Lumens from amps drawn

Newb issue #1:

I see a lot of amp draw measurements, but isn't it watts that drive lumens?

If so, then would double the batteries at half the amps but double the volts produce the same lumens?

I'd like to know how to do the right calcs for the different types and numbers of battteries.

On a tangent, I tested an alkaline at 1.5+v and 4.2mA and a NiMH at 1.3+v and ~3.7mA (both Duracells) using a cheap & handy HF MM. Why trying both in a new iTP A3 it sure looked like the rechargable produced a stronger light on every level. Why?

Thanks in advance for any and all help.

Amps across the emitter what we are interested in.

Most reading on here are from the cell, this is not an accurate way to tell but the easiest.

Two cells in series = double the voltage, so half the current to get the same power.

Two cells in parallel = no difference in voltage or current, the capacity will be increased

Amp draw at the tailcap will only tell you the approximate run time, depending on the capacity of the cell. It does not always directly translate to emitter lumens.

Well known drivers have been tested at different voltages so knowing these you can be reasonably accurate on current to LED. Direct drive allows you to be very accurate also.

Welcome to BLF, GottaZoom! :)

I have fun here and I'm sure you will too.

I tried measuring the amps across the emitter by removing the bezel and reflector, turning the light on High, and touching my probes to the + and - solder points.

Only thing that accomplished was t make me see spots for the next 10 minutes, from trying to stare at the emitter while it was on High. Aarrrrgh

Lol, lucky it wasn't a fused meter then or at least a tripped cell if protected

LOL, so when I read people with fixes that increase the draw they're trying to shorten the run time? OK, they're probably looking at direct drive.

Thanks y'all for the welcome and the funnies, too.

When people are trying to increase the tailcap current draw, they're trying to reduce the overall resistance in the circuit. The circuit = the batteries, contact resistance, wiring, driver board, springs, and LEDs. Reducing the circuit's resistance means maximizing the output that the light is capable of. So people try thicker wires, batteries with lower internal impedance, unprotected batteries, etc.

BTW, it's current through the LED, not across it. Voltage is measured across the LED.

http://flashlightwiki.com/Brightness_Bins

Is there a reference voltage for the chart . . specifically table mA/lumen specs are at what voltage?

GottaZoom asked

"On a tangent, I tested an alkaline at 1.5+v and 4.2mA and a NiMH at 1.3+v and ~3.7mA (both Duracells) using a cheap & handy HF MM. Why trying both in a new iTP A3 it sure looked like the rechargable produced a stronger light on every level. Why?"

It can be true that a lower voltage battery can "push" more Amps through an emitter than a higher voltage battery. In the case of your alkaline batteries vs NiMH the short answer is lower internal resistance. All batteries have an internal resistance that can affect how much current they can produce. NiMH rechargable batteries can have a lower internal resistance than Alkalines. Voltage across a resistance produces a current represented by I = E/R. I know this might sound strange, but you can also think of a current through a resistance produces a voltage, In the case of current draw from a battery, the discarge current going through it's internal resistance produces a voltage drop. So at a certain high enough current draw, i.e. driving a high power LED, it is possible that a lower voltage battery with a lower internal resistance can actually deliver more current to a high power LED, therefore driving it higher and brighter!

You could also measure this effect with a volt meter across the terminals of the battery. The battery with the lower internal resistance will have less of a voltage drop under load. In this case a 1.5V battery(open circuit) looses to 1.2V battery (open circuit) because under a high load the NiMH battery is able to maintain a higher voltage under that load.

Off Topic,

With all this talk of mAh ratings of various 18650 batteries (2200mAh, 2400mAh etc.), I was curious to find out what the mAh rating of a D Cell Alkaline battery was, so I looked it up. Wikipedia says that a good D Cell is rated at 15,000 to 20,000 mAh!

Without knowing the math ...I know the answer .. internal resistance . the alkaline is holding the light back from it's potential .. the lower resistance in the nimh lets it take more of what it wants ..of course this shortens the run time ...you just made the light brighter and pulled more amperage .

http://www.cree.com/led-components-and-modules/tools-and-support/document-library

..Now it all sounds to much like math..gotta run :P

Aloha and welcome to BLF GottaZoom!

That I do not know, but it raises a different question do all the manufacturers use the same reference voltage or is it different for each manufacturer and each emitter model. I guess the chart raises more questions than it answers, sorry.

I think you are spot on with an often confused topic. Efficiency, which is power In divided by power out is the truest measure. The biggest loser is generally heat caused by resistance and comes from every part of a circuit. Hot batteries, too small wiring, switches, spring contacts, a poor match of battery voltage vs driver voltage. Even an over driven led will produce more excess heat than extra light. Some people may think that direct drive is the most efficient method but this is not necessarily true and more often than not isn't. When you build a light, it's your light so you get to decide how it's done. There are many right ways, and some wrong ways that work, but not as well. One thing is true, V=I*I*R.

Wrong! Sorry, it's V= I x R and P = I x I x R.

This means that for any given circuit, doubling the current increases power loss in that circuit Fourfold. Design wise this means that any LEDs run in parallel would be less efficient than the same LEDs run in series. That is overly simplistic but true nonetheless. A buck driver is generally more efficient than a boost driver simply because the half of the circuit that supplies the power is running at a lower current. I could go on and on but there are others who can cover this more concisely. I encourage you to read as much as you can from lots of different posts as different viewpoints reach different shaped minds. Go ahead and make something. It will probably work and if it doesn't , when you figure out why not you will have learned even more. You don't have to mod an exspensive light to start with. Almost any light can be reworked and for me it has been less stressfully starting with something that was junk in the first place. Post your work and read. Read about the cool lights, the disasters, different ways to do the same thing. And when you start, get yourself a decent magnifying lamp 'cause this sh**s small.

I'm not sure I'm understanding your question, so if my answer is not what you were looking for, sorry.

The way a diode works is like this: You apply a forward voltage across it. Starting from zero, no current flows through the diode. As you ramp up the voltage, a very small current (microamps) starts to flow. When you reach a certain voltage, which is around 3 V for an XM-L LED for example, all of a sudden the diode starts to conduct lots of current (amps), assuming the power source is capable of producing unlimited current. The diode behaves more or less like a piece of wire once you reach that voltage. If you increase the voltage too much, e.g. 3.5 or 4.0 V, without any limiting devices like a series resistor, the current through the diode flies through the roof and if the diode isn't capable of dissipating the generated heat, poof.

If you do use a series resistor, you can flatten out the curve, so to speak. As you ramp the voltage beyond 3 V, the diode behaves more or less like a piece of wire, but the resistor prevents the current from flying through the roof.

So what this means is that you don't apply a certain voltage across the diode and then independently push a certain amount of current through it. The voltage across the diode determines the amount of current through it.

See the first graph on this page: http://www.kpsec.freeuk.com/components/diode.htm

That graph is for a general purpose diode. As I mentioned, for an XM-L LED, the "knee" is around 3 V instead of 0.7 V.

So when you ask if there's a reference voltage, I'm not quite sure what you mean, but given a specified current through the diode, you can look at the diode's data sheet to find out what the voltage must be.

Aside: one of the reasons that the XM-L is such a good LED, compared to older ones, is that its forward voltage is lower. What that means is that it takes less voltage before the LED starts to conduct current, and therefore it operates at a lower voltage at a given current, and that means lower power draw at a given current.

That's interesting. I didn't know alkalines had that much energy capacity. For the sake of comparison, let's compare energy capacity. A high-end 18650 is about 3 Ah. Wikipedia says a good D cell is 15 - 20 Ah. Let's say the D cell is 1.5 V, and the 18650 is 3.7 V.

Energy capacity is voltage * current * time, or voltage * amp-hour rating. Therefore:

18650 = 3.7*3 = 11 Wh (watt-hours)

D cell = 1.5*20 = 30 Wh. Wow 3x the 18650.

Okay let's compare to a typical nimh D cell. The first hit I find with Google says that an Energizer D cell is 2.5 Ah.

D nimh = 1.2*2.5 = 3 Wh. Wow, that's all? Is that correct?

Okay I looked at more nimh D cells. Seems like 10 Ah is more typical. So that would be 12 Wh. Pretty close to an 18650. I guess that's why most people here prefer 18650s.

Typo: it's P=I*I*R. :) I know you meant that.

One other thing I'd like to mention: in your series/parallel example, it's not that the LEDs themselves are any more or less efficient depending on the series or parallel configuration; it's the drive circuitry that's less efficient when LEDs are in parallel.

Here is a link the a PDF file of The characteristics of a Cree XM-L

http://www.kosmodrom.com.ua/brand/cree/XLamp%20XM-L.pdf

On page 5 there is a graph of Lumen output vs forward current. If you will notice the y axis is relative % of rated lumens (which is 280 lumens at 700ma) Yes, Cree only rates the XM-L at 280 lumens. At 3000ma, the XM-L puts out 325% of 280 lumens = 910. Why do they rate it at only 280 lumens then? I suppose its because it is most efficient at that lumen output. Look at the graph, after 700ma the curve starts to get flatter, meaning that the LED is getting less and less efficient. In the real world, efficiency is most important, especially in a battery operated device. Only enthusiasts such as ourselves would go for maximum lumen output.

BTW, a resister is not an efficient way to limit the current through an LED, because the current that goes through the resister will create heat and is wasted energy. The drivers used are more sophisticated than that and use PWM, etc. PWM is turning the LED on and off very quickly so that during say 1 second interval it is off 50% of the time and on 50% of the time so on average it is running at half brightness.

Now having said this, I am wondering, why would a driver use PWM?? It seems it would be more efficient to run the LED in lower current modes for the lower brightness modes that PWM the 3000ma where the LED is least efficient. Now that I think about it PWM is just a cheap and dirty way of providing a lower Lumen mode, a better way is current regulation? (probably more espensive?) I know there are driver experts here, maybe someone has the correct info