I was taking some measurements on a simple circuit consisting of a 3.7 volt battery (an old one used for vaping and such), a 20mA LED that has a listed voltage of around 2.3 V, and a resistor of about 100 Ohms (though when I measured it with the multimeter it was 102-ish, ranging from 99.5 or so to 102.1).

So, I measure the voltage on the breadboard across my circuit. I get about 2.96 V, which I kind of expect given that the battery is a bit old and has seen some use. So far, so good.

Now I measure the current through the circuit. From Ohm's Law I expect to see about 30 mA, give or take.

(Because I = V/R and I = 2.96/102, approximately 0.029 A, but I am using a less-than-high-precision multimeter, so anything around 27 mA to 30 mA would be expected).

Then I measure the current. I get 1.9 mA in a couple of different measurements. The LED lights up when I do this, the meter is part of the circuit. Maybe the meter is just old and messed up.

I know that LEDs are not perfect conductors, but that amount of resistance from it seems a little high. So I try something else. I break out a 10k resistor (measured at 10.65 kΩ) with a different meter. My hand calculation says I should see 0.28 mA, since the voltage is 2.98 (again). This time the meter doesn't even pick up the current reading. I sort of expected that but with another meter I get 1.9 mA.

And then I try something else. Using the 10k resistor I measure the voltage drop across it, and across the diode. The diode shows a drop of 2.46 V and the resistor a drop of 0.48 V -- it adds up perfectly, and I suppose it should. In this case, the "effective resistance" of the diode would be 2.46 V / I and here again I get about 1.9 mA. That would mean the diode is acting like a resistor of 1.3 kΩ or thereabouts.

Anyhow, I was wondering what the explanation might be for this behavior. I suspect it is because diodes don't behave in a linear fashion, so while they are rated for a given current that isn't going to follow Ohm's law at all (nor do I expect them to).

Or, I could be working with crappy meters, which are these school-issued ones that are a bit on the aged side. AFAICT they work ok tho, as the values they give for currents and the like seem reasonable in other contexts (I should probably try them with a bunch of different known resistors tho, just to see if the current readings are way out of whack, maybe there's a blown fuse or something).

  • 2
    $\begingroup$ Wouldn't the Electrical Engineering Stack Exchange site be a better home for this question? $\endgroup$ Oct 17, 2022 at 23:39
  • 1
    $\begingroup$ Actually, if you ask on the EE stack exchange the question probably will be closed as a duplicate. There's already lots of answered questions there about a circuit that consists of a battery, a resistor, and an LED all in series. Go there, search for "LED" and "resistor", and you'll find gobs of them. $\endgroup$ Oct 17, 2022 at 23:47
  • $\begingroup$ @SolomonSlow, good point. It would be discouraging to OP to have a question migrated to EESE only to be closed there as a duplicate after migration. Vote to close here then? $\endgroup$ Oct 17, 2022 at 23:53
  • 1
    $\begingroup$ P.S.: No. The LED is not acting like a resistor. It's acting like a diode. See hibp.ecse.rpi.edu/~connor/education/EI_archives/exp08.pdf Compare figure A-2 (describes the ideal diode behavior that you may have been expecting) with figure A-3 (describes the behavior of a real diode). You said you measured a voltage drop of about 2.46 V across your diode. That probably corresponds to the "knee" in the plot where the current starts to go up fast. $\endgroup$ Oct 17, 2022 at 23:54
  • 1
    $\begingroup$ I didn't know the EE stack exchange existed... d'OH. $\endgroup$
    – Jesse
    Oct 18, 2022 at 0:11

2 Answers 2


I'm posting this despite my conviction that:

(1) this would be better homed on EESE and

(2) it is almost certainly a duplicate there (and possible here too)

(Because I = V/R and I = 2.96/102, approximately 0.029 A, but I am using a less-than-high-precision multimeter, so anything around 27 mA to 30 mA would be expected).

You can't use Ohm's law here since the circuit has a non-linear element, i.e., the LED, in series with the resistor.

But you can make a first approximation by assuming the nominal voltage across the LED (the voltage across when the current through is nominal) and then see how well the solution 'fits' using Ohm's law for the voltage across and current through the resistor.

For example, if the LED has a voltage across of 2.3V (at nominal current of 20mA), and the total voltage across the series connected LED - resistor combination is ~3V, then the voltage across the resistor is ~0.7V.

Now apply Ohm's law for the resistor and find the series current to be:

$$ \frac{0.7\mathrm{V}}{102\Omega} \sim 6.9\mathrm{mA}$$

This result isn't consistent with the assumption of a 20mA series current and so you must then iterate.

Generally, one assumes a nominal value of LED current and a total voltage across the series combination, and then one chooses the resistor value that is consistent with this.

  • 1
    $\begingroup$ Thanks so much. I know this probably had a duplicate somewhere but when I look this kind of thing up it never seems to quite answer the question I had. In any case I suspected the nonlinear behavior was the issue. I use LEDs in circuits for students a lot because you basically have an indicator light that tells you stuff is working. But it does mess up readings. $\endgroup$
    – Jesse
    Oct 18, 2022 at 0:18

$\rm resistance,\,\Omega = \dfrac{\rm voltage,\,V}{\rm current,\,\rm A}$

Here is an LED characteristic graph of current against voltage.

enter image description here

For each current (or voltage) you can define a resistance for the LED.
In this example at a current of $1\,\rm mA$ the resistance of the LED is $1.92/0.001 = 1920\,\Omega$ and at a current of $2\,\rm mA$ the resistance of the LED is $2.04/0.002 = 1020\,\Omega$.

Suppose you have a $10\,\rm V$ power supply and consider that the LED is bright enough when a current of $2\,\rm mA$ is passing through it, how do you link the power supply to the LED so that the LED is not destroyed by having too much current passing through it?
You add a series resistor which when a current of $2\,\rm mA$ is passing though it has a voltage of $10- 2.04= 7.96\,\rm V$ across it. Think of it as the "excess" voltage being across the resistor.
So the resistance of the series resistor should be $7.96/0.002=3980\, \Omega$.
In fact resistors normally come with preferred values and the nearest such value is $3900\,\Omega$ with the current through the diode then being slightly over $2\,\rm mA$.

If the choice was to have the LED shining less brightly using a current of $1\,\rm mA$ then the resistance of the series should be $8080\,\Omega$ with the nearest preferred value being $8200\,\Omega$ with the LED running a little less brightly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.