Check me on this

Timothybil

Flashlight Enthusiast
Joined
Nov 9, 2007
Messages
3,662
Location
The great state of Misery (Missouri)
I have a couple of those cheap 3AAA undercounter COB lights. High-Low-Off. My understanding is that to get the Low light lever they just switch some resistance into the circuit,Since by Ohm's Law, current is equal to voltage divided by resistance, adding resistance would reduce the current, and therefore produce less output.

The question I have, does this still draw the same amount of energy from the cells at either level?There is less current for the LED, but the difference is what was used to move through the resistance, as I see. Am I right or wrong?
 

peter yetman

Flashlight Enthusiast
Joined
Mar 23, 2014
Messages
5,100
Location
North Norfolk UK
I used to wonder the same thing.
Leds are not a linear resitive load. As you get to the higher voltages the current rises disproportionately.
If you consider the Malkoff Hi/Lo ring, which is just a switch with a resistor in it, the runtimes are increased when the resistor is in the circuit.

Hope that makes sense, I'm pre-Cornflakes and we've run out of milk.
P
 

Timothybil

Flashlight Enthusiast
Joined
Nov 9, 2007
Messages
3,662
Location
The great state of Misery (Missouri)
I used to wonder the same thing.
Leds are not a linear resitive load. As you get to the higher voltages the current rises disproportionately.
If you consider the Malkoff Hi/Lo ring, which is just a switch with a resistor in it, the runtimes are increased when the resistor is in the circuit.

Hope that makes sense, I'm pre-Cornflakes and we've run out of milk.
P

You are right. The extra resistance drops the voltage to the LED.
 

DIWdiver

Flashlight Enthusiast
Joined
Jan 27, 2010
Messages
2,725
Location
Connecticut, USA
The resistor is added in series with the battery and the LED. In a series circuit, the current is the same in every element.

Thus, lower current in the LED means lower current in the battery, which means less power is being drawn from the batteries, which means longer run times.

You could dim the light by placing a resistor in parallel with the LED. However, this would be phenomenally stupid, as it would increase the load on the battery, making the run time shorter. Nobody clever enough to bring a product to market would be this stupid. I don't say that to insult anyone, but rather to say "NEVER GONNA HAPPEN!"

Now, to go beyond the OP's question: As archimedes points out, there is now power coming from the battery that doesn't go to the LED, but to the resistor instead. This power loss represents a lower efficiency of the system in transferring power from the battery to the LED. The math here is quite clear and unambiguous, given only the information presented.

However, due to the highly non-linear behavior of the LED, this may very well NOT cause a lower efficiency in converting battery power to light. In fact, it may well improve it.

Say your LED draws 100 mA on high power and 50 mA on low, and say that on high it generates about 30 lumens. Because the lumens/mA of LEDs tends to fall with increasing current (and vice versa), at half the current you are likely to get a little more than half the light, so maybe 15.5, maybe 16 lumens. It's not a big effect, but it's quite often real (improvements in lm/W are much larger than in lm/mA).

Since you cut the current in half, your batteries should theoretically last twice as long, and as shown above, you're likely to get more than half the light. Overall, that's improvement in efficacy.

But it gets even better. Because batteries (especially alkalines, and even worse, carbon/zinc) have lower capacity at higher loads (and vice versa), cutting the current in half will give you more than twice the run time, compounding the improvement in efficacy.

Each of these effects does have a point of diminishing returns. But it is typically at least a factor of 10 below the rated current of the LED, and for the battery it's where the run time is starting to be a significant fraction of the shelf life of the battery, so it could be weeks for a NiCad, years for alkaline or lithium primary cells.
 

archimedes

Flashaholic
Joined
Nov 12, 2010
Messages
15,780
Location
CONUS, top left
Yes ... as you note, "all else being held equal" (which it never is)

Thanks for the adding the more detailed discussion above.
 
Last edited:

Lynx_Arc

Flashaholic
Joined
Oct 1, 2004
Messages
11,212
Location
Tulsa,OK
There is another type of LED light circuit that is out there too.... called a linear regulator which is essentially a variable electronic resistor that adjusts the throttling of power with the depletion of battery keeping the current to the LED constant until it drops below a threshold upon which the regulator does not have any headroom to operate then typically it goes into direct drive mode. The advantage of a linear regulator is constant light output till the batteries deplete too much then slow fade to nothing in light. The disadvantage is shorter runtime as static resistor loads the power to the LED drops as the battery voltage drops thus throttling power in the light increasing runtime.
Another advantage of a linear regulator is lights that are poorly designed that are essentially direct drive or overdriving LEDs on high mode a linear regulator can lengthen the life of the LEDs by elimination of overdriving them.
I'm no expert at detecting linear regulators on lights sometimes you have to measure current to the LED over time to detect if there is regulation or use a light box. For multi mode lights with a linear regulator you will still get the advantage of lower drive current reducing loss of power with higher internal resistance battery sources but the fade to nothing from the start static resistor advantage is lost.
 
Top