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I am working on some circuit lab homework this week and am a bit stumped by something I just can't seem to find anywhere in my Circuit Analysis I book.

The lab has me construct the following circuit on my breadboard:

schematic

simulate this circuit – Schematic created using CircuitLab

I then mark down the ammeter and voltmeter readings, taking away one resistor each time until down to one. Using this information, my worksheet asks:

When the LED has current flowing through it, a voltage can be seen across it. Thus, the LED will have a resistance. Calculate the resistance of the LED for each of the four cases and record your results in Table III

My first thought is R=V/I. But the result just doesn't seem right to me.

Example data I have measured from this circuit with all 4 resistors: 1.586V, 186.2 microAmps. Using these measurements, the LED has a resistance of 8517.72 Ohms? That can't be right can it?

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marked as duplicate by PeterJ, Daniel Grillo, nidhin, brhans, MarkU Feb 15 '16 at 9:00

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  • \$\begingroup\$ You will find that the "resistance" of the LED will actually change as the current through the LED is set by the resistors and the Voltage drop is a physical property of the semiconductor junction in the diode. With 186.2uA the LED is likely to not be illuminated at all, in reality this is because the resistors chosen are much too large for the LED \$\endgroup\$ – crasic Jan 27 '16 at 2:05
  • \$\begingroup\$ The LED does indeed light up, although it is extremely dim. Yes the resistor values are pretty silly for a single LED, but that is what the lab instructions call for. \$\endgroup\$ – Seth Duke Jan 27 '16 at 2:35
  • \$\begingroup\$ There is a way to calculate the series resistance of an LED, but it is rather... involved. For more information, see this. Does the assignment have you do anything after this? It might be there to demonstrate that Ohm's law by itself doesn't fit a diode. \$\endgroup\$ – helloworld922 Jan 27 '16 at 2:56
  • \$\begingroup\$ The follow on step in the lab is: You should be able to get a close approximation of the current flowing in this circuit by including your calculated resistance for the LED as a part of the overall resistance: I = V(battery)/R(resistors)+R(led) \$\endgroup\$ – Seth Duke Jan 27 '16 at 3:13
  • \$\begingroup\$ Since it is a series circuit, your current used to calculate LED resistance is in fact the system current, while you will confirm your measurement its effectively a tautology. I think this is a poor way to teach about diodes as they specifically do not behave like a linear resistance and their equivalent Ohmic resistance is a function of the current! However its possible the following steps demonstrate this. \$\endgroup\$ – crasic Jan 27 '16 at 4:27
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Well, you're on your way to finding out that the resistance of a diode is not linear! But you have the right equipment and know the right formula R = V / I (ohms law). Measure the voltage and the current and then calculate the resistance of the LED. You will find that it changes!

Remember - if you replaced the LED with a fixed resistor at that calculated value, you should get the same current and voltage reading.

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In a normal resistor, the resistance is constant, and independent of the voltage you use to measure it.

In a non-linear device (such as a diode), the resistance depends dramatically on the voltage used to measure it -- meaning that if you measure using different voltages (or currents), you will get different answers.

In the case of a diode, the I-V curve is exponential, and you will find that the resistance decreases nearly inversely with current.

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