I have always been told that LEDs need a resistor in series when powered from a battery.

If this is true, how is it possible to power an LED from a coin cell battery without a resistor? I have tried this, and it works without burning out the LED.

  • \$\begingroup\$ The grand kids soon realised it was not magic!! \$\endgroup\$ – vince Dec 21 '15 at 12:37
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    \$\begingroup\$ Small button batteries have quite a lot of internal resistance and are close to the forward voltage of the LED. \$\endgroup\$ – pjc50 Dec 21 '15 at 12:41
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    \$\begingroup\$ Have you ever measured the short circuit current of a button battery? \$\endgroup\$ – Andy aka Dec 21 '15 at 12:53
  • \$\begingroup\$ Another way to look at it - is that such batteries can only deliver a small amount of current, even if they were short-circuited. The resistor is there to prevent too much current from burning-out the LED. Example: if your battery can only deliver 13mA (max), that's about what a 330 ohm resistor would do connected with a +5v source. The spec on the LED limits the current. \$\endgroup\$ – Brad Apr 3 '17 at 16:28

That's because your battery is a voltage source and a resistor at the same time.


No battery is an ideal voltage source. This is easy to demonstrate: measure the voltage of a coin cell battery with a voltmeter. For your typical lithium coin cell, it should be around 3V. Now connect the LED to it, and measure the voltage again. The voltage should be around the forward voltage of the LED.

This works because real batteries have some internal resistance. A fresh coin cell might be approximately modelled by adding a series resistor. With an LED connected, it might look something like this:


simulate this circuit – Schematic created using CircuitLab

It's the internal resistance of the battery, R1 here, which limits the current. Just like a discrete resistor, this internal resistance drops a voltage proportional to the current through it according to Ohm's law. It also makes the battery warm.

The internal resistance also increases significantly as the battery discharges, and this is why a fresh alkaline battery might measure 1.65V, whereas a dead one might still measure 1.4V, but is unable to provide enough current to work as intended.

There's another way to demonstrate a battery's internal resistance: short a coin cell's terminals with a wire and note that it does not explode. The internal resistance of the battery limits the current through the wire (which has nearly zero resistance), and thus, limits the rate at which electrical energy is converted to heat.

Now try the same experiment with a lithium ion battery (which has a very low internal resistance) and a heavy gauge wire (which also has a very low resistance) and you will create a sizable explosion, or at least a fire.

On second thought, don't try that second experiment.

enter image description here


This... became a bit longer, but it nicely describes how LEDs and resistors work together without any math...

Have a look at the violet curve of this plot, and ignore the rest for a moment.

enter image description here

The curve sketches (i.e. the slope is just a number chosen by me, not real data) the voltage-current-characteristic of an LED with a nominal current of 20mA at 3.2V.

  • It's perfectly fine to apply exactly 3.2V to that LED. It will sink exactly 20mA, and everything is fine.
  • Apply 3.3V, and the LED will sink 40mA, which is definitely too much and will burn it.
  • Apply 3.1V, and he LED will sink just 10mA, which will not harm the LED, but it will be a little dim.

Now, consider the green curve, which could be another LED of the same type with some part spread, or even the same LED at higher temperature.

  • At 3.2V, the LED will sink about 30mA, which may not burn the LED instantaneously, but may damage it slowly.
  • At 3.3V, it sinks 60mA (immediate burn)
  • At 3.1V, it sinks about 15mA, which is fine again.

So, if the nominal LED voltage is above the source voltage, you don't need a resistor. But make sure this is still valid in worst cases where the LED will sink 20mA at a lower voltage.

Now, let's think about a higher source voltage of 3.7V and a resistor of 25Ohm, which is perfect for the LED:


simulate this circuit – Schematic created using CircuitLab

Let's play an intellectual game: Replace the LED by a potentiometer, and turn it. The voltage and current measured at the indicated node are drawn as blue line in my diagram above. (You can keep the LED and replace the resistor to get the violet/green curve, too)

The working point is where the blue line crosses the curve of the LED, since there current and voltage are the same for both games.

Now, the working point of the "bad" LED is about 3.16V and 23mA, which should still be fine.

Now, a coin cell usually has a nominal voltage of 3.0V, though you can usually measure 3.1V when it's full. Most bright LEDs have a nominal voltage of 3.2V, but even if it's lower than the battery voltage, the inner resistance of the battery protects the LED as shown above - even if it doesn't have the exactly correct value.

This also means: If you connect this LED to a set of two AA batteries with a very low internal resistance, your LED may blow up.


LEDs are current driven devices, and although they do not obey ohms law, they behave resistance like when driven with voltages near their forward voltage. So there are multiple ways to limit the current through an LED

  • with a resistor
  • with a linear current regulator (essentially a dynamic resistance)
  • a pwm signal smoothed by an inductor (probably most efficient)
  • by applying a voltage that is near enough to the forward voltage that the flowing current is small enough (see the LEDs datasheet for a graph)

What is likely happening here is (one of) two (or both) things:

  • The button cell has high internal resistance so that the voltage drops to a level that causes small enough current to flow
  • the LEDs current at the voltage the button cell can provide is small enough (common for white LEDs and some blue and UV ones)
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    \$\begingroup\$ Hmmm do not obey ohms law... thinking about this one.... \$\endgroup\$ – Andy aka Dec 21 '15 at 12:52
  • \$\begingroup\$ @Andyaka: nasty little rebels these droid^H^H^H^Hiodes ... \$\endgroup\$ – PlasmaHH Dec 21 '15 at 12:55
  • \$\begingroup\$ Pretty sure they still follow Ohm's Law, but there are a lot of other variables that come into play as well \$\endgroup\$ – DerStrom8 Dec 21 '15 at 13:23
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    \$\begingroup\$ Plasma is correct. Technically Ohm's law only applies to devices (really materials) that obey a linear voltage/current relationship, i.e. the resistance is independent of the current. It has been generally applied to the expression R=V/I, but that really only defines a value of R for a given V and I. R, in general, will vary with the magnitude of I, such as for a diode. \$\endgroup\$ – Barry Dec 21 '15 at 13:39
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    \$\begingroup\$ @derstrom8: About everything with formulas but the hardest quantum level stuff in physics is about useful approximations. Ohms law is not useful for LEDs, no one attaches a resistance value to them, or to BJTs. In most nontrivial circuits you rather talk about a complex (frequency dependent) impedance than about simple resistance. I think when we talk about such formulas as not being obeyed, it really means to not be useful to describe such a situation. \$\endgroup\$ – PlasmaHH Dec 21 '15 at 14:38

No, an LED does not always require a resistor. An LED, as its name implies (Light Emitting Diode) is a diode. So as long as you put a high enough voltage at its terminals to overcome the forward bias it will start conducting and thus emit light. What the resistor does is limit the current that will flow through the diode and thus stop it from "burning". It is a safety device. If the current provided by the battery you have is small enough or if the period that the current is provided is short enough to not cause the diode to "burn" then you do not need a resistor.

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    \$\begingroup\$ Actually, overcurrent kills LEDs rather quickly - you can't really create a safe short overcurrent pulse by plugging the LED in a circuit manually. \$\endgroup\$ – Dmitry Grigoryev Dec 21 '15 at 14:11
  • \$\begingroup\$ All diodes powered by a constant voltage source need a series resistance. Otherwise, how will you limit the current? \$\endgroup\$ – jose.angel.jimenez Dec 22 '15 at 8:29

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