Watts Volts Amps some explaining

Copy pasted from a topic that drifted of topic because of the confusing nature!

Thanks for the great posts guys and feel free to ask/add :wink:

I think defenitions of constant voltage and constant current should be added to the first post.

I was in a hurry and just copy / pasted the first explanation I found that compared it to water flow, I didn’t even completely read it honestly.

Although I think you are basically saying the same thing as what the quote said.

Water pressure / Speed (same thing in the case of electronics 101 for beginners) = Voltage.

Amps / Current = Hose size

Total water output = watts

Sure talking in water speed may be easier for some to understand but it is also less correct and doesn’t translate as easily to what is actually going on in an electrical circuit. For example you can have water pressure in a hose that is capped off but you can not have any water speed.

A good video example of your giant 1.5v battery is this :

That is only a 2.2v “battery” in the video but with buttloads of current capability.

Like I said before, I copy pasted the explanation because these people needed a basic understanding on electrical circuits 101.

Trying to explain the forward voltage, LED efficiency, coefficient of heat ect is most certainty electronics 102.

That said in the quote they are talking about incandescent light bulbs, NOT LED’s.

With an incandescent light bulb watts is king since they are virtually unaffected by other factors. So yes, the light output would be the same with an incandescent light bulb in the above examples.

In a perfect world so would LED’s but this is not a perfect world and people need to learn how things are supposed to work before they can grasp how they actually work.

The 101 course always comes before the 102 course.

Thanks for making this Miller.

Good job, Miller.

Yes. A valuable clarification I could have noted originally is that a car-sized single-cell 1.5V alkaline battery still would not light up a Cree LED. Despite having tons of power. Nor would one of these supercapacitors, in theory (depending the voltage they provide).

So to me, in a way I see voltage as ‘height of the water flowing downhill’. I know we all hate the water analogy, but the fact is, if that analogy had never been used in high school, I never would have gotten it.

So to me, it’s back to the old ‘water wheel’. To make a water wheel go, you need “head” (height or drop of the water), and “flow” (speed of the water). With a water wheel, it’s not necessarily about the sheer amount of water.

The more “head” you have, the more distance the water drops, so it’s about height or steepness of decline. If you have a pond up on the hill, and it drains to a pond down at the bottom, the steepness of a hill is a factor… that’s like voltage. (Like producing “pressure” of the water.)

I say this because, if you have one pond that is lower than the other, the water in the lower pond will not flow into the water of the bigger pond, even if there is much more water in the lower pond.

Water still always flows downhill. It is very difficult to get water to flow uphill.

So, if you had a car-sized battery at 1.5V, it still would not light up a Cree LED that needs 3 or so volts to light up. That would be like water flowing uphill, I guess.

So the higher and steeper that hill, the higher the voltage. (And at some point, the height and steepness of the hill can become dangerous.)

So, you have a chemistry of a battery, and that voltage will always be the same (in a certain range). So, the “height” of the hill is already set. The only thing you can modify is the width of the water opening, to a degree. But, in our world, you get to stack multiple “hills” on top of each other, so they form one great big hill :slight_smile: . This is called batteries being in Series.

Or you can have multiple hills side by side, draining off their water at the pond on the top of their hill, all the same height draining into the same pool. This is called batteries in Parallel. Can be the same amount of water in the end, but the way the water flows is different. Obviously, hills stacked on top of each other is more dramatic, and has more potential… to a point.
^These were further thoughts since writing the above :wink: .

Crazy video, BTW, Texas Ace. That was a good thing you pasted… and it spurred me on to more thoughts, which apparently have been helpful, gratified to hear, and you’re welcome, BrianK and any lurkers who felt the same… you know who you are.
I hope we get past chemical battery technologies soon… this is what’s holding back most electrical innovation, particularly with vehicles and flashlights.

The sad thing: beyond my explanations above, I have little grasp of electrical “electricity 102”, to borrow a phrase from above. For instance, I just can’t wrap my mind around Power Factor; I’ve tried. Perhaps when the analogies to our 3D macro world run out, I don’t have much aptitude. There’s probably a youtube video that could help me on that, though, if I weren’t too busy researching other more-pressing stuff.

And I guess it’s better to say it here than in the original thread: I thought it was funny that the begging to stop writing about this subject only came after it was clear that there wasn’t going to be more discussion on it (there, anyway). I even pre-said it in my last post there (“[resolved, so] complaints are belated”), and it didn’t do any good. I’ve seen that a lot on forums, and elsewhere, and haven’t figured that out. Like, an argument can break out (even though that’s not what this was), then it ends, and THEN people rush in to beg the people who were fighting to “stop!”. Weird.

Can I make things even more complicated?

I skimmed through and didn’t see any mention to cell internal resistance. Please excuse me if there was a reference.

The highest capacity cell (of it’s size) is not always the one to deliver the most amps output. In fact with 18650’s there is a bit of a push me/pull me effect for mid capacity high current and high capacity mid current.

For regular edc’s with average output then a high capacity cell is best. A wow light is best with high current cell which is a combination of capacity and ‘C rating’, these cells have low internal resistance. Generally speaking the lower the better. Having lower internal resistance will mean the cell will hold a higher voltage when under load, thereby giving more watts to the driver/led.

Larger capacity cells like the D size l-ion may not have very high ‘C ratings’ but that is offset by the high capacity eg -

18650 cell, 2400mah, 10 c rating
2.4ah * 10 = 24 peak amps (dead short) it will quickly reduce as the cell heats up and internal resistance changes

32650 cell, 6000mah, 5 c rating
6.0ah * 5 = 30 highest possible amps (more than the 18650 even with lower c rating)

Hope I confused things a little more. :disguised_face:

Note - the above numbers were pulled out of my head, happy to be corrected

Keep it simple. Internal cell resistance is just like having another resistor in the curcuit.

Ditto^, for the people that this discussion was aimed at the internal resistance is way over what they need to know. They simply want / need to know how batteries work together in series vs parallel since that is all that 95% of them will ever do with electronics.

That said once you grasp how electricity works and interfaces with your daily life, lots of things start making a lot more sense.

Sorry. I missed the target

Not to worry, you didn’t do anything wrong. The info is mostly correct as well but to hash out those small differences would seriously derail the thread for the target audience and honestly the vast majority of them will never care about all the things that would be discussed after going down that road.

DB Custom wrote:

. . .

A few rare lights have the ability to run one cell or two cells to the same 3V emitter, and this is what’s causing the confusion. When fed 2 cells this driver goes into Buck drive with constant current, halving the voltage to the emitter so it doesn’t blow. These types of drivers are usually only around 86% efficient, making heat in the conversion process to eliminate the excess voltage. So there isn’t a doubling effect due to the heat losses. This kind of driver is what’s causing the confusion, no doubt.

. . .

DBC's post (most not included above) does a nice job of explaining some of the key concepts we deal with in this hobby.

To the best of my understanding, the bold text is slightly off though. For the type of buck drivers we typically use, heat is a by-product of the work being done by the components used in the bucking process. The higher the current, the greater the amount of heat. At moderate currents, it is difficult to detect any heat in some buck drivers.

These buck drivers reduce voltage to the emitter is by using an inductor to absorb excess voltage. The emitter takes the voltage that it needs and the remainder is absorbed by this inductor. The inductor stores the excess voltage by building a magnetic field. A chip ("buck converter") monitors the process and when the inductor is getting saturated, it cuts battery power to the LED. The inductor then releases it's stored energy to the LED. The magnetic field in the inductor begins to collapse as the stored energy leaves. This triggers the buck converter to reconnect the battery to the LED and the cycle repeats.

I too am an electronics novice. The above is just my gleaning of what appears to be happening by looking as circuits and working with them as a hobbyist.

Once again. I don`t know english language enougth to make this myself.
Somebody please add defenitions - what is constant voltage, what is constant current, what is current–voltage characteristic and how does it looks for leds.
Without this first post looks like some explanation that starts in the middle and ends not far away from it.

In linear drivers using the 7135 chip(nanjg 105C) the heat waste is a design feature needed to drop the supply voltage to led voltage, it’s supposed to do that. For a buck or boost driver any heat waste is not intended or desired but comes as a penalty for driver inefficiency. Better parts means more energy for use by the led. Space is the biggest handicap for boost/buck as better usually means bigger.

Linear just means same current into the driver as out(less what the driver itself uses).

A boost driver takes low voltage and high current from the cell and changes it to higher voltage lower current to the led. Input voltage needs to be below led voltage in lowest mode or it goes direct drive.

A buck driver takes higher voltage lower current from the cell and alters it to lower voltage higher current to the led. Input voltage needs to be ~1V above led voltage in highest mode or it falls out of regulation.

All 3 are most efficient when the change necessary is minimal.

I have a question I’ve been meaning to ask, but before that I need to preface it with a statement that needs confirmation: the current corresponding to a particular forward voltage is the MAXIMUM current that the LED will take @ that voltage, right? In other words, at that forward voltage, the LED can be powered by anything equal to or less than that max current.

If so, is it possible to design a driver for 1x18650 and 3V LEDs, that instead of bucking the voltage, feeds the full cell voltage to the LED (minus any voltage losses), BUT varies the current to provide constant WATTAGE to the LED, thereby providing constant output? At any output level, as the cell voltage drops, the driver draws more current to keep wattage constant, until the cell can no longer supply the required forward voltage needed at that output level. Then the driver drops to the next lower output level, which requires a lower forward voltage, and the cycle continues.

Would this kind of driver be more efficient than a buck driver?
How would the current variation be achieved?
Would the means of varying the current introduce additional losses that would negate any gains, assuming there are any to begin with?

Forward voltage and forward current are tied together. It’s more accurate to say that to get a given current through the led you need to supply the forward voltage for that current. Current flows as a result of having enough voltage. This is why as the battery voltage drops, the current through the led drops, at least in direct drive applications. With 7135 drivers the battery usually has a surplus of voltage which the 7135’s burn off as waste in order to maintain constant 350 or 380 mA output. When voltage falls below the emitter Vf for the number of chips used then current starts to drop.

If nobody will do this, there will be tons of such questions (#15) in this thread.

Thanks, Rufusbduck. I understand that. What are you thoughts regarding my statement and questions?

You’re describing the way a buck regulator works but there’s not enough difference in the cell voltage and the led voltage for the driver to operate, it’s called headroom. At low currents with low Vf it would work for awhile but not for the full battery cycle. There are buck/boost drivers that do both as needed but they are proprietary to certain brands and have yet to be either reverse engineered or designed from scratch to accommodate our penchant for custom modes. They are also limited to fairly low output levels by BLF standards.

Kiriba-ru, I would think constant voltage and constant current are definitions in and of themselves, or are you referring to that as opposed to pwm control? We’re not university profs with experience in clear, textbook analysis, just a bunch of hacks doing our best to share what we know. Since your not in such great command of the language maybe you could be less critical of that lack in others, eh? We each understand and learn different ways, it makes sense that different explanations will make sense to different people. If all else fails, the curious reader can poke around the forum and do a little searching on their own. It’s all here several times over. I find for myself I tend to learn more this way if not always what I was anticipating.

I don’t think I am. This is what I’m talking about: Say the LED requires 2W at a Vf of 2V. Instead of taking 4V and 0.5A from the cell and bucking it down to 2V and 1A to feed to the LED, the driver feeds 4V and 0.5A directly to the LED. Then when the cell is down to 3V, it feeds 3V and 0.67A, and so on.

I know about voltage overhead.