I have a simple circuit arrangement of a 9V battery - 1k Ohm resistor - LED. This circuit has a current of 7.5 mA (measured between the LED and the battery). If there's voltage drop of 2V in the LED, does it mean that my LED has a (2 % 7.5)k Ohm resistance?



4 Answers 4


Diodes are not ohmic. Resistors have a linear relationship between current and voltage. That is the relationship described by Ohm's law which you used in your question. Diodes look like this:

We normally model them as a simple voltage drop when they are forward biased and an open circuit when they are reverse biased. If more precision is desired, we sometimes model the forward biased state with a voltage drop and a resistance. However, what you have computed is not the same as this resistance. Your calculation fits a line between the origin and your present circuit's operating point, giving a much different number.

To say that your LED has that resistance is misleading as if you change the current, the voltage will not change proportionately.


You should not consider LEDs (and other diodes) as having resistance, as they do not obey Ohm's Law. The voltage across a lit LED varies only slightly as the current varies. It is best to say that an LED has a (nearly) fixed voltage drop, which varies with the colour of the LED.


Since resistance is defined as:

$$ R= \frac{E}{I}$$

In the case you've mentioned, by citing a particular forward voltage (Vf) dropped across an LED with a particular forward current (If) through it, we have

$$ R= \frac{Vf}{If} = \frac{2V}{0.0075A} =267 \text{ ohms}$$

However, since an LED's Vf will stay fairly constant while its If varies since the LED isn't ohmic, its resistance won't remain constant, and its calculated resistance will only be valid at the particular Vf and If cited.

For example, if the LED you you were referring to developed a Vf of 2.1 volts with an If of 15 mA through it, then its resistance under those conditions would be:

$$ R= \frac{Vf}{If} = \frac{2.1V}{0.015A} = 140 \text{ ohms,}$$

which, as you can see, is quite different from the first case.

Following is an interesting plot of forward voltage and resistance as a function of forward current for a randomly picked (junkbox) 20 mA red LED.

The current through, and voltage dropped across the LED were measured, and the resistance was calculated as discussed earlier.

Enjoy! :)

enter image description here


A better way to look at the resistance of the LED is to look at the total current and voltage for the circuit.

R = V / I = 9 / .0075 = 1200 ohms

Since the resistor is 1000 ohms, the LED must be providing 200 ohms of resistance. As mentioned though, this value is not constant like a normal resistor.

Also, assuming the battery is actually putting out 9 V, the LED voltage drop could not be 2 V. This would make the current (9 - 2) / 1000 = 7 mA.

  • \$\begingroup\$ That's not really a better way because, as you stated, it doesn't work. If the point of the exercise is to determine the resistance of the LED with 7.5 mA though it, the right way to do it is to force 7.5 mA through the LED, then to measure the voltage across it, and then to use Ohm's law to calculate the resistance under those conditions. \$\endgroup\$
    – EM Fields
    Commented Oct 7, 2015 at 10:31
  • \$\begingroup\$ I didn't say it was a good way. As has been mentioned, resistance is not something you usually look at for LEDs. But this is a conceptual way to figure out the resistance at that particular point on the LED's IV curve. I disagree with "it doesn't work" though. It does work, but you have to recalculate if anything changes. \$\endgroup\$ Commented Oct 7, 2015 at 14:26

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