Regarding the circuit below, when the switch is opened very fast according to the formula V(t) = L (di/dt) how can we interpret why the inductor current changes direction?:

enter image description here

During the switch is closed, the current flows in steady state from the upper terminal of the inductor to the lower terminal of the inductor. But the moment the switch is opened the current "wants" to flow the opposite direction. By "wants" I mean something causes it to flow that way. How can we relate this and the formula V(t) = L (di/dt) to explain what is happening more clearly?


Does the inductor polarity really reverses here:?

enter image description here

Edit 2:

This is regarding an answer:

enter image description here

  • 3
    \$\begingroup\$ Polarity of what? Voltage polarity does reverse. Current polarity (direction) does not. \$\endgroup\$ – Curd Feb 20 '18 at 11:35
  • \$\begingroup\$ I think what you mean by polarity in Figure 2, the inductor acts as a source so in my Figure 2 the polarity is shown wrong(?). But in Figure 1 how can we talk about polarity since the L acts like a short. So actually voltage does not reverse, but inductor starts acting as a voltage source. How can something does not exist get reversed? \$\endgroup\$ – pnatk Feb 20 '18 at 11:36
  • \$\begingroup\$ i was talking about voltage polarity. \$\endgroup\$ – pnatk Feb 20 '18 at 11:40
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    \$\begingroup\$ yes, in the second picture voltage polarity is shown wrong. The right terminal is more positive than the left one. In the first picture there is still voltage across the inductor (it's no a short; just a power sink; similar to a resistor) so it makes very well sense to talk about voltage polarity. \$\endgroup\$ – Curd Feb 20 '18 at 11:52
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    \$\begingroup\$ The inductor wants the current to flow in the same direction it did before the change. \$\endgroup\$ – Chu Feb 20 '18 at 13:02

This phenomenon confuses most beginners due to the whole, "A complete circuit must exist for current to flow!" thing. I find it is A LOT easier to understand if you add in the parasitic impedance around the inductor.

Closing the circuit

With an IDEAL switch, change your circuit and add a VERY LARGE impedance across the inductor, as shown below.


simulate this circuit – Schematic created using CircuitLab

That impedance exists in real life, though perhaps not with the value I chose in this example. It comprises of, amongst other things, air resistance, PCB resistance, and capacitive coupling across the inductor.

In a theoretical ideal circuit that impedance is infinite, but it still factors in, read on.

Now re-examine the events.

If the switch is initially closed long enough, a steady state current, in this example 15A, exists in the inductor along with a magnetic field. At this point, since an ideal inductor has no resistance, there is zero voltage across the inductor. The current is dictated entirely by R1 which has the full supply voltage across it.

When the switch opens, the magnetic field maintains the current in the coil in the same direction till it decays.

Now the only place for that current to come from is through the parasitic impedance. That means I-Coil, 15A, is flowing up through Z1. By simple Ohm's Law you can see that there will be a LARGE voltage drop up across Z1.

That puts the top of L1 at a very negative voltage, -15 terravolts with the indicated values.

Again, in a theoretical ideal circuit, the impedance, as I mentioned earlier, is infinite. That makes the peak -15 x Infinite = -Infinite Volts. See, it still factors in.

In reality of course something somewhere will break down before those voltages will be reached. Either an arc will form, insulation will breakdown, or some component will fail.

Of course, since Z1 is large, the LZ time constant is very small, so the voltage spike is very short.

Reality Check

In the real world, the impedance will actually be across the switch.

As the contacts break an arc will be formed due to the high negative voltage on the right side of the switch coupled with the initially very short contact distance. The arc acts like a switch, ionizing the air between the contacts, forming a much smaller impedance across the switch. The voltage on the right of the switch will then be at some less negative level low enough to just maintain that arc. Since LZ is less, the current will take much longer to decay.

In fact, with the wrong kind of switch, the arc resistance can be so low as to allow the voltage on the inductor to rise back up positive close to the supply voltage. At that point the arc will be maintained indefinitely, driven by the supply. Or at least till the whole thing melts or catches fire and falls apart.

The energy from the inductor will be released in the heat of the arc and cooking the switch, and may even blow your power supply.


simulate this circuit

If the switch is actually a transistor of some type, it will, in all likelihood, be destroyed. This is why you will see folks on here continually reminding people to add fly-back diodes to any circuit driving significantly inductive loads, like relay, solenoid, motor and transformer drivers. Even the inductance of long wires to a purely resistive load can cause the same destructive effect if the switching edges are fast enough and the currents are high.

Adding A Diode

In your second example..

You have added a low resistance path, the diode, for the current to continue to flow when the left end of the inductor tries to go negative.

That is, when the switch opens the current initially tries to pass through the parasitic impedance. This drives the left end negative as before. However, when the voltage reaches about -0.7V, the diode will turn on and carry the current and hold the left end of the inductor at close to -0.7V.

The voltage on the right side of the inductor is dictated by the voltage drop across the resistor, which initially, with 15A still flowing, is still 15V.

Before the switch opens there is zero voltage across the ideal coil. When the switch first opens, the initial voltage across the coil is -15.7V


simulate this circuit

  • 9
    \$\begingroup\$ Lovely description. You have a great, patient way of explaining things to people and can think down to beginner's level without being patronising to them or making them feel stupid. Your posts are a pleasure to read. \$\endgroup\$ – DiBosco Feb 20 '18 at 13:52
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    \$\begingroup\$ My pleasure. I think sometimes people forget how hard it is when you start out and how easy it is to get confused. It's not always easy thinking down to the level of when you started on electronics (or anything complicated really) and surprisingly difficult at times trying to explain things and theories that you use every day when designing. Blimey, when I try to explain to people what I do ... :D \$\endgroup\$ – DiBosco Feb 20 '18 at 14:00
  • \$\begingroup\$ @Trevor_G I got the first part but here: i.stack.imgur.com/zUOmS.png Does the current pass through the power supply? It is not so clear in that schematics how the negative voltage across L is created since there is nothing in parallel to L. What are + - pins at Z1 terminals? Can you tell how does the current loop through in that circuit when the switch is opened? Thanks \$\endgroup\$ – pnatk Feb 21 '18 at 17:02
  • \$\begingroup\$ I will try to re-draw that circuit and ask you what I mean in more detail \$\endgroup\$ – pnatk Feb 21 '18 at 17:09
  • \$\begingroup\$ @panicattack Yes the current will come through the power supply when the switch is arcing. There is ALWAYS something in parallel with the inductor, or at least some path for current to flow, though in an ideal circuit, that resistance is infinite, it still exists. The +- is just part of that voltage activated switch.. you can ignore them, just understand the arc works like a switch. \$\endgroup\$ – Trevor_G Feb 21 '18 at 17:35

The current stays flowing in the same direction. It's the voltage that changes, to form an arc across the switch if necessary, to 'try to' keep the current flowing at its previous value.

The simple equation \$V(t)=L\frac{di}{dt}\$ shows us that if the current tries to change value in zero time, the voltage will be infinite. In practice, a very small time for the change generates a very large voltage.

  • \$\begingroup\$ Is it possible to show it mathematically? I mean a formula for the current so that we can see in such equation to keep the current constant V must go to infinity ect? \$\endgroup\$ – pnatk Feb 20 '18 at 10:54
  • \$\begingroup\$ @panicattack yes \$\endgroup\$ – Neil_UK Feb 20 '18 at 11:07
  • \$\begingroup\$ Very large NEGATIVE voltage.... \$\endgroup\$ – Trevor_G Feb 20 '18 at 13:30

During the switch is closed, the current flows in steady state from the upper terminal of the inductor to the lower terminal of the inductor. But the moment the switch is opened the current "wants" to flow the opposite direction.

Your learning is flawed. Current continues to flow in the same direction until all the magnetic energy in the inductor is depleted by the generation of a spark across the switch.

It's the voltage across the inductor that reverses in order to try and keep the same current flowing.

How can we relate this and the formula V(t) = L (di/dt) to explain what is happening more clearly?

If you apply a constant voltage to an inductor, current ramps up at so many amps per second and the formula that relates the circuit parameters is: -

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

So when current is rising positively with time the applied voltage has a certain polarity. If current stops ramping up and "holds" at some DC value, the voltage across the inductor must fall to zero due to the same formula. If current then starts to fall, the voltage on the inductor must reverse as a natural outcome of the maths.

  • \$\begingroup\$ Please see my edit. Why do you say the inductor voltage reverses? In my edit the current keeps flowing across the inductor same direction in both Figure 1 and Figure 2, which means if I measure the potential difference between the left terminal and the right terminal of the inductor is always positive regardless of the switch is open or closed. Where am I wrong? Voltage seems not reversing across the inductor (?) \$\endgroup\$ – pnatk Feb 20 '18 at 11:28
  • \$\begingroup\$ You are mistaken. For current to continue flowing through the diode there has to be voltage pushing it through and this can only happen when the inductor voltage reverses. You have shown + and - symbols indictaing voltage but these are the worng way round else how could current flow through the diode? The voltage reverses and the inductor (with stored energy) acts like a generator to keep current flow as near to constant as it can. The voltage produced by the inductor is generated internally so it now pushes current out of the coil. \$\endgroup\$ – Andy aka Feb 20 '18 at 11:41
  • \$\begingroup\$ Ok let me ask the same to you as to the user above In Figure 1 how can we talk about polarity since the L acts like a short? So actually voltage does not reverse, but inductor starts acting as a voltage source that I see. But how can something does not exist get reversed? You guys saying inductor voltage reverses as if it has polarity when switch is ON. There is no polarity so what is reversing here. \$\endgroup\$ – pnatk Feb 20 '18 at 11:45
  • \$\begingroup\$ L doesn't act like a short unless the current is totally constant. Because the current is starting to reduce as energy is being taken from the magnetic field, the rate of change of current (di/dt) IS negative hence the terminal voltage it produces is negative. \$V = L\frac{di}{dt}\$. If di/dt is 1000 amps per second and falling it numerically equals -1000 hence, V = -1000 x L. If L is 1 henry then V = -1000 volts. \$\endgroup\$ – Andy aka Feb 20 '18 at 11:51

Theoretically after the switch is opened no current is flowing at all which would cause an infinite voltage Peak across the inductor. Practically the sudden change of current causes a high voltage peak (at the former negative terminal of the inductor) causing a short arc which closes the circuit again and disspates the energy stored in the inductor.

The current won't change directon (as Neil_UK noted). The voltage across the inductor will have, however, opposite polarity as before the switch was opened.

That's why the formula should be \$v = -L\frac{di}{dt}\$ (i.e. with negative sign).

The polarity of the voltage does change.
The polarity (i.e. direction) of the current does not change.

The inductor changes its behaviour from power sink (power goes into inductor before switch is opened) to power source (power goes out of inductor after switch is opened).

In order to model more accurately (avoiding an infinite voltage peak) what's happening you have to model the arc (which will be quite difficult) or change the problem (e.g. adding a freewheeling diode).

  • \$\begingroup\$ Please see my edit. Voltage across the inductor reverses? But in my edit current keeps flowing in same direction so the polarity of the inductor voltage does not reverse. Im a bit confused. \$\endgroup\$ – pnatk Feb 20 '18 at 11:31
  • \$\begingroup\$ @panic attack: voltage polariy DOES reverse. The voltage polarity shown in your right picture is wrong. Current comes out of the positive terminal of a power source. \$\endgroup\$ – Curd Feb 20 '18 at 11:55
  • \$\begingroup\$ @panic attack: "But in my edit current keeps flowing in same direction so the polarity of the inductor voltage does not reverse." The fact that current keeps going the same direction does not mean that voltage does not reverse. \$\endgroup\$ – Curd Feb 20 '18 at 13:17
  • \$\begingroup\$ @panic attack: Here's a mechanical analogy: if you accelerate a mass (inductance) you need a certain force (voltage) acting on it. If you suddely try to slow down the mass its momentum (current) is still pointing in the same direction but the force (voltage) acting on the mass reverses. First the mass is resisting, later the mass is pushing. \$\endgroup\$ – Curd Feb 20 '18 at 13:22
  • \$\begingroup\$ Ya you might want to reword that to state the inductor peak is large minus... Intuitively peak is though of as positive, or high, if not qualified. \$\endgroup\$ – Trevor_G Feb 20 '18 at 13:29

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