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When a current flow in an inductor is interrupted, what limits the voltage rise?
Can a Zener diode that protects a switch against inductance when the switch opens, affect turn on speed of the valve when you close it again?Inductance and “kick back” voltagesQuestions About InductorsBehavior of RLC circuit elements at t = 0Solving a circuit containing a resistor and inductor in parallelWhere does inductor current go when the RL circuit switch opens?How does a negative current start in the inductor in ac circuit?Energy from an inductor breakdownChoosing inductor to limit inrush currentWhy does duty cycle seem relevant when calculating inductor current in boost converter circuit?Inductor Saturation Current and Temperature Rise Current
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When the current flowing in an inductor is interrupted the voltage will rise until (typically) there is a flashover or insulation breakdown in the switch. Assuming extremely good insulation what would limit the voltage rise in practice? I can't really believe it would head into the megavolt region.
inductor
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
$endgroup$
add a comment |
$begingroup$
When the current flowing in an inductor is interrupted the voltage will rise until (typically) there is a flashover or insulation breakdown in the switch. Assuming extremely good insulation what would limit the voltage rise in practice? I can't really believe it would head into the megavolt region.
inductor
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
$endgroup$
5
$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
1
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51
add a comment |
$begingroup$
When the current flowing in an inductor is interrupted the voltage will rise until (typically) there is a flashover or insulation breakdown in the switch. Assuming extremely good insulation what would limit the voltage rise in practice? I can't really believe it would head into the megavolt region.
inductor
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
$endgroup$
When the current flowing in an inductor is interrupted the voltage will rise until (typically) there is a flashover or insulation breakdown in the switch. Assuming extremely good insulation what would limit the voltage rise in practice? I can't really believe it would head into the megavolt region.
inductor
inductor
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
asked May 28 at 8:49
Dirk BruereDirk Bruere
6,21253367
6,21253367
5
$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
1
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51
add a comment |
5
$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
1
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51
5
5
$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
1
1
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
TL;DR: It also depends on the actual setup.
It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is $E=1 over 2 L I^2$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.
If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.
Marko Bursic has already mentioned self capacitance in his answer.
You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.
For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.
At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).
BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.
In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.
On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):
The answer is that the vacuum doesn't impede the electrons at all.
In a perfect vacuum (except for the test charge), the charged
particle's behavior can be calculated by just three factors: 1) its
mass, which gives it inertial forces, 2) its electric charge, which
makes it responsive to an electric field, and 3) the electric field
acting on the charged particle. Then the acceleration of the particle
is just the equilibrium between the inertial and electric forces.
[...]
This has been mistaken to be a measure of resistance, which seems to
become infinite at low pressures, but this is not correct. At higher
pressures, the breakdown voltage increases steadily with pressure (off
the right side of Figure 2), and so does the resistance. So there is a
direct relationship between breakdown voltage and resistance in that
range. If we erroneously take that as a hard and fast rule, and we
observe the breakdown voltage shooting up at very low pressures, we
conclude that the resistance must be increasing at very low pressures,
asymptotically approaching infinite resistance at a pressure
characteristic for that gas. But here we have to remember that Paschen
was studying breakdown voltages, and made no mention of resistance. If
we make direct measurements of the resistance, we find that it varies
directly with the pressure, and continues straight down to nothing at
no pressure, without the sudden deviation at the threshold discovered
by Paschen. This can be confirmed by putting an ammeter into the
circuit, and finding the resistance by the volts divided by the amps.
So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).
That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.
Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel
In slide #29 (page 25 of the PDF) you get (emphasis mine):
Intrinsic Dielectric Strength Limits
- The breakdown of Gasses at STP is dependent on type.
- Air : ~3 kV/mm (75 V/mil)
- He : ~0.37 kV/mm (9.3 V/mil)
- SF 6 : ~9 kV/mm (222 V/mil)
High Vacuum breakdown is surface and configuration
dependent but is in the range of 20 to 40 kV/mm (500 to 1000
V/mil)
- Ignoring surface effects, Liquids and Solids bulk properties are
generally similar in the same range of 15 to 20 kV/mm (375
V/mil to 500 V/mil).
So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
$endgroup$
add a comment |
$begingroup$
self capacitance
simulate this circuit – Schematic created using CircuitLab
edited May 28 at 11:02
Warren Hill
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
$endgroup$
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
add a comment |
$begingroup$
If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.
However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.
Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of $fracI_indC_stray$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.
If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, $0.5CV^2$, is (ideally) equal to the original energy stored in the inductor, $0.5LI^2$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.
If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
$endgroup$
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
|
show 1 more comment
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.
If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.
These may be much more limiting than any (tiny) capacitance in the inductance.
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
$endgroup$
add a comment |
$begingroup$
Voltage across inductor will rise to v =L.di/dt
di/dt is determined by how fast the transistor switch switches off and cuts the current off.
answered May 28 at 12:05
JamesJames
1174
$endgroup$
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
add a comment |
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5 Answers
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5 Answers
5
active
oldest
votes
active
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$begingroup$
TL;DR: It also depends on the actual setup.
It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is $E=1 over 2 L I^2$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.
If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.
Marko Bursic has already mentioned self capacitance in his answer.
You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.
For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.
At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).
BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.
In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.
On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):
The answer is that the vacuum doesn't impede the electrons at all.
In a perfect vacuum (except for the test charge), the charged
particle's behavior can be calculated by just three factors: 1) its
mass, which gives it inertial forces, 2) its electric charge, which
makes it responsive to an electric field, and 3) the electric field
acting on the charged particle. Then the acceleration of the particle
is just the equilibrium between the inertial and electric forces.
[...]
This has been mistaken to be a measure of resistance, which seems to
become infinite at low pressures, but this is not correct. At higher
pressures, the breakdown voltage increases steadily with pressure (off
the right side of Figure 2), and so does the resistance. So there is a
direct relationship between breakdown voltage and resistance in that
range. If we erroneously take that as a hard and fast rule, and we
observe the breakdown voltage shooting up at very low pressures, we
conclude that the resistance must be increasing at very low pressures,
asymptotically approaching infinite resistance at a pressure
characteristic for that gas. But here we have to remember that Paschen
was studying breakdown voltages, and made no mention of resistance. If
we make direct measurements of the resistance, we find that it varies
directly with the pressure, and continues straight down to nothing at
no pressure, without the sudden deviation at the threshold discovered
by Paschen. This can be confirmed by putting an ammeter into the
circuit, and finding the resistance by the volts divided by the amps.
So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).
That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.
Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel
In slide #29 (page 25 of the PDF) you get (emphasis mine):
Intrinsic Dielectric Strength Limits
- The breakdown of Gasses at STP is dependent on type.
- Air : ~3 kV/mm (75 V/mil)
- He : ~0.37 kV/mm (9.3 V/mil)
- SF 6 : ~9 kV/mm (222 V/mil)
High Vacuum breakdown is surface and configuration
dependent but is in the range of 20 to 40 kV/mm (500 to 1000
V/mil)
- Ignoring surface effects, Liquids and Solids bulk properties are
generally similar in the same range of 15 to 20 kV/mm (375
V/mil to 500 V/mil).
So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
$endgroup$
add a comment |
$begingroup$
TL;DR: It also depends on the actual setup.
It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is $E=1 over 2 L I^2$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.
If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.
Marko Bursic has already mentioned self capacitance in his answer.
You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.
For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.
At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).
BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.
In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.
On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):
The answer is that the vacuum doesn't impede the electrons at all.
In a perfect vacuum (except for the test charge), the charged
particle's behavior can be calculated by just three factors: 1) its
mass, which gives it inertial forces, 2) its electric charge, which
makes it responsive to an electric field, and 3) the electric field
acting on the charged particle. Then the acceleration of the particle
is just the equilibrium between the inertial and electric forces.
[...]
This has been mistaken to be a measure of resistance, which seems to
become infinite at low pressures, but this is not correct. At higher
pressures, the breakdown voltage increases steadily with pressure (off
the right side of Figure 2), and so does the resistance. So there is a
direct relationship between breakdown voltage and resistance in that
range. If we erroneously take that as a hard and fast rule, and we
observe the breakdown voltage shooting up at very low pressures, we
conclude that the resistance must be increasing at very low pressures,
asymptotically approaching infinite resistance at a pressure
characteristic for that gas. But here we have to remember that Paschen
was studying breakdown voltages, and made no mention of resistance. If
we make direct measurements of the resistance, we find that it varies
directly with the pressure, and continues straight down to nothing at
no pressure, without the sudden deviation at the threshold discovered
by Paschen. This can be confirmed by putting an ammeter into the
circuit, and finding the resistance by the volts divided by the amps.
So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).
That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.
Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel
In slide #29 (page 25 of the PDF) you get (emphasis mine):
Intrinsic Dielectric Strength Limits
- The breakdown of Gasses at STP is dependent on type.
- Air : ~3 kV/mm (75 V/mil)
- He : ~0.37 kV/mm (9.3 V/mil)
- SF 6 : ~9 kV/mm (222 V/mil)
High Vacuum breakdown is surface and configuration
dependent but is in the range of 20 to 40 kV/mm (500 to 1000
V/mil)
- Ignoring surface effects, Liquids and Solids bulk properties are
generally similar in the same range of 15 to 20 kV/mm (375
V/mil to 500 V/mil).
So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
$endgroup$
add a comment |
$begingroup$
TL;DR: It also depends on the actual setup.
It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is $E=1 over 2 L I^2$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.
If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.
Marko Bursic has already mentioned self capacitance in his answer.
You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.
For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.
At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).
BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.
In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.
On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):
The answer is that the vacuum doesn't impede the electrons at all.
In a perfect vacuum (except for the test charge), the charged
particle's behavior can be calculated by just three factors: 1) its
mass, which gives it inertial forces, 2) its electric charge, which
makes it responsive to an electric field, and 3) the electric field
acting on the charged particle. Then the acceleration of the particle
is just the equilibrium between the inertial and electric forces.
[...]
This has been mistaken to be a measure of resistance, which seems to
become infinite at low pressures, but this is not correct. At higher
pressures, the breakdown voltage increases steadily with pressure (off
the right side of Figure 2), and so does the resistance. So there is a
direct relationship between breakdown voltage and resistance in that
range. If we erroneously take that as a hard and fast rule, and we
observe the breakdown voltage shooting up at very low pressures, we
conclude that the resistance must be increasing at very low pressures,
asymptotically approaching infinite resistance at a pressure
characteristic for that gas. But here we have to remember that Paschen
was studying breakdown voltages, and made no mention of resistance. If
we make direct measurements of the resistance, we find that it varies
directly with the pressure, and continues straight down to nothing at
no pressure, without the sudden deviation at the threshold discovered
by Paschen. This can be confirmed by putting an ammeter into the
circuit, and finding the resistance by the volts divided by the amps.
So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).
That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.
Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel
In slide #29 (page 25 of the PDF) you get (emphasis mine):
Intrinsic Dielectric Strength Limits
- The breakdown of Gasses at STP is dependent on type.
- Air : ~3 kV/mm (75 V/mil)
- He : ~0.37 kV/mm (9.3 V/mil)
- SF 6 : ~9 kV/mm (222 V/mil)
High Vacuum breakdown is surface and configuration
dependent but is in the range of 20 to 40 kV/mm (500 to 1000
V/mil)
- Ignoring surface effects, Liquids and Solids bulk properties are
generally similar in the same range of 15 to 20 kV/mm (375
V/mil to 500 V/mil).
So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
$endgroup$
TL;DR: It also depends on the actual setup.
It can be useful to see the problem from an energy point of view. Since the energy stored in the inductor is $E=1 over 2 L I^2$, when you "instantly" bring the current to zero that energy has to go somewhere. It cannot vanish.
If you model the current interruption as an extremely rapid increase of resistance at the point of the circuit where the switching element is placed, what you get is that the voltage across the inductor begins to raise rapidly. When does it stop raising? It depends on the inductor setup and its surrounding environment.
Marko Bursic has already mentioned self capacitance in his answer.
You say "assuming extremely good insulation", but "extremely good" is still "not perfect". There is resistance and capacitance toward anything surrounding the inductor. You can get an increase of leakage currents due to voltage increase or displacement current into nearby object due to any tiny parasitic capacitance, whose "reactance" becomes low when the rise time is small.
For example If you have a regular, mechanical switch, you may get an arc between the contacts that dissipates the energy into heat, ionization of the air and generation of EM waves.
At high enough voltages, electrons can be extracted from the metal and an electric arc can be formed in vacuum, too (there are also switches that operate using a vacuum arc).
BTW, unless you do want to get high voltage spikes, that's why you put overvoltage protection devices or snubber circuits in parallel to inductors whose current may be interrupted. So, in practice it is you (the circuit designer) that wants to limit the voltage rise.
In a sense, controlled generation of inductor overvoltage spikes is what is done in some step-up DC-DC converters: you "interrupt" the current in an inductor in order to get an increase in voltage, then you "dump" the increased voltage into an output capacitor to "store" it for the load. Of course the switching controller is key to useful operation.
On a related note: it can come as a surprise, but vacuum has not "infinite resistance", i.e. it doesn't impede the flow of current at all! It simply has no free charges to support currents. Once a charge is extracted from a nearby object things change dramatically. This is explained in detail in this article by Charles Chandler. Excerpts (emphasis mine):
The answer is that the vacuum doesn't impede the electrons at all.
In a perfect vacuum (except for the test charge), the charged
particle's behavior can be calculated by just three factors: 1) its
mass, which gives it inertial forces, 2) its electric charge, which
makes it responsive to an electric field, and 3) the electric field
acting on the charged particle. Then the acceleration of the particle
is just the equilibrium between the inertial and electric forces.
[...]
This has been mistaken to be a measure of resistance, which seems to
become infinite at low pressures, but this is not correct. At higher
pressures, the breakdown voltage increases steadily with pressure (off
the right side of Figure 2), and so does the resistance. So there is a
direct relationship between breakdown voltage and resistance in that
range. If we erroneously take that as a hard and fast rule, and we
observe the breakdown voltage shooting up at very low pressures, we
conclude that the resistance must be increasing at very low pressures,
asymptotically approaching infinite resistance at a pressure
characteristic for that gas. But here we have to remember that Paschen
was studying breakdown voltages, and made no mention of resistance. If
we make direct measurements of the resistance, we find that it varies
directly with the pressure, and continues straight down to nothing at
no pressure, without the sudden deviation at the threshold discovered
by Paschen. This can be confirmed by putting an ammeter into the
circuit, and finding the resistance by the volts divided by the amps.
So an inductor placed in an "ideal" vacuum could have its terminal voltage increase enormously, but once the vacuum breaks down, because even a small quantity of electrons is extracted from nearby objects (e.g. the terminals of the inductor itself, for example), a large current may be produced in the vacuum (hence an arc).
That's enough for the concept: the rest boils down of how good are the insulators surrounding the inductor and how big is the extraction potential of electrons from the materials on which the inductor voltage spike is applied.
Some interesting data on this point can be found in this lengthy slide collection from NASA site (it is a set of presentations by several authors); the first part is: High Voltage Engineering Techniques For Space Applications by Steven Battel
In slide #29 (page 25 of the PDF) you get (emphasis mine):
Intrinsic Dielectric Strength Limits
- The breakdown of Gasses at STP is dependent on type.
- Air : ~3 kV/mm (75 V/mil)
- He : ~0.37 kV/mm (9.3 V/mil)
- SF 6 : ~9 kV/mm (222 V/mil)
High Vacuum breakdown is surface and configuration
dependent but is in the range of 20 to 40 kV/mm (500 to 1000
V/mil)
- Ignoring surface effects, Liquids and Solids bulk properties are
generally similar in the same range of 15 to 20 kV/mm (375
V/mil to 500 V/mil).
So here you go: even in vacuum the best you can get is a dielectric strength of 40kV/mm. Can this lead to MV voltages? It depends on the setup, but probably something will break down before that happens.
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
edited May 28 at 10:24
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
answered May 28 at 9:02
Lorenzo DonatiLorenzo Donati
17.2k44577
17.2k44577
add a comment |
add a comment |
$begingroup$
self capacitance
simulate this circuit – Schematic created using CircuitLab
edited May 28 at 11:02
Warren Hill
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
$endgroup$
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
add a comment |
$begingroup$
self capacitance
simulate this circuit – Schematic created using CircuitLab
edited May 28 at 11:02
Warren Hill
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
$endgroup$
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
add a comment |
$begingroup$
self capacitance
simulate this circuit – Schematic created using CircuitLab
edited May 28 at 11:02
Warren Hill
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
$endgroup$
self capacitance
simulate this circuit – Schematic created using CircuitLab
edited May 28 at 11:02
Warren Hill
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
edited May 28 at 11:02
Warren Hill
3,7951127
edited May 28 at 11:02
Warren Hill
3,7951127
edited May 28 at 11:02
Warren Hill
3,7951127
3,7951127
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
answered May 28 at 8:58
Marko BuršičMarko Buršič
10.8k2812
10.8k2812
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
add a comment |
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
4
4
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
$begingroup$
You may want to refer to winding resistance too so as to damp any resultant ring.
$endgroup$
– Warren Hill
May 28 at 11:07
add a comment |
$begingroup$
If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.
However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.
Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of $fracI_indC_stray$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.
If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, $0.5CV^2$, is (ideally) equal to the original energy stored in the inductor, $0.5LI^2$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.
If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
$endgroup$
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
|
show 1 more comment
$begingroup$
If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.
However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.
Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of $fracI_indC_stray$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.
If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, $0.5CV^2$, is (ideally) equal to the original energy stored in the inductor, $0.5LI^2$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.
If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
$endgroup$
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
|
show 1 more comment
$begingroup$
If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.
However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.
Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of $fracI_indC_stray$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.
If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, $0.5CV^2$, is (ideally) equal to the original energy stored in the inductor, $0.5LI^2$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.
If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
$endgroup$
If you have a non-physical pure inductor, as in a thought experiment, or in a SPICE simulator, then nothing limits the rise in voltage. Why stop at megavolts, you could go to zettavolts and beyond.
However, a physical inductor has dimensions, and that means capacitance. The capacitance makes the rate of voltage rise finite, and ultimately limits the peak voltage if nothing else does.
Establish a current in the inductor, then open the switch. The current starts flowing into the stray capacitance, and charging it up, at an initial rate of $fracI_indC_stray$ volts per second. With a few pFs stray capacitance, and 1 amp flowing, that's going to be pretty quick.
If something breaks down, the insulation, air between slowly opening switch contacts, then the current will divert to that path, and the voltage will stop rising. If nothing breaks down, then the current will continue to flow into the stray capacitance, and the voltage will rise such the energy stored in it, $0.5CV^2$, is (ideally) equal to the original energy stored in the inductor, $0.5LI^2$. That's the first quarter cycle of an LC resonance. After that, the circuit will continue to ring, losing energy through dissipation and EM radiation.
If the stray capacitance is enhanced, for instance by the 'condensor' fitted across the points in an old-style car ignition, then the rate of voltage rise will be slower. In this case, the voltage rise rate is slowed to allow the mechanical points to open to an adequate gap before being stressed with the several hundred volts needed for the coil to break down the spark plug.
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
edited May 28 at 19:16
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
answered May 28 at 9:26
Neil_UKNeil_UK
81.3k285188
81.3k285188
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
|
show 1 more comment
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
If nothing breaks down, wouldn't it end up resonating in this LC tank?
$endgroup$
– Huisman
May 28 at 9:59
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
$begingroup$
@Huisman absolutely, I was wondering whether to add that to my answer.
$endgroup$
– Neil_UK
May 28 at 10:08
1
1
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
@Huisman (shameless plug): you can observe the resonance phenomenon in a LTspice simulation of a Zener snubber circuit in this answer of mine.
$endgroup$
– Lorenzo Donati
May 28 at 10:29
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
$begingroup$
What does Quantum Electrodynamics say about this?
$endgroup$
– David Tonhofer
May 28 at 23:17
1
1
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
$begingroup$
@DavidTonhofer I'm not a physicist, but I have some doubts QED can be usefully applied here. These are not quantum phenomena (well, extracting an electron from a metal is, but that's not the key point here), the lumped element model of an inductor is all Maxwell's equations at work, albeit in a very simplified way (quasi-static fields hypotesis, etc. etc.)....
$endgroup$
– Lorenzo Donati
May 29 at 16:21
|
show 1 more comment
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.
If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.
These may be much more limiting than any (tiny) capacitance in the inductance.
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
$endgroup$
add a comment |
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.
If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.
These may be much more limiting than any (tiny) capacitance in the inductance.
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
$endgroup$
add a comment |
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.
If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.
These may be much more limiting than any (tiny) capacitance in the inductance.
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
$endgroup$
Mainly the breakdown voltage of whatever is breaking the current - the air-gap in the mechanical switch (circuit breaker) as it opens (~3 kV/mm for dry air), forward voltage of any flyback diode (if there is one in the circuit), or breakdown or avalanche voltage of a transistor turning off.
If it's a mechanical switch, remember that it's a race between the air-gap as it opens, and the energy in the inductor dissipating. By the time the gap between the switch contacts is only a millimetre wide, there will only be a 3000 Volt spark across it even if there is still energy left in the inductor to support that, as it's magnetic field collapses.
These may be much more limiting than any (tiny) capacitance in the inductance.
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
edited May 29 at 12:11
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
answered May 29 at 12:00
DaveBoltmanDaveBoltman
45626
45626
add a comment |
add a comment |
$begingroup$
Voltage across inductor will rise to v =L.di/dt
di/dt is determined by how fast the transistor switch switches off and cuts the current off.
answered May 28 at 12:05
JamesJames
1174
$endgroup$
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
add a comment |
$begingroup$
Voltage across inductor will rise to v =L.di/dt
di/dt is determined by how fast the transistor switch switches off and cuts the current off.
answered May 28 at 12:05
JamesJames
1174
$endgroup$
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
add a comment |
$begingroup$
Voltage across inductor will rise to v =L.di/dt
di/dt is determined by how fast the transistor switch switches off and cuts the current off.
answered May 28 at 12:05
JamesJames
1174
$endgroup$
Voltage across inductor will rise to v =L.di/dt
di/dt is determined by how fast the transistor switch switches off and cuts the current off.
answered May 28 at 12:05
JamesJames
1174
answered May 28 at 12:05
JamesJames
1174
answered May 28 at 12:05
JamesJames
1174
answered May 28 at 12:05
JamesJames
1174
1174
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
add a comment |
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
Down markers - You may not like it but that's the way it is! You obviously don't know your Horowitz & Hill!
$endgroup$
– James
May 30 at 14:06
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
$begingroup$
I don't think downvoters disagree with the truth you are saying. The key point is that you are not trying to answer the question of the OP. Please review the help center on how to answer a question successfully.
$endgroup$
– Lorenzo Donati
May 30 at 15:32
add a comment |
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$begingroup$
self capacitance
$endgroup$
– Marko Buršič
May 28 at 8:55
1
$begingroup$
Lets have some numbers; 10 milliHenry and 10picoFarad (perhaps those old auto ignition systems) and 1 amp in the inductor. 0.5 * L * I ^2= 0.5 * C * V^2, and we solve for V = I * sqrt(L/C); thus V = 1 * sqrt(10 milli/10pico) or about 30,000 volts.
$endgroup$
– analogsystemsrf
May 29 at 2:38
$begingroup$
Mainly the breakdown voltage of whatever is breaking the current - air gap in a switch (circuit breaker), avalanche voltage in a transistor. I think these would be much more limiting than any (tiny) capacitance in the inductance.
$endgroup$
– DaveBoltman
May 29 at 11:51