I couldn't find the resistance values for an LED so I tried to measure it and ended up much more confused than before:

I have a plain (red) LED and initially I used a multimeter to measure the resistance but it didn't show any values. I tried 2 multimeters with the same results; also, I tried to connect it the other way around with no improvement. (I do know that the LED isn't broken, it does emit light in a circuit).

So I went another route and added it to a (series) circuit to measure it from there. I have a 3V battery and a 100 ohm resistor, if I measure the current with just those 2 I get 30 mA, which checks out: $$ I = \frac{V}{R} = \frac{3} {100} = 0.03A $$

But when I add the LED to this circuit (also in series) I measure a current of about 11 mA. Meaning $$ R = \frac{V}{I} = \frac{3} {0.011} = 273\Omega $$ That would be the total resistance and since I know that the resistor has 100 ohm the LED should have a resistance of 173 ohm. This seems like a lot, from reading online I usually see that the default value is 75.

So to double check I swapped the 100 resistor with a 200 ohm one but that confused me a lot: If I measure the current without the LED I get 15mA which aligns with the calculations $$ I = \frac{V}{R} = \frac{3}{200} = 0.015A$$

but when I add the LED the current drops to about 6mA. Calculating the resistance from there I get $$ R = \frac{V}{I} = \frac{3}{0.006} = 500\Omega$$ So the resistance of the same LED would be 300.

I also tried another approach by measuring the Voltage drop and the results got even more confusing: The voltage dropped by 1.9 for the LED and 1.1 for the resistor; this applies to both the 100 and 200 ohm resistor; I had the same numbers for the 2 resistors which would get me very different numbers using Ohm's law to calculate the resistance.

I know I am doing something wrong but I don't get where my mistake is. Am I messing up the measuring with the multimeter, or am I misunderstanding the formulas, or do I get something more fundamental wrong?

This is how I measured the current with a 100 ohm resistor 100ohm measure Thanks so much to anyone reading all of this!


3 Answers 3


I couldn't find the resistance values for an LED so I tried to measure it

An LED isn't a resistor, and won't behave linearly the way you're expecting it to - your mental model is flawed and Ohm's Law doesn't apply to LEDs.

With a resistor, the formula V = I × R applies - a change in one of these three values will directly impact the others in some way. Increase the voltage with a static resistance and the current rises too.

The Voltage vs Current graph for a resistor looks something like this - the slope is controlled by the resistor's value (but this graph has no scale, so can be used for all resistor values).

A diagonal line with 1:1 relationship between increasing voltage and current

The fact that this graph is a straight line is what permits the application of Ohm's Law, emphasis mine:

Ohm's law states that the electric current through a conductor between two points is directly proportional to the voltage across the two points

An LED (Light Emitting Diode) can be conceptualised just like a diode - hypothetically, zero current will flow through it in one direction (i.e: the voltage drop is infinite), and a fixed voltage drop will present in the other direction. It is of course more complicated than that, but let's stick with the basic / perfect model for now.

If you have a diode with a known forward voltage drop, Vf = 2.1v, you'll get a Voltage vs Current graph looks something like below... (ignoring various realities of life)

A simplified graph showing zero current until 2.1v at which point the current rises rapidly

The fact that this line is bent, means that Ohm's Law does not apply, and you cannot directly model a diode as a resistor. It's just not possible.

This is also why a current limiting resistor is so important - once you reach the threshold voltage, the current through an LED / Diode will rapidly increase, possibly destroying the part.

LEDs and Diodes don't have a "resistance value", because they aren't resistors.

Would you have a source to read up more on the relationship between diodes and their relationship with current/volt and resistance? I've been reading introductory electronics books and they all gloss over it.

With regards to resistance - as above, it's not that they "gloss over it" so much as that isn't a thing / doesn't make sense.

If you're interested in reading into this further, then you can approximate a diode / LED using the Lumped-Elemenet Model.

Transistor provided a link in the comments, 'Resistance' of an LED, which helps to clarify this a little. Because a part of the graph is straight-ish, you can model it as a resistor... and because there is an apparent voltage offset away from zero, you can model that as a voltage source. However, this is an approximation, and the single LED shown in Transistor's article has two potential solutions depending on which area you're interested in (with more if you're interested elsewhere).


If you have a handful of resistor values, you can start to plot a real graph that approximately matches what I've shown above - with the LED in circuit, take a current reading (or voltage over a resistor and calculate the current) for a number of different resistor values, and plot them on a graph... then turn the LED around (reverse bias) and do the same... see what you get, and keep an eye on the LED's brightness too.

The datasheet for the selected LED should have a graph that looks something like below (from the Wurth 151051RS11000 datasheet, a Red 5mm LED). Here, the line is dashed above ~30mA, as that's the part's rated Continuous Forward Current (If)... sustained current above this will damage the LED.

V/I graph for 151051RS11000

  • \$\begingroup\$ That is super helpful, thank you! Only question: Would you have a source to read up more on the relationship between diodes and their relationship with current/volt and resistance? I've been reading introductory electronics books and they all gloss over it. \$\endgroup\$ Apr 19 at 17:56
  • \$\begingroup\$ I've edited my answer a bit to address this... but fundamentally, LEDs / diodes aren't resistors, and they don't have an inherent resistance. This line of thinking doesn't make sense, sorry. \$\endgroup\$
    – Attie
    Apr 19 at 21:28
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    \$\begingroup\$ "With a resistor, the formula V = I / R applies" – I think you mean V = I R or I = V / R. \$\endgroup\$ Apr 20 at 1:56
  • \$\begingroup\$ Whoops, thanks @TannerSwett! \$\endgroup\$
    – Attie
    Apr 20 at 11:31

A red LED at reasonable current levels has a voltage drop of about 1.8V or 1.9V.

So with a 3V supply and 100Ω series resistor we would expect a current of about (3-1.8)/100 = 12mA

With 200Ω series resistor we would expect 6mA. Which is what you measured.

Referring to 'resistance' of a nonlinear device like an LED does not make much sense because you can't really use it to predict the current under different conditions.

A better model is a fixed voltage source. And an even better model is Shockley diode with series resistance.

There is a characteristic called dynamic resistance that does make sense in some contexts where you look at the change in voltage across the LED for the change in current. For example, you changed the current by about 5mA (from 6 to 11mA) and the LED voltage will have changed a bit. You can measure this. The dynamic resistance will be \$\Delta V\$/\$\Delta I\$. However it is useless for calculating the LED current.

Below I've simulated the two situations simultaneously (source + resistor + LED) with lots of meters to show the currents/voltages. Hopefully the clutter does not obscure how simple the situation is.


simulate this circuit – Schematic created using CircuitLab

Here you can see that the voltage across the LED (using the Circuitlab model for a red LED) changes from 1.788V to 1.9V when the current is increased by 4.94 mA, so the dynamic resistance is about 22Ω. You will find that dynamic resistance is not only useless for calculating the total LED voltage or current, but varies a lot with the current, in fact it's more-or-less inversely proportional to the current.

In any case, you can see that the fixed-voltage-source model gives numbers that are in the ballpark, but the simulation model gives numbers that are even closer to what you actually measure.

In most cases we use the Vf numbers from the datasheet, since that is what is provided. Usually we get limits on Vf at one or two currents and a 'typical' (which means maybe not real, certainly not guaranteed) variation of Vf with current, and that is all we have to work with for design purposes. If it is inadequate to reasonably accurately predict the current that's a hint that the LED Vf near the desired current is too high for the supply voltage and we need to increase the supply voltage or use a different LED.

  • 1
    \$\begingroup\$ +1000 points for simulating. @user3553471 You can grasp a lot more about this concept by playing around in a simulator like CircuitLab (click the link within the answer). Change values with a couple clicks and re-run the simulation all within a few seconds. \$\endgroup\$
    – Ste Kulov
    Apr 19 at 15:17

A LED is not a resistor so it won't register with a multimeter to a resistance value. As a first order model, an LED has only a forward voltage (Vf) drop. If you look at the I-V characteristics of a real LED, you will notice it has both a forward voltage drop, a knee in the I-V graph, followed by a rise in voltage with current. The last part would be the series resistance of the LED.

To get a better model for calculating the current in a circuit with an LED, use I = (V-Vf)/R.

Here are some examples of LEDs and their I-V graphs: enter image description here

Image from If voltage determines the brightness of a LED, how does amps affect it?

  • \$\begingroup\$ I get that but then why do the numbers in the calculations go so weird? \$\endgroup\$ Apr 19 at 13:53
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    \$\begingroup\$ @user3553471 It gets weird because your formulas are weird and do not match the circuit. If you have 3V battery, 2V over LED and 1V over the 100 ohm resistor, that is exactly what the meter shows, 10mA. \$\endgroup\$
    – Justme
    Apr 19 at 14:28
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    \$\begingroup\$ @user3553471 Because you’re still trying to calculate the resistance of the LED. For a single operating point, it will have a value. At a lower or higher current, it will have a different. A resistance value isn’t suitable for an LED. A forward voltage is. \$\endgroup\$
    – winny
    Apr 19 at 14:31

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