Why Don't Paper Circuits Need Resistors for LEDs?

Paper circuits are a popular type of project that use 3V coin cell batteries and copper tape instead of AA batteries and a breadboard to build a circuit. However, they don't need current-limiting resistors with LEDs! If you've ever built circuits with LEDs before, you know current-limiting resistors are a must in order to avoid burning out your LED. This video explains the concept of battery internal resistance and how it applies to these two different type of circuits.
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[@Noxoreos]

I'm sorry, but I feel like this will lead many people to a wrong understanding, since the info is not quite right, or at least misleading. I'll try to explain why that is:
While the internal resistance of the coin cell might be the reason for the LED to not burn out that fast, I would not conclude, that the missing resistor in the AA battery circuit is the reason why the LED fails. The coin cell doesn't do very well to the LED either, when operated for a longer duration. Also when you short the coin cell, it drops it's voltage significantly, so the internal resistance of it is just enough to drop the voltage, so the LED can be operated without breaking it immediately. It is acually still slightly outside of the safe operating range for a 5mm LED and that will damage the LED over time. It depends on the color of the LED though, because not all operate at the same forward voltage. In a paper circuit the duration is not long enough to see the LED burn out in most cases.
It is still a good reminder that the battery itself can drop it's voltage due to internal resistance, which is often overlooked and the main reason why a battery can't keep up the nominal voltage when the current draw is too high.
One thing to watch out for: Many coin cells provide up to 4V. I have checked a few coin cells of mine and they are all marked with 3.6V and rest at a voltage of 3.9V when measured without any load. To find out what the actual current draw (and voltage) would be, it is necessery to measure it's voltage, while connected to a load with a similar impedance. Measuring the battery voltage without a load doesn't tell anything and reading the nominal voltage on it is also not enough to find out if it's safe to use or not.
When a circuit is calculated, then the battery resistance is assumed to not exist at all in most cases, but it is still possible to drive leds without resistors when the correct voltage is supplied.
Every LED has it's own internal resistance and power rating. That means that the current through that LED is actually a result of the applied voltage (assuming that the connected power supply can deliver any required current draw without dropping the voltage). When the combination of applied voltage and the internal resistance of the LED exceeds the LEDs power rating (or current rating, which essentially is the same thing, because they are proportional to each other), only then the LED burns out (theoretically - in reality they can still fail when manufactured poorly).
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So saying "the LED burns out because of the missing resistor" is only half of the truth (or actually wrong - kind of). Yes, adding a resistor with the right properties prevents the LED from burning out, but you don't necessarily need to use a resistor to drop the voltage if you can achieve this effect by other means, which there are plenty of: Buck converters, normal diodes, zener diodes, additional leds in series, a chopped signal and a capacitor, half charged batteries and many other ways.
For example you could even use five AA batteries (assuming they don't exceed 1.5V each) and connect four typical 5mm LEDs in series that have a forward voltage at around 1.7V. If the total voltage of 7.5V is split by four, that would mean each LED is driven at 1.875V, which is completely safe and you don't need a resistor.
I think you could even use two NiMH AA batteries, because they have a nominal voltage of 1.2V.
I recently built a night lamp with multiple 5mm LEDs and one high power led all in series which are driven at 12.9V using a boost converter connected to a 3.7V lipo battery. I have not used any series resistors for the leds in there and the lamp still didn't fail after multiple nights of operation (all night). The used lipo battery can easily provide about 40A of current at 3.7V, because it is made to be used on a 2" freestyle drone.
Furthermore, while I know that everyone calls it a "current limiting" resistor. It is actually not limiting at all. It drops the voltage across it, which results in the current being reduced, but it does not actually "limit" the current and the voltage still matters. A zener diode would be a better candidate for this term, because it drops as much voltage as it needs to (before it burns out).
It also highly depends on the charging status of the AA battery. A full AA battery actually provides up to 1.65V and two of them would be way above the safe operating range of one typical 5mm led.

[@Science.Buddies]

Hi - you are correct that there's actually a lot more nuance to this than discussed in this video. Our resources are primarily intended for K-12 students/parents/teachers using related projects on our site, who the vast majority of the time are going to be facing one of these options - coin cell with no resistor, or AA battery pack/Arduino with resistor, as opposed to the other various options for designing a circuit with no resistor. Your explanation is more appropriate for a more advanced level of understanding.

[@tarunsharma5254]

damn.. this is very good info...God blesw you mate 🎉 cheers

[@Noxoreos]

The info about internal resistance is useful and correct, but the conlusion drawn here is actually wrong.

[@tonywright8294]

Well who’d of thought it ?