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I'm aware of that, in a circuit, inductive loads need to be protected with a flyback diode because if the circuit gets open the magnetic field will collapse forcing a current flow in the opposite direction.

What I don't understand is that last part: opposite direction, let me explain:

Let's take the following flyback diode example at normal operation (circuit closed): enter image description here

Suppose now that the circuit opens. My understanding of what a current in the opposite direction should be is the blue line, and the actual current flow that must happen (otherwise the flyback diode needs to be inverted) is shown in red:

enter image description here

So, why is the actual current flow (red) described as in the opposite direction while it is going in the same direction as before?

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    \$\begingroup\$ Whoever said "opposite direction" meant that the current through the diode is flowing bottom to top, while the current through the load is flowing top to bottom. \$\endgroup\$
    – The Photon
    Jan 9, 2021 at 19:31
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    \$\begingroup\$ The current flow through the speaker continues in the same direction after the transistor opens. It does not reverse. \$\endgroup\$
    – JRE
    Jan 9, 2021 at 19:31
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    \$\begingroup\$ Inductors "try" to maintain current flow through them in the same direction (and same magnitude). It's what they do. The voltage across them may reverse. \$\endgroup\$ Jan 9, 2021 at 19:34
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    \$\begingroup\$ Your red arrows are correct. Your understanding is correct. Your confusion is understandable. I can't explain what someone might have meant by saying "opposite direction." Yes, the current will decay over time. The series resistance of the inductor will convert energy into heat. The forward voltage drop on the diode is also an indication that the diode is converting energy into heat. \$\endgroup\$
    – user57037
    Jan 9, 2021 at 20:41
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    \$\begingroup\$ One thing to mention explicitly: while the current keeps going through the speaker in the same direction, the voltage across the speaker will change sign - so there's something that loosely "reverses direction" in a sense. This might be part of the confusion, given that current usually flows from high voltage to low voltage, but this is a situation where that's not the case since the inductor (speaker) uses up stored energy to push current upwards in voltage - or, otherwise stated, the reversed voltage is what allows the current through the inductor to decrease. \$\endgroup\$ Jan 10, 2021 at 4:02

3 Answers 3

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The loudspeaker is an inductor; so you have to create a good notion about its behavior. Here is a simple explanation.

Both capacitors and inductors are accumulating elements. They accumulate energy; so they can be considered as sources... rechargeable sources.

The capacitor can be thought of as a voltage source containing potential energy (as a tensioned spring). So, when discharges, it retains its polarity and passes a current in an opposite direction (the spring returns).

The inductor can be thought of as a current source containing kinetic energy (like an inert moving object). So, when it discharges (step 4 in Fig. 2 below), it reverses its polarity and passes a current in the same direction (the object continues moving in the same direction).

You can investigate the inductor behavior by this attractive Flash movie. This is an exe file with embedded Flash player (it is absolutely safe since it is uploaded on my site of circuit-fantasia.com); so play and have fun. Here are two typical steps:

inductive_differentiator_charging

Fig. 1. The inductor is charging

inductive_differentiator_discharging

Fig. 2. The inductor is discharging

Also, you can find an explanation of this clever circuit trick in another answer.

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    \$\begingroup\$ "(it is absolutely safe since it is uploaded on my site of circuit-fantasia.com)" -- umm, yeah, I don't it works like that. While you obviously trust yourself, for the reader, you're still just another unknown on the internet. \$\endgroup\$
    – ilkkachu
    Jan 10, 2021 at 10:34
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    \$\begingroup\$ @ilkkachu, I see... But unlike most contributors here, I am not anonymous. Everywhere I have posted on the web, I have done it openly because I don't feel the need to hide anything. I use the username "Circuit fantasist" (dreamer) not to hide but to imply that I have a different view of circuits... but I give my real data in my profile pages. If you want, you can visit my blog Circuit Stories that I started last month. Adobe put me in an awkward situation (my site is made on Flash in 2000, then I did not know it would happen today). \$\endgroup\$ Jan 10, 2021 at 10:51
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    \$\begingroup\$ @Circuitfantasist you are not only asking us to trust your good intentions, which many of us are probably willing to do, but also asking us to trust that you are an security expert and that your site hasn't been hacked by anyone who has injected malicious code. \$\endgroup\$
    – BrtH
    Jan 10, 2021 at 16:06
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    \$\begingroup\$ @BrtH, Good sense of humor:) Do you understand that I can give guarantees only for myself? I have two antivirus programs, I do not open suspicious emails with attachments... What else would you suggest I do? \$\endgroup\$ Jan 10, 2021 at 16:35
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I'm aware of that, in a circuit, inductive loads need to be protected with a flyback diode because if the circuit gets open the magnetic field will collapse forcing a current flow in the opposite direction.

And the reason why it doesn't make sense is because it is wrong; current continues to flow in the inductive load in the same direction until all the magnetic energy is spent. At that point, current has fallen to zero amps but, importantly, it never reversed direction.

What I don't understand is that last part: opposite direction, let me explain:

You don't need to explain because anyone sensible will know that current continues to flow in the same direction until the magnetic stored energy is spent.

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  • \$\begingroup\$ An attempt to answer the OP's question would be nice. \$\endgroup\$
    – Voltage Spike
    Jan 11, 2021 at 23:26
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There is one comment or two which cleared my doubt. My real concern was the change in voltage polarity without change in current direction. The change in polarity made me assume that the current would flow in opposite direction as the “+” lead indicate higher potential energy. It’s a bit confusing the way the concept is explained.

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