I've reviewed the questions:

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?

  • 1
    \$\begingroup\$ LED Vf always varies with current. Look at any LED datasheet from a reputable manufacturer. If it has not got a Vf versus I graph it is not a reputable manufacturer :-). Alkaline batteries when new have Voc of over 1.6V. When almost dead they are about 1V. No white LED will operate usefully (for 99.9% of all values of useful) at 2V. To allow it to work over the whole battery range use 3 cells plus a resistor OR a red LED. I would ALWAYS design so as to use a resistor. The "spread" otherwise experienced is far too wide in practice to be useful if the design is meant to be repeatable. \$\endgroup\$
    – Russell McMahon
    Commented Dec 13, 2012 at 4:34
  • \$\begingroup\$ You have a current limiting resistor; it's just built into the batteries. \$\endgroup\$
    – Lenne
    Commented Jul 23, 2017 at 14:01

3 Answers 3


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.

  • \$\begingroup\$ My meter shows the two cells in particular as having a voltage of 3.08V. I suppose two possibilities exist, 1. the cells aren't new and replacing them could yield sufficient voltage to deliver excess current; or 2. The LED's actual forward voltage varies from some value below 3.08 to 3.2 or more. \$\endgroup\$
    – JYelton
    Commented Dec 11, 2012 at 22:02
  • \$\begingroup\$ @JYelton another possibility is your meter is not calibrated / accurate ;) \$\endgroup\$
    – vicatcu
    Commented Dec 11, 2012 at 22:35
  • \$\begingroup\$ Very true! Still, the project is for a toy (not quite a throwie) which is a one-off. Obviously if the LED doesn't last, it's not a big deal. \$\endgroup\$
    – JYelton
    Commented Dec 11, 2012 at 22:45

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.


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.

  • \$\begingroup\$ This is a pretty good explanation of what's going on, but I'm curious what an experienced EE would do to make it simple, functional, and safe. In other words, what would you change/implement? \$\endgroup\$
    – JYelton
    Commented Dec 12, 2012 at 17:49
  • \$\begingroup\$ I would talk to sales/marketing/management about what they are trying to achieve. Then I would measure the circuit voltage and current over time. It may not need an changes at all: even not fully discharging the cells may not matter if it only gets used a few seconds every month. For other uses a flashing circuit will give you both better cell life and better lighting. \$\endgroup\$
    – david
    Commented Dec 17, 2012 at 5:05

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