Inductor turn-off and flyback diodes

[bigmessowires]

Inductor turn-off: Can you help me understand the theory behind what happens when the current through an inductor is abruptly stopped? Specifically, I'm confused about why the voltage that appears across the inductor is reversed, relative to the original direction of current flow. As a practical matter I could just drop in a flyback diode to handle the voltage spike, and not worry about it WHY it works, but I really want to understand it better.

Take a look at this diagram from the Wikipedia page about flyback diodes: http://en.wikipedia.org/wiki/File:FlybackExample.GIF

In figure 1, the switch is closed and current flows through the inductor, eventually reaching a steady state. The upper terminal of the inductor is at a positive voltage relative to the lower terminal.

In figure 2, the switch is opened, cutting off the power supply. According to Wikipedia and every other source, this creates a large NEGATIVE voltage across the inductor, making the upper terminal negative relative to the lower terminal. But if the "goal" of the inductor is to resist changes in current flow, and it wants to keep current flowing in the same direction as it was when the switch was closed, then shouldn't it create a large POSITIVE voltage? That would keep the current moving as it was before. By creating a large negative voltage, the inductor is actually inducing current to flow in the opposite direction as before.

In figure 3, my confusion is compounded. This figure shows current flowing the wrong way through a potential gradient across the inductor. Huh?

[c4757p]

But if the "goal" of the inductor is to resist changes in current flow, and it wants to keep current flowing in the same direction as it was when the switch was closed, then shouldn't it create a large POSITIVE voltage? That would keep the current moving as it was before. By creating a large negative voltage, the inductor is actually inducing current to flow in the opposite direction as before.

No! I'll drag out the old hydraulic analogy here: Imagine you have water flowing through a pipe (current), and inside that pipe is a flywheel (inductor). The pressure (voltage) is lower at the "ground" end of the pipe (the end out of which the current is flowing). Now, stick your hand over it. The pressure at that previously lower-pressure end will now become massively positive as the flywheel tries to push your hand out of the way.

Likewise, if you cover the higher-pressure (positive) end, where the current is flowing in, the flywheel will attempt to keep pulling the water through, creating suction (negative pressure/voltage) at the top.

If you prefer math to pipes, the equation for an inductor is:

v = L di/dt.

If the current suddenly falls, that is a negative change in current (di/dt). If di/dt becomes negative, v will also be negative.

[c4757p]

In figure 3, my confusion is compounded. This figure shows current flowing the wrong way through a potential gradient across the inductor. Huh?

The current is flowing the wrong way? Conventional current flows from high (positive) to low (negative) potential, which is exactly what is happening.

[bigmessowires]

The current is flowing the wrong way? Conventional current flows from high (positive) to low (negative) potential, which is exactly what is happening.

Yes, notice how figure 3 shows current flowing through the inductor from the - side to the + side? That's where I went cross-eyed. :-)

Your hydraulic analogy makes good sense, as does the V = L di/dt math. But I can't seem to reconcile these with the numbers in my brain. Take your hydraulic analogy. If I cover the outflow with my hand, then the pressure there rises to a large positive value. But now I've got the outflow at a positive pressure relative to the inflow. Why doesn't that force water to flow down that pressure gradient, reverse its previous direction, and exit at the intake?

[c4757p]

The current is flowing the wrong way? Conventional current flows from high (positive) to low (negative) potential, which is exactly what is happening.

Yes, notice how figure 3 shows current flowing through the inductor from the - side to the + side? That's where I went cross-eyed. :-)

Right, because it is pushed out of the +ve side and pulled into the -ve side.

Your hydraulic analogy makes good sense, as does the V = L di/dt math. But I can't seem to reconcile these with the numbers in my brain. Take your hydraulic analogy. If I cover the outflow with my hand, then the pressure there rises to a large positive value. But now I've got the outflow at a positive pressure relative to the inflow. Why doesn't that force water to flow down that pressure gradient, reverse its previous direction, and exit at the intake?

You've just discovered the LC tank, where the flexibility of your hand is the capacitance

In the real world, either the capacitance (whether a true capacitor, or stray capacitance) will dominate, giving this oscillation that you just described, or something will break down or arc, causing conductance to dominate and the "maintain the same current" behavior to succeed. Yes, a pure, ideal inductor being switched to a pure, ideal open circuit is mathematically inconsistent. Try doing it in a SPICE simulator, which allows you to use perfect components - it'll flip you the bird.

[geraldjhg]

the current continues flowing in the same direction THANKS to the flybak diode
the voltage signals are just old fashioned way of anlyzing it
the current ramps up during the on time and ramps down via the flyback
during the off time
and whats usefull is the current stored and or delivered by the inductor
voltage is just needed to set up current when swith is on

[free_electron]

In simple terms:

Basically a current flowing through an inductor cannot be stopped. if you interrupt the current flow the magnetic field in the inductor starts to decay. This induces current in the wire.

The current keeps on flowing in the SAME direction during the field collapse. as it was flowing before.
So this is why you get the apparent sign reversal for voltage.

If you do not provide a pathway for this current you will get a very high voltage. Ohm's law is to be obeyed at all times. ! current x resistance = voltage . since the open loop resistance of air is very large you get a very large voltage.

The flyback diode closes the loop allowing the continuation of the current.

[AG6QR]

I think all the explanations have been correct, but I've always thought of it in the way free_electron wrote it.

An inductor is something that absolutely refuses to allow an instant change in current.  It will do whatever it has to in order to keep that current flowing the same way through the inductor.  It will build up an enormous voltage if needed.  But if there's a flyback diode, it can keep the current flowing through itself by only producing a small voltage.

In the wikipedia diagrams, the current always flows through the inductor in the same direction.  That's what inductors do.

Incidentally, a capacitor is analogous to an inductor, except voltage is swapped with current.  The capacitor won't allow an instant change in voltage, and will source or sink whatever current is required to prevent an instant voltage change.

There are certainly more precise descriptions and equations which you'll need to learn and understand, but the first intuitive thing to grasp is that an inductor won't allow an instant current change, and a capacitor won't allow an instant voltage change.

[Psi]

The current keeps on flowing in the SAME direction during the field collapse. as it was flowing before.
So this is why you get the apparent sign reversal for voltage.

Now you guys are confusing me, hehe

Could you explain how current in the same direction = voltage in different polarity?
I seem to be missing something

[free_electron]

simple.

The moment you disconnect the input , the inductor becomes the SOURCE of energy.
look at the direction of the current arrow.

 For a source : the tip of the current arrow always points to the positive

So. draw your battery and your inductor.
Draw the arrow showing the direction of current.
Remove the battery ( erase it. )
The arrow does not change. now place + where the tip of the current arrow is. ( your inductor is now no longer a consumer , it delivers the current )
the minus is then opposite side.

[Stonent]

Here's a few good videos. The last one specifically addresses inductive spiking.





Inductive spiking and diodes.