.........Many lights have a tail switch so when it's removed it's difficult to power the head, select the mode and then make a current sweep to see how it works. I'm beginning to think it may be easiest to discharge a battery to compare to a fresh cell but even then you need a way to measure the current and voltage.
Has anyone come up with a way to do this?...........
Yes, what you do is forget about measuring the voltage and only concern yourself with how long a battery lasts compared to how much light it puts out. The light is proportional to current. If that output stays pretty level though out the discharge of the battery, then it has a good current regulated driver circuit.
Let me help explain the relationship of voltage, current in an LED flashlight which might help you understand that you really don't need to concern yourself with voltage if wanting to measure efficiency.
An LED is a current driven device. Light output is proportional to the current going through it. The voltage across the LED is called the forward voltage drop. That voltage drop is around 3.00 volts for Cree LEDs in a lot of flashlights. It is a very narrow band of voltage from which the LED just starts to conduct current to the point of reaching its maximum current. Thus its a lot easier to control the light of an LED with current regulation and not voltage regulation.
A 3.00V cree LED can be driven by a number of battery configurations from single AA on up to 4xAA or Lithium Ion voltage. The driver is designed to convert a battery configuration to the proper level to drive the LED. The relationship is via power. Power in from the battery(s) = power consumed by the LED + power loss of the LED driver circuit. Hopefully the driver circuit is very efficient such that the majority of the battery power makes it to the LED.
Power is voltage x current. Example: A single AA light is drawing 1 amp at 1.30 volts from its battery. That's 1.30 watts. The LED in the flashlight has a 3.00V drop across it. What is the current going through the LED? We will use power in = power out to the LED minus about 10% loss in the driver circuit. Thus 1.30 watts from the battery = 1.17 watts at the LED with 0.13 watts loss in the driver. Using the equation of P=VI, we can work backwards with 1.17w/3.00v = 0.390 amps through the LED. If this was a 2xAA flashlight and the LED was the same brightness as the example, then we can assume that the current from the batteries will be half since the voltage is twice with the extra battery.
It's just a lot easier to use a data logging light meter and makeshift light box, measure the light output over time and graph it in Excel. Efficiency would then be those lights that have the biggest output x run time product..............or area under the curve. Efficiency is a combination of a good driver (ie: not use a resistor like you mentioned), and an LED that is efficient such as the newer Cree LEDs. Typically the more you spend on a light, the better the driver and thus its efficiency. What is nice is when I come across a light that isn't too expensive and yet is very efficient. I don't use 2xAA lights but have tested a lot of single AA lights. Most of these lights will run all the juice out of a battery. But not all will be equal in the efficiency area. The hobby in it for me being an electronics guy is with a good well regulated efficient driver.
See the last two graphs comparing an alkaline, an L91, an NiMh, and a 14500 lithium ion in the same flashlight. Note how the L91 kind of chokes with a heavy load whereas the lithium ion excelled. And then on a lighter load the L91 did better. Some batteries simply do not like a heavy load on them and their capacity is higher with a more gentile load......ie: an alkaline battery.
http://www.candlepowerforums.com/vb/showthread.php?417347-Fenix-LD11-Single-AA-Light-Performance-Run-Time-Graphs
Sorry for the long post. Hope this was helpful
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