Do I need a current limiting resistor when powering an LED below its forward voltage?

Question

I've reviewed the questions:

• Why can you use an LED below voltage?
• Is a current-limiting resistor required for LEDs if the forward voltage and supply voltage are equal?

But combining them begs another question:

If an LED (example: a 3.2 \$V_F\$ white LED) lights satisfactorily when powered using two alkaline cells (3V), is there a need for a current limiting resistor? Assuming the battery source never varies above the \$V_F\$ of the diode, can too much current ever be applied?

If it is recommended to always have a current limiting resistor, as in Martin's answer to the second question, does that mean the best option is to use three cells (4.5V) and add the resistor?

Answer 1

Have you tested the two alkaline batteries? I'd bet they are slightly above 1.5V each, bringing your total voltage up to the LED forward voltage. However, when the LED starts conducting, its low impedance pulls the battery voltage down slightly, and the battery itself acts as the current limiting resistor.

As to the second part of your question, I can't say whether its always better to add the resistor and use more cells. If you're making an LED throwie, then it definitely isn't. If you want a consistant runtime that doesn't rely on hitting a bullseye between battery voltage and current draw, then probably.

Answer 2

Note that the forward voltage of an LED as specified by the manufacturer is only an average value. Different parts may vary in their actual forward voltage at their specified current.

Therefore, it is not a good idea to blindly try to exactly hit the specified forward voltage (3.3V, for instance). To be safe, you could apply the correct forward current and measure the actual voltage for your specific part. This will give a pretty good value for the 'correct' voltage to use.

If that's not feasible, some safety margin can be considered to make sure to not overload the LED in even the worst case (specimen with relatively low actual Vf being used in high temperatures). Powering a white LED (Vf typically 3.2-3.3V) with 3.0V should be enough of a safety margin. - On the other hand, your LED ist probably not expected to live those 50-100k hours, so overdriving it a little should not be an actual problem.

Answer 3

Martins answer to the second question refers to "using a voltage source" and "using a current source". You are not using a voltage source, and you are not using a current source: you are using a two-cell alkaline battery.

Given that you've got a 3.08V circuit, I expect that (1) The battery will discharge rapidly and wastefully initially, then (2) The light will drop out while the battery is still almost fully charged.

In between (1) and (2) you should get a brief region where most of the battery discharge is going to power the LEDs (rather than heating the cells or just switched off). If the cell discharge curve exactly matches the LEDs voltage curve, this could be most of the energy in the cells, but I doubt I would be that lucky. At 3.08V with alkaline cells, I would expect the crossover to be fairly short, and up near the fully-charged end of the discharge curve (but it's the kind of thing you could measure)

If you use 3 cells and a resistor, you waste energy in the resistor, but I think you will use far more of the discharge curve of the cells, and not leave them almost fully charged when they dip below the LED voltage. So you may pay less for cells, even though you use three at a time instead of 2.

Obviously, the main problems with the simple circuit are blowing up the LED or exploding the cells, but clearly that hasn't happened with your experiment, which supports my idea that you are operating up near the top voltage of the cells.