Does a diode really follow Ohm's Law?

This really is not a black and white question and many folks will argue it does not follow "Ohm's Law", and depending how you argue it, they can be right.

However, the truth is the resistance of a diode changes depending on the applied current or voltage. As such, you can not simply look up the resistance of a diode and use "Ohm's Law" to determine the relationship between voltage and current by the good old V=IR formula like you can with a resistor. From that argument, no a diode, or more accurately, semiconductor, does not seem to follow Ohm's Law.

However, if you have a circuit with a diode in it, biased at voltage V or with a bias current of I, the resistance of the diode under those conditions is still a constant. That is, Ohm's formula still applies when the diode is in a steady state. If you are trying to calculate the output impedance of your circuit in that state, that is important to know, while acknowledging the impedance will be different when the circuit is in a different state.

In fact, I would go as far as to argue that a diode always follows Ohm's formula. Yes V=IR. However, in the case of the diode R follows a rather complex equation that includes V or I as variables..

That is for a diode

\$V = I.R_D\$ Where
\$R_D = F(I,V)\$
\$V = I.F(I,V)\$

So yes, mathematically, it does follow Ohm's formula, just not in a form that is much use to you except under very specific static conditions.

For those that argue "Ohm's Law does not apply if the resistance is not constant" I am afraid that is a misquote by Maxwell. Ohm's intent with that was that the resistance should be constant with time under stable excitation conditions. That is, the resistance can't change spontaneously with no change in the applied voltage and current. The truth is, nothing has a fixed resistance. Even your humble quarter watt resistor will change resistance when it warms up and as it ages.

If you think this is just he opinion of one man, you would be right, his name is
Georg Simon Ohm

Chances are you have never actually read his work, or if you read German, the original version. If you ever do, and, at 281 pages or antiquated English and electrical terminology, I warn you, it is a very hard thing to read, you will discover that he indeed covered non-linear devices and, as such, they should be included in Ohm's Law. In fact there is a whole Appendix, some 35 pages, devoted entirely to the subject. He even acknowledges there were things to still be discovered there and leaves it open for further investigation.

Ohms Law states.. according to Maxwell..

"The electromotive force acting between the extremities of any part of a circuit is the product of the strength of the current, and the resistance of that part of the circuit."

That however is only part of Ohm's thesis and is qualified in Ohm's words by the statement, "a voltaic circuit... which has acquired it's permanent state" which is defined in the paper, and I paraphrase, as any element whose resistance is dependent on the applied voltage or current or anything else must be allowed to settle into it's balanced condition. Further, after any change in the excitation of the circuit as a whole, a rebalance must occur before the formula is effective. Maxwell, on the other hand qualified it as, R must not change with V or I.

That may not be what your were taught in school, or even what you have heard quoted or read from many reputable sources, but it is from Ohm himself. The real issue is many people perceive or understand only a very simplified interpretation of Ohm's thesis, penned by Maxwell, that has been, possibly mistakenly, propagated over the decades since the great man actually performed his work as "Ohm's Law".

Which of course leaves you with a paradox.

The fact is Ohm simply stated, once it settles into a stable state the voltage across the circuit is the sum of the current times the resistances of the parts.

^{simulate this circuit – Schematic created using CircuitLab}

\$E = I.R1 + I.R2 + I.R3\$

Where R3 is whatever resistance the diode settles into. As such, it does not matter whether R3 is a diode or not. Which of course is correct. Maxwell, on the other hand, implies that since the circuit contains a non-linear element, the formula does not apply, which of course, is wrong.

So do we believe what Maxwell wrote was an error in oversimplification and go with what Ohm really said, or do we throw away what Ohm really said and go with Maxwell's simplification which leaves non-linear parts out in the cold?

If you believe a diode does not fit your mental model of Ohm's Law, then your model of Ohm's Law is actually Maxwell's Law. Something that needs to be qualified as being a subset of Ohm's thesis. If you believe a diode does fit the model then you are really quoting Ohm's thesis.

As I said, it is not black and white. In the end, it does not really matter since it changes nothing.

Diodes do not follow Ohm's Law. But. At any given current level, you can measure the change in voltage (\$\Delta V\$) for small changes in current (\$\Delta i\$), and get a local equivalent resistance called dynamic resistance. Graphically, this is simply the slope of the voltage/current curve for the diode, or \$Rd=\frac{\Delta V}{\Delta i}\$. This is often useful for describing how a diode in a circuit will behave at a given current level.

Your friend is simply describing the behavior of a standard (silicon, non-Schottky) diode, whose v-i curve is an exponential which is essentially essentially zero (for a graph which uses mA as the current axis) and which starts visibly rising at about 0.6 volts and which will normally hit very high currents by about 0.7 volts. That is, the dynamic resistance is very high at low currents and after (about) 0.6 volts rapidly drops. This means that, if you have a forward-biased diode driven by a variable voltage and fixed resistor, over quite a range of voltages the diode forward voltage will be pretty close to 0.6 or 0.7 volts.

Diodes do not follow ohms law. As you can see in your quoted passage, Ohm's law specifically states that R remains constant. If you try to calculate R from V/I while looking at a diodes IV curve, you will see that as you increase the voltage, "R" will change.

Your electrical engineer friend is incorrect. Saying that "the resistance varies to keep a constant Vdrop" is completely meaningless. In this case, the "resistance" is literally just V/I, which is changing. If you allow R to have any value in V=IR, the equation becomes useless because you cannot predict anything.

In your situation, you would not see a voltage drop. Both sides of the device would be at the same positive voltage (relative to the - terminal of the power supply)