11
\$\begingroup\$

In a previous question, it was brought to me that LEDs do not obey the Ohm's law. (See Calculate expected voltage around a resistor)

Simply put: how is that?

What makes them behave so differently? How should we treat them in a circuit and calculations?

Are there other components with similar behavior?

\$\endgroup\$
  • 5
    \$\begingroup\$ The non-ideal diode model has an exponential term in it. More importantly, Kirchoff's laws are satisfied, and those always apply. \$\endgroup\$ – Matt Young Apr 16 '13 at 19:31
  • \$\begingroup\$ @MattYoung just for clarity, the ideal diode has an exponential term, and the threshold model is just a very rough simplification \$\endgroup\$ – clabacchio Apr 16 '13 at 19:40
  • \$\begingroup\$ Try applying a certain variable voltage to water. What you'll find is that the resistance changes with voltage. Air also doesn't obey Ohm's Law - you've got gigantic voltages floating in the air. But there's almost no current until the voltage reaches a certain level. What you observe then is a spark in the form of a lighting. Ohm's law applies only to resistive materials - by definition. What does not obey Ohm's Law is not a resistor. \$\endgroup\$ – Jonny B Good Apr 16 '13 at 19:45
  • 5
    \$\begingroup\$ You have just discovered the principle of linear vs non-linear devices. Expect non-linear behavior from all the semiconductors. \$\endgroup\$ – gbarry Apr 16 '13 at 20:48
15
\$\begingroup\$

Ohm's law applies to resistance. All resistive aspects of a device will behave according to OHm's law.

If you invert your question you see that every thing that behaves according to Ohm's law must be a resistor. There is only so much that one can do with pure resistance. So logically the anything that doesn't behave according to ohms law isn't a resistor. Or any thing that isn't a resistor won't behave according to ohms law.

I believe that is called a Tautology.

In circuit design we have many different devices all having unique properties to be able to implement different things/functions.

\$\endgroup\$
  • 3
    \$\begingroup\$ I think your anwswer needs to be highlighted more as it is the only correct one (at the moment I am writing this). Ohm's law is empirical and was originally derived from obserwing the behaviour of wires of different length. Water doesn't obey Ohm's Law, air doesn't - only conductive materials do, and even then not always. \$\endgroup\$ – Jonny B Good Apr 16 '13 at 19:42
  • 2
    \$\begingroup\$ If I had a black box, and ran a current through it - then measured the voltage across it - I could calculate it's resistance at that point in time. It' doesn't matter what is in the black box. \$\endgroup\$ – Brad Apr 16 '13 at 19:55
  • 1
    \$\begingroup\$ Exactly - like I said: A diode does not have a "fixed resistance". I could however argue that it does have a known resistance for a given current. The question was about if it obeys ohms law, and it does. It just doesn't have a constant resistance. \$\endgroup\$ – Brad Apr 16 '13 at 20:18
  • 3
    \$\begingroup\$ @Brad, it doesn't obey Ohm's law - period. For Ohm's law, V and I are proportional, i.e., the ratio is constant. \$\endgroup\$ – Alfred Centauri Apr 16 '13 at 22:16
  • 1
    \$\begingroup\$ You are describing the fact the the response is nonlinear - and I agree it is not. However that was not the question. By this logic a "variable resistor" would also disobey ohms law. Ohms law defines the relationship between resistance, current and voltage as an equation - proportions to each other. It merely states a change in one will require at least a change on one other to remain valid. You are insisting R must maintain constant while only V and I change for a device. This would describe a device which is linear and purely resistive - but not the only one which would apply. \$\endgroup\$ – Brad Apr 16 '13 at 22:46
16
\$\begingroup\$

They do - they just do not have a "fixed" resistance. If you look at it from a standpoint of having a fixed forward voltage drop (which they sort of do - depending on operating region) - look at them more as having a fixed voltage across them. Therefore, as different currents go through them, their voltage will stay (relatively) constant, but the resistance will change.

This is a simplistic answer - but I think you're talking at this level.

\$\endgroup\$
  • 5
    \$\begingroup\$ Well, it is a diode, which is a semiconductor which inherently means it does not have a fixed conductance, like a normal conductor. The properties of this (and other) semiconductors are complex. They do different things in different operating regions. It's resistance is more of an artifact of it's operation at any specific point - as opposed to a fixed quantity. See "Voltage-Current Characteristic" here: en.wikipedia.org/wiki/Diode \$\endgroup\$ – Brad Apr 16 '13 at 19:44
  • 3
    \$\begingroup\$ This is not a good answer. If a circuit element obeys Ohm's law, the voltage is proportional to the current, i.e., the voltage across is a linear function of the current through - full stop. Moreover, this answer conflates the notion of resistance, V/I, and dynamic resistance, dv/di. See, for example, youtube.com/watch?v=QF6V74D2hbY \$\endgroup\$ – Alfred Centauri Apr 16 '13 at 21:42
  • 3
    \$\begingroup\$ I don't agree. Ohm's Law does not assert that resistance cannot be a function. To deny this means that, for instance, a potentiometer or rheostat do not obey Ohm's Law, because someone can turn the knob. \$\endgroup\$ – Kaz Apr 16 '13 at 23:16
  • 3
    \$\begingroup\$ @Kaz, linearity and time invariance are distinctly different. You're conflating the two. It they were the same, we wouldn't need to separately specify, for example, linear time-invariant system. A variable resistor is, at any moment of time, a resistor with resistance that is constant with respect to the voltage across it or the current through it. \$\endgroup\$ – Alfred Centauri Apr 17 '13 at 0:42
  • 2
    \$\begingroup\$ @AlfredCentauri But so is a diode, at any moment in time. It's like a rheostat, except that the daemon inside the diode which turns the knob is looking at the forward voltage rather than time. \$\endgroup\$ – Kaz Apr 17 '13 at 0:49
4
\$\begingroup\$

Simply put, because the're not resistors but p-n junctions, and because of that their V-I ratio is exponential.

It doesn't mean that you can't calculate their current, just that it's not as simple as for resistors. For instance, you can treat them with a threshold model, with a fixed voltage drop. Then the current will be set by external resistors or active components.

The LEDs are diodes, so that's the obvious similarity. Also the base-emitter junction of a bipolar transistor is a diode, and behaves similarly. The only difference with diodes is that their threshold voltage is higher due to the different materials and doping.

\$\endgroup\$
3
\$\begingroup\$

A lightbulb, on first examination may not appear to obey ohms law. Measure its resistance with a multimeter and it might be 5 ohms. Connect it to a power supply capable of illuminating it and measure current and voltage and its resistance will have considerably risen (maybe 20 or 30 ohms). Its still a resistor but its resistance changes with power delivered to it.

A light dependent resistor is another example - its resistance changes with incident light - it's still a resistor and obeys ohms law - but it takes a little bit more than a linear volt-current graph to figure things out.

\$\endgroup\$

protected by W5VO May 5 '13 at 21:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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