Voltage and lumens

[021411]

I missed the boat and just need some clarification..

(Hypothetical) Say you have a drop-in LED module that is rated 3-14v and is rated for 200 lumens. So let's say you drop in two CR123's into the light for a total of 6v from the batteries total. Does this give you 200 lumens or do you have to max out the module to achieve that number? In this case it would be 14v (again hypothetical numbers and situation). Or is runtime the only thing that will be changed?
I'm confused there.

Thanks.

 

[winston]

I feel a little presumptuous answering this question, and hope that one of the experts will correct me if I go wrong here...

Electrical current, rather than voltage, is the key value when determining lumens. So light output in your hypothetical emitter assembly would depend on how the driver for the LED dealt with regulation of different input voltages. The most common method of regulation would yield 200 lumens for any voltage in the 3-14V range, as the driver would maintain the current at a constant level by stepping-up or stepping-down the input voltage to a suitable level.

I hope this helped a little.

-Winston

 

[021411]

Thanks. Makes perfect sense. I never really thought about the driver doing most of the work. Didn't even know it existed in lights. I guess I know what does the "regulating" now. :thumbsup:

 

[xiorcal]

Short answer, preceding long answer:
You will get 200lm from 2 CR123s if it uses a relatively recent SSC or Cree LED. You may or may not get 200lm if it uses a luxeon LED, or generic brand LED, depending on the LED's requirements, and the ability of your batteries.

Long answer:
The lumen output of an LED depends on the power that is going through it. Power = Volts * Current (I). With LEDs, as the current is increased, the voltage also rises. I think it's the same with most other electronic things, but don't quote me.

Being a drop-in I will presume it is a single LED. A Cree or SSC will only want < 3.7V to run maxed out. A Luxeon III/V, or something else will want significantly higher voltage (Not so sure of these numbers ~8V at .7A for 120lm).

If the drop-in you have is rated 3-14V, then it has a driver in it, because that is what allows a vatiation in input range, without frying the LED with excessive power. Drivers will match the input in that range to output the voltage the LED wants. This is done by sucking extra current, and effectively turning it into voltage.


Luxeon example:
The next limitation is whether the high current is possible.
If, for instance the LED is an overdriven lux V, and wants to run at (pulling numbers out of my --- here) 14V, and you put in 6V (2*cr123), then the driver has to pull that extra current out of the batteries. (6V at 2.3x the current the LED wants at 14V), and converting it to (14V at the current the LED wants).

[The 2.3x is derived from the multiple of 6V required to make 14V.]
In reality, drivers aren't 100% efficient, some are about 80%, some more, some less, so you would need batteries capable of giving current 2.9x greater than a luxeon wanted at 14V, which is possible around 1A.

Luxeon example Battery requirements:
This would be 2.9A. If your batteries are in series to make the 6V, that would mean both batteries would only give as much current-providing-ability as one on its own.

If you were using Li-ion batteries, that would mean both batteries would have to be 1450mAh (@2C discharge) to meet the requirements for 200lm. They don't make li-ion batteries that high in energy, so you would be more likely to get 150 or 130 something lm.

If you were using primary Lithiums, they may have a higher drain rate (I don't know this) so if they could both drain at 2.9A each, you could get 200lm.


Cree/SSC example:
A much simpler example, because these LEDs are more efficient and lower V than luxeons. If you put 6V in, you will require less than 1A at 3.7V for 200+ lm. That means you need less than ~650mA at 6V, which is well within 2C for Lithium-Ion, and well within Lithium primary cell discharge rate.