# Estimating LED lumen output using before and after bat. volts.

So I go on a bike ride for about 50 minutes last night with both LED lights on high. Two 18650s in parallel power two LEDs. Voltage at start of ride, 4.13V, voltage at end of ride, 3.61V. A smart charger tells me my batteries have about 2200 mAhr of capacity. Now if I assume that about half the energy of the battery goes into light should I be able to estimate the average lumens given off by the LEDs?

Is 50% conversion efficiency a close number to use?

I’m not sure that is possible. With so many variables factoring into led . I you’ll have to factor in the resistance of the springs, the drain of the driver, the size of the lead wires to the led itself, the cooling rate of mcpcb, the cooling rate of the housing, the glass if its AR coated or not, etc… From what I understand, measuring total lumens requires the use of a white box. You may be able to calculate the number to within a couple lumens that way,but I doubt you’d get anything accurate. I’m no expert, but that’s just my personal view on the matter.

If you really want to know, other members have devices to test lumen output. Maybe one could help you out if you mailed it to them.

The lights and batteries are attached to a bike helmet so it is not so portable. The lights are supposed to have the Cree XM-L T6 LEDs. If I use the data sheet and measure the current should that give me the Lumens?

Data sheet,

http://www.cree.com/~/media/files/cree/led%20components%20and%20modules/xlamp/data%20and%20binning/xlampxml.pdf

If that is correct, can I then work backwards and estimate the efficiency of the light?

Thanks for all the help!

Since you have a smart charger, you can discharge the battery right after use and find out the remaining juice in mAh. Subtract this with your known battery capacity (e.g 2200mAh) and get the actual mAh used.

Divide this with your time used in hour and you get the average current. I think it’s safe to assume driver can get about 70% efficiency so multiply by 0.7 to get emitter current.

Cross check this emitter current with the XML spec and this gets you the emitter output in lumens.

Note this does not account for optics losses, final brightness will be even lower.

The good news is the lens on the light allows me to get a good magnified view of the LED. From a review of LEDs on budgetlightforum,

it looks like I have the XM-L T6. Will experiment and try and work backwards to estimate efficiency.

Thanks all!

Edit. From the picture all I guess I can tell is I have a XM-L and not the particular version.

Edit. On the other hand the color of the light seems to be cool white which would favor it being a T6?

Edit. Cool white models of XM-L could be T6, T5, or U2, all powerful emitters.

Some of my rule-of-thumb observations:

My budget flashlights I measure at around 40 lumen /watt. If it is lower than that you have a very bad flashlight, if it is higher it is quite good. High end flashlights with good drivers, reflectors, AR-coated lenses and such can be up to 80 to 100 lumen/watt.

Mine is certainly a budget light. And I think I was running about 1.5 amps at say 3.8 volts, 5.7 watts, or by your rule about 230 lumens times 2 lights.

More lumens!

Some numbers,

Battery at rest 4.10V
Voltage across battery, LED on high 3.83V
Current out of battery LED on high 1.23A
Voltage across wires that go to LED 2.98V
Current in wires going to LED 1.19A

Assuming numbers are close and the LED is in fact a XM-L T6 using the data sheet from Cree the lumens at .7A are 280 but at nearly 1.2 amps the lumens increase by 160% to about 450. The voltage verses current curve from the Cree data sheet show my numbers 1.19A and 2.98 volts are off the curve a bit?

It looks like the battery is producing electrical energy at a rate 3.83Vx1.23A = 4.7W
LED is using energy at a rate 2.98Vx1.19A = 3.55W so the electronics seem fairly efficient.

The data sheet says that we can expect up to 100 lumens per watt efficiency with the XM-L family which would give me more like 355 lumens using 3.55W above.

As mentioned data sheets don’t account for loses.

As for any LED, it won’t reproduce the exact V vs I curve per spec, temperature for one will affect the correlations. So normally test value not matching the spec is expected.

However something is off with your actual current measurement, doesn’t match your real world test, you said in 50 minutes the battery would drop to 3.6V at rest which is probably well under 20% remaining. (If it were a phone it would say 0% and shut down at this point).

If the light actually pull 1.23A per battery, running for 50 minutes, you’d have at least still half. But your battery is almost depleted - meaning this light pulls way more than 1.2A, probably closer to 2A.

Thank you! Something does not add up! Using the multi-meter (with a new battery) and shorting out the test leads I get .2 -.3 ohms resistance. So with roughly an amp going through both of them we drop about .2—.3 volts across the leads during testing throwing off the current measurements?

More Lumens.

Edit, the old 9V battery in the multi-meter measures 8V, time to re-test.

Measured led voltage and current doesn’t include power used in the driver itself but I don’t think you need battery capacity anyway just measure led voltage and current at a few other times during discharge and get an average for led current on high. Compare that with T6 output at that current and guess at transmission losses. Also, led current is not battery current, you need to disconnect one of the emitter wires to run it through the DMM or use a clamp meter instead.

I wanted all the numbers just for an idea of how efficient the electronics were. Between the battery pack and the light is a clicky switch and from there two wires lead to the LED. For current through the LED one wire was unsoldered. If I can find my other multi-meter I can measure both current and volts and let the battery run and take a series of measurements. Next time I will use low resistance leads for the current measurement.

Thanks!

You could probably just extend the wires so the driver fits in your hand, then turn on the light and see how long the driver takes to get hot. Losses are converted to heat. While it’s not scientific, it might mean more to you than a calculation with errors.

I did try and power both lights with one driver and the electronics got very warm fast but with the electronics powering a single light the heat produced did not seem so bad. I thought by measuring power produced by battery minus power used by the LED (if measured accurately) would give a good estimate of power used by electronics. Will try and do some more accurate measurements.

Thanks!

After more careful testing on the garage workbench, 2 batteries at start, 4.17V, thick short wires for current measurement, connections soldered, power 1 LED.

Time voltage to LED Current through LED

8:00 3.05V 1.53A
8:01 3.03 1.49
8:05 3.01 1.45
8:15 2.98 1.41
8:30 2.96 1.33

Batteries voltages at finish 3.94V

Using the data sheet table “Relative Flux vs. Current (TJ = 25 °C)”, at the start it looks like the LED is putting out almost 560 lumens, after 30 minutes that drops to about 490 lumens and the LED housing is getting toasty. No where near the claim of 1600 lumens, but two of them working together still light up the road OK.

The one data sheet “Electrical Characteristics (TJ = 25 °C)” must be for a particular submodel of the XM-L family as I cannot believe all the LEDs in the XM-L family have the same electrical characteristics, similar but not the same. For a given voltage the LEDs in the XM-L family that give off more light will have more current going through them in general?