LL TAKE THAT ADVICE IS BAD AND WILL BLOW UP YOUR LED! D8
Diodes are non-ohmic and when they enter the forward-conducting state their conductivity shoots through the roof. (In the ideal case, the conductivity of a forward-biased diode is infinite, but in practical applications it's not but it may as well be.) You have to regulate the current through a diode or it'll pass however much your power supply can give it until it pops. This is typically done using a current-limiting resistor in series with the LED, or with a current-regulated supply (which is what your lab power supply is capable of doing).
The CLR approach requires you to know Vf, the forward voltage of the diode where it enters the forward-conducting state, and is calculated thus: (Vsupply - Vf)/I = R where V is voltage, I is current (specifically the current you want to allow the diode to conduct), and R is resistance.
This is much simpler with a current-regulated supply since you can just define the current limit and just make sure Vsupply > Vf so it has enough voltage to forward-bias the diode. The diode will very quickly hit the current limit at that point and your power supply will regulate it accordingly.
So, I’ve done plenty with LEDs, and never had this issue. I tie them directly into the 5v rail on my pc psu, and they don’t consume excess current. I’ve done battery powered LEDs on cosplay as well, and never had an issue of it slurping excessive amperage. I’ve always regulated voltage, not amperage. And this is what I see done in commercial products as well when repairing electronics. So, I don’t understand why people are saying they’ll take infinite current? Am I missing something here? Like even my electric razor has an led just slapped on to the two terminals to indicate charging, no current regulation at all. You seem to know what you’re talking about, so any insight into this discrepancy would really be appreciated.
@@haroldsaxon1075 "I tie them directly into the 5v rail on my pc psu, and they don’t consume excess current." Bullshit.
@@0x8badf00d I already got the answer to my question elsewhere. There are two ways to drive an led, constant voltage or constant current. Constant voltage driven LEDs do not require a blast as set voltage regulates the current. This is what I am used to. Constant current driven however does require a blast to keep voltage within a target range as left to itself it will burn out.
@@haroldsaxon1075 Hey sorry I didn't see your comment earlier. I'm terrible at checking the notifications panel as of late. Something about not enough spoons lol.
I've never deliberately used the constant voltage method of controlling a bare LED. If an LED works with constant voltage control, it usually either has integrated current limiting circuitry (some fancy addressable LEDs are like this) or has current limiting circuitry in series with it. Diodes alone pretty exclusively operate in the current controlled regime and require additional circuitry (either via current-limiting resistor or constant current source) in series with the diode for it to work via constant voltage control.
In the typical case, if you power a diode--any diode; not just LEDs--with a voltage supply that exceeds Vf, the forward voltage of the diode's PN junction (this is based on both the diode chemistry and the diode's construction), the conductivity of the diode essentially becomes that of the semiconductor materials used to construct the diode. This conductivity increases exponentially with applied voltage.
In the ideal case, we generally assume the diode's conductivity rapidly shoots off towards infinite when Vsupply > Vf, and ensure that the diode has some kind of current regulation in series with it to make sure the diode doesn't suck enough current such that I x Vf = P doesn't exceed the diode's maximum power rating. An ideal constant voltage supply can supply infinite current, and an ideal diode can conduct infinite current when forward-biased, so we have to regulate current through the diode somehow!
In the practical case, the conductivity is obviously not infinite. That can only happen in superconductors, and diodes are... not that lol. So a diode that is forward conducting will happily chew up as much current as the power supply can or will give it, or as much current as its semiconductor chemistry will conduct at the voltage provided, whichever is lower. If Vsupply approximately equals Vf, the diode may actually work without current regulation, as the supply voltage may not be high enough for the diode's chemistry's conductivity to exceed the diode's current rating! But as Vsupply increases, that conductivity increases exponentially, and you can very rapidly exceed the diode's current limit and cook it. This is not something you want to depend on, and you should always make sure you have current regulation in series with your diode to keep it happy. :)
As an additional useful nugget of knowledge, specifically pertaining to LEDs: more current = more lumens. Reason for this is you have more electrons jumping the transition gap across the PN junction and relaxing into lower energy states, and releasing their lost energy as photons of light roughly around the characteristic wavelengths of the diode's chemistry. :)
Another nugget of trivia: ALL diodes are light emitting. The ones that aren't by design are just (really shitty) IR diodes in disguise. Radiant heat is infrared light, after all, and the energy lost by electrons that jump the gap even in small signal and rectifier diodes has to go somewhere. ;)
@@calyodelphi124 It's late at night when i saw this, so im going to keep this short. So, Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points, right? And diodes have a variable resistance depending on the forward voltage, right?
Here's a chart i.stack.imgur.com/Z9STp.png
This means there is a safe operating voltage for any diode, including LEDs. Current increase as forward voltage increase. Neither number can simply run away to infinity while the other remains constant, they are inexorably linked and inseparable variables.
Here's another link with actual LEDs and the operating specifications. 3.0-3.2 volts is your safe range for them.
www.amazon.com/JABINCO-Circuit-Assorted-Science-Experiment/dp/B0827KYRFH/ref=asc_df_B0827KYRFH/?tag=hyprod-20&linkCode=df0&hvadid=416655650137&hvpos=&hvnetw=g&hvrand=12374622667303135318&hvpone=&hvptwo=&hvqmt=&hvdev=m&hvdvcmdl=&hvlocint=&hvlocphy=9028279&hvtargid=pla-870425772422&psc=1&tag=&ref=&adgrpid=94181909936&hvpone=&hvptwo=&hvadid=416655650137&hvpos=&hvnetw=g&hvrand=12374622667303135318&hvqmt=&hvdev=m&hvdvcmdl=&hvlocint=&hvlocphy=9028279&hvtargid=pla-870425772422
Watching a Chemist work a lab bench power supply to power LEDs will go right into the stash of material if I ever get to hold a lecture about basic electrical engineering. BEAUTYFULL !
You're toasting them. It's actually quite surprising that they don't burn off. In most things in electronics you get voltage specification, so you have motor that needs 12v, you give it, and it draws what it needs. LEDs are different because they have specified current, 40mA for this in the video. That is because, as you could see, at 11.5V they drawing almost nothing at 11.55V 40mA (440mW), and at 11.6V 1A (12W) dying out of heat, and when they heat up, this ideal 11.55V is changing, so exponential dependence with some temperature drift.
What you need is constant current supply, and you had one from the beginning of the video, just use those knobs on the left side, regulate roughly 40mA while shorting supply, and you have ideal supply for the LEDs. (V have to be also high enough, but it will never be too high with so small A). Now when you give them 40mA they will draw how much voltage they need.
The supply that you bought is also a constant current, but in terms of voltage (what was presented for single one, for 2 and 3 in series it's ok), and also in terms of current (800mA, is little bit more than 40mA), it's too powerful.
Yes thats the way to go
I'm screaming at the screen like a grandfather. "I'm getting too caught up with this 40mA, 900mA should be fine". TBH, constant current regime is always a bit hard to grasp for outsiders, and the threshold voltage marking feels like we could use the given voltage to control them.
Pretty sure he could run 22 stars on 900mA
@@RaunienTheFirst you run LEDs in series not in parallel....
Meaning that you give the string voltage until 40mA flow then you stop at whatever voltage that Happens to be!!!
Edit: spelling
Everyone seems to have the led wiring under control.
I'm just going to note that steel has pretty poor reflectivity in uv ranges.
A Teflon reaction vessel will capture more of the bounce light.
3:52 - fun or not so fun anecdote: one or two years ago in the nearby uni they used UV lamps to disinfect room after lectures. Staff forgot to turn off the lamp for a lecture and students got severe burns. Surprisingly the students were blamed because they didn't notify the staff during the lecture (even though burns show up some time after the exposure).
Of course the students got blamed, because if the Uni admitted the blame, they'd be liable for any and all damages incurred due to the exposure. (Immediate medical costs, and long term ones as well.)
@@The_Keeper yup, they didn't help them at all, even worse they still required presence on the lectures even though some of the students had to be hospitalised
@GsaUce Rug look up Silesian Medical University in Poland, that's a case from 2021 I think. The staff there is notorious for mobbing and other such practices. I don't really know what happened later as I didn't follow the case.
"surprisingly"
Man, I work in a lab with a lot of diode lasers and boy was this...interesting to watch. I think other people have mostly covered the issues and suggestions, but I'll throw one more in: LEDs and laser diodes like this with low current ratings are often very sensitive to shock discharges. Whenever we handle laser diodes, we make sure to ground ourselves well to avoid accidentally frying them. We have these conductive bracelets that we literally plug directly into the ground pin of an outlet. I'll also add that the things other people have said are basically why we are willing to spend a few hundred dollars on current sources specifically marketed as laser diode drivers :) they're purpose-built to precisely drive these low amounts of current through LEDs or laser diodes.
I would reach out to Big Clive for technical electronics, especially lighting. He usually does Collab's with stem people.
I was thinking exactly the same thing! I've been watching a lot of his videos and this seems exactly up his alley!
YES! I was thinking he needed an explosion containment pie tin nearby
Fourth vote for Clive
"You don't win the nobel prize for solving easy problems"
Sometimes you win it for inventing entirely new problems as well.
LEDs basically entirely block current until the operating voltage, then from there they have very little resistance. You want a big resistor in series and then the LED will be polite and take what it needs
Yes that would be fine. Big resistor means roughly 300 ohms here to drop 12 V at 40 mA, that way half the power is dissipated in resistor and half in LED.
This is the best explanation I've ever seen of how LEDs work.
Kind of? If you put a big resistor in there it's going to limit how much the LED can politely take. Throwing in a 150ohm res will let the LED draw ~45mA. Putting in a 300ohm resistor will drop that by half and the LED will only draw 23mA and will be a lot dimmer.
☝️☝️☝️☝️This. I've never used UV LEDs, but if they're anything like regular LEDs, you *always* put a resistor in the circuit with them, that's how you keep them from blowing up.
Google "LED resistor calculation". Based on the voltage of the power supply and the rated current of the LED, you can work out what size resistor you need.
you want a constant current source, which delivers 40mA
"You can't pump 30 amps through it and expect it to be fine"
I'm stealing this quote
Got the notification on my subs list, couldn't find the video in the videos tab, BUT i WAS able to find that the playlists was updated, and I found this! YEAH SCIENCE!
Can't wait for a Styropyro collab and UV-C lasers for the Cubane series.
Oops no more reagents! Vapourized by the 60000000W laser they designed together.
😂
basically, an LED acts like an open circuit until you hit a voltage threshold, the forward voltage. Then it behaves like a constant voltage source (with a small internal resistance) with a voltage equal to its forward voltage. So any voltage greater than its forward voltage would cause it to draw huge current (obviously a real LED has a tiny internal resistance blah blah). Thus you need to add a series resistor to limit the current. Say you need 40mA of current to drive the LEDs and the LEDs have a forward voltage of 6V, and you want to power it from 12V, you will need a resistor that drops (12V - 6V) = 6V while passing 40mA, which is (12V-6V)/(0.04A) = 150ohms. I hope this makes sense!
This.
Doing the math will save you so many headaches down the line, and there is a lot of reference out there to consume. I'm sure that Will or Backyard Scientist could help.
You don’t need a current limiting resistor if you use a constant current supply. That said, a string of 40 mA LEDs requires a 40 mA constant current supply (LED driver), not a 500 mA one. What I don’t get is why he isn’t using that bench power supply in CC mode instead of ordering stuff online…
@@mass1s it's not clear to me that his power supply has a constant current mode. It looks old enough to be of a vintage were CC mode was only in high end models.
This video has to be one of the best advertisement for the company that manufactured these leds: pumping 1A inside a 40mA rated LED without instantly frying it is quite a feat, plus they are UVC or UVB which have been notorious for being fragile !
Looking at the specs of your LEDs I'm also somewhat suspicious about the announced output. Typical efficiency for LEDs in that range tends to be below 4%, and yours have about 8% efficiency based on the nominal output/(nomimal voltage*amps) . My guess would be that the nominal output is exagerated but the only way to check that is with an optical power meter.
best advertisement that they sold him the wrong product?
I'm a PhD student doing research on UV LED growth and I just want to say that the main challenge of high efficiency UV LEDs right now is partly due to light extraction (i.e. using UV transparent packaging), but mostly due to the difficulties in growing high quality AlGaN p-n junctions and quantum wells. P-type doped AlGaN is extremely challenging, as the acceptor state of the Mg dopant is deeper in the band gap than with p-GaN and thus getting sufficient amounts of holes in the material is extremely challenging, especially at higher Al compositions. Surface roughness is also a major issue, as we need very good interfaces in order to get good heterostructures for AlGaN quantum wells and quantum barriers. There is also the issue of defects in the material and how to get rid of them, as they result in nonradiative recombination. You kind of just missed the point on my entire PhD's work by saying that light extraction is the sole issue lol.
The major reason it's took so long for progress in UV is mostly due to the fact that research in AlGaN growth is extremely costly. Each growth I do is hundreds if not ~$1000 in materials and consumables (and I do maybe 4 growths every day). This is why there are only very few research groups actively pursuing UV growth and it's mostly companies that are doing the research.
To be fair, no one outside of a person working on UV LED growth would know any of this. Even reading this is extremely difficult because of the large amount of context required for every part of the process.
(I personally don't understand about 1/2 of this what with p-n junctions and quantum wells. Its just magic words from a magic man)
I couldn’t agree more. Or less. It just sounds like brilliant word salad to me.
Thanks for understanding this for me!
@@giggabiite4417 If you just replace the words “quantum (anything)” with “weird stuff” then it’s a little easier.
Also wanted to say exactly that! Btw whats up with these quantum waves, aigan(or something like that) and p-thingys?
Yeah, you'll want to use the timer to cut power to the driver. The driver has a voltage range and a current rating, and will provide whatever voltage it can within that range to make that current flow. So if you open a switch in its output line, it will instantly swing to it's maximum voltage, trying to overcome the new infinite resistance of the open switch.
and that's exactly what's fried the first LEDs
No need for timers at all. Just properly drive the LEDs and they wont overheat.
I can't wait for my UVC backlit TV in 10 years.
8:14 Resistor. You need a resistor. Diodes of any sort have an exponential current versus voltage curve. Adding a resistor makes it much easier to control the current.
8:47 No, the 40mA number is very important. That power supply has a current limit8ng knob. Use it.
As an electronics hobbyist who dabbles in chem, this was extremely entertaining but also a little frustrating. Imagine watching someone perform a distillation but they know gas volume and temperature are linked so they crank up the heat and hope their apparatus manages itself. You can steadily see their glassware show signs of cracking and heat stress and you know it's going to all fall apart soon but they are just happy to be getting product.
I know electronics are complicated, but just like chem it's an applied set of knowledge on physical operations - the systems can be reliable but for a while you're going to be breaking things and I think that's the best way to learn. Fun to see you try something new!
Right when I was about to start being productive. Can't say no to a new cubane episode
8:49 Watching and this kept tickling the back of my "danger, will robinson!" instincts.... why are you operating an unprotected PCB on top of a flexible conductive surface (foil)? I literally can't think of an easier way to have an unpredictably timed catastrophic short circuit...
Those aluminum substrate PCBs are conductive anyway. It's also made my heart skip a beat when I saw the aluminium foil, but it's not an issue with LED modules (the lack of proper heat dissipation is, though).
A lot of people are already talking about other issues, so I'll throw my electrical knowledge at another thing you mentioned:
If you don't want to burn up a bunch of expensive components it's relatively inexpensive to get circuit breakers to put in your test circuit when you set the whole thing up for the first time. it helps ease the worry of frying something you don't want to replace and also helps minimize the risk of a fire, arcs or bad results. The basics are:
-get the right size, the rated current of your breaker should be ~120% of the load listed (take the mA and multiply by 1.2x, that's about the right size).
-don't mix AC and DC, breakers are one or the other just buy the one that matches where you're putting it (You can probably ask any of the electrical people in the discord or something if you're unsure)
-wire the circuit up with the breaker open (not passing current) and in a way so that the breaker opening again will stop all current
This way you can close the breaker once the whole circuit is made, be less at risk of shocks and arcs when doing so and if it's sized right and you're a little lucky it should save your expensive components. Better safety for your parts, and more importantly better safety for the user!
when's the last time you've seen a 50mA "circuit breaker"? maybe you could use a very small fuse, but even then, the exponential runaway behavior of LEDs will probably kill it before the fuse can even warm up. fuses and breakers can protect against *thermal* damage where it's ok if it takes a few milliseconds to kick in since the component will have barely heated up. but they certainly can't protect a semiconductor device from electrical breakdown, we're talking microseconds or nanoseconds then!
remember, the magic smoke is often just a symptom, the device fails first, then a high-current failure mode causes the heat and lets out the smoke. a breaker will stop the smoke but not necessarily the failure
I don't think they make circuit breakers below like 10 amps. Although they certainly do make small *fuses*.
@@thewhitefalcon8539 I think they make smaller (down to a few amps) breakers for individual appliances and stuff, but regardless that's still way too high so your point still stands.
His PSU comes with a current limit. He just never ever used it.
I have nothing to say except CUBE, and that this episode reminds me of how when you badly overdrive a red LED it will try becoming green before exploding
From an electrical engineer, put a resistor in series such that the difference between the supply and the nominal forward voltage of the LEDs divided by the resistance gives the recommended current. Then heat sink the LEDs and adjust the supply so that in steady state the LEDs get their rated current. If they are all in series, this is much easier.
LEDs work best with current regulation and that is what a resistor does crudely with a fixed supply voltage. Make sure the resistor is rated for the power going through it.
(Vsupp-Vled)/R=I
Presistor=R*I^2
If you do these things, a supply with a constant current mode is not necessary.
Firstly, congrats for 'giving it a go'.
Having said that, I suggest you grab another full set of UV LEDs as I suspect those ones will be pretty screwed (you've thrown over 2000% of their rated current through them)
As others have already said in their responses, 40mA MEANS 40mA and anything more starts 'frying' them.
Your benchtop power supply is not accurate enough on its current setting
It's pretty simple to create a constant current circuit - 2 * transistors and 2 * resistors, but the suggested LM317 style is even easier. 1 * LM317 and 1 * 33 ohm resistor will give you 37.8mA. The power supply can be any voltage between about 15V and 40V so your 19V laptop supply will be fine.
LEDs are most likely failure cause is overheating. Not the temp of the heatsink, but the temp of the tiny silicon die itself.
The hotter they are, the more current they will consume so it's a BAD idea to run them in parallel - due to manufacturing tolerances, one will consume a TINY bit more current than the others in parallel. This will make that one a little hotter, so it will consume even MORE current -> RUD (RAPID UNSCHEDULED DISASS$EMBLY!)
You CAN safely place LED in series (but you'd need to increase the voltage of the power supply accordingly)
The MOST 'finicky' LEDs IMO are LASER LEDs. These will 'let out the magic smoke' if you even THINK about over driving them! (e.g. the cheap red ones you can buy from Jaycar will die in less than a millisecond if they're over driven by only 2mA! Don't ask me how I know ). From what I know, the LASER LEDs die by burning the mirrors at each end of the cavity LONG before the die itself is fried.
The LEAST 'finicky' LEDs IMO are the infrared LEDs like those found in your TV remote. These are normally rated around 50mA or so, but are often driven at over 1000mA (1A) to make them 'bright' enough for your TV to sense. The key is that they're pulsed off/on rather than CW (Constant Wave) so they don't get enough TIME to overheat and die.
In your UV case, you're MOSTLY looking at CW operation. (Pulsing LEDs involves timescales of microseconds to millseconds while you're talking about multiple seconds!)
In short, do a simply google search on "LM317 constant current" and run with that
Chemists getting the current 2 orders of magnitude different. And buying worse power supplies while having a perfectly working one. I still love it.
Nail curing lamps might be a good alternative. My lab uses them for most of our photoreactions under UV light. They're like ten bucks off of amazon. They usually have around 5 mW/cm^2 and a lambda max around 365 nm
Can confirm, am in the same lab and used the same lamps successfully.
They seem just too long a wavelength though, which is a shame because yes they are really quite cheap and available. We need shorter than 350 nm
NO. Those nail and acrylic curing LEDs are all ~400nm and will be useless for his purposes as he rightly suspects above. HE NEEDS THEM SWEET SWEET HIGH ENERGY 300nm CANCER RAYS.
@@ExtractionsAndIre this is probably insane but if you do it and get away with it youd be a legend
But did you consider cooling the LED enough to decrease the wavelength
The wavelength difference isnt big enough for this to be completely infeasible (I think) but hey it will be insane if you do it and sounds very doable if you have some kind of temperature buffer
@@aduantas it's the nail salon industry. They've spent decades sitting around huffing solvent and epoxy fumes thinking a surgical mask will protect them. They don't care about your 'science' and 'health & safety'. They're gonna look fabulous in a casket.
Original Cubane ran 64-bit. Make sure you're using 64-bit chemistry not 32-bit
He's still using analog chemistry
Obviously, 64 is a cube and 32 isn't.
I have transcended the anger and frustration, and embraced the chaos of this video. This is Tom's baptism of Magic Smoke™. Now it's a matter of time until he combines electronics and chemistry and turns to the dark arts of electrochemistry!
God bless electrochemistry!
You need a constant current source or at least resistor between power supply and LED... and a (cheap) multimeter for mA measurement. The LED has Uled = 12V and Iled = 0.04A and the supply e.g. Us = 15V . Than you calculate the voltage drop for a resistor Ur = Us - Uled = 15V - 12V = 3V , with that you can calculate the needed resistor value R = Ur/Iled = 3V / 0.04A = 75Ohm. The power rating (what the resistor NEEDs to dissipate) is Pr = Ur * Iled = 3V * 0.04A = 0.12W. So you need 75Ohm resistors to limit the current to 40mA for a 12V led at a power supply voltage of 15V, and the resistor should have 0.12W at least (0.25W is a common value btw )
This.
I would go for a 5W ceramic one, just to be safe. They also are able to run hot all day long and can dissipate heat way better.
Hey, I feel your pain man. Trying to do electronics with almost zero experience is basically impossible because you don’t know what knowledge you are missing. My suggestion is to ask for a circuit design or even someone to build the circuit for you. Best of luck man!
To be fair, I think it would be about the same to try to jump into chemistry without any background -- especially considering all the varied naming conventions for things.
E&F x Medhi collab
@@blackroberts6290 OMG he should see if Electroboom would do a crossover yes! Good thinking! i would pay to see that!
3:48 fun fact, they did this at a Hype Beast fashion show not knowing what kinds of tubes they were using and sunburned everyone's eyeballs causing excruciating pain the night after
Funny chemistry man my beloved
00:00 Intro
00:59 *LED history*
02:41 UV LEDs
05:01 *AliExpress haul* - LEDs
06:52 *Power*
07:45 bench power supply test
09:00 *Ebay haul* - driver & timer
10:43 *Heat sinks*
12:13 *Unexpected explosion*
12:59 now we've got one
13:37 *Driver* - earth?
14:05 oop sparks
14:28 officially cooked
15:40 *Timer relais*
16:13 highlighter demo
16:36 *Reaction chamber setup*
18:00 Outro
"Feels wreckless to not have one..." says the man who runs Ex&F. I think that almost warrants a Can Test.
I admire your tenacity to expand beyond your comfort zone. To admit you are underprepared in a field, yet willing to dive in, make mistakes, learn from them, and learn the correct action are the hallmarks of a great scientist.
Whatever uni you went to, I'm sure they're proud to count you amongst their alumni. Good on ya, mate!
Sincerely: a fellow chem PhD student in the USA.
Just limit the current at 40 mA and set the voltage above the drop off. You can measure the drop off voltage with a simple multimiter in the diode setting.
To be clear, a lot of bench power supplies can do this, but not all. If yours does, it will automatically reduce voltage until it is pushing no more than 40ma (or whatever you set the amp limit to), so even if you set the voltage to something absurdly high it'll be safe. The power supply will use whichever limit is lower, though, voltage or amps.
@@oasntet Isn't being able to regulate both voltage and current the whole point of bench top power supplies like that one?
@@Pr3Va1L That one can, but some don't have CC regulation. Even in the case of some of the ones that have CC regulation, it's often difficult to actually get it dialed in to anything specific. The behavior of the pot at the end of its stroke and the tiny knob can easily make it difficult to set a current limit like 40mA on a supply where the scale maxes out at 5 or 10 A.
Yeah you need to short the output through a multimeter’s current input and then fiddle with the current limit knob until the multimeter reads 40 mA. Can’t rely on the shitty built-in analog meter.
the multimeater might not be able to pick up a forward voltage as big as 12V
an entire video on achieving literally nothing except for blowing up a bunch of LED's and power supplies. not even touching any chemistry. mind blown, keep up the great work, looking forward to the next one.
Haha yeah, I miss chemistry
My brother in Christ you need a resistor to drop the current to 40mA
The GameCube intro was sped up just enough that I didn't immediately realize what was wrong but I knew _something_ was. Bravo.
"we have to leave ourselves some room for error"
mY bRoThEr In ChRiSt , you should've bought 5 !
Also, you're drawing high current because you're powering them with more than 12-14V.
One is none, two is one.
Get redundant in a big way.
I like when you go on a tangent and then realize and the video cuts to the next thing, you must be aware of it cuz you always leave in a perfect second or two as you are actually realizing it as you make the video, I love it lol
i love how everyone gets distracted by him overdriving the leds but noone is mentioning that he is literally doing all of this on a piece of aluminum foil
I was so confused to why.. It's useless as a heat sink because it's so thin and it's pointless to have a reflector behind the leds. If it was meant to compensate for not connecting the ground wire to the Aliexpress power supply to reduce the risk of electrocuting himself it achieves the opposite.
@@bjornSE not only that, its an esd hazard, espeically with expensive new gens parts like those leds ^^
im guessing cause hes doing all of this on his chemistry bench
I think I figured it out! It's a diy anti-static mat! PSA: the purpose of an anti-static mat is to prevent electrons from stampeding to ground through sensitive electronics and works by providing a slow path to ground for high energy (thousands of volt) electrons. Aluminum foil provides a low resistance path for any electrons to go wherever they feel like so it's not the best choice.
@@contomo5710 Haha, I saw your reply after I posted mine! :)
@@bjornSE this one is more a static mat then :P aluminum would oppose a lot of capacitance unless grounded, i doubt he grounded it
You don't see many constant current drivers at such low current since it is much easier (at very low current) to use a resistor to limit current. You use a constant current driver at higher currents where you'd need a giant heat-sinked resistor that would cost a lot of money (and waste tons of power). At 40mA wasting a tiny bit power is fine.
I'd recommend putting a cooling fan over the top of the can as well. Keeping my fingers crossed there'll be some positive results on the [2+2] step soon!
That power supply you have looks like it does constant current, you can (with the LED disconnected) turn the voltage to about 15 volts, then turn the current knob down until the voltage drops. After that you can hook up the LED and turn the current knob up until the desired current. this way the LED can run at as high or as low of a voltage as it wants (largely dictated by LED temperature) and the LED current is controlled by the power supply.
It's 115am in NZ. And I need to be asleep. But this latest Cubane update can't be ignored.
Lots of people say its painful to watch, but as someone that doesn't know much about electronics myself (or any of the other mistakes done), the journey tend to be the most enjoyable part of these series. Keep it up Tom!
It feels like eye sand for a couple days when you get UV exposure, also sun burns. Aluminum reflects and glass absorbs pretty well. Made some UV cabinets for 2020 and cooked me and my lab mate a bit.
Doesn't matter how many chips you burned, you got the job done.
You’re looking good bro. Keep doing whatever you’re doing.
love your chemistry adventures, keep it up dude.... after watching the video maybe stay away from electronics lol....
I love the genuine "I don't know what im doing, let me know how to fix anything". it feels real, and the janky electronics is funny.
The LED star needs 40mA and not a mA more (ok, it depends what the data sheet says, but 10-50mA is all I would accept). I am surprised they would survive 900mA for long at all. The volt drop of the LED is determined by semiconductor physics.
Gotta love the Aphex Twin background music
I died inside when you gave those 40ma leds a full amp, poor things
LEDs: "Professional installation required."
Tom: "Didn't say what kind of professional, let's go!"
I looked at the listing for those LED stars and the three different voltage versions seem to be different combinations of series and parallel chips. I don’t see any balancing components on the boards, so the 12-14V and the 6-7V versions probably have parallel LEDs that are not going to share current equally. You would be best to avoid those and only use the 24-28V 20mA kind. Since this is such a low current, you don’t really want a power led driver. You want something closer to the circuit you would use to drive an indicator LED, but at a bit higher voltage. The easiest way would be a 36V DC power supply and a ~500 ohm resistor. You should be able to run a bunch of stars off one power supply this way (each with their own resistor of course). I recommend at least 1 watt rated resistors. Technically you would only be at 0.2W, but normal 1/4 watt resistors are going to get really hot that close to their rating.
cubane is stripping this man of his sanity. tom has become the cube and by the emperors light the cube will sustain.
Love the channel! Since I simulate electronics for a living maybe I can provide a chemistry-ish analog for you that doesn't offend too many people. Internally the LED as two regions, an n and p region put together. The n region is electron rich and the p region is electron deficient. These two regions can be modeled as electron and hole "gases". The concentration of these two "gasses" are determined by the internal voltage but also by the external voltage you apply. The internal and external voltage is exponentially related to the electron and hole concentrations in the two regions. At equilibrium two regions form a depletion region where the electrons and holes "react" resulting in a continuous carrier distribution and barrier between them. In non-equilibrium conditions (when voltage is applied) this depletion zone can be modified and is eventually eliminated with enough applied "positive" bias. With no depletion region the electrons and holes flow into each other's region and recombine (a sort of reaction) producing light and heat. In direct bandgap semiconductors light is more efficiently produced than heat (the reverse is true for indirect bandgap semiconductors). This "reaction" is still exponentially dependent on the applied voltage.
A direct chemical analog for this behavior might be difficult for the equilibrium conditions but non-equilibrium dynamics should make sense in terms of reaction kinetics in chemistry. You have two reactants electrons and holes forming light. There is a reverse reaction from light into electrons and holes but the light can leave the diode while electrons and holes cannot (for lasers there is a third reaction possible, stimulated emission). Regardless, all this should point out the fact trying to understand this "reaction" in terms of voltage is difficult. You want to provide a constant rate of electrons and holes that the device is rated for turning into light. In this case its 40 mA. The device will drop a certain amount of voltage at that current rating (something around the built in potential). So, you could linearize the LEDs behavior at that bias and setup the circuit such that the LED is modeled as a resistor with the voltage from your power supply at 40mA. Other people recommended using a series resistor to help provide the proper bias for the LED. That would work but you could also use a current mirror or some other type of constant current source. These type of electronic circuits are not too difficult to make. A current mirror is two transistors and a resistor. But you also should be able to find a proper LED driver or driver circuit online. The bottom line is you need to deliver the proper current because the "reaction" is exponentially dependent on voltage. Hooking an LED directly up to a voltage source isn't the best of ideas, although this setup might work for your purposes if these LEDs are robust enough... I'm an engineer so I understand the pragmatic argument for things "just working"
learning so much about LEDS. Give me back my oldschool UV tubes.
There was a night club in Hong Kong that used a bunch of UV C lights.. didn't go well.
It was a huge event/exhibition- I remember reading about the fallout. The designers decided they liked the aesthetics of UVC lamps, so they put up a ton of those 6ft Osram UVC tubes and wrecked everyone's eyes and skin.
@@ephjaymusic I think I heard about it on ua-cam.com/users/Bigclive
As an electronics nerd, I was screaming internally from the moment you decided to run the LEDs at 900mA.
My Chemistry Teacher once told me that cubane is not possible because of the angles. Let's prove her wrong
Sigma supply it and its made on pilot scales
It's possible but it doesn't like existing. That's why it's so good for making explosives from it.
Bruh! That's so misinformed! Thermodynamic instability doesn't necessarily mean kinetic instability. Cubane is a good example, [Co(NH3)6]3+ in acid solution is another good example.
@HammerTh yeah and when he makes it he’ll definitely try to nitrate it 😂😂
@@iFloxy Octa-nitro-cubane would be a definite milestone if it can even exist.
E&F out here pondering his Cubane
DUDE NOOO! Your using the powersupply wrong! It has a knop vor Current limmiting.... drive it with CC! Not with CV!!
I don't know what a cubane is or why you want it but I am so invested in it's synthesis
Wooooo! New Cubane episode!
Best wow!
Muchest yes!
Now with 10^40% more cancerous photons!
Cancer light, best light!
Photons, phoward!
@@lairdcummings9092
Fact: Ionizing photons kick off electrons.
Fact: This requires physical work.
Conclusion: Cancer photons are photons that work for you and the whole family.
@@AtlasReburdened better living (?!) through physics!
As someone with a background in Electrical Engineering, this video is insanely funny to watch.
I always love more of Tom!
A note about LEDs; there's a direct correspondence between the energy of the emitted photons and the forward voltage drop across the diode. As you demonstrated with the sharp current response from your power supply, the forward voltage drop of each of these 4-in-series diode arrays is ~12V, so the voltage you would use to calculate the current draw with Ohm's law is ((applied voltage)-12). The manufacturer gives the upper limit as 14V, and a maximum current draw of 40mA, so (2V)/(40mA)=50Ω; when you drive it at ~19V, the current was (19-12(V))/50Ω=140mA
Generally speaking with LEDs, you adjust the voltage until they're drawing their rated current. You can probably get away with over driving them some for a while but they won't last near as long. That driver you hooked up to them will output whatever voltage it can in its range to force the circuit to the 900 milliamps it's rated for. Which is significantly higher than what you showed for the LEDs. They may last a little while but you would almost be safer using your power supply and cranking up it's voltage and using the current limiting knob to hold it somewhere around 50 milliamps.
Thank you, this explains what people are screaming about ballasts for LEDs. I’ve always voltage regulated LEDs and could not fathom current regulating them instead on variable voltage.
@@haroldsaxon1075 I'm glad I was able to help. :)
10:36 really showing your Aussie side. Love it, great video as always cobber x
As an amateure electrician I would recommend giving the LEDs a little bit less than their stated voltage. Although LEDs can work for a long time, sellers sometimes exaggerate the amount of Volts they can comfortably handle so they will be brighter, but this will severely decrease their lifespan
Sufficient cooling also effects lifespan significantly
ssry f english 2 language
I think you mean current rather than voltage?
@@woosix7735 well, aren't they both equally important?
Dig around the indoor gardening forums lots of folks build their own lights so they are probably the best place to find the proper design criteria. I am no electrical egi'nerd but electrified sharp metal is very dangerous, especially with no ground. Take the time to debur or round over any edge that could cut you, I have a slice on my hand right now from just lifting a BBQ. A cut can be an open door to your heart for the angry pixies that come out of the wall.
These do require the very low current of about 40 mA per star.
mechanical engineering reporting in with "I dunno if that's right, but der blinkenlights are blinkenlighting so maybe it's right enough for cube time"
The 40mA is critical, if this is specified as the maximum current the LEDs can stand. LEDs are non linear devices, which do not follow Ohms law. Once the voltage across an LED reaches its forward bias voltage, any further attempt to increase this voltage will result in an exponential increase in current through the device, resulting in its destruction. The current through low power LEDs is usually limited using a series resistor, but this solution is inefficient for high power LEDs, where the current is better controlled using an adjustable constant current supply.
When driving an LED, the primary concerns are running it below its maximum current rating, and ensuring that it can dissipate sufficient heat to avoid exceeding its maximum operating temperature. It is recommended that power LEDs are mounted on a metal heat sink, usually aluminium, using a heat sink compound such as that used to mount computer CPUs. . (Note, it's important to avoid putting heatsink compound on the LED window, as it tends to react and frost it. )
When using a constant current supply with more than one LED, , the LEDs should be connected in series. In this circuit configuration, each LED will pasx the same amount of current, and the supply voltage will automatically adjust to that required, so that each LED in the chain is operating at its stated forward bias voltage for the operating current. This is conditional on the constant current open circuit supply voltage exceeds the sum of the LED forward voltages in the series chain.
Being such compact devices, LEDs are very unforgiving, when driven with excessive current. They over heat very quickly and burn out, so acquit heat sinking is essential.
To avoid excess heat dissipation in the constant current power supply, it is better to use a switch mode constant current converter than a linear constant current supply.
An approximate calculation, suggests the forward voltage for the LED emitting 420nm UV should be around 3.9 volts.
( Vf=hc/(e*wavelength), assuming no charging pumping. This equation is only a guide to the bandgap voltage.)
If your high power LEDs are 3watt, it suggests that each package has two LEDs in series, given a effective forward voltage of 7.8 volts.
As these are rated at 3 watts, it suggests the forward current maximum is approximately 0.38 amps, so maybe the actual current is not 40mA, but 400mA. Certainly 40ma seems rather low for an power LED.
A quick view of the LED data sheet would confirm these assumptions.
P.S, I froze the picture of the LED data sheet, but the data on the segment of the sheet I could see did not make much sense. It is difficult to reconcile the 3 watt rating with chips taking 40mA. There are four LEDs on each die, but it is not clear how they are wired.
To avoid the instantaneous destruction of the LEDs, due to excess current, it is a good idea to fit a series power resistor when performing the initial tests. A 2.5 ohm half watt resistor would provide about a volt of supply rail stabilisation over the rated Vforward of the LED package.
Tom does it right. Doesn't post often so when he finally does, it's like Christmas morning
Keep up the effort! Chromium coated (which seems to be the case here) is actually quite an optimal (low cost) choice for this range of wavelenght but keep in mind that very small oil film can absorb it quite well so try to clean the can with acetone before use (and close it during use of course)
amazing job king not only an incredible chemist but an amazing electrician too
turn the psu to 12v and the current limit to 30 mA.
that is it >w<
also
led is an diode, their current grows exponentially over the voltage.
edit again:
their current also increase over temperature.
This very good. PLEASE make more.
Absolute joy
I'm not an electrical engineer but I would put a small resistor (around 600 ohms or lower9 before the LED to limit the current, and I would also keep a multimeter measuring the current just to make sure you're running them at with the proper current. Also are you supposed to see the blue glow from those LEDs or it's just the camera picking UV?
It's just the camera and some extra light having its energy lowered by the plastic. And yes I would also recommend using a resistor. Since these are voltage dependant leds keeping the current regulated and the voltage variable is important.
@GsaUce Rug they're not though
Just an FYI on the current thing, when you read an LED specsheet the most important thing is the voltage, because an LED doesn't draw current until it hits a threshold voltage and by that point it is stable (ish)
Now a normal source you will find in a lab(or in your socket) is a voltage source, that means it produces a fixed voltage. The current produced by the source is determined by whatever is connected to it. So in the LED problem the source produces the voltage, the current output is exactly what the LED needs.
To be honest with the setup of this video, I was expecting a little more explosions and fire. ;P
Probably the series I'm most excited for next parts
LEDs aren't light bulbs, they don't regulate the current that passes through them. You need to put a current limiting resistor inline with them. The value of that resistor should be roughly (supply voltage - total voltage drop of LEDs in circuit) / (desired current in amps).
If the total voltage drop of the LEDs exceeds your supply voltage, current won't flow.