At 1 mA draw maybe 
At any serious draw (1A and up) D cells do 4 Ah at most.
The best visual representation I have seen is by by googling the "discharge curve" for various batteries. It shows voltage vs runtime for a given battery type and often times with different rates of discharge in the same graph; .1C, 1C, 10C, etc. Nimh cells suffer less voltage sag at higher discharge rates than alkaline cells and can provide a more constant voltage over each cycle. Alkaline cells are rated for more watthours but to get that extra energy you must discharge them at a low rate, like .1C where C is the rated capacity if the battery. You can also look at graphs of how other things like temperature affect cell life.
You're right, v=i*r,p=v*i and thank you. I WAS talking about the circuit in general not the LEDs. There is much more to know about this topic than can be conveyed in a few paragraphs that it's important to keep reading and asking questions.
One reason Cree may rate their LEDs based on current value is there is a much broader range of current values than voltage values. Xpg and XML have similar forward voltages but not input currents.
Quite correct. That's why the driver I designed for my light actually reduces the constant-current reference in order to dim. No secondary PWM. It's a LOT more efficient, actually. It's just more difficult/expensive to do. At very low levels, however, even I start PWMing the constant current source - but at that point, the constant current is already regulated to 500mA, so I'm PWMing 500mA, not 3 amps. The reason for this is that the circuit is setup for 3 Amps, and it starts to get really inefficient when you try to convince it to then output only a few hundred mA.
PPtk
PPtk
LOL, I'm just now starting to get v x a = w and how to measure the first two (plus continuity and battery charge) on the mm! Haven't tried resistance, yet.
Thanks to you and others with the kindly instruction.
I've been playing with a bunch of old incans that stopped working . . most were bulbs and corrosion and parts forced out of place . . they're basically junk, so I just may try something with those.
I have an old Brinkman 2 x AAA (similar to a AAA micro-minimag) I'd love to see running l-ions and making a LOT more light than a NiteIze drop in when I feel good enough to tear it apart . . I want it to survive, though.
This may partially explain why one of my lights shows an usually high lumen rating only for strobe mode and not SOS or other power levels. The other ratings all seem much more realistic and I wondered why.
Well, besides the usual BS we sometimes see.
Thanks, but I think you circled my original and later question . . if watts measure the consumed power, wouldn't a lumen rating be the product of consumed watts thereby have both a volt and amp specification? I assume the table shows relative amps for a given voltage, but if so what is it?
And as you thought it through, what is the most efficient way for a driver to deliver the watts consumed.
You are right, Watts = Volts x Amps (P = E x I) but from Ohm's Law, I = E/R or stated another way E = I x R.
Doing the Algebra, If P = E x I, and E = I x R, substituting I x R for E, we can express Watts as P = (I x R) x I. After multiplying this becomes P = I squared x R.(sorry I do not know how to do the exponent 2 on the keyboard). So it could seem that Power could be expressed without the Voltage. This is not true, the voltage is accounted for, it is just not obvious. Think of it this way, When the XM-L is giving off 910 lumens it is because there is enough voltage present to push 3 amps thru it.
Cree rates their LEDs at a baseline standardized current that results in 1watt power consumption. From there, they go on to state the maximum approved output levels. If you read more of the fine print you will see that for energy star certification and maximum diode life the recommended forward current is well below the maximum value given. I think(and this is purely speculation on my part) that the rating at low current value is for industrial use where the need to compare the output of different LEDs at 1watt power consumption(apples to apples) is necessary. It is not intended as as maximum value. The graphs show how many more lumens are produced at higher currents, which is another area where Cree LEDs excell. They are tooting their horn. As much as we enjoy being at the forefront of people in the know abou LEDs we as modders represent a very small segment of led sales. I happen to think we are important in getting the word out but still, we don't buy diodes in lots counted by the thousand.
Two cells in parallel = no difference in voltage, but amperage doubles. If the power is regulated, then there’s longer capacity, but direct it would push more amperage, since it’s available. Correct or not?
Probably not unless a single cell could not supply the amperage. A smaller cell is less likely to be able to keep up and would "sag" thus reducing current draw. The amperage doesnt double. The available capacity doubles, amphours is a measure of capacity.
I think LED mfrs commonly state a lumens/Watt rating because people want to know "how much light will I get out of these things compared to the incandescent, fluorescent, or metal halide lights I'm currently using?" With those other types of lighting, the bulb is often directly attached to AC mains voltage (I think. I'm not sure about metal halide). Therefore, lumens/W = lumens/(Amp*120 V) (or whatever the AC mains voltage is in your country). Even though the mfrs might state a lumens/W rating, it's essentially the same as a lumens/Amp rating since the voltage is always the same. Just divide or multiply one or the other by 120.
With an LED, the lumens are a result of electrons passing through the diode. I don't remember the details well, but it has something to do with electrons dropping down a quantum level and releasing energy in the form of visible light. That is, the light is produced by current through the LED. But as I said earlier, for any given current through the LED, the forward voltage across the LED is known. So if you know the current, you know the voltage and therefore wattage. Specifying lumens/Watt is again almost the same as specifying lumens/Amp, but in the case of an LED, to get one from the other, you'd multiply or divide by the LED forward voltage, which is not fixed at 120 V but you'd look it up on a voltage vs. current graph for that particular LED. The voltage is not fixed for a range of currents. For each current, the voltage is different. It might be something like 2.0 A corresponds to 3.00 V, and 2.1 A corresponds to 3.01 V, and so on, or something like that. It's clearer if you look at the graph I included a link to in one of my earlier comments.
2 cells in parallel = no difference in voltage, when no current is flowing. When current flows, the battery voltage will drop due to the battery's internal resistance. At high currents, we all know that battery voltage sags. This is due to the internal resistance. Adding a 2nd battery in parallel is essentially placing two resistors in parallel, where the resistor represents the battery's internal resistance. 2 resistors of equal value in parallel, results in a resistance of half the original value. So putting two batteries in parallel means you're halving the effective battery internal resistance. So at high currents, the battery voltage will sag less. But I think I know what you meant: putting two batteries in series doubles the (open circuit) voltage, and putting them in parallel does not.
Available current (amperage) doubles, but that doesn't mean that actual current doubles. That is, adding a 2nd battery in parallel won't force 2x the current through a circuit. What will happen is that if the circuit was trying to pull X Amps from a single battery, but the battery's internal resistance wouldn't let it, then adding the 2nd battery in parallel would allow the circuit to pull more than X Amps, possibly up to double. This is because of the halved internal resistance.
edit: actually, what Rufusbduck said is more accurate: the amp-hour capacity doubles, not the amperage, but it's still true that because of reduced internal resistance, the combined battery pair will be able to provide about double the current for any specified amount of voltage sag.
I agree with the above :) so a DD LED may need a bit of consideration if more cells are added in parallel
Oh yeah, lots of consideration. :)
Have a look at the forward voltage curve of the XM-L, at the top of page 4 of this data sheet: http://www.cree.com/~/media/Files/Cree/LED%20Components%20and%20Modules/XLamp/Data%20and%20Binning/XLampXML.pdf
It's been confirmed that the XM-L can go beyond 3 A if it is heatsinked well. What is the forward voltage beyond 3 A? Someone may have charted those numbers, but extrapolating from that curve, it should be around 3.6 V. Beyond a certain current, the output starts to drop and the LED will eventually go poof.
Now a single 18650 is around 4.2 V open-circuit. If you direct-drive an XM-L, it will immediately drop to that 3.6 V, and the current through the XM-L will be whatever current corresponds to that voltage, according to the (extrapolated) curve. maybe around 4-5 A.
If you parallel a second 18650 with the first one though, suddenly you have half the internal resistance, so the current could be double, for a very short time before the LED goes poof!
I think that graph helps when looked at with page 5 and the flashlight wiki. . if we take page 4 we can compute watts consumed and then cross-reference it to page 5 and the table to show the lumens produced from the watts used. At least if I understand what I think I read.
Using the flashlight wiki table for the T6: 100% is about 2W and 290 lumens and 325% is about 10W and 950 lumens (give or take).
NOW, what's less clear to me is what is happening with a 2 cell setup that is delivering 8+v. Is there is a resister to limit flow to something around 3v?? Or would a 10W flow come to the LED in just a different combination of Volts & Amps (say 1.5A at double the volts compared to a 1 cell)?
Thanks!
Power companies sell energy by the watt, not the amp. For industry, lumens/watt allows a straightforward expense calculation. How much light do we need and how much will it cost. Sales reps and ceos ore not generally electrical engineers and just want to know how much it will cost. Lumens is a measure of light output and the specific ratio relative to amperage is different for each led type and does not lend itself to comparison whereas lumens per watt allows for different currents and is a more easily compared measure of efficiency.
2watts consummed = .75 A x 2.66V
10watts consummed = 3A x 3.33V
For a 2-cell setup you would need some sort of current limiter or the led would fry. A resistor for this would be cheap but about the least efficient method along with a linear driver used in a standard fashion. A linear driver acts as a resistor to burn off extra voltage. A buck driver takes a voltage higher than what is needed and steps it down to the desired level and converts some of the excess voltage into supply current. How well it does this is the measure of the drivers efficiency. Whatever voltage the driver fails to convert ends up as waste heat.
Matching the supply voltage to the led voltage is relatively easy with mains power and voltage regulators. With flashlights, of whatever type, we are constantly struggling with the mismatch between battery voltage and led voltage and the fact that batteries do not maintain the same voltage over the entire discharge cycle.
Choosing the leds(how many), the batteries(what voltage), the driver(boost, buck, linear, or direct drive). For me, this represents the most crucial compromise for any build and is the sum of all the constraints placed on the build. Actually making the thing once I have decided what "it" is pretty straightforward even though I do more cursing during the build than over a bar napkin circuit diagram.
^^^^^^^^^^^^^^^ Excellent post above. ^^^^^^^^^^^^^^^^^^