LED circuit seems to defy Ohm's law. Of course it’s something else

Ok, I promise I have checked and rechecked values here.

I needed to make an LED circuit for a Christmas display using one white LED inside a Santa Claus figure to replace the incandescent light bulb because the whole thing was so old it fell apart.

Ok, piece of cake. The white LED had minimal info, 20mA 5mm. I checked it with a component checker that had a readout that it had Vf (I take to mean forward voltage) = 2.82 and C = 36pF.

I am using two AA batteries in series for (measured) 3.11v when fresh.

Calculated 3.11/0.020 (3.11v/20mA) required a 155 ohm resistor. However when I set up the circuit the LED was way too dim, and the current measured was around 2mA.

I experimented with some resistors and ended up with (and this is all measured with two different voltohm ammeters) 4 niminal 10 ohm resistors for a resistance of 2.9-3 ohms, 3.11 volt supply voltage. Measuring current I measured 16.6mA and voltage drops of 0.05v drop on the 4 parallel 10 ohm resistors, and 3.06v drop across the LED. The LED is an appropriate brightness. But none of this adds up. Should be over 1 amp.

Have been running this setup for three days now, still glowing brightly. But it makes no sense. Anyone know what is happening here?

• You measured your LED's Vf at a very low forward current, and so you got only 2.82V. If you find & read the datasheet for your LED, you'll see that its Vf at its recommended operating current is substantially higher. Vf for an LED is not a constant. Oct 3 at 17:21
• Set R = (Vsupply-Vled)/Irequired Oct 3 at 17:25
• Re "I am using two AA batteries in series" - could you use three batteries? Would make entire circuit much more stable and predictable.
– Arvo
Oct 4 at 7:37
• @Arvo Not doubting you, but why that big of a difference? Oct 4 at 10:58
• @RockPaperLz-MaskitorCasket Batteries don't have identical and constant voltage; for 2 ones resulting voltage (3.0-3.3V) is too near to LED Vf and this may require new resistor value for another batch of batteries. 3 batteries give some headroom. (And with 2 NiMh rechargeables (2*1.2V=2.4V) LED will almost not light at all :))
– Arvo
Oct 4 at 12:17

To get 20mA with a 3.11V drop is 155 ohms just like you calculated.

But you did not add in the fact that from the 3.11V, there is already a 2.8V drop over the LED like you measured, so that leaves only 0.31 drop over the resistor.

In reality the LED might have even larger voltage drop with 20mA current that what your tester says. And in fact your next measurements prove it.

So anyway, if you measured 0.05V over 2.5 ohm resistance, that's 20mA, almost spot on with your measurement of 16mA. Which means, your resistor perfectly obeys Ohm's law, and nothing in the circuit defies any laws or equations of physics. It leaves 0.05V for resistor, 3.06V for the LEDs, and that adds up to equal the battery voltage of 3.11V.

• There may be a voltage drop over the internal resistance of the battery too Oct 4 at 9:53
• @psmears Yes, but so what, that does not affect anything as that is internal to the battery. The voltages and currents were measured at the points of a closed circuit, and KVL and KCL apply, even if in reality there is a black box called battery and it has internal ideal voltage source and source resistance. Oct 4 at 9:59
• It means the battery voltage under load will be less than 3.11V which will have a knock-on effect on the other calculations... Oct 4 at 10:02
• @psmears The battery already is measured under load, so the effect is already taken into account on all calculations, so measured LED and resistor voltages must add to measured battery voltage. KVL and KCL are satisfied. Yes, unloaded battery voltage will be higher, but also it means LED and resistor are disconnected and have no voltage or current. Oct 4 at 13:07
• Yes! Thank you and all the others that helped. It’s been a long time and my mind got stuck on diodes being “short” when forward biased, but of course, the voltage drop alone tells me it isn’t. So I was just not putting it together. Thanks for your help! Oct 4 at 21:11

People often misconstrue Ohm's Suggestion as some sort of promise that all things are linear, i.e. voltage proportional to current. Many real things do not behave that way. They are non-linear.

LEDs are at the top of the list of non-linear things.

Their data sheet should show the voltage/current curve that they do follow in their working range. Note that it is quite steep.

Most people are accustomed to constant-voltage devices or voltage-varying devices. Supply them a voltage and they do their thing. LEDs are a constant-current or current-varying device. You were annoyed that the only data they gave you was 20mA, which you perceived as useless - no, that's the spec. You need to push <=20mA through that LED at whatever voltage that happens. Weird!

So your goal was to come up with a simple resistor regulator that would - well, you were trying to engineer it to hit a voltage, but you needed to engineer it to hit a current at the voltage drop (battery minus LED operating voltage at the current you wanted).

The trick with LEDs is they will always behave at target current, but the voltage that happens at will wander around based on temperature, age, binning etc. Giving LEDs a constant voltage is risky since the voltage-current curve is so steep. So using a resistor as a current regulator, especially with LED voltage near battery voltage, is fraught with risk.

• LEDs are not constant-current devices. The datasheet-specified current is the current at which they produce the datasheet-specified brightness; the objectively-measured brightness is usually pretty linear with the current across the LED. In many applications, a much lower brightness is acceptable or even desirable. Oct 4 at 19:35
• @Russell yeah, I meant if you want a particular setting you go for a particular current not voltage, but sure. OK. Oct 5 at 5:53

Natural "laws" may be divided into three groups:

1. Those which define a quantifiable trait in terms of other quantities. Ohm's law is an example of this. Note that quantities may often be defined numerically in some circumstances where the numbers don't have any relation to any real-world phenomenon.

2. Those which specify the behavior of an idealized form of something. Many laws related to the behavior of gases fall into this category. To the extent that a substance behaves as an ideal gas, given any three of pressure, volume, temperature, and number of molecules, one can compute the fourth. Unlike resistance, which is meaningful only as a derived quantity, there are independent ways of measuring the pressure, volume, temperature, and molecular count of a quantity of gas.

3. Those which are believed to specify universal behaviors which are common to everything, such as laws of universal gravitation. Such laws may seem to be affected by curvature of space time, but within the appropriate reference frame they are believed to hold absolutely.

Many objects behave as though they have a "resistance" property such that, if one computes the property for some particular combination of applied voltage and measured current, or applied current and measured voltage, one can predict the current at other amounts of voltage, or the voltage at other amounts of current. Not all objects will behave in such fashion, and expecting them to do so would be analogous to expecting that applying two atmospheres of pressure to an ice cube which is sitting in open air would cause its volume to shrink by 50%. Because ice is not a gas, should hardly be surprising that its behavior deviates from the Boyle's law even more than most real-world gases.

No, LED isn’t ‘defying Ohm’s Law.’

As you may know, the LED current vs. forward voltage is highly non-linear: small changes in voltage near the 'knee' point have a large impact on current.

The Vf quoted in the data sheet is the forward drop at rated current (e.g. 20mA). What you measured with the ‘component checker’ (I assume the diode setting on your meter) isn’t going to give you the same forward voltage (Vf) as the data sheet, because the meter is sensing at a much lower current, and thus, gives a lower indicated Vf.

However, the LED will still give off light at much lower currents and somewhat lower voltages (plus, your eye is more sensitive at lower intensities.) That's why you still see a dim LED with 155 ohm load resistor. You might even be able to see it with the tester, too.

This nonlinear LED V-I characteristic I mentioned wouldn't be so much of a problem if it were tightly controlled. It isn't. Instead, LED V-I characteristics vary due to both manufacturing tolerance and operating conditions.

These variances pose a challenge when using a barely-enough voltage for the LED together with a low-value dropping resistor: with such a small voltage drop, it's very hard to 'tune' the resistor to hit the exact LED current.

The 'barely-enough voltage' is certainly the case for two "AA" alkalines (nominal 1.5V), which together add up to about 3.4V (2 x 1.7V) when fresh, then degrade down to about 2V (2 x 1.0V) when depleted. Meanwhile, the forward voltage (Vf) manufacturing spread for a white LED might be +/-10% for a 3.1V Vf LED. So Vf could be as high as 3.41V or as low as 2.79V.

Taken together then, 2 AA's working working voltage range vs. LED forward Vf is very narrow for a reasonable brightness.

The solution? Many 'fairy lights' that you might buy off Alibaba or at the craft store use just two AAs, and use a low value series resistor (10 ohms or so). These let the battery internal resistance roughly regulate current. This works surprisingly well with very long runtimes. That's kind of what you've hit upon.

Another solution is to use three AAs and a higher value resistor. This will be less sensitive to Vf variation and battery voltage than the two-cell approach.

A more involved solution would be to use a boost constant-current LED driver. This would have the benefit of longer run time, since these work as 'joule thieves' as the batteries go ever lower. In fact, you could probably find this pre-made as a fairy light battery pack, and reuse the module for your Santa.