Open Circuit Inductor - Electronic Circuits Projects Diagrams Free

Styx · 25th April 2006, 07:27 PM

If you open-cct an inductor while current is flowing an arc will form to maintain the current-flowing (to satify V=Ldi/dt).

However...
IF the inductor is in a perfect vacuum (with no atoms of gas) and the coils were such that they were not touching each other (dont even need any insulation around wire).

If you got a current established and THEN open-cct the inductor what will happen?

since there is no "atmosphere" to ionise an arc cannot form, or can it?

as far as I can figure out one of two things will happen

1) infinite voltage would develope across the terminals
2) the copper at the terminal ends would start to vapourise under the extreamly high electro-field (not infinite) and the liberated copper atoms/electrons would facilitate an arch.

#2 is what I am leaning towards but what kind of voltage would be needed?

**Nigel Goodwin** · 25th April 2006, 07:57 PM

Have you considered that a valve (known as a VACUUM tube to our colonial cousins!) passes current through a vacuum, and indeed needs the vacuum in order to work.

Also, if you apply too high a voltage it will arc between the electrodes.

Styx · 25th April 2006, 07:58 PM

but that has a heating element to liberate electrons from it

**Nigel Goodwin** · 25th April 2006, 08:04 PM

Quote:

Originally Posted by Styx

but that has a heating element to liberate electrons from it

In normal use yes, but a cold valve will arc over if you apply too much voltage anyway.

I don't even know if a vacuum is any better an insulator than dry air, dry air takes about 12,000V to jump 1cm - I don't see as a vacuum would take much more?.

Optikon · 25th April 2006, 08:12 PM

Quote:

Originally Posted by Nigel Goodwin

Quote:

Originally Posted by Styx

but that has a heating element to liberate electrons from it

In normal use yes, but a cold valve will arc over if you apply too much voltage anyway.

I don't even know if a vacuum is any better an insulator than dry air, dry air takes about 12,000V to jump 1cm - I don't see as a vacuum would take much more?.

But in the vacuum case, _what_ is the mechanism that will conduct current. This is, I believe, the essence of styx question. Obviously, an infinite voltage will not develop. So if it does arc and conduct, is the current displacement current?

Styx · 25th April 2006, 08:12 PM

Quote:

Originally Posted by Nigel Goodwin

Quote:

Originally Posted by Styx

but that has a heating element to liberate electrons from it

In normal use yes, but a cold valve will arc over if you apply too much voltage anyway.

I don't even know if a vacuum is any better an insulator than dry air, dry air takes about 12,000V to jump 1cm - I don't see as a vacuum would take much more?.

ahh but it would. In air there are atoms around floating. When one passes into a strong electrical field it will liberate an electron (ie ionising) THIS ionised air then provides a path for the arc to flow.

This is how lightning works, it has to ionise a path to the ground.

Anyway I found part of the solution.

Work Function
Either heat (in the form of a valve element) or a strong electrical field will liberate electrons. THUS with an increaing electrical field from an open-cct inductor (in a vacuum) electrons fro mteh copper will start to be liberated creating an electron-cloud around each terminal. when the cloud from one meets the cloud from the other an arc will form[/url]

Optikon · 25th April 2006, 08:15 PM

Quote:

Originally Posted by Styx

Quote:

Originally Posted by Nigel Goodwin

Quote:

Originally Posted by Styx

but that has a heating element to liberate electrons from it

In normal use yes, but a cold valve will arc over if you apply too much voltage anyway.

I don't even know if a vacuum is any better an insulator than dry air, dry air takes about 12,000V to jump 1cm - I don't see as a vacuum would take much more?.

ahh but it would. In air there are atoms around floating. When one passes into a strong electrical field it will liberate an electron (ie ionising) THIS ionised air then provides a path for the arc to flow.

This is how lightning works, it has to ionise a path to the ground.

Anyway I found part of the solution.

Work Function
Either heat (in the form of a valve element) or a strong electrical field will liberate electrons. THUS with an increaing electrical field from an open-cct inductor (in a vacuum) electrons fro mteh copper will start to be liberated creating an electron-cloud around each terminal. when the cloud from one meets the cloud from the other an arc will form[/url]

The essence of how a vacuum tube works?? I've never used them as they aren't relevant these days (even for audio IMO) but the physics is fascinating nevertheless.

phalanx · 25th April 2006, 09:37 PM

Quote:

Originally Posted by Styx

Anyway I found part of the solution.

Work Function
Either heat (in the form of a valve element) or a strong electrical field will liberate electrons. THUS with an increaing electrical field from an open-cct inductor (in a vacuum) electrons fro mteh copper will start to be liberated creating an electron-cloud around each terminal. when the cloud from one meets the cloud from the other an arc will form[/url]

You found the right information. There is actually considerable thought put into high voltage vacuum feedthroughs because you can get arcing at much closer distances then would normally be expected.

zevon8 · 25th April 2006, 11:04 PM

hmm.. you got me thinking about an old piece of gear I worked on years ago, a high potential test set that had Kilovac vacuum relays to switch the high voltage circuits. I had always figured as you did that the vacuum was used to up the insulation value and stop arcing. Seems to be the case, heres more grist for the mill:

http://relays.tycoelectronics.com/ki...tro/vacuum.asp

G8RPI · 28th April 2006, 02:58 PM

The current does not have to flow through an arc in this case.
While current must flow when the circuit is broken, if the insulation and switch are perfect (impossible), the current flow will charge up the self capacitance of the coil. Obviously if you have high inductance and low capacitance you will acheive very high voltages, this is why something else (your switching transistor

) normally breaks down.

G8RPI.

ljcox · 29th April 2006, 04:42 AM

If you assume that an arc will not occur, then the insulation resistance of the winding will break down. Either way, the current will continue to flow until the energy stored in the magnetic field has been dissipated.

G8RPI · 29th April 2006, 08:37 AM

Hi,
I disagree, current flows when a capacitor is charging and a capacitor stores energy (1/2CV2). The energy stored in the magnetic field of a coil does not have to get dissipated, it can be stored. The reasion this mechanism is not recognised is that the self capacitance is so small that the voltage exceeds the insulation breakdown limits for practical switching devices. The olrginal question asked what happened with a perfectly insulated coil and switch.

ljcox · 29th April 2006, 11:26 AM

Quote:

Originally Posted by G8RPI

Hi,
I disagree, current flows when a capacitor is charging and a capacitor stores energy (1/2CV2). The energy stored in the magnetic field of a coil does not have to get dissipated, it can be stored. The reasion this mechanism is not recognised is that the self capacitance is so small that the voltage exceeds the insulation breakdown limits for practical switching devices. The olrginal question asked what happened with a perfectly insulated coil and switch.

Yes, I see what you mean. However, if you assume that the inductor and switch have "perfect" insulation & there is no arcing, the voltage will tend to infinity but will be limited by the interwinding capacitance. Thus there will be a damped oscillation and the energy will be dissipated in the coil resistance. If the coil was wound with a superconducting wire, it would oscillate for ever.

However, in practice, there is not such thing as perfect insulation.

eblc1388 · 29th April 2006, 04:25 PM

Assumption:
Current is flowing in an isolated coil in space. The connection at both ends is suddenly removed so the coil become truely isolated.

Quote:

Originally Posted by G8RPI

the current flow will charge up the self capacitance of the coil.

OK. There is some form of energy stored. Then what next?

After some hours later, i.e. steady state, what happens?

This "special capacitor" is unlike a normal capacitor consists of two plates with opposite charges. The "capacitor" now changed into a single electrode fromed by the coil made up of metal. No potential difference can exists on different part of a metal object because it is now isolated in space, with no current flowing. So the stored energy immediately during the disconnection may well have become a static charge on the isolated coil or have gone all together and lost as heat in oscillation.

Would there be a net charge on the coil? If so, does the polarity of this charge related to the direction of the original current direction?

ljcox · 29th April 2006, 11:47 PM

Quote:

Originally Posted by eblc1388

Assumption:
Current is flowing in an isolated coil in space. The connection at both ends is suddenly removed so the coil become truely isolated.

Quote:

Originally Posted by G8RPI

the current flow will charge up the self capacitance of the coil.

OK. There is some form of energy stored. Then what next?

After some hours later, i.e. steady state, what happens?

This "special capacitor" is unlike a normal capacitor consists of two plates with opposite charges. The capacitance is distributed through out the coil since one turn lies next to those beside it and above the ones under it, etc. The electric field is due to the potential difference across the coil due to its resistance and the current flowing. The "capacitor" now changed into a single electrode fromed by the coil made up of metal. No potential difference can exists on different part of a metal object because it is now isolated in space, with no current flowing. There is a current flowing, ie. the current that was flowing at the instant the circuit was opened. And there is a potential difference across the coil since it has resistance. So the stored energy immediately during the disconnection may well have become a static charge on the isolated coil or have gone all together and lost as heat in oscillation. As I wrote previously, it will oscillate and the energy will be dissipated as heat due to the resistance of the coil.

Would there be a net charge on the coil? No. When an LC circuit oscillates, the energy is transferred from the magnetic field to the electric field and vice versa. So eventually, all the energy will be dissipated. There cannot be a static charge left since the coil is a conductor. If so, does the polarity of this charge related to the direction of the original current direction?

The situation is difficult to visualise since the capacitance is distributed. I don't know how it would be analysed mathematically. I assume you would have to treat the coil like a transmission line.

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25th April 2006, 07:27 PM	(permalink)
Styx	Open Circuit Inductor If you open-cct an inductor while current is flowing an arc will form to maintain the current-flowing (to satify V=Ldi/dt). However... IF the inductor is in a perfect vacuum (with no atoms of gas) and the coils were such that they were not touching each other (dont even need any insulation around wire). If you got a current established and THEN open-cct the inductor what will happen? since there is no "atmosphere" to ionise an arc cannot form, or can it? as far as I can figure out one of two things will happen 1) infinite voltage would develope across the terminals 2) the copper at the terminal ends would start to vapourise under the extreamly high electro-field (not infinite) and the liberated copper atoms/electrons would facilitate an arch. #2 is what I am leaning towards but what kind of voltage would be needed? __________________ Nothing is impossible. Once a problem is realised, the rest is just details

25th April 2006, 07:57 PM	(permalink)
Nigel Goodwin	Have you considered that a valve (known as a VACUUM tube to our colonial cousins!) passes current through a vacuum, and indeed needs the vacuum in order to work. Also, if you apply too high a voltage it will arc between the electrodes. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

25th April 2006, 07:58 PM	(permalink)
Styx	but that has a heating element to liberate electrons from it __________________ Nothing is impossible. Once a problem is realised, the rest is just details

25th April 2006, 11:04 PM	(permalink)
zevon8	hmm.. you got me thinking about an old piece of gear I worked on years ago, a high potential test set that had Kilovac vacuum relays to switch the high voltage circuits. I had always figured as you did that the vacuum was used to up the insulation value and stop arcing. Seems to be the case, heres more grist for the mill: http://relays.tycoelectronics.com/ki...tro/vacuum.asp

28th April 2006, 02:58 PM	(permalink)
G8RPI	The current does not have to flow through an arc in this case. While current must flow when the circuit is broken, if the insulation and switch are perfect (impossible), the current flow will charge up the self capacitance of the coil. Obviously if you have high inductance and low capacitance you will acheive very high voltages, this is why something else (your switching transistor ) normally breaks down. G8RPI.

29th April 2006, 04:42 AM	(permalink)
ljcox	If you assume that an arc will not occur, then the insulation resistance of the winding will break down. Either way, the current will continue to flow until the energy stored in the magnetic field has been dissipated. __________________ Len

29th April 2006, 08:37 AM	(permalink)
G8RPI	Hi, I disagree, current flows when a capacitor is charging and a capacitor stores energy (1/2CV2). The energy stored in the magnetic field of a coil does not have to get dissipated, it can be stored. The reasion this mechanism is not recognised is that the self capacitance is so small that the voltage exceeds the insulation breakdown limits for practical switching devices. The olrginal question asked what happened with a perfectly insulated coil and switch.