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I recently started to play with online circuit simulator and in a very simple circuit I can't understand the behavior

I have a voltage source connected through a current limiting resistor to a diode and an inductor in parallel. As far as I know, a diode should behave like a short circuit when its anode is connected to the positive terminal of the voltage source. In this simulator something very strange happens: as I close the switch, a lot of current passes through the diode (and very small amount through the inductor) and after couple of seconds no there is significant drop of the current through the diode until it stops completely. Why is that?

enter image description here

link for my circuit in the simulator: Link (click on the swtich to close it and watch the simulation)

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    \$\begingroup\$ Without having looked at the simulation, I'm going to venture a guess by saying it's accurately simulating the release of magic smoke from said diode. \$\endgroup\$
    – Mels
    May 23, 2017 at 8:50
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    \$\begingroup\$ Oh, not the case. The circuit in the simulator contains a current limiting resistor that wasn't mentioned in the question. \$\endgroup\$
    – Mels
    May 23, 2017 at 8:52
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    \$\begingroup\$ Edited to reflect that + screenshot of schematic and waveforms. \$\endgroup\$
    – Enric Blanco
    May 23, 2017 at 9:06

5 Answers 5

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As others have pointed out, a diode is not a "perfect" short (or open) circuit. However, if you understand its "limitations," then you can use the idealized behavior every were, except the limitations area.
For your particular circuit, you have to know that an inductor initially appears as an open circuit , and then as a short circuit after reaching steady state. What this means is that initially, your circuit behaves as if only the resistor and diode (in series) are connected to the supply. So the diode is forward biased and acts like a short.
As the inductor reaches steady state, the voltage across it goes to zero, and therefore the voltage across the diode goes to zero. Since the diode needs at least .6V to be forward biased, it stops conducting when the voltage on the inductor goes below .6V. At this point, the circuit behaves as if only the resistor and inductor (in series) are connected to the supply.
I hope you can now see that your simulator is showing the correct circuit behavior.

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Initially, the inductor resists the change of current, making the diode the path of least resistance and causing it to carry most of the current. When the magnetic field in the inductor builds up, the voltage across it decreases as it allows more current to pass. The diode has a forward voltage drop (typically 0.6V) to account for, so it won't conduct any current after the voltage across the inductor falls below the diode's forward voltage.

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    \$\begingroup\$ Also, when you turn that circuit off, you are likely to fry the diode with back EMF. \$\endgroup\$
    – Simon Richter
    May 23, 2017 at 13:49
  • \$\begingroup\$ @Meis, but why the current is trying to go through the inductor anyway, if a diode should behave like a short circuit when the voltage on it is greater than the forward voltage? \$\endgroup\$
    – yanivps
    May 24, 2017 at 15:30
  • \$\begingroup\$ @yanivps, because a diode does not behave like a short circuit. The idea that a diode acts like a short-circuit in one direction and an open circuit in the other is a significant simplification of its actual behavior. \$\endgroup\$
    – Solomon Slow
    May 24, 2017 at 16:58
  • \$\begingroup\$ @SimonRichter Couldn't many power diodes (including the 1N400x series) handle the short voltage spike considering the fact that they have a nonzero capacitance? \$\endgroup\$
    – Caleb Reister
    May 26, 2017 at 7:53
  • \$\begingroup\$ The voltage spike wouldn't be that short, considering that there's nothing in the circuit to dissipate the back EMF from the (pretty sizable) inductor. I haven't done the math, but I'd venture a guess that one would have to bring quite a beefy diode to avoid a reverse breakdown. \$\endgroup\$
    – Mels
    May 29, 2017 at 10:13
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Yes the previous posters are right. To further clarify, a diode is not a short circuit but a threshold device, it starts conducting whenever voltage across it (when oriented properly to conduct) is greater than some value, typically 0.6V (but may differ for special types).
So it behaves like this whenever the voltage is lower than 0.6 V no current will flow and when the voltage is over this threshold current flows.

The inductor responds to sudden changes in current in a different way, it exhibits something called impedance, that is a way to say that while it has a resistance R it also has an inductance L, a component that is directly dependent on frequency.

So an inductor when suddenly connected or disconnected from a voltage supply reacts by spiking up voltage for a brief while and the current is initially almost zero, only to settle a brief moment later with smaller currents and voltages approaching zero.

The diode in the circuit sees this increase in voltage (while the current is still almost zero in the coil) and it closes, letting the spike flow through it, reducing also the excessive voltage on the coil and thus the large current in the diode that flows for a very short time.

A very common arrangement usually called a SNUBBER is what you will find in some switching relays or even solid state devices. Its function is to stop the excessive voltage spike from breaking coil insulation by conducting temporarily the large voltage spike and then to close as the voltage on the coil returns close to zero. I merely translated the above equations and observations in layman terms, hope it helps.

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    \$\begingroup\$ In the future you should split answers into paragraphs so that they'll be easier to read. And a switch is not a tap: when you open a switch, the current stops flowing. // (This time, I have edited it for you.) \$\endgroup\$
    – Oskar Skog
    May 23, 2017 at 16:37
  • \$\begingroup\$ +1 for mentioning that a diode is not an actual short circuit. \$\endgroup\$
    – Oskar Skog
    May 23, 2017 at 16:57
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    \$\begingroup\$ It may be worth adding a little bit about the I-V characteristic of a typical diode along with the Shockley equation. \$\endgroup\$
    – Caleb Reister
    May 26, 2017 at 7:49
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When enough time has passed, the inductor behaves as a short circuit and bypasses the diode. That means \$V_d=0\$ and no current flowing through the diode.

OTOH, at the precise time of switching, the inductor will try to keep its current (that happens to be zero because it is not energized). Because of that, it briefly behaves like an open circuit and all current flows through the diode, which will be forward biased until the voltage in the inductor falls below \$V_f\$.

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For an inductor,

\$V = L\dfrac{di}{dt}\$

In any steady state condition, there is no change of current against time hence, the voltage across the inductor MUST be zero.

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