That would be synonymous with a break in the circuit. No electrons available means no conductor there. It’s not like an electron gets in his yugo and heads off to work. For current to flow, it flows everywhere in the circuit simultaneously so if there’s a break then mr yugo doesn’t leave his driveway, let alone get to Starbucks for Java.
I hate auto correct.
P = V × I (volts by amps)
Leds are generally killed because of overheating to the point of melting; they “get fried”, sounds familiar?
If cooling is adequate enough for it to stay within safe “thermal envelope”, nothing should go wrong in this respect.
For a led, overvoltage usually means crazy high overcurrent so they get nuked by the excessive power dissipated through them. Now please bear in mind that, if we feed them with “weak” power supplies like, for example, relatively high internal resistance batteries, this may not happen because a lot of voltage will be dropped in the batteries: inneficiently lots of “wasted” power in them.
No overcurrent? No overvolting, simple as that because what the driver does is tuning the voltage output for a certain voltage drop through a precisely known value & perfectly linear shunt resistor be measured and succesfully compared with a reference.
All of this is as far as I understand/know.
Cheers ^:)
P.S.: Wow! Guess I had not refreshed the page in a while… 
P.S.: fixed the Power = Voltage × Intensity Rufusbduck, it was a literal translation of what my brother taught me as a kid (Potencia = Voltaje × Intensidad).
The problem is that the thermal resistance between the die junction and thermal pad is great enough that the die will blow before the pad can transfer excessive heat away. Poofing from over voltage is virtually instantaneous but thermal transfer isn’t. the thermal pad doesn’t even have time to warm the copper. It nukes from the inside out.
“I” refers to current not intensity.
Don’t forget thermal runaway, as the die overheats and breaks down current continues to rise. This is why drivers are current controlled devices rather than voltage controlled. Changes in the temp of the die would otherwise allow a constant voltage to supply a rising current. Drivers fix the current regardless of changes in the die. FET drivers just happen to operate right on the cusp of the point of failure, a smidgen more current and no super cooled star could save the led. It probably would survive longer, with a thermal resistance of 2.5C/W junction to thermal pad at 20W output the die would be 50C hotter than the thermal pad. At some point the thermal resistance of the copper mount will allow localized heating to overcome forced cooling but I’ve no idea at what point. I do know that even mounted on a massive sink LEDs still have a very limited range above spec’d voltage. I don’t know enough about semiconductors to know how excessive voltage affects them on a molecular level which is where they operate. At some point I went beyond what I know and passed into the realm of BS and lost the way. In any case, I doubt there’s any way any of us could successfully power a 3V led directly from a 6 or 12 v source. That’s assuming no artificial limits on cell current. Personally I think even 3 10180’s in series could summon enough current for the time it took to poof any 3V led in any flashlight we can lay our hands on.
Over volting by a little (half to one volt) does lead to over amp as the voltage is within range of design specs. Over volting by ~400% is instant poof.
This thread doesn’t seem to be a simple explanation any more. I guess that’s the nature of electricity. It can start off with simple terms but very quickly gets highly technical.
CC/CV is usually seen in relation to the charge cycle.
Watts Up With The Water?
I hate to be a wet blanket. I mean this is flowing so well…
But to pump up the volume, I can’t resist floating a drop of my favorite Smart Guy, Richard Feynman.
Forewarned: You might want a life preserver to swim in Feynman’s pool!
But if you want to have this electromotive force concept soak in thoroughly, I can think of no more pure, clean, highly potable source than this:
”
The Feynman Lectures on Physics”:The Feynman Lectures on Physics
Specifically, Electromagnetism, The Flow of Wet Water, The Flow of Dry Water (hey… “Feynman”!!), and Electricity in the Atmosphere.
I would highly recommend anyone who wants to Know Things, pay careful attention to everything you can read by Feynman. It’s not like any normal human being can absorb all of it, but it’s a useful goal. (And it beats the ever-loving dog snot out of pocket monster hunting!!!)
Dim
Great Link
Thanks for reminding how wonderful the Feynman Lectures are.
Thank you The Miller for putting this all in one spot here,
You are certainly one of the best and always so very organized, and quit noticeably extremely forward thinking,
This was something I really wanted to finish reading but forgot where in that huge thread it was, so after many query search’s I finally lucked out.
So just wanted to say thanks The Miller, very good reading for sure…
Been reading for the past 2 hours and still can’t quite figure out something. If I have 2 18650 batteries, one rated (10A) & one rated at (20A) which would be better for my triple ? The 20A would run brighter, and the 10A would run longer ? Or would they both run the same with 1 draining faster ? Can’t quite grasp this because I keep getting vaping stuff and most sites discribe the various kinds, benefits, drawbacks and such.
Could one of you more knowledgeable folks help me understand this as far as my lights are concerned?
+1
I would add Chaos Theory and Benoit Mandelbrot and his Mandelbrot Set. 
This is all from understanding Complex Numbers.
From my limited knowledge and I hope others will correct my misunderstanding I will venture an answer.
If you are using only 1 amp a higher capacity battery would give you longer run time. Some of my flashlights run at 5 amps and what I am after is longer run times at high current.
You might want to visit HJK’s excellent 18650 Battery Comparator and see how different batteries deliver current.
http://lygte-info.dk/review/batteries2012/Common18650comparator.php
The amp rating on the battery really means nothing when it comes to how long it will run something. It simply means that you can draw the stated current without the battery being damaged.
Now higher amp rated batteries will generally provide more power to FET driven LED’s but that is because they have lower internal resistance.
That really depends on what you mean by “better”. Assuming the circuit can flow the full 20A, it will be brighter depending on the point on the output curve that current hits vs. the 10A point. It’s not likely the Datasheet will rate output at those levels, so you’ll have to rely on HKJ’s (etc. here — you know who you are) testing to estimate the brightness at those Currents.
OTOH, if you want the light to last, you’d need to look at the Capacity of the two 18650s to estimate how long. The 20A would deplete that Capacity (to some agreed-upon level) ~twice as fast as the 10A, assuming equal Capacity (and that the circuit — including the battery — can pass that much).
The real problem comes in when you understand that the 10A and 20A are just the rating at which a large percentage of those particular 18650s can source Current without damage. If your circuit will flow 20A, it will probably get close to that out of the 10A battery, meaning don’t get used to having it around. OTOH, sinking as much Current as your battery can source puts your circuit in the category of a “dead-short rig”, which seems extreme.
Taking away the assumptions, what are the rated Capacities of the 18650s and how much Current can your circuit actually flow?
Thanks for the assist…Think I understand now. I’ve seen that battery comparator a while back. I’ll take a closer look this time.
Thanks guys….
I see (here and elsewhere) things posted sometimes that are incorrect, but because I don’t want to start a fight or don’t have the time to type out long responses I usually try to steer clear, but this thread has real promise it would seem.
I’ve only made it part way through the first page, so I don’t know if it’s been covered, but much of the confusion about drivers and LEDs stems from a misunderstanding of how LEDs actually behave, or operate, IMO.
An LED can most simply be described, IMO, as acting like a gate or bridge. There is a certain amount of voltage (V) that is required in order for the diode to allow current to pass through the gate or cross the bridge. Until you reach that voltage, the gate remains closed and no current crosses the junction. As you increase the voltage (V) you reach a threshold where the gate begins to open, as voltage continues to increase you are opening the gate further and further. How much electricity flows across the LED is determined by the makeup of each individual LED. Current is measured in amps (A). The wider the gate is opened (meaning the higher the (V), the more (A) can travel through or across the junction.
We control the amount of current going across an emitter by opening or closing the gate with drivers. Imagine the gate as voltage (V), and current (A) as individual electrons passing though the gate.
A Constant Current driver doesnt pay any attention to how far the gate is opened or closed (voltage), so long as a certain number of electrons pass through the gate (A).
A Constant Voltage driver doesn’t care how many electrons (A) pass though the gate so long as the gate stays at a certain spot in its opened or closed range (V).
This analogy is of course oversimplified and imperfect, and there is a better analogy I’m thinking of, but it’s a lot longer, so hopefully this will suffice and explain the relationship between LEDs and drivers more clearly.
-Michael
” As you increase the voltage (V) you reach a threshold where the gate begins to open, as voltage continues to increase you are opening the gate further and further. How much electricity flows across the LED is determined by the makeup of each individual LED. Current is measured in amps (A). The wider the gate is opened (meaning the higher the (V), the more (A) can travel through or across the junction”
That was helpful….understood!