Inductor behavior (in general and in SMPS)

[derGoldstein]

I've been trying to understand the behavior of inductors in different circuits and there's a particular property that I'm having trouble grasping: the magnetic field collapsing generates a gradually increasing voltage across the inductor.

When current is initially passed through an inductor, the inductor resists the change in current across it, starting with a high impedance that gradually decreases as the magnetic field is formed. When the field is fully formed, the circuit is in steady state, and as long as the current passing through the inductor doesn't change, it will just behave as the length of wire in it.
(I'm assuming that I'm correct so far, please tell me if I'm not)
At this point I disconnect the power from the inductor, and this is where my understanding fails: The inductor again resists the change in current across it, and as the magnetic field collapses, it generates a current across the inductor that will keep increasing in voltage until it's high enough to overcome the resistance between the two ends of the inductor and close the circuit. If the inductance is very high the voltage will increase enough to arc through very high resistance such as air.
(here again I'm assuming that the last paragraph was correct, even if I don't understand why it's so)

If this is the case, couldn't a buck circuit output a higher voltage than the input if the resistance of the load is high enough (or if the load is removed completely)? Does the capacitor prevent this from happening due to its behavior of resisting the change in voltage? If so, wouldn't the voltage across the capacitor in a buck circuit that's not connected to a load keep increasing with no limit? Could this happen if the regulation circuit fails?

I'd be grateful to anyone who could elucidate.

[uncle_bob]

Hi

An inductor stores energy. It and the capacitor are two "ideal" elements that do so.

The inductor stores energy in a magnetic field.

When you attach it to a power source, it "soaks up" energy from the source.

When you remove it from a power source (or change the source), it dumps that energy.

There are basic rules that dictate how fast it will soak up energy and how much energy it will store. The same rules help to estimate how it will dump that energy when the source is removed (or changed).

An inductor wants to maintain a constant current. If something external tries to change that, it resists the change by either storing more energy or by dumping some of the energy it has stored.

So far so good?

Bob

[rstofer]

One simple equation says it all:  V_L = L di/dt  The voltage across an inductor is proportional to its inductance times the rate of change of current.  If you could shut off the current instantaneously, di/dt would be infinite and the voltage would  rise to infinity.  There are practical limitations that keep this from happening but that's what the math says.  Ignition coils come to mind...

https://en.wikipedia.org/wiki/Inductor

[derGoldstein]

An inductor stores energy. It and the capacitor are two "ideal" elements that do so.
The inductor stores energy in a magnetic field.
When you attach it to a power source, it "soaks up" energy from the source.
When you remove it from a power source (or change the source), it dumps that energy.
There are basic rules that dictate how fast it will soak up energy and how much energy it will store. The same rules help to estimate how it will dump that energy when the source is removed (or changed).
An inductor wants to maintain a constant current. If something external tries to change that, it resists the change by either storing more energy or by dumping some of the energy it has stored.
So far so good?

I've seen the comparison between the inductor and the capacitor in so many places, and it always seems like a very poor equation. The inductor and the capacitor are two passive (solid-state) energy storage devices, but the similarities don't extend much further.
An ideal capacitor will keep its charge indefinitely if there's no electrical connection between its terminals (and the real-life version will "leak" very slowly), and won't ever output a voltage higher than the one that flowed into it (I understand that some ceramic capacitors can generate voltage transients, but I mean the ideal model). It's very similar to a battery in the mathematical sense.
An inductor will only store and discharge energy under specific, dynamic conditions, and it doesn't have a non-energized storage state. Also, unlike the capacitor, the magnetic field has "other options". If there's a secondary winding, the energy can be discharged through it instead. If there's a change in the arrangement of ferromagnetic matter around it then the energy can be discharged mechanically. In DC terms, it's nothing like a battery.

The capacitor's behavior is much more intuitive. It can be compared to a latched spring, or just a volume of mass that's raised from the ground to a higher position like a shelf. In a way you can also compare it to linear momentum (or, if you want to think of it as a stationary version, then angular momentum).
You put energy in and you take energy out the same way, and in between the energy "stays there".

One simple equation says it all:  V_L = L di/dt  The voltage across an inductor is proportional to its inductance times the rate of change of current.  If you could shut off the current instantaneously, di/dt would be infinite and the voltage would  rise to infinity.  There are practical limitations that keep this from happening but that's what the math says.  Ignition coils come to mind...

So if you have an energized inductor which you disconnect from the circuit, does the voltage instantly rise to theoretical infinity, or does it rise gradually? If the change is instantaneous, how can any additional semiconductors in the circuit (like a freewheeling diode) survive the reaction?

[rstofer]

One simple equation says it all:  V_L = L di/dt  The voltage across an inductor is proportional to its inductance times the rate of change of current.  If you could shut off the current instantaneously, di/dt would be infinite and the voltage would  rise to infinity.  There are practical limitations that keep this from happening but that's what the math says.  Ignition coils come to mind...

So if you have an energized inductor which you disconnect from the circuit, does the voltage instantly rise to theoretical infinity, or does it rise gradually? If the change is instantaneous, how can any additional semiconductors in the circuit (like a freewheeling diode) survive the reaction?

Because nothing changes in zero time.  Even if the current is switched with a mechanical switch, there will be enough voltage generated to arc across the switch contacts for a while. The magnetics aren't quite as perfect as the math.  But I do remember making "shock boxes" from audio transformers and a battery.  Closing the switch was no big deal but when it opened, it stung!  As to freewheeling diodes, they only see a 0.6 or 0.7V drop.  Some resistive device in the circuit accounts for the rest of the energy.

Mostly we deal with low levels of inductance and limited switching speeds.  It all works out fine.  But the math is what it is - if it were possible to switch in zero time, the voltage would be infinite.  Neither can happen in the real world.

Really, it's the same story with a capacitor.  The current is C dv/dt.  If the rate of change of voltage is high, so is the capacitor current.  The current causes heating and is a limiting factor on how fast computers can go.  Smaller feature size with shorter interconnects means lower capacitance which means less heat for a given switching frequency.  One way to limit dv/dt is to use lower and lower logic voltages.

[Molenaar]

The voltage upon disconnection does not rise to infinity due to the resistance of the coil. If the series resistance would be zero, it would be ideal and indeed go to infinity.

[BobsURuncle]

The energy stored in a capacitor is proportional to the square of the voltage between the plates
The energy stored in an inductor is proportional to the square of the current running through the coil.

Ei =  0.5*L*I^2
Ec = 0.5*C*V^2

When you remove an inductor from a circuit, via a switch for example, the current path is interrupted, but energy must be conserved so the voltage will rise sufficiently to keep the current flowing till the stored energy is dissipated through resistive elements in the wire and circuit.  With a simple switch it will arc across the poles: voltage will = di/dt:  the faster the current drops the higher the voltage.

[uncle_bob]

An inductor stores energy. It and the capacitor are two "ideal" elements that do so.
The inductor stores energy in a magnetic field.
When you attach it to a power source, it "soaks up" energy from the source.
When you remove it from a power source (or change the source), it dumps that energy.
There are basic rules that dictate how fast it will soak up energy and how much energy it will store. The same rules help to estimate how it will dump that energy when the source is removed (or changed).
An inductor wants to maintain a constant current. If something external tries to change that, it resists the change by either storing more energy or by dumping some of the energy it has stored.
So far so good?

I've seen the comparison between the inductor and the capacitor in so many places, and it always seems like a very poor equation. The inductor and the capacitor are two passive (solid-state) energy storage devices, but the similarities don't extend much further.
An ideal capacitor will keep its charge indefinitely if there's no electrical connection between its terminals (and the real-life version will "leak" very slowly), and won't ever output a voltage higher than the one that flowed into it (I understand that some ceramic capacitors can generate voltage transients, but I mean the ideal model). It's very similar to a battery in the mathematical sense.
An inductor will only store and discharge energy under specific, dynamic conditions, and it doesn't have a non-energized storage state. Also, unlike the capacitor, the magnetic field has "other options". If there's a secondary winding, the energy can be discharged through it instead. If there's a change in the arrangement of ferromagnetic matter around it then the energy can be discharged mechanically. In DC terms, it's nothing like a battery.

The capacitor's behavior is much more intuitive. It can be compared to a latched spring, or just a volume of mass that's raised from the ground to a higher position like a shelf. In a way you can also compare it to linear momentum (or, if you want to think of it as a stationary version, then angular momentum).
You put energy in and you take energy out the same way, and in between the energy "stays there".

One simple equation says it all:  V_L = L di/dt  The voltage across an inductor is proportional to its inductance times the rate of change of current.  If you could shut off the current instantaneously, di/dt would be infinite and the voltage would  rise to infinity.  There are practical limitations that keep this from happening but that's what the math says.  Ignition coils come to mind...

So if you have an energized inductor which you disconnect from the circuit, does the voltage instantly rise to theoretical infinity, or does it rise gradually? If the change is instantaneous, how can any additional semiconductors in the circuit (like a freewheeling diode) survive the reaction?

Hi

The inductor does not go to infinite voltage in zero time. The capacitor does not go to infinite current in zero time. Same issue on both devices.

The real capacitor does not store energy forever and ever. Neither does the real inductor. Again the same thing on both devices. Ideal inductors would work just as well as ideal capacitors. Superconducting coils are an "ideal" inductor. Short the terminals and the field stays "forever".

A multi winding inductor is not that different than a multi plate capacitor, but that gets a bit more complex.

Since they are both ideal non- realizable elements, they are purely defined by math. One converts current to stored energy, the other converts voltage to stored energy. That's the only difference.

Bob

 

[derGoldstein]

Thanks everyone, I'm starting to alter my intuition of how the inductor behaves, and how the equations defining the devices are useful comparisons.

So in practice, in a buck converter circuit, suppose these 3 conditions occur:
1) the regulation circuitry fails (while the mosfet is off)
2) the load is almost zero
3) the output capacitor is (potentially) too small

Now we're left with the inductor, the freewheeling diode, and the output capacitor. For the sake of this example let's say there's nowhere for the current to flow back "upstream" through the switching mosfet. Is it then possible for that circuit to output voltage higher than the input voltage?

Suppose I'm trying to prevent this exact scenario from happening using only passives. Would adding more capacitance to the output along with a bleed resistor help?



uncle_bob:

Since they are both ideal non- realizable elements, they are purely defined by math.  One converts current to stored energy, the other converts voltage to stored energy. That's the only difference.

Ok, I'm beginning to see the usefulness of the comparison in the equations, but there are some properties which I don't find to be good comparisons:

The inductor does not go to infinite voltage in zero time. The capacitor does not go to infinite current in zero time. Same issue on both devices.

Right, so in practice we change "infinity" to "as high as the series resistance and other imperfections permit". Ok, I can see this one.

The real capacitor does not store energy forever and ever. Neither does the real inductor. Again the same thing on both devices. Ideal inductors would work just as well as ideal capacitors. Superconducting coils are an "ideal" inductor. Short the terminals and the field stays "forever".

This is the comparison I find impractical. If we consider the real-world behavior, assuming that we open the circuit after we charge the devices, the *effective* property of the capacitor is to keep the charge, while the effective property of the inductor is to discharge by any means. The inductor's charge is by nature transient.
Neither the resistance of copper nor the resistance of teflon is infinite, so we can, in theory, call them both "conductors". But for any practical purpose, we'd classify copper as a conductor, and teflon as an insulator. In practice, the charge of a capacitor does "stay put" for most practical purposes, it's objectively a fairly good energy storage device.

A multi winding inductor is not that different than a multi plate capacitor, but that gets a bit more complex.

Multi-plate capacitors are kind of like capacitors in parallel, while something like an electrolytic capacitor has one pair of plates that are spooled, but I see what you mean.

[uncle_bob]

Thanks everyone, I'm starting to alter my intuition of how the inductor behaves, and how the equations defining the devices are useful comparisons.

So in practice, in a buck converter circuit, suppose these 3 conditions occur:
1) the regulation circuitry fails (while the mosfet is off)
2) the load is almost zero
3) the output capacitor is (potentially) too small

Now we're left with the inductor, the freewheeling diode, and the output capacitor. For the sake of this example let's say there's nowhere for the current to flow back "upstream" through the switching mosfet. Is it then possible for that circuit to output voltage higher than the input voltage?

Suppose I'm trying to prevent this exact scenario from happening using only passives. Would adding more capacitance to the output along with a bleed resistor help?



uncle_bob:

Since they are both ideal non- realizable elements, they are purely defined by math.  One converts current to stored energy, the other converts voltage to stored energy. That's the only difference.

Ok, I'm beginning to see the usefulness of the comparison in the equations, but there are some properties which I don't find to be good comparisons:

The inductor does not go to infinite voltage in zero time. The capacitor does not go to infinite current in zero time. Same issue on both devices.

Right, so in practice we change "infinity" to "as high as the series resistance and other imperfections permit". Ok, I can see this one.

The real capacitor does not store energy forever and ever. Neither does the real inductor. Again the same thing on both devices. Ideal inductors would work just as well as ideal capacitors. Superconducting coils are an "ideal" inductor. Short the terminals and the field stays "forever".

This is the comparison I find impractical. If we consider the real-world behavior, assuming that we open the circuit after we charge the devices, the *effective* property of the capacitor is to keep the charge, while the effective property of the inductor is to discharge by any means. The inductor's charge is by nature transient.
Neither the resistance of copper nor the resistance of teflon is infinite, so we can, in theory, call them both "conductors". But for any practical purpose, we'd classify copper as a conductor, and teflon as an insulator. In practice, the charge of a capacitor does "stay put" for most practical purposes, it's objectively a fairly good energy storage device.

A multi winding inductor is not that different than a multi plate capacitor, but that gets a bit more complex.

Multi-plate capacitors are kind of like capacitors in parallel, while something like an electrolytic capacitor has one pair of plates that are spooled, but I see what you mean.

Hi

Your basic issue is still not looking at the two devices (inductor and capacitor) as working on voltage versus current.

To put a capacitor in "storage mode" you open circuit it.

To put an inductor in "storage mode" you short circuit it.

If you short a capacitor .... zonk ... all energy gone.

If you open an inductor ... same ... all energy gone.

Current for one, voltage for the other.

Bob

[Chris Mr]

If one could manufacture a capacitor that had no other properties (like series inductance, series resistance, leakage etc) I have no doubt there would be a Nobel prize in the offing.

Equally, if one could make an inductor that had no inter-winding capacitance or series resistance one would have similar accolades bestowed.

We go around drawing lines (wires) on schematics as if they have zero resistance when in the end they are made of copper rather than superconductors - the very same substance they make transformers out of which have copper losses - and then have to debug our 'perfect' drawings.

As it is we have to live in the universe we have which is way more complicated, and lovely, where ones mind has to grapple with many more things than those which are optimal.

It's the arty part of electronics; to get ones head around the not so obvious things that are going on 

[derGoldstein]

Your basic issue is still not looking at the two devices (inductor and capacitor) as working on voltage versus current.

To put a capacitor in "storage mode" you open circuit it.
To put an inductor in "storage mode" you short circuit it.
If you short a capacitor .... zonk ... all energy gone.
If you open an inductor ... same ... all energy gone.

Current for one, voltage for the other.

I understand the comparison in mathmatical terms, but in practical terms those four statements are: true, false, true, true.
Even if you took a toroidal inductor, charged it, and then magically made the winding vanish completely, leaving a closed magnetic circuit, the field would not stay put for any "useful" amount of time (because of the imperfection of the ferromagnetic material).

The steady state of the two devices, when implemented in the real world, are not equally "steady", that's all I'm saying

[uncle_bob]

Your basic issue is still not looking at the two devices (inductor and capacitor) as working on voltage versus current.

To put a capacitor in "storage mode" you open circuit it.
To put an inductor in "storage mode" you short circuit it.
If you short a capacitor .... zonk ... all energy gone.
If you open an inductor ... same ... all energy gone.

Current for one, voltage for the other.

I understand the comparison in mathmatical terms, but in practical terms those four statements are: true, false, true, true.
Even if you took a toroidal inductor, charged it, and then magically made the winding vanish completely, leaving a closed magnetic circuit, the field would not stay put for any "useful" amount of time (because of the imperfection of the ferromagnetic material).

The steady state of the two devices, when implemented in the real world, are not equally "steady", that's all I'm saying

Hi

You are flipping back and forth between open circuiting an inductor and complaining that they are not ideal. One is a fundamental theoretical issue the other is simply a function of how much money you want to pay. If you took both an ideal capacitor and an ideal inductor, they would both store energy forever.

Bob

[Chris Mr]

Bob has it - there is no steady state.

Nothing is perfect (least of all me) even though the mathematics persuades us that there is perfection

[derGoldstein]

You are flipping back and forth between open circuiting an inductor and complaining that they are not ideal. One is a fundamental theoretical issue the other is simply a function of how much money you want to pay. If you took both an ideal capacitor and an ideal inductor, they would both store energy forever.

The original question was about open-circuiting an inductor. The post that you just replied to concerned the practicality of the two components as energy storage devices. If this were a multi-threaded forum, I'd split the two up, but it's not, so I have to alternate.

I understand that an ideal version of both devices would store the energy forever. More fundamentally, an ideal, closed magnetic circuit will remain there forever. The same goes for an ideal, closed electrical circuit.

As for it being a function of how much money you want to spend, this is exactly the same as saying that, if the voltage is high enough, teflon is a *practical* conductor. As in, it would, under certain real-world conditions, make sense to fabricate a circuit where the traces were made out of teflon, because its conductivity isn't zero, and therefore it's a conductor. For the absolutely overwhelming majority of real-world purposes, this would not happen.

So yes, it's a function of how much money you want to spend. A 1 farad capacitor charged to 10V holds 0.015Wh of energy. You can calculate the amount of energy lost over 1 minute while it's just standing there after being charged, it's not going to be much in terms of percentage. In the real world, I can buy this capacitor for $40 retail. How much would it cost for someone to construct a device that stores 0.015Wh for 1 minute using an inductor? How large would that device be?

In theory, an inductor is an energy storage device, and in theory, teflon is a conductor. In practice, you wound't use an inductor to store energy, and in practice, you wouldn't use teflon as a conductor.

[Chris Mr]

An inductor IS an energy storage device - while the energy is moving (alternating) - Like a flywheel

A Capacitor is a storage device - while the energy is static - like a bucket of water on a ladder, waiting to fall

Magnetic fields collapse; capacitors leak

[uncle_bob]

You are flipping back and forth between open circuiting an inductor and complaining that they are not ideal. One is a fundamental theoretical issue the other is simply a function of how much money you want to pay. If you took both an ideal capacitor and an ideal inductor, they would both store energy forever.

The original question was about open-circuiting an inductor. The post that you just replied to concerned the practicality of the two components as energy storage devices. If this were a multi-threaded forum, I'd split the two up, but it's not, so I have to alternate.

I understand that an ideal version of both devices would store the energy forever. More fundamentally, an ideal, closed magnetic circuit will remain there forever. The same goes for an ideal, closed electrical circuit.

As for it being a function of how much money you want to spend, this is exactly the same as saying that, if the voltage is high enough, teflon is a *practical* conductor. As in, it would, under certain real-world conditions, make sense to fabricate a circuit where the traces were made out of teflon, because its conductivity isn't zero, and therefore it's a conductor. For the absolutely overwhelming majority of real-world purposes, this would not happen.

So yes, it's a function of how much money you want to spend. A 1 farad capacitor charged to 10V holds 0.015Wh of energy. You can calculate the amount of energy lost over 1 minute while it's just standing there after being charged, it's not going to be much in terms of percentage. In the real world, I can buy this capacitor for $40 retail. How much would it cost for someone to construct a device that stores 0.015Wh for 1 minute using an inductor? How large would that device be?

In theory, an inductor is an energy storage device, and in theory, teflon is a conductor. In practice, you wound't use an inductor to store energy, and in practice, you wouldn't use teflon as a conductor.

Hi

Pardon me, energy storage is *exactly* what you use inductors for in switched mode power supplies. As long as you believe are useless for that purpose, there is no need to get into the power supply part of this. The fact that they store energy is so fundamental to the switcher question that you have to get it straight first.

Bob

[Chris Mr]

Dear component supplier,

Please supply four of your "ideal" inductors...


Constant current - from whence (Shakespeare!)


LC starts life with no energy, then it gets a flick from outside, oscillates in a decaying fashion as it circulates the energy around, then goes back to starting state.

[derGoldstein]

Pardon me, energy storage is *exactly* what you use inductors for in switched mode power supplies. As long as you believe are useless for that purpose, there is no need to get into the power supply part of this. The fact that they store energy is so fundamental to the switcher question that you have to get it straight first.

I never once said that inductors do not store energy. The single claim that I made which in any way contradicted or argued with any of your points was that comparing the manner in which capacitors and inductors store energy was, for the purposes of understanding them, not a practical one.
The only thing I disagree with is the *simile*. I never contradicted the any of the facts, equations, ideal functions of the devices, real-world behaviors, or practical uses.

[derGoldstein]

I think I have a better way of explaining the problem I have with the simile.

Suppose you take a ball and throw it into the air. You applied force to move the mass of the ball away from the mass of the earth. The mass of the ball will now strive to get back as close as it can to the mass of the earth, and will, to a certain limit, apply force to do so (displacing the air away from its path). In throwing the ball into the air you've "charged" it with kinetic energy. I would not, however, say that as a result, the ball is an "energy storage device". I'm not saying that energy *isn't* stored in the ball, I just wouldn't classify the ball as an energy storage device.
Now I'll modify the situation slightly. I'll take the same ball and put it on a shelf. I've applied the same force, but "latched" the energy by not allowing it to fall back down. To release this energy, I would only have to push the ball very slightly, which would cause it to roll off of the shelf and once again apply force to get back to the ground. While I still wouldn't call the ball an energy storage device, it's now a bit closer to that description, subjectively speaking.

My classification or subjective description of the ball does not in any way change how it behaves. I'm merely arguing that I would describe it differently.

[Chris Mr]

And what if the same ball were spinning around inside a bucket on a string?

[uncle_bob]

Pardon me, energy storage is *exactly* what you use inductors for in switched mode power supplies. As long as you believe are useless for that purpose, there is no need to get into the power supply part of this. The fact that they store energy is so fundamental to the switcher question that you have to get it straight first.

I never once said that inductors do not store energy. The single claim that I made which in any way contradicted or argued with any of your points was that comparing the manner in which capacitors and inductors store energy was, for the purposes of understanding them, not a practical one.
The only thing I disagree with is the *simile*. I never contradicted the any of the facts, equations, ideal functions of the devices, real-world behaviors, or practical uses.

Hi

It is not a "cute trick" used to compare the two devices. They are uniquely coupled in what they do. That actually is deliberate. If it were not true, we would use a different pair of parts for which is was true.  The way in which they are coupled is what makes the math work. Understanding that coupling is a very fundamental part of circuit theory. Without it, the rest of the analysis should not / would not / could not make sense.

Bob

[Chris Mr]

Bad explanation 

Ball inside bucket, bucket attached to string, someone is holding the string in their hand and making the whole thing turn either round horizontally, or vertically.

Now the ball isn't moving with respect to the bucket, but the bucket is moving.  If you push the ball outside the bucket there is energy stored in the ball that can be captured - like the ball on the shelf falling off due to gravity, this way due to cetripedal force.

[Chris Mr]

Sorry Bob, not your explanation, mine 

[derGoldstein]

It is not a "cute trick" used to compare the two devices. They are uniquely coupled in what they do. That actually is deliberate. If it were not true, we would use a different pair of parts for which is was true.  The way in which they are coupled is what makes the math work. Understanding that coupling is a very fundamental part of circuit theory. Without it, the rest of the analysis should not / would not / could not make sense.

I see that, I can see the mirroring of their properties in the equations, I see how they complement each other and work as LC filters or oscillators, and that if the math were different then those mechanisms wouldn't work.
I was just making a subjective observation about how they're explained. Maybe I should have said that a capacitor seems (to me) more like a battery and therefore more intuitive to understand (to me).

Ok, I retract the observation. I see what you mean about having to understand both as energy storage devices in order to understand how power circuits work. I really did come away from this with a better understanding of their interaction. Thanks for your patience.