Formula to Estimate Lumens from Amps?

[recDNA]

I once read a rough formula to determine LED lumens of an XP-G R5 from a reading of amps at the tailcap. Can anybody refresh my memory?

 

[csshih]

I believe it would depend largely on the efficiency of the driver,
If I recall correctly, a buck driver is less efficient than a boost driver-- with a boost driver having ~80-90% efficiency, and a buck having ~70-80%?

I might be wrong -- hoping someone more knowledgeable will chime in here.

 

[recDNA]

I believe it would depend largely on the efficiency of the driver,
If I recall correctly, a buck driver is less efficient than a boost driver-- with a boost driver having ~80-90% efficiency, and a buck having ~70-80%?

I might be wrong -- hoping someone more knowledgeable will chime in here.

And in direct drive?

 

[Mr Happy]

It depends on the specific emitter. If you know the emitter, just look at the manufacturer's data sheet. There will be a nice graph there plotting light output against emitter current. Bob's your uncle.

 

[mdocod]

Depends on the batteries and configuration of batteries being used.

If you have [email protected] that will be dimmer than [email protected] assuming a regulator where all things are staying equal in efficiency with nice lower overhead.

 

[csshih]

And in direct drive?

oh, direct drive? specs should all be in the datasheet.

 

[mdocod]

I'll second that: In direct drive, (or in series with a simple resistor), you should be able to estimate based on the datasheets.

Eric

 

[recDNA]

It only goes up to 1.5 amps and the vertical axis is luminus flux - not lumens.

I'm running an R5 at 3 volts and 2.25 amps. Doesn't fit anywhere on graph.

Does anybody know the approximate efficiency with buck or boost drivers? My next question will be which is used in the Dereelight 1S R5

 

[gcbryan]

You're running it over spec and that's why you can't find it in the spec sheets. Assuming an inefficient driver maybe the emitter is only seeing 1.5A so you could use that.

I don't know if the spec sheets have been updated to list the max current draw from 1A to 1.5A. If so use that figure. It was 325 lumen for R5 at 1A. I seem to have seen 450 lumens as possible with 1.5A but I'm not sure about that.

FYI-As far as I know, luminous flux is lumens.

 

[recDNA]

You're running it over spec and that's why you can't find it in the spec sheets. Assuming an inefficient driver maybe the emitter is only seeing 1.5A so you could use that.

I don't know if the spec sheets have been updated to list the max current draw from 1A to 1.5A. If so use that figure. It was 325 lumen for R5 at 1A. I seem to have seen 450 lumens as possible with 1.5A but I'm not sure about that.

FYI-As far as I know, luminous flux is lumens.

Oh thanks - duh. I was very surprised by the amps reading too. I'll have to do it again but I don't see how I could have messed it up. It's pretty straight forward. One probe in the amps DC socket of the multimeter the other in the common -. One probe on the negative end of the batteries in the flashlight the other on the shiny metal of the body where the tailcap would normally complete the circuit. I will try again and report it if I get a different result.

My other results were reasonable though -

1.33 amps Mini with IMR123

.9 amps P20C2 R5 with 2 X CR123 primaries.

 

[csshih]

I'm running an R5 at 3 volts and 2.25 amps.

? that's not physically possible with the current forward voltage ranges of the XP-G.

if you did have a LED that did that, you'd get.... 500-600 lumens.. goodbye, MC-E! :nana:

 

[recDNA]

? that's not physically possible with the current forward voltage ranges of the XP-G.

I'll check again right now - but if the driver in only 70% efficient isn't it possible? Can you see any error in my procedure? Of course normally I use two hands to hold the probes and get the same reading it isn't a short circuit or anything. The flashlight is on although you can't see it I can clearly.

 

[65535]

V*I is your supply Wattage.

V*I*Eff. is the power to the LED from the driver.

You'll need to find power curves for the LED. But you should be able to just divide the wattage by the Vf and determine current. Which will allow you to determine the luminous flux from the datasheet. Then you'll need to factor for reflective losses in the optics/reflector and window.

 

[csshih]

huh... I'm slightly confused, are we running direct drive?

how many batteries are in there, 1 primary?

 

[recDNA]

huh... I'm slightly confused, are we running direct drive?

how many batteries are in there, 1 primary?

2 X L91 AA lithium primaries. I'm told under load they run at 1.5 volts each so 3 volts. Well under the 4.2 volt max of this pill.

 

[Mr Happy]

I'll check again right now - but if the driver in only 70% efficient isn't it possible? Can you see any error in my procedure? Of course normally I use two hands to hold the probes and get the same reading it isn't a short circuit or anything. The flashlight is on although you can't see it I can clearly.

I'm not sure what's happening there, but one thing to watch out for is the accuracy of the meter reading. Inexpensive meters on the DC A range are calibrated to read pure DC, but a light with a boost driver does not draw pure DC, it draws a complex DC+AC wave form. You cannot necessarily believe what the meter says under those circumstances.

 

[recDNA]

V*I is your supply Wattage.

V*I*Eff. is the power to the LED from the driver.

You'll need to find power curves for the LED. But you should be able to just divide the wattage by the Vf and determine current. Which will allow you to determine the luminous flux from the datasheet. Then you'll need to factor for reflective losses in the optics/reflector and window.

The curve I found was forward current vs luminus flux not power vs luminus flux and I don't know the Vf.

 

[csshih]

oh! then it's a regulated dropin. oops, I thought you meant a direct drive.

alright, let's try some quick calculations with the currently supplied readings, assuming they are accurate.

3V * 2.28A = 6.84 watts, assuming 70 % driver inefficiency = 4.788 Watts to the emitter.
Cree supplies that the emitter should be giving out... 432.6lumens @ 4.748Watts.. closest figure that I could find.

that, of course, does not account for the heat of the emitter (which decreases output)

 

[recDNA]

It's pretty bright. Who knows. I wouldn't be surprised if it's over 300 lumens otf when first turned on.

I mentioned direct drive because my Mini on an IMR123 is 1.33 amps at the tailcap (so that IS direct drive) so I wanted to estimate that one too.

Pretty cool though. The Javelin LOOKED ultra-bright to me. The odd thing is I think the driver for the 1S R5 is 1.2 amps so how is it running at 2.28 or even 1.5?

I want to be sure it's clear I really LIKE this drop in. I'm not complaining about it!

 

[recDNA]

I'm not sure what's happening there, but one thing to watch out for is the accuracy of the meter reading. Inexpensive meters on the DC A range are calibrated to read pure DC, but a light with a boost driver does not draw pure DC, it draws a complex DC+AC wave form. You cannot necessarily believe what the meter says under those circumstances.

You're probably right but other readings I've done on other flashlights have been the same as posted values in several threads. BTW, where's the love for my meter? Hasn't let me down yet!