Role of diode in a circuit with coil

[MikeMl]

If the OP claims to understand water flow, then look up WATER HAMMER, which is analogous to inductive kick...

 

[crutschow]

In my mind, I use this analogy with water, which works just fine with wires, resistor, capacitors, diodes etc. Inductors are kind off, an odd man out.

Actually the analogy with water works for inductors also, if you remember that water has weight (mass which corresponds to inductance). Thus when the water is flowing in a pipe and you remove the pressure source, the water will still keep flowing due to its inertial mass, until friction (resistance) slows it down.

If the water is flowing rapidly in a pipe and the faucet is suddenly shut, you may hear a clank from the pipe (water hammer), which is a result of this inertia. This is similar to the voltage spike that occurs when you open an inductor carrying current.

 

[Ross Craney]

That's not true. Just place a short across an inductor which is carrying current and monitor the current flow (you can easily do this in SPICE). The current will continue to flow but goes to zero due to IR loss in the resistance of the coil, which absorbs energy from the magnetic field and causes it to collapse. If you have a superconducting coil, the current will continue forever without any power required, which is why they use superconducting magnets in the big particle accelerators.

If you interrupt that current, then you get a voltage spike as the field attempts to keep the current flowing.

Maybe the field collapses because with a short circuit across the inductor there is now no current flowing through it & hence no field.
Since the induced voltage is of the opposite polarity wouldn't it be fair to say that it doesn't attempt to keep the current flowing but in fact opposes it.

 

[dknguyen]

When you have the inductor connected to a voltage source driving current through it, yes, the polarity of the induced voltage opposes current flow and that's why it takes time for the current in an inductor ramp up.

When I said "the inductor tries to keep current flowing" I am referring specifically to the situtation during flyback when the inductor has no source connected to it. In that case the induced voltage has the same polarity as a source supplying the current and in that case the inductor tries to keep current flowing and is the reason why it takes current time to ramp down.

So yeah, I guess it depends on which side of the coin you're talking about. The inductor opposing current flow when the switch is closed and trying to keep current going when the switch goes open are two sides to the same coin of "the current flow in an inductor must be continuous".

When we start talking about voltage spikes and things like that, my mind kind of shuts off on the aspect of the inductor opposing current flow causing it to slowly ramp up so I just tend to say the inductor without qualifying that the inductor doesn't always try to keep current flowing. What the inductor does always try to do is to keep current flowing through it the same (whether it's zero amps or 100A).

 

[unclejed613]

Maybe the field collapses because with a short circuit across the inductor there is now no current flowing through it & hence no field.
Since the induced voltage is of the opposite polarity wouldn't it be fair to say that it doesn't attempt to keep the current flowing but in fact opposes it.

which brings to mind another definition of inductance: "inductance opposes CHANGES in CURRENT flow."

 

[The Electrician]

Maybe the field collapses because with a short circuit across the inductor there is now no current flowing through it & hence no field.

You are not understanding what happens. When you short the inductor (which previously had a current established in it by some source), the current continues, through the short. It's only because the wire in the inductor has some resistance that the current decays at all. Because the short doesn't allow any voltage external to the inductor to be developed, the time for the current to decay will be the longest possible. In fact, the ratio of the inductance to the resistance of the wire (L/R) is the intrinsic time constant of the inductor.

If the inductor were wound with superconducting wire, the current that existed at the time the short was applied would continue to flow indefinitely.

 

[unclejed613]

..........
When I said "the inductor tries to keep current flowing" I am referring specifically to the situtation during flyback when the inductor has no source connected to it. In that case the induced voltage has the same polarity as a source supplying the current and in that case the inductor tries to keep current flowing and is the reason why it takes current time to ramp down.
........

When we start talking about voltage spikes and things like that, my mind kind of shuts off on the aspect of the inductor opposing current flow causing it to slowly ramp up so I just tend to say the inductor without qualifying that the inductor doesn't always try to keep current flowing. What the inductor does always try to do is to keep current flowing through it the same (whether it's zero amps or 100A).

and as it does so in a de-energized coil, it does so into a very high ("ideally" infinite, but nothing's ideal) impedance, hence the high voltage that develops.

 

[unclejed613]

You are not understanding what happens. When you short the inductor (which previously had a current established in it by some source), the current continues, through the short. It's only because the wire in the inductor has some resistance that the current decays at all. Because the short doesn't allow any voltage external to the inductor to be developed, the time for the current to decay will be the longest possible. In fact, the ratio of the inductance to the resistance of the wire (L/R) is the intrinsic time constant of the inductor.

If the inductor were wound with superconducting wire, the current that existed at the time the short was applied would continue to flow indefinitely.

it would also have to be an ideal diode or a superconducting short.... and once you release the short, you generate a very large voltage spike

 

[The Electrician]

it would also have to be an ideal diode or a superconducting short....

When I say "short", I mean an ideal short; otherwise it's not a short, it's a low value resistor.

and once you release the short, you generate a very large voltage spike

You just don't do that if you have a superconducting coil carrying a large current, or you get damage. People who use superconducting magnets know better than to do that.

 

[Roff]

Post content removed.

 

[Ross Craney]

When you have the inductor connected to a voltage source driving current through it, yes, the polarity of the induced voltage opposes current flow and that's why it takes time for the current in an inductor ramp up.

When I said "the inductor tries to keep current flowing" I am referring specifically to the situtation during flyback when the inductor has no source connected to it. In that case the induced voltage has the same polarity as a source supplying the current and in that case the inductor tries to keep current flowing and is the reason why it takes current time to ramp down.
So yeah, I guess it depends on which side of the coin you're talking about. The inductor opposing current flow when the switch is closed and trying to keep current going when the switch goes open are two sides to the same coin of "the current flow in an inductor must be continuous".

When we start talking about voltage spikes and things like that, my mind kind of shuts off on the aspect of the inductor opposing current flow causing it to slowly ramp up so I just tend to say the inductor without qualifying that the inductor doesn't always try to keep current flowing. What the inductor does always try to do is to keep current flowing through it the same (whether it's zero amps or 100A).

When the source voltage has been removed there is no magical supply to keep the current flowing. There is ONLY a generated magnetic field which upon collapse generates a voltage in the REVERSE polarity to the original source & decays at a rate set by the load as seen by the inductor.

 

[crutschow]

When the source voltage has been removed there is no magical supply to keep the current flowing. There is ONLY a generated magnetic field which upon collapse generates a voltage in the REVERSE polarity to the original source & decays at a rate set by the load as seen by the inductor.

First you say there's nothing to keep the current flowing and then you say the field decays by the load as seen by the inductor. Well, the inductor does not see the load unless there is current in the load. A resistive load is meaningless unless there is current flowing. You can't dissipate power in a load without current.

You keep mentioning REVERSE polarity as if that's something unexpected or unusual. It's not. It just the result of the magnetic field attempting to keep the current flowing in the same direction. Think about the polarity of an inductor. Say you put a positive voltage on the left terminal to move the current left to right. When you remove the voltage and short the inductor, the current will still keep moving left to right from the magnetic field, but now the inductor is acting like a battery with the left terminal negative and the right terminal positive. No magic there.

The magnetic field simply keeps the current flowing in the load until all the magnetic field energy has been dissipated. Why is that so hard to understand?

 

[dknguyen]

When the source voltage has been removed there is no magical supply to keep the current flowing. There is ONLY a generated magnetic field which upon collapse generates a voltage in the REVERSE polarity to the original source & decays at a rate set by the load as seen by the inductor.

I don't think you understand what I'm saying and your wording is not correct either. When the inductor is connected to the source, the voltage across the inductor actually opposes that of the source (as the inductor produces a voltage that opposes the flow of current in an attempt to keep it the same level). When the magnetic field collapses because you disconnected the source, the voltage across inductor reverses and it is now the SAME polarity as the source that was disconnected. That's because the inductor is now behaving like a source and using the energy stored in it's magnetic field to try to maintain the current flow.

Maybe the field collapses because with a short circuit across the inductor there is now no current flowing through it & hence no field.

The field only collapses if it requires energy to maintain the current flow. With a short-circuit (a true short-circuit) no energy is required to maintain the current flow so the field does not collapse because no energy is being drawn from it. Anything in the path of the current flow that dissipates energy will cause collapse of the field though, whether it be a the forward voltage drop of a diode or the smallest resistance.

Since the induced voltage is of the opposite polarity wouldn't it be fair to say that it doesn't attempt to keep the current flowing but in fact opposes it.

Yes, when a source is connected to the inductor it opposes the current flow resulting in a continuous ramp up of current rather than an instaneous step-change. But not when the source has been disconnected and the inductor tries to maintain current flow through it by trying to behave as a source through the use of energy stored in it's magnetic field. In this case it is not opposing current flow but is actually assisting it. However, unless the current flow is through a short-circuit that dissipates no energy, it is a losing battle as the inductor will finally run out of the energy that it has stored in it's magnetic field. THe result is a ramp down of current rather than a step-drop in current.

You dont force current through an inductor
Inductors don't try to keep current flowing.
When a voltage is removed from the inductor the magnetic field collapses & in doing so generates a voltage in reverse polarity to the original voltage.

Okay, but what does that generated voltage do? Well the level that the voltage is just happens to be whatever is required to maintain the current level as best it can with the limited energy that has been stored in it's magnetic field. In other words, that voltage will be just high enough to drive current through an alternate path around the inductor. So yes, the inductor does try to keep to current flowing.

The diode is to provide a short circuit to this induced voltage.

Indeed it is, but then you contradict yourself in the next statement.

The presence of the diode actually increases the induced voltage as it causes the rate of magnetic collapse to increase.

A short circuit means zero voltage. In this case, the diode only serves as an approximation of a short-circuit so it's near zero voltage. So, no the diode doesn't increase the induced voltage. This statement is just wrong.

The diode is there to clamp the induced voltage and does so by providing the inductor with an "easy path" where it can try to maintain current flow. The "easy path" means that a only a low voltage is required to drive the current. So the flyback current stays in a controlled area while not requiring the generation of a large voltage spike to drive it through alternate paths which would damage components. There's few current paths that are easier than a short-circuit and the diode approximates one well in this case.

 

[unclejed613]

You just don't do that if you have a superconducting coil carrying a large current, or you get damage. People who use superconducting magnets know better than to do that.

by definition, isn't ANY current in a superconductor generally unlimited or approaching infinity? if R=0, and E>0, then any condition of E/R=∞ ... or since R=0, then E can never be >0??

 

[Ross Craney]

is now the SAME polarity as the source that was disconnected. That's because the inductor is now behaving like a source and using the energy stored in it's magnetic field to try to maintain the current flow.

Why do you keep saying it wants to maintain current flow , it is simply a coil of wire. The field collapses in the opposite direction & "generates" a voltage of the opposite polarity. Go & read electromagnetism 101

That source in my sentence, is not the source that was disconnected. It's also not a discrete source either. It's a model of what the inductor is doing and the inductor uses the energy stored in it's magnetic field to maintain the current flow when the actual source has been removed.

Why oh why does this inductor feel a need to replace the current it has been deprived of?

The field only collapses if it requires energy to maintain the current flow.

The field collapses because its got no bloody current to maintain it - this isn't rocket science

Well the level that the voltage is just happens to be whatever is required to maintain the current level as best it can with the limited energy that has been stored in it's magnetic field. In other words, that voltage will be just high enough to drive current through an alternate path around the inductor.

The resultant voltage is determined 'basically" by the magnetic field strength , the rate of field collapse & the number of turns of the inductor

The diode is there to clamp the induced voltage and does so by providing the inductor with an "easy path" where it can try to maintain current flow. The "easy path" means that a only a low voltage is required to drive the current. So the flyback current stays in a controlled area while not requiring the generation of a large voltage spike to drive it through alternate paths which would damage components. There's few current paths that are easier than a short-circuit and the diode approximates one well in this case.

Since you say the induced voltage is the same polarity as the applied voltage wouldn't the diode also provide an "easy path" for tha applied voltage ? This would mean no magnetic field , no induced voltage & hence no argument. End of my input.

 

[crutschow]

Well Mr. Craney we've tried our best to educate you, but you obviously have never actually done any experiments with an inductor, and are hung up on a misinterpretation of the information you received in electromagnetism 101 about inductance. But as my old grade school teacher used to say, you can lead a horse to water but you can't make him drink.

 

[Roff]

by definition, isn't ANY current in a superconductor generally unlimited or approaching infinity? if R=0, and E>0, then any condition of E/R=∞ ... or since R=0, then E can never be >0??

A superconductor is just a wire (or whatever) with zero resistance. The current through it can be limited by other (non-superconducting) components, e.g., a resistor.

 

[dknguyen]

When the source voltage has been removed there is no magical supply to keep the current flowing. There is ONLY a generated magnetic field which upon collapse generates a voltage in the REVERSE polarity to the original source & decays at a rate set by the load as seen by the inductor.

That reverse polarity you just described causes the inductor to behave like the "supply" that you just said didn't exist. Granted, it's not magical, but rather an energy limited supply. Being energy limited, it will be unable to maintain a constant continuous level of current through the "load as seen by the inductor".

What is especially interesting is that while you actually do say "When a voltage is removed from the inductor the magnetic field collapses & in doing so generates a voltage in reverse polarity to the original voltage", you do not seem to realize that this has the effect of resisting change in current flow or as I prefer to say it "tries to maintain current flow". BTW, that reverse polarity voltage is only required if the current path is dissipative which will draw energy from the magnetic field which in turn causes it to collapse. If the path is a short-circuit, no induced voltage is required to drive the current through the loop and therefore no energy is draw from the field and the field is maintained even as the current flows.

Why do you keep saying it wants to maintain current flow , it is simply a coil of wire. The field collapses in the opposite direction & "generates" a voltage of the opposite polarity. Go & read electromagnetism 101

I've done more than read it, but you you don't seem to realize that "generating a voltage of the opposite polarity" means maintaing the current flow (within the limits of the energy stored by the magnetic field). In fact, the reason that voltage is generated is to drive that current. If no voltage is required to keep that current going like in a short-circuit, no voltage is generated.

Why oh why does this inductor feel a need to replace the current it has been deprived of?

The reason is that changing currents produce changing magnetic fields and changing magnetic fields induce electrons to move. Disconnecting the source counts as a change in current, change in current = changing magnetic field = induced voltage = electrons to flow. And the way all the polarities work out, these electrons continue to flow in the same direction they were flowing before the source was disconnected. That induced voltage is whatever is required to resist this change in current (ie. continue to drive those electrons) or as I like to say "try to maintain current flow". That's why.

But this does bring up something very interesting that may seem contradictory at first. I claim that inductors try to maintain current flow during flyback (aka resist changes in current, in this scenario resist a drop in current). But the very mechanism I explained above that causes this behaviour requires an actual change in current which seems to contradict what happens in when the flyback current loop is a short-circuit. In these loops, no energy is required from the magnetic field of the inductor to keep the current flowingso the inductor is actually successful at at maintain the current at the same level. But the very phenomena of maintaing current flow requires a drop in current! This is why...
In a sense, it's almost like energy is oscillating between the current flow and the magnetic field because current will drop, induce a magnetic field which causes the inductor to bring the current level back up again. The inductor reacts instantly to it though and it just all happens at a frequency so high you can't even observe it. It's not so hard to believe when you consider that most people accept that the inductor instantly reacts to resist a change in current. If it didn't there would be a brief surge of current as the inductor "set things up".

An instantaneous circuit analysis does model of the circuit does show the inductor as a current source during flyback. If the current path is dissipative, In the next instant the current source will be of a different, lower value. "But a current source is supposed to be constant!". Maybe in most usages, but what really makes the behaviour of a current source is the voltage adjusting itself to cause a certain amount of current flow whether that current is cosntant or not which is what an inductor does during flyback (you do have to take into account the limited energy reserves of the inductor however).

Would you prefer it better if I worded it as resisting the changing of current flow? Because "resisting" and "trying it's best to maintain the current flow with the limited energy it has" really means the same thing, except one is easier to understand because it refers specifically to the flyback scenario we are talking about, and not the universal behaviour of the inductor. It's the same as lifting a weight that's too heavy for you- you're resisting the force of gravity, or you're trying to lift the weight and overcome the force of gravity as best you can with the strength you have- by no means does it actually mean you are ultimately able to lift it. It definately does slow down how fast the weight falls though in the same way inductor current ramps down in a dissipative current loop.

The field collapses because its got no bloody current to maintain it - this isn't rocket science

Wrong, as I said above, the field only collapses because energy is being drawn from it in an effort to "maintain current flow"/"resist current changes". If it's a true short-circuit across the inductor like a superconducting coil the field does NOT collapse when the source is removed (assuming nothing else is also drawing energy from the field by some other means) as no energy is being drawn from the magnetic field. THat energy has to go somewhere and in this case it has nowhere to go and stays in the magnetic field. It really isn't rocket science which is strange why you don't seem to get it. Conservation of energy at work.

The resultant voltage is determined 'basically" by the magnetic field strength , the rate of field collapse & the number of turns of the inductor

You seem to have a habit of providing physical explanations while not realizing they result in the circuit behaviour I am describing. Rate of field collapse is dependent on how much energy is being drawn from the field to resist the change in current flow. THat rate is dependent on the power dissipation of the flyback current loop.

Since you say the induced voltage is the same polarity as the applied voltage wouldn't the diode also provide an "easy path" for tha applied voltage ?

Again, you are having trouble understanding what it is I actually write because that is what I said. See my sentence that says "The diode is there to clamp the induced voltage and does so by providing the inductor with an "easy path" where it can try to maintain current flow"?

This would mean no magnetic field , no induced voltage & hence no argument.

To see why this is wrong you have to look no further than any old DC electromagnet. An inductor's induced voltage is whatever it needs to be resist current flow (whether opposting the current flow of a source, or trying to maintain it when the source is removed). In this example, diode is an "easy path" but not a perfect one and still requires voltage to drive current through it so there is an inducved voltage.

As a result of this current-source behaviour (or more accurately, limited-energy current source behaviour), an inductor can still have a magnetic field even when there is no induced voltage. Think about a circuit of just a source-resistor-inductor that's been allowed to reach steady state. THe inductor will be a short so it has no induced voltage, but it still has current flowing through it which means it still has a magnetic field. Another example is an inductor in flyback trying to resist the change in current through a short. It requires zero volts to drive current through a short so it has zero induced voltage, but current is still flowing and so the inductor still has a magnetic field. If what you were saying was correct, then DC electromagnets wouldn't work continuously. There is no induced voltage across them the electromagnet winding in steady state and yet it still a magnet. The voltage applied is to overcome the wiring resistance.

 

[dknguyen]

For people actually trying to learn something from this thread and getting confused by all the back and forth, I am going to summarize what Ross and I are talking about and why we are contentious on a few points and how closely they actually agree. Regardless of how it sounds like, our standpoints are a lot more similar than it may seem (well, except for this first point, but he may have just had an momentary lapse since it's obviously not right and has not come up since he mentioned it).

FALSE: short-circuits or low resistance current paths cause inductors to have a higher induced voltage and increases the rate of magnetic collapse
(low resistance = low voltage so this one obviously doesn't add up.)

TRUE HALF THE TIME/FALSE HALF THE TIME - inductors don't try to maintain current flow
(I can see why he might argue this because it can be misinterpreted. In general, it's only true half the time. What you can be sure of is that inductors resist changes in current all the time. There are two aspects of this behaviour though: This means that when trying to increase current through it, the inductor opposes the current flow. When trying to decrease the current through it, it tries to maintain the current flow. Hence it resists CHANGES in current, not the actual the current itself.)

SEMI-TRUE: that the inductor's field collapses because no current is there to maintain it
This is a finer point that has to do with conservation of energy as well as theoretical and real scenarios. The way that a inductor tries to resist a drop in current (in plainer terms, "maintain current flow when you try to decrease current through the inductor") is by using the energy that was stored in it's magnetic field when current was increasing through it before power was disconnected. If maintaining the current flow requires no energy like in a short-circuit around the inductor, this magnetic energy doesn't go anywhere and the magnetic field just hangs around. It's a lot like a flywheel on a car that's being left alone. As soon as you start sucking energy from it though (like putting a resistor in the current path or even using the magnetic field to move something), the magnetic field will be reduced, eventually to the point where it just dissapears. In that case, to maintain the field you actually do need a an energy source (or as Ross called it, a "current") to maintain the field indefinately.

So Ross's claim that the inductor field collapses because there is no current to maintain it is only true if there is actually energy being dissipated from the magnetic field of the inductor. No dissipation = no collapse. Granted, in real world devices, there's always some dissipation due to inherent resistance in the wire as well as the coils of the inductor or even as using the magnetic field to perform work so the field should always collapse...eventually. Though not wrong in most cases, the explanation is at too high a level to properly explain what is going on so that it applies in all cases.

 

[Ross Craney]

Well Mr. Craney we've tried our best to educate you, but you obviously have never actually done any experiments with an inductor, and are hung up on a misinterpretation of the information you received in electromagnetism 101 about inductance. But as my old grade school teacher used to say, you can lead a horse to water but you can't make him drink.

Read your first post. You say exactly what I've been saying. I'm not sure when you changed your mind. The question in the opening post is simple & concise. It does not reqire theoretical circuit analysis or computor simulations to answer. You have already provided an excellent answer in post 2 other than the maintain current flow scenario. But I am beginning to think that is just semantics. Somehow you have become more confused as the topic has progressed.