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I am trying to understand the current direction during non-steady states of inductor.

The circuit I am testing is very simple :

    (+)---Resistor--+----L---+---SWITCH-----(-)
                    |        |
                    |        |
                    |        |
                    +--LED---+

I am using the fastald.com simulator to see whats going on. And it appears like direction of the current changes according to the LED's polarity. for eg. If I connect LED's negative to the switch side, then the current will flow right to left from L/inductor. Opposite also happens if I connect LED's positive to switch.

My understanding is that L doesn't allow instantaneous change in current, and this pretty much appears instantaneous change by twice the amount of I, ie. from -I to +I or from +I to -I.

Whats going on ? is the simulator no good ?

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2 Answers 2

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I can't comment on the usefulness of Falstad- I don't use it and have seen some very silly results from others using it, but it's not necessarily the fault of the simulator. It could also be an issue of a model or the way a simulation is set up, or a misunderstanding of what the circuit itself does in reality.

As to what happens in simulation, it depends on the LED model. A reasonably accurate LED model has some capacitance so some current will flow immediately upon the switch closing, about V/R for very fast rise time, regardless of polarity.

Then the voltage across the LED will ramp up as the capacitance charges (assuming a relatively large value ideal (no capacitance) inductor. In one polarity the LED will conduct when the voltage reaches a couple volts or the inductor will conduct so the voltage falls again (ideally asymptotic to 0V across the inductor). The steady-state current in either direction is V/R.

Here is an LTspice simulation with real diode model and ideal R/L models.

enter image description here

You can see the current in either polarity jumps up to close to 10mA (not quite, because of the 10ns rise time) and in one polarity the behavior of the inductor dominates (the LED never conducts) and in the other the LED conduction dominates for the first few microseconds, then the inductor starts to hog some of the current. After 10us or so the circuit is very close to being in steady-state with virtually all the 10mA flowing through the inductor.

The initial conduction due to to the LED capacitance is very short with the values and models I've chosen- less than 100ns, so the 10ns rise time is significant.

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The inductor is like a inertia. When the switch is closed, there is a steady current of,

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

At this time, the inductor has a stored electric energy is

$$ W = \frac{LI^2}{2} $$

Once the switch is opened, the inductor will try to keep its current value. And the same current \$I\$ is sent through the LED, from right to left.

  • If the LED cathode is on the switch side, the current direction on the LED is reversed. If \$W\$, the accumulated energy on \$L\$, is enough to destroy the LED's junction, the LED will break down.
  • If the LED anode is on the switch side, the current direction to the LED is forward. The LED will be turned on and will be dimmed as the energy of \$L\$ is consumed for lighting.

In theory, the LED current should be a steady value, not depending on the LED direction. Because if the LED is wired in opposite, the reverse voltage should rise up to breakdown voltage.

But all real inductors have a quality factor. It makes the situation different. An ideal inductor has an infinity value of \$Q\$, but you need to check the real \$Q\$ value used on your simulation model.

The \$Q\$ factor is mainly used to analyze RF circuits. But here, if the LED is reverse wired, the voltage rising speed is very high, and the skin effect cannot be ignored.

Thus, the currents of two situations become different. The current of forward-wired LED will be greater than reverse-wired LED.

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