How do you collapse a magnetic field quickly?
The Power of Soup
I've been designing (slowly) over a couple of years an electro-magnetic
mass driver. Basically it's a coil-gun: a machined barrel with multiple
coils and sensors mounted on it. When all the coils fire, they attract
the payload simultaneously. Before each coil is a pair of IR sensors
that shine criss-cross beams through the barrel, sensing when the payload
passes. At this time, I want each coil to sense the payload in turn, and
as fast as possible turn off it's field. If the coil's field doesn't
collapse fast enough, it will slow it's progress, or worse capture the
payload. It should allow it to continue towards the other coils which are
still energized.
From my old college days, I remember that a coil that is energized sets up
a magnetic field around it (duh), but when power is cut there is a decay time
for the field as current still flows because of back EMF (?). I'm assuming
that I'll have to short the coils out through some kind of power resistors,
say 5-10 ohms at 50-100 watts at least. I really haven't done any power
calculations past the coil heat dissipation, because I don't know how to
calculate the strength of the field, and how much power it's going to try
and dump back into the driving circuitry.
I'm also afraid of arcing and such across the power transistors, coil over
heating (I'm planning on using thermocouples to monitor internal temp and
pumped coolant to control it), and speeds too high for the sensors near
the end of the barrel to see.
Well, any help, insight, criticisms or "you're going to friggin kill yourself
because:" on my plan/problem would be appreciated.
-Ken
--
Ken "It's only a small fire" Griest
pos...@world.std.com - The Power of Soup - http://world.std.com/~posoup
"I want to conquer the world! Give all the idiots a brand new religion!"
cooper says: Sorry I considered killing you yesterday...
Alan "Uncle Al" Schwartz
The equation of state for a magnetic field is the same as that of an
ideal gas, where volume times pressure (volume times field strength) is
energy.
To collapse the field >rapidly<, a low inductance/impedence bleed circuit
is not enough. To blast away the field you must reverse the coil current
flow with the proper half-life. Thus slapping in a big, charged
capacitor against the original coil polarization would be interesting,
and probably somewhat messy.
Cry "CHEEBLES!" and loose the gerbils of war.
--
Alan "Uncle Al" Schwartz
Uncl...@ix.netcom.com ("zero" before "@")
http://vvv.com/adsint/freehand/uncleal/
"Quis custodiet ipsos custodes?" The Net!
(Uncle Al has been Officially convinced to "voluntarily" shut down his
homepage. You can own his complete 529 essay collection. Surf by
before it dies!)
David K. Bryant
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>coils and sensors mounted on it.
Every boy's dream.... especially the boys over the hill
from me at Lawrence Livermore Labs.
>From my old college days, I remember that a coil that is energized sets up
>a magnetic field around it (duh), but when power is cut there is a decay time
>for the field as current still flows because of back EMF (?). I'm assuming
>that I'll have to short the coils out through some kind of power resistors,
>say 5-10 ohms at 50-100 watts at least. <snip> I don't know how
>much power it's going to try and dump back into the driving circuitry.
Try this idea.... You have your main motive coils and their drivers.
Include another coil and driver set which is about 1/3 the size of
the main set. Drive this second coil in the opposite direction to
nullify the effect of the main coil as it is collapsing. Also, if
you "slug" is a magnet you can reverse the field on the main coils
and push the slug from behind as it's being pulled from the front.
James P. Meyer
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>coils and sensors mounted on it.
I admire your effort, but wonder if you are spending your time
wisely. Mass drivers, coil guns, rail guns, and similar devices already
exist. And in much more efficient forms than the one you're working on.
How much time have you spent in the library looking up references to
projects similar to yours?
Jim
David K. Bryant
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>coils and sensors mounted on it. When all the coils fire, they attract
>the payload simultaneously. Before each coil is a pair of IR sensors
>that shine criss-cross beams through the barrel, sensing when the payload
>passes.
Move the detectors closer to the breech. It will give you more
time to work. Like timing an engine before TDC.
>At this time, I want each coil to sense the payload in turn, and
>as fast as possible turn off it's field. If the coil's field doesn't
>collapse fast enough, it will slow it's progress, or worse capture the
>payload.
Throw cold water on it. Works for dogs.
>Well, any help, insight, criticisms or "you're going to friggin kill yourself
>because:" on my plan/problem would be appreciated.
You're gonna poke your eye out with that thing!
> Ken "It's only a small fire" Griest
Ahhh, words to live by.
Kevin McMurtrie
Soup) wrote:
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>coils and sensors mounted on it. When all the coils fire, they attract
>the payload simultaneously. Before each coil is a pair of IR sensors
>that shine criss-cross beams through the barrel, sensing when the payload
>passes. At this time, I want each coil to sense the payload in turn, and
>as fast as possible turn off it's field. If the coil's field doesn't
>collapse fast enough, it will slow it's progress, or worse capture the
>payload. It should allow it to continue towards the other coils which are
>still energized.
>
>From my old college days, I remember that a coil that is energized sets up
>a magnetic field around it (duh), but when power is cut there is a decay time
>for the field as current still flows because of back EMF (?). I'm assuming
>that I'll have to short the coils out through some kind of power resistors,
>say 5-10 ohms at 50-100 watts at least. I really haven't done any power
>calculations past the coil heat dissipation, because I don't know how to
>calculate the strength of the field, and how much power it's going to try
>and dump back into the driving circuitry.
>
>I'm also afraid of arcing and such across the power transistors, coil over
>heating (I'm planning on using thermocouples to monitor internal temp and
>pumped coolant to control it), and speeds too high for the sensors near
>the end of the barrel to see.
>
>Well, any help, insight, criticisms or "you're going to friggin kill yourself
>because:" on my plan/problem would be appreciated.
>
>-Ken
>--
> Ken "It's only a small fire" Griest
> pos...@world.std.com - The Power of Soup - http://world.std.com/~posoup
> "I want to conquer the world! Give all the idiots a brand new religion!"
> cooper says: Sorry I considered killing you yesterday...
Just cut the power. Having the coil kick back is the fastest way to get
rid of the field other than reversing the power. The passing object and
the next energized coil should absorb enough inductive energy as to not
blow the transistors.
I'd dump the sensors. If you know how much the object wieghs and how much
you can pull then you know the constant rate to increase the switching
frequency. You might want to play with some stepper motors. They are a
circular version of what you are building.
The Power of Soup
James P. Meyer <jim...@acpub.duke.edu> wrote:
> I admire your effort, but wonder if you are spending your time
>wisely. Mass drivers, coil guns, rail guns, and similar devices already
>exist. And in much more efficient forms than the one you're working on.
But you can't just go and buy one, and building is most of the fun!
> How much time have you spent in the library looking up references to
>projects similar to yours?
Say about 3-4 years off and on?
I wanted to thank everyone who replied with suggestions, and it's opening
up new avenues for research. The idea of channeling the back EMF into the
successive coils is a good one, and it may work. I can also see that
there will be as much a time delay in _increasing_ the fields of the next
coils as there will be in simply turning the current one off and having
it decay naturally. If this is so the only viable options center around
throwing the excess energy away, probably at heat.
I've taken into account that I probably don't want a ferrous barrel to
eliminate just making a big electromagnet. I was thinking about using
machined, polished and oiled aluminum for work-ability and friction
reduction. I'm now faced with calculating the forces acting upon
something such as a ball bearing. Calculating a B field is easy,
done that in college, but determining the forces that act at a distance
on an object of unknown magnetic permeability takes a little more thought.
A magnet has a certain field around it, granted. A non-magnetic, but
ferrous object like a nail is attracted with a certain force which I would
assume to be something on the lines of the inverse square of distance.
Some other notes, all the coils will be energized at the same time, to
maximize the forward vector of acceleration, and considering that once
the object is moving, it's probably better to have the coils already running
because of the inductive delay that comes with turning them on.
Also, a push-pull configuration won't work unless I use a polarized object
such as a magnet, or a self-contained battery powered electromagnet,
or if possible an statically charged ball bearing or the like. Currently
I'm just planning on using a pull-pull configuration to simplify controller
electronic design.
James P. Meyer
>I've taken into account that I probably don't want a ferrous barrel to
>eliminate just making a big electromagnet.
You don't want a metalic, conductive, barrel of *any* material. If
you wind a coil of wire around a conductive core (barrel), you have just
inserted a "shorted turn" into the coil. The result will be that when you
try to change the current in the coil, the shorted turn represented by the
barrel will have a current induced in it that will try to oppose the
original change in the coil current. It's all "transformer action" if you
want to look it up.
However, you can make this shorted turn stuff work *for* you instead
of against you. Make a barrel out of fiberglass and epoxy or some kind of
plastic. Wind your turns on that. The barrel doesn't have to be strong
like a regular gun barrel, because there's no pressure *inside* the
barrel. Line the barrel with a thin sheet of teflon to reduce friction if
you want to get fancy.
Then wind some heavy wire for your coils. You don't need many turns,
I'd say about 20 turns per coil would be a good place to start. Start
with one coil in the beginning. Put the first coil down near the
"breech". Then make a "slug" out of copper. Make it a close fit, not too
snug, for the barrel. Make the slug about 4 or 5 times as long as it is
big around. The length of the slug and the coil should be about the same.
Put a stop in the barrel so that when the slug is inserted, it is
positioned against the stop and about 2/3rds of the way into the first
coil.
Then get a 300 volt power supply, a 10,000 ohm 10 watt resistor, an
electrolytic capacitor with a rating of around 1,000 microfarads at 300
volts, and a heavy-duty push-button switch. Wire it all up like this:
The power supply minus terminal goes to the cap's minus terminal and
one side of the coil (either side is ok). The plus terminal of the power
supply goes to one end of the 10 watt resistor. The other end of the
resistor goes to the cap's plus terminal and one side of the switch. The
other side of the switch goes to the side of the coil that is left over.
Put a slug in, turn the power supply on and wait about 30 seconds for
the cap to charge up. Then press the switch.
You should get a bang from the switch and the slug should be shot out
of the end of the barrel.
If you get that far, it's just a matter of adding more coils to the
gun and replacing the push button with a series of timed electronic
switches to finish the job. If you want more theory about what's
happening, you can either look it up or ask me here.
Here's a question or two for you to think about.....
Why do you suppose it might be a good idea to *not* completely close
off the "breech" of your gun? Why might you want to consider making the
barrel with a lot of small holes cross-wise in it?
Hint... What's going to happen to the air ahead and behind of your
slug?
Jim "Don't poke your eye out." Meyer
Kevin McMurtrie
Soup) wrote:
>In article <31537334...@news.duke.edu>,
>James P. Meyer <jim...@acpub.duke.edu> wrote:
>> I admire your effort, but wonder if you are spending your time
>>wisely. Mass drivers, coil guns, rail guns, and similar devices already
>>exist. And in much more efficient forms than the one you're working on.
>
>But you can't just go and buy one, and building is most of the fun!
>
>> How much time have you spent in the library looking up references to
>>projects similar to yours?
>
>Say about 3-4 years off and on?
>
>I wanted to thank everyone who replied with suggestions, and it's opening
>up new avenues for research. The idea of channeling the back EMF into the
>successive coils is a good one, and it may work. I can also see that
>there will be as much a time delay in _increasing_ the fields of the next
>coils as there will be in simply turning the current one off and having
>it decay naturally. If this is so the only viable options center around
>throwing the excess energy away, probably at heat.
>
>I've taken into account that I probably don't want a ferrous barrel to
>eliminate just making a big electromagnet. I was thinking about using
>machined, polished and oiled aluminum for work-ability and friction
>reduction.
No, it mustn't conduct electricity or it will be REALLY slow. The only
way you could use aluminum would be to cut it lengthwise and insert a
ceramic or plastic insulator in the gap to stop conduction in the same
direction as the coil winding. I would suggest using a disposable plastic
tube that can be slid into the coils.
>I'm now faced with calculating the forces acting upon
>something such as a ball bearing. Calculating a B field is easy,
>done that in college, but determining the forces that act at a distance
>on an object of unknown magnetic permeability takes a little more thought.
>
>A magnet has a certain field around it, granted. A non-magnetic, but
>ferrous object like a nail is attracted with a certain force which I would
>assume to be something on the lines of the inverse square of distance.
>
>Some other notes, all the coils will be energized at the same time, to
>maximize the forward vector of acceleration, and considering that once
>the object is moving, it's probably better to have the coils already running
>because of the inductive delay that comes with turning them on.
Just compensate for the delay and run only one or two at a time. You have
to compensate anyway in turning them off. You'll then be able to
concentrate the power.
Tom Sheridan
|> To collapse the field >rapidly<, a low inductance/impedence bleed circuit
|> is not enough. To blast away the field you must reverse the coil current
|> flow with the proper half-life. Thus slapping in a big, charged
|> capacitor against the original coil polarization would be interesting,
|> and probably somewhat messy.
I think what you want is a 'snubber'. :-|
It's a capacitor/diode combination that the arcs'n'sparks guys use when they turn off large currents, like New York.
But if I was you i wouldn't listen to people like me who think a milliamp ia a huge current :-)
|> Cry "CHEEBLES!" and loose the gerbils of war.
^^^^^
'let slip', surely?
--------------------------------------------------------------------------
Tom Sheridan My views are my own.
GPS Correlation with views of GPS is random.
Oldham, UK.
Wenzel
comments about discharging an inductor's field rapidly. The energy in the
field is one-half the inductance times the current squared. The fastest
way to dissipate the energy is to dump it into a high voltage "supply"
through a diode. The initial current when discharge starts is fixed so
the higher the voltage, the more power flows out of the magnetic field.
Select a voltage near but below the breakdown of your transistors or
other components.The high voltage "supply" could be a high voltage
capacitor with a shunt regulator or I suppose you could design a
switching converter to return the energy to the low voltage supply. (Good
luck with that one!) A resistor and diode across the coil is pretty good
but by holding the voltage high during the entire discharge, the rate of
energy flow is maximum. If you use a resistor and diode then select a
resistor which gives you the highest acceptable voltage when the full
coil current flows in it. The power in the resistor is one-half the
inductance times the full coil current squared times the number of
discharges per second. If that is unclear, please say so and I will
clarify! I have no comment about the relative safety of this project,
however!
wa4...@vnet.ibm.com
>In article <Dor7L...@world.std.com>, pos...@world.std.com (The Power of
>Soup) wrote:
>
>No, it mustn't conduct electricity or it will be REALLY slow. The only
>way you could use aluminum would be to cut it lengthwise and insert a
>ceramic or plastic insulator in the gap to stop conduction in the same
>direction as the coil winding. I would suggest using a disposable plastic
>tube that can be slid into the coils.
>
>to compensate anyway in turning them off. You'll then be able to
>concentrate the power.
>
some random thoughts.
Would it be possible to use the natural LC circuit formed by the coil
capacitor combination to produce the pulse and the damping waveform?
This will require 'tuning' the circuit, but might be an interesting
approach. You probably want to stay away from electrolytic capacitors,
since the voltage on the capacitor should reach a significant negative
value on the second half cycle.
Use caution in the type of plastic tubing that is used. Certain plastics
can absorb a significant amount of electromagnetic energy. PVC seems to
be rather bad about this for radio frequencies (e.g., Don't build a loading
coil or trap based on a PVC form; it may melt or even catch fire!).
What properties should the slug have? Shouldn't it be constructed so as to
minimize eddy currents? I would think a powdered iron slug might be the
best, while copper may be one of the worst.
Dave
P.S. Don't shoot yourself in the foot with the railgun.
P.P.S. Opinions expressed here are mine, and not my company's.
Steve Jackson
>
> If the coil's field doesn't
> collapse fast enough, it will slow it's progress, or worse capture the
> payload. It should allow it to continue towards the other coils which are
> still energized.
>
Condiser using a 'H' Bridge Driver for each coil. These are commonally used
to drive stepper motors. With a H driver you can reverse pulse each coil for
a very precise period. This should force the coil off very quickly, and with
fine tunning, you can even ditch any residual magnetism. A 'H' bridge will
also assist you in dealing with the back-emf. Remember that a stepper motor
can be thouht of as a set of electro-magnets that have to be controlled very
quickly.
--
Steve Jackson
mailto:st...@dater.demon.co.uk
Terry Harris
>Condiser using a 'H' Bridge Driver for each coil. These are commonally used
>to drive stepper motors. With a H driver you can reverse pulse each coil for
>a very precise period. This should force the coil off very quickly,
I don't think so, the current in the coil will continue to flow in the
same direction till the stored energy is removed. How quick this
happens depends on the voltage the current works into. The flywheel
(or parasitic MOSFET) diodes in an 'H' bridge route the current into
the supply rails but you don't need to turn on the 'reverse' half of
the bridge to do this, you don't need that half of the bridge at all.
You can collapse the current just as quick with a zener diode the
same voltage as the supply (and faster with higher voltages if the
diode and drivers can stand it). Less efficient (assuming you can do
something with the energy pumped back into the supply) but a lot
simpler.
Cheers Terry...
John Wadley x3293 JWPE 8211
: In article <Doot3...@world.std.com>, pos...@world.std.com (The Power of
: Soup) wrote:
: >I've been designing (slowly) over a couple of years an electro-magnetic
: >mass driver. Basically it's a coil-gun: a machined barrel with multiple
: >coils and sensors mounted on it. When all the coils fire, they attract
: >the payload simultaneously. Before each coil is a pair of IR sensors
: >that shine criss-cross beams through the barrel, sensing when the payload
: >passes. At this time, I want each coil to sense the payload in turn, and
: >as fast as possible turn off it's field. If the coil's field doesn't
: >collapse fast enough, it will slow it's progress, or worse capture the
: >payload. It should allow it to continue towards the other coils which are
: >still energized.
: >
: >From my old college days, I remember that a coil that is energized sets up
: >a magnetic field around it (duh), but when power is cut there is a decay time
: >for the field as current still flows because of back EMF (?). I'm assuming
: >that I'll have to short the coils out through some kind of power resistors,
: >say 5-10 ohms at 50-100 watts at least. I really haven't done any power
: >calculations past the coil heat dissipation, because I don't know how to
: >calculate the strength of the field, and how much power it's going to try
: >and dump back into the driving circuitry.
: >
: >I'm also afraid of arcing and such across the power transistors, coil over
: >heating (I'm planning on using thermocouples to monitor internal temp and
: >pumped coolant to control it), and speeds too high for the sensors near
: >the end of the barrel to see.
: >
: >Well, any help, insight, criticisms or "you're going to friggin kill yourself
: >because:" on my plan/problem would be appreciated.
: >
: >-Ken
: >--
: > Ken "It's only a small fire" Griest
: > pos...@world.std.com - The Power of Soup - http://world.std.com/~posoup
: > "I want to conquer the world! Give all the idiots a brand new religion!"
: > cooper says: Sorry I considered killing you yesterday...
: Just cut the power. Having the coil kick back is the fastest way to get
: rid of the field other than reversing the power. The passing object and
: the next energized coil should absorb enough inductive energy as to not
: blow the transistors.
: I'd dump the sensors. If you know how much the object wieghs and how much
: you can pull then you know the constant rate to increase the switching
: frequency. You might want to play with some stepper motors. They are a
: circular version of what you are building.
--
000000000011111111112222222222333333333344444444445555555555666666666677777777
John Wadley |This post is strictly my opinion, not TI's.
a18...@pooh.msp.sc.ti.com |
Texas Instruments, Inc. |
James P. Meyer
>I've taken into account that I probably don't want a ferrous barrel to
>eliminate just making a big electromagnet.
You don't want a metalic, conductive, barrel of *any* material. If
Dan Mills
jim...@acpub.duke.edu "James P. Meyer" writes:
>pos...@world.std.com (The Power of Soup) wrote:
[Chop]
>
> If you get that far, it's just a matter of adding more coils to the
>gun and replacing the push button with a series of timed electronic
>switches to finish the job. If you want more theory about what's
>happening, you can either look it up or ask me here.
>
> Here's a question or two for you to think about.....
>
> Why do you suppose it might be a good idea to *not* completely close
>off the "breech" of your gun? Why might you want to consider making the
>barrel with a lot of small holes cross-wise in it?
>
> Hint... What's going to happen to the air ahead and behind of your
>slug?
>
How about using a short section of copper pipe as your projectile? This
would eliminate the problem with the air? Is there any reason why this would
be a bad idea?
--
********************************************************
* And on the first day the lord said...... * *
*......Lx1, Go! and there was light! * Dan Mills *
********************************************************
David R Mulligan
>
> Dan Mills <Dmi...@abcde.demon.co.uk> wrote:
>
> >> Hint... What's going to happen to the air ahead and behind of your
> >>slug?
> >>
> > How about using a short section of copper pipe as your projectile? This
> >would eliminate the problem with the air? Is there any reason why this would
> >be a bad idea?
>
> ago for a hollow nerf football that you could throw really far and it
> didn't wobble when it flew through the air?
>
> Jim
There's a very good reason. Copper! You'd have to use an iron or
nickle based object.
James P. Meyer
James P. Meyer
> > > How about using a short section of copper pipe as your projectile? This
> > >would eliminate the problem with the air? Is there any reason why this would
> > >be a bad idea?
> >
>
> There's a very good reason. Copper! You'd have to use an iron or
> nickle based object.
You didn't read my first message, did you?
Copper will work fine if you build and operate the rifle the way
I described it there. Iron is *not* required.
Jim
Allan Duncan
>
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>coils and sensors mounted on it. When all the coils fire, they attract
>the payload simultaneously. Before each coil is a pair of IR sensors
>that shine criss-cross beams through the barrel, sensing when the payload
>passes. At this time, I want each coil to sense the payload in turn, and
>as fast as possible turn off it's field. If the coil's field doesn't
>collapse fast enough, it will slow it's progress, or worse capture the
>payload. It should allow it to continue towards the other coils which are
>still energized.
>
>From my old college days, I remember that a coil that is energized sets up
>a magnetic field around it (duh), but when power is cut there is a decay time
>for the field as current still flows because of back EMF (?). I'm assuming
>that I'll have to short the coils out through some kind of power resistors,
>say 5-10 ohms at 50-100 watts at least. I really haven't done any power
>calculations past the coil heat dissipation, because I don't know how to
>calculate the strength of the field, and how much power it's going to try
>and dump back into the driving circuitry.
>
>I'm also afraid of arcing and such across the power transistors, coil over
>heating (I'm planning on using thermocouples to monitor internal temp and
>pumped coolant to control it), and speeds too high for the sensors near
>the end of the barrel to see.
>
>Well, any help, insight, criticisms or "you're going to friggin kill yourself
>because:" on my plan/problem would be appreciated.
Well, your college days should best be forgotten, and a text book
consulted - the best way to _sustain_ the field is to short the coils!
(Low loss through only the coil resistance allows the current to continue
to flow).
The way to quickly quench the current is to allow a large back emf to
develop - in the limit, opening a switch and quenching the arc, but
normally a small capacity high voltage capacitor is placed across it.
You also need a diode in series, or the capacitor
will discharge back into the coil and you will have the field back
again, but with opposite polarity (you might want that if the projectile
was magnetised), and a resistor across the cap to bleed off the charge
at a suitable rate.
This is a variation of your TV's horizontal deflection cicuit, or a
capacitor-discharge ignition in reverse.
Allan Duncan a.du...@trl.telstra.com.au (+613) 9253 6708, Fax 9253 6664
Photonics & Reference Standards Section
Telstra Research Labs, Box 249 Rosebank MDC
Clayton, Victoria, 3169, Australia (a world renowned Lab in past times)
Dan Mills
ski...@interlog.com "David R Mulligan" writes:
>James P. Meyer wrote:
>>
>> Dan Mills <Dmi...@abcde.demon.co.uk> wrote:
>>
>> >> Hint... What's going to happen to the air ahead and behind of your
>> >>slug?
>> >>
>> > How about using a short section of copper pipe as your projectile? This
>> >would eliminate the problem with the air? Is there any reason why this would
>> >be a bad idea?
>>
>> None that I can think of. Didn't I see A TV commercial not too long
>> ago for a hollow nerf football that you could throw really far and it
>> didn't wobble when it flew through the air?
>>
>> Jim
>
>There's a very good reason. Copper! You'd have to use an iron or
>nickle based object.
We are using Eddy curent effects! A conductive material is needed a ferrous
one is not!
Dan Mills
>pos...@world.std.com (The Power of Soup) wrote:
>
>>I've taken into account that I probably don't want a ferrous barrel to
>>eliminate just making a big electromagnet.
>
[Snip]
>
> If you get that far, it's just a matter of adding more coils to the
>gun and replacing the push button with a series of timed electronic
>switches to finish the job. If you want more theory about what's
>happening, you can either look it up or ask me here.
>
The electronic switch is decidedly non trivial. The peak current will be
measured in KA. You can not use Thyristors as dI/dT is to great.
Would Thyratrons be a possibility?
Just another thought : The projectile needs to be a Non magnetic conductor?
How about creating a highly ionised plasma in the base of the tube? How
about discharging a cap. through a thin wire forming a conductive metal gas?
The timing and the design of the magnetic circuit would be critical. You
could also use a electro-static lense to focus the output pulse. This would
probably work better in space!
Just me thinking improbable thoughts <g>.
William W Janssen
>
>In article <31557935...@news.duke.edu>, jim...@acpub.duke.edu (James P. Meyer) writes:
>|> pos...@world.std.com (The Power of Soup) wrote:
>|> "breech". Then make a "slug" out of copper. Make it a close fit, not too
>|> snug, for the barrel. Make the slug about 4 or 5 times as long as it is
>|> big around. The length of the slug and the coil should be about the same.
>
>
>Huh? Wouldn't you want to make the slug out of a ferrous metal?
>
>
>--
>The above comments are my own and not my employer's. Have a great day!
I thought the desire was to collapse the magnetic field QUICKLY.
I also understood that a "slug" or shorted turn would SLOW the collapse.
Where did I go wrong.
Bill K7NOM
Don Abernathy
James P. Meyer
> > If you get that far, it's just a matter of adding more coils to the
> >gun and replacing the push button with a series of timed electronic
> >switches to finish the job. If you want more theory about what's
> >happening, you can either look it up or ask me here.
> >
> The electronic switch is decidedly non trivial. The peak current will be
> measured in KA. You can not use Thyristors as dI/dT is to great.
> Would Thyratrons be a possibility?
Thyratrons would be one way. Another would be a triggered spark
gap. That's a gap that is spaced far enough apart so that the voltage
won't make it break down until a smaller spark close to one electrode
makes a cloud of plasma and triggers the large gap. It's like a
thyratron but without the hydrogen.
>
> The timing and the design of the magnetic circuit would be critical. You
> could also use a electro-static lense to focus the output pulse. This would
> probably work better in space!
Good idea. Next time the mothership comes to pick you up and
take you back to where ever it came from, you should try it. 8-)
Jim
fin...@where.com
just use a diode across the coils ...this would also protect any solid
state devices switching on the coils ...make sure diode is installed
opposite to the source polarity..Try it ....
James P. Meyer
> |> "breech". Then make a "slug" out of copper. Make it a close fit, not too
>
> Huh? Wouldn't you want to make the slug out of a ferrous metal?
No. The expanding magnetic field from the coil will induce a
current in the copper. The current in the copper will generate a
magnetic field of its own. The polarity of the two magnetic fields will
be such that they will oppose each other and shove the slug out of the
coil. This action will be the same for either polarity of electric
current in the coil and will also work for AC currents in the coil.
The original post that started this assumed an iron slug and
coils that were always "on" and needed to be turned off rapidly. That
scheme has many practical problems. Causing a coil to have a brief pulse
in it only when the slug in in the coil is much easier to do and results
in a much more efficient gun.
Jim "No iron need apply." Meyer
James P. Meyer
> >|> "breech". Then make a "slug" out of copper. Make it a close fit, not too
> >
> >Huh? Wouldn't you want to make the slug out of a ferrous metal?
> >
> I thought the desire was to collapse the magnetic field QUICKLY.
> I also understood that a "slug" or shorted turn would SLOW the collapse.
You're mixing the original iron slug gun with my improved copper
slug gun. You only pulse the coil when the slug is in the coil. That
kicks the slug out and after the slug is out, you don't care what
happens to the field. That's the beauty of getting rid of the iron. In
my design, the slug is always repelled by the coil so there's no need to
"collapse the field".
Jim
Dwight Elvey
|>
|> I've been designing (slowly) over a couple of years an electro-magnetic
|> mass driver. Basically it's a coil-gun: a machined barrel with multiple
|> coils and sensors mounted on it. When all the coils fire, they attract
|> the payload simultaneously. Before each coil is a pair of IR sensors
|> that shine criss-cross beams through the barrel, sensing when the payload
|> passes. At this time, I want each coil to sense the payload in turn, and
|> as fast as possible turn off it's field. If the coil's field doesn't
|> collapse fast enough, it will slow it's progress, or worse capture the
|> payload. It should allow it to continue towards the other coils which are
|> still energized.
|>
Hi
Looking at all the suggestions of using zeners or tuned
circuits, it seems like people are missing the obvious.
It can be done mechanically without actually dumping
the energy of the coil. All one has to do is throw a megnetic
shunt into the field path that will be shorter or less
resistance to the field than the guns path. Doing it this
way, one could even save some of the energy. As a matter
of fact, one need not even use electro magnets at all
with big enough motors one could use a bunch of permanent
magnets. If one was trting to throw a large weight, one
could store enough energy in some flywheels to really
kick the load off.
IMHO
Dwight
Allan Duncan
>adu...@rhea.trl.OZ.AU (Allan Duncan) wrote:
>just use a diode across the coils ...this would also protect any solid
>state devices switching on the coils ...make sure diode is installed
>opposite to the source polarity..Try it ....
Oh no he didn't - he said the bit at the end.
>>From article <Doot3...@world.std.com>, by pos...@world.std.com (The Power of Soup):
>>>
>>>I've been designing (slowly) over a couple of years an electro-magnetic
>>>mass driver. Basically it's a coil-gun: a machined barrel with multiple
>>>coils and sensors mounted on it. When all the coils fire, they attract
>>>the payload simultaneously. Before each coil is a pair of IR sensors
>>>that shine criss-cross beams through the barrel, sensing when the payload
>>>passes. At this time, I want each coil to sense the payload in turn, and
>>>as fast as possible turn off it's field. If the coil's field doesn't
>>>collapse fast enough, it will slow it's progress, or worse capture the
>>>payload. It should allow it to continue towards the other coils which are
>>>still energized.
>>>
Dwight Elvey
|> James P. Meyer wrote:
|> >
|> > Dan Mills <Dmi...@abcde.demon.co.uk> wrote:
|> >
|> > >> Hint... What's going to happen to the air ahead and behind of your
|> > >>slug?
|> > >>
|> > > How about using a short section of copper pipe as your projectile? This
|> > >would eliminate the problem with the air? Is there any reason why this would
|> > >be a bad idea?
|> >
|> > None that I can think of. Didn't I see A TV commercial not too long
|> > ago for a hollow nerf football that you could throw really far and it
|> > didn't wobble when it flew through the air?
|> >
|> > Jim
|>
|> There's a very good reason. Copper! You'd have to use an iron or
|> nickle based object.
Not true. Copper could work as long as the field was changing
fast enough. The changing field will set up a current in the
copper that would tend to push away from the magnetic field.
So instead of pulling the projectile you push it with the
field. Switching coils in sequence will still make it move.
Dwight
Mark Kinsler
>In article <Doot3...@world.std.com>, pos...@world.std.com (The Power of
Soup) writes:
>Hi
> Looking at all the suggestions of using zeners or tuned
>circuits, it seems like people are missing the obvious.
>It can be done mechanically without actually dumping
>the energy of the coil. All one has to do is throw a megnetic
>shunt into the field path that will be shorter or less
>resistance to the field than the guns path. Doing it this
>way, one could even save some of the energy. As a matter
>of fact, one need not even use electro magnets at all
>with big enough motors one could use a bunch of permanent
>magnets. If one was trting to throw a large weight, one
>could store enough energy in some flywheels to really
>kick the load off.
Which gives you a mechanical catapult. That energy has to come from
somewhere.
I shall stick my oar in here a bit further: Most of the devices of the
railgun variety don't use any iron at all. That's because it takes a
while to change the magnetic field through a piece of iron, and the delay
is unacceptable in a device in which things have to happen fast.
So the idea is to use the interaction between electromagnets,
neither of which has an iron core. They also have only a few turns of
wire at most. This cuts down the inductance and thus the time needed to
change the magnetic field therein. The current requirements are
therefore increased drastically. The things aren't very easy to make.
Electric cannon were first proposed in the 1880's, and for all the work
that's been done on the things we haven't had notable success. This is
not to say that the problem isn't soluble, but it sure isn't a trivial
problem.
I think that the problem with launching something into orbit from any
kind of cannon is that the projectile's speed would be highest where the
air is thickest. That's why rockets are better. Pity: I'd like to see
that sometime.
M. Kinsler
gonna build an HO-gauge railgun
William Murphy
Soup) wrote:
>
>I've been designing (slowly) over a couple of years an electro-magnetic
>mass driver.
<snip>
I recall a student or group of students at MIT making a device like
you are describing. You might want to research what they did and the
problems they encountered.
Smooth Sailing
William Murphy
David K. Bryant
>Soup) wrote:
>>
>>I've been designing (slowly) over a couple of years an electro-magnetic
>>mass driver.
> <snip>
I was cleaning out the garage today and I found it!!!
I have a NASA tech Brief that discusses a Superconducting
Magnetic Projectile Launcher. Want a copy?
Dwight Elvey
|> adu...@rhea.trl.OZ.AU (Allan Duncan) wrote:
|> just use a diode across the coils ...this would also protect any solid
|> state devices switching on the coils ...make sure diode is installed
|> opposite to the source polarity..Try it ....
|> > >
|> > >I've been designing (slowly) over a couple of years an electro-magnetic
|> > >coils and sensors mounted on it. When all the coils fire, they attract
|> > >the payload simultaneously. Before each coil is a pair of IR sensors
|> > >that shine criss-cross beams through the barrel, sensing when the payload
|> > >passes. At this time, I want each coil to sense the payload in turn, and
|> > >as fast as possible turn off it's field. If the coil's field doesn't
|> > >collapse fast enough, it will slow it's progress, or worse capture the
|> > >payload. It should allow it to continue towards the other coils which are
|> > >still energized.
Much deleted
Hi
Allen is incorrect. Using a diode will not quickly
dump the current in the coil. The rate that the energy
is removed from the coil is related to the resistance that
the coil feeds the current into. The diode will look like a
low ohms resistance to the energy in the coil and have a long
time constant. The bigger the resistance the quicker the coil
will dump its current. Using a diode with a series resistor
will have a better effect. The value of the resistor should
be chosen to keep the voltage flyback of the coil below the
breakdown of the switching device. Also, since I would
assume that you're coil will store a large amount of energy,
the resistor and diode should be heave in size.
Dwight
Mustafa Soysal MS57
payload, IR beams, and coils...
Messing with your design aside, you need to look at the properties of
the magnetic and electric equipment involved, and then you optimize with
them to get the maximal change in d'phi'/dt. Note that you must
consider the magnetic and electrical properties of your payload as well.
To get the best accelaration, you should actually calculate the forces
from the previous/current coils and upcoming coils to see what sort of
transfer function you need to apply to each coil to maximize the
integral area under the force over location.....Your transfer function
for each coil could be the same for all, or a precalculated one for each
depending on the mass of your payload and what you want to do with it.
To further widen the area of possibilities, you could make each coil
adaptive to produce a maximum accelaration, if that's your objective
rather than a controlled output. Please note that as the speeds
increase, you will have to deal with the effects of turbulences etc, so
you may have to pull thru a range of speed quickly to achive better
results.
This could yield you to improved designs etc. and that in turn opens new
possibilities on the electrical/physical side...there are zillions of
possibilities...
brian whatcott
>
>In article <Dp9so...@news2.new-york.net>, fin...@where.com writes:
>|> adu...@rhea.trl.OZ.AU (Allan Duncan) wrote:
>|> just use a diode across the coils ...this would also protect any solid
>|> state devices switching on the coils ..
> Allen is incorrect. Using a diode will not quickly
>dump the current in the coil. The rate that the energy
>is removed from the coil is related to the resistance that
>the coil feeds the current into. The diode will look like a
>low ohms resistance to the energy in the coil and have a long
>time constant. The bigger the resistance the quicker the coil
>will dump its current. Using a diode with a series resistor
>will have a better effect. The value of the resistor should
>be chosen to keep the voltage flyback of the coil below the
>breakdown of the switching device.
Hmmm...Duncan advocates a diode
and Elvey advocates a diode and as high a resistor value as will not pop the
coil driver.
It turns out that the power transfer theorem covers this case adequately:
maximum power is transferred when the load and the source impedences are
equal. That's how a given amount of energy is dissipated fastest.
The source is the coil, the load is the diode resistor combination.
Brian
Harry H Conover
:
: It turns out that the power transfer theorem covers this case adequately:
: maximum power is transferred when the load and the source impedences are
: equal.
Sure, maximum power transfer occurs when you match the impedances, but
this has nothing to do with the rate at which the field is collapsed.
The rate of field collapse -- d/dt(magnetic flux) is directly proportional
to the voltage induced in the coil. For maximum collapse rate, you want
this voltage to be as high as possible. This implies that you dump the
stored energy into a resistor of high value.
:That's how a given amount of energy is dissipated fastest.
No, that's how you maximize the coupling efficiency -- a different
thing entirely.
Harry C.
Terry Harris
>> Allen is incorrect. Using a diode will not quickly
>>dump the current in the coil. The rate that the energy
>>is removed from the coil is related to the resistance that
>>the coil feeds the current into. The diode will look like a
>>low ohms resistance to the energy in the coil and have a long
>>time constant. The bigger the resistance the quicker the coil
>>will dump its current. Using a diode with a series resistor
>>will have a better effect. The value of the resistor should
>>be chosen to keep the voltage flyback of the coil below the
>>breakdown of the switching device.
>///
>
>Hmmm...Duncan advocates a diode
>and Elvey advocates a diode and as high a resistor value as will not pop the
>coil driver.
>
>It turns out that the power transfer theorem covers this case adequately:
But that theorem assumes a constant source voltage which a colapsing
field in an inductor isn't. It doesn't cover the case at all.
Cheers Terry...
Harry H Conover
: brian whatcott (in...@intellisys.net) wrote:
: :
: : It turns out that the power transfer theorem covers this case adequately:
: : maximum power is transferred when the load and the source impedences are
:
: Sure, maximum power transfer occurs when you match the impedances, but
: this has nothing to do with the rate at which the field is collapsed.
:
: The rate of field collapse -- d/dt(magnetic flux) is directly proportional
: to the voltage induced in the coil. For maximum collapse rate, you want
: this voltage to be as high as possible. This implies that you dump the
: stored energy into a resistor of high value.
:
: :That's how a given amount of energy is dissipated fastest.
In retrospect, the quickest way of dumping an inductive field is with
an infinite resistance or open circuit. Unfortunately, this will develop
a voltage spike with potentially enough voltage to jump or breakdown
any switching device. A high resistance is the reasonable compromise.
By contrast, a low resistance will result in very slow collapse of the
magnetic field. Certain type of time delay relays exploit this principle
by placing a shorted turn of heavy gauge wire around the relay coil.
Harry C.
Tom Bruhns
: In retrospect, the quickest way of dumping an inductive field is with
: an infinite resistance or open circuit. Unfortunately, this will develop
: a voltage spike with potentially enough voltage to jump or breakdown
: any switching device. A high resistance is the reasonable compromise.
A resistor will result in an exponential decay of current, assuming
only inductance and resistance in the model. You can get a quicker
decay if you clamp with a diode to the maximum allowed voltage (e.g.,
a rail at just below the breakdown voltage of the switching device).
You are dumping energy back into that rail, so you might consider how
to use it and not waste it. All this is modified somewhat when you
take into account parasitic capacitance.
You can even use a pair of switches, if you want: one to pump energy
into the coil and one to remove energy. The second takes the place of
the diode suggested above, to a high voltage rail. You must turn that
second one off at exactly the right time, when no energy is left in
the inductor (i.e., the current has dropped to zero). Again, this is
modified by parasitic capacitance...
The problem is very similar to those encountered designing switching
power supplies.
--
Cheers,
Tom
to...@lsid.hp.com
Daniel P. Engel III
>Huh? Wouldn't you want to make the slug out of a ferrous metal?
I don't know who asked the original question about a rail gun, but
would it be possible to build one that pushed, rather than pulled the
mass? I've never studied them, so maybe it's a dumb idea, but if a
copper "ring" was used as the payload, then wouldn't a rising magnetic
field induce a current that would cause an opposing magnetic field.
Then, you could turn the coils ON as the payload passed, and not worry
about collapsing the field--only about establishing it at the proper
rate for maximum work out of each coil. Then, the acceleration from
the fields (given a set of payload properties) would be entirely a
function of their rate of growth, and not of their absolute strength.
I don't know how the coils themselves would interact. Perhaps there
would be an issue of preventing the field from one coil from inducing
a current in the next (or vice-versa), but it seems like it's worth
thinking through on paper.
Just my humble opinion.
--
Dan Engel (den...@hom.net)
brian whatcott
>: :
>: : It turns out that the power transfer theorem covers this case
adequately:
>: : maximum power is transferred when the load and the source impedences are
>: : equal.
>:
>: Sure, maximum power transfer occurs when you match the impedances, but
>: this has nothing to do with the rate at which the field is collapsed.
>:
>: The rate of field collapse -- d/dt(magnetic flux) is directly proportional
>: to the voltage induced in the coil. For maximum collapse rate, you want
>: this voltage to be as high as possible. This implies that you dump the
>: stored energy into a resistor of high value.
>:
>: :That's how a given amount of energy is dissipated fastest.
>
>an infinite resistance or open circuit. Unfortunately, this will develop
>a voltage spike with potentially enough voltage to jump or breakdown
>any switching device. A high resistance is the reasonable compromise.
>
>magnetic field. Certain type of time delay relays exploit this principle
>by placing a shorted turn of heavy gauge wire around the relay coil.
>
> Harry C.
>
>
(which doesn't seem to have made it to the newsgroup yet;) he reminds me
that you need the complex conjugate of the source Z for max power transfer.
It is obvious to most that a low sink impedance prolongs the field.
It is not so obvious that a judiciously chosen C+R should sink the field
energy fastest.
Evidently, a diode+resistor is a non linear element which is not so easy to
handle analytically (but like so many others, can be handled as an iterated
model in a few lines of Basic code).
I fancy that if a diode must be used, a C+R sink will do the job best.
But there's a bright cybernickel just waiting for the person with a
plausible (intuitive) explanation of why an arc discharge dumps the mag field
energy slower than either of the two networks mentioned above.
Brian
Paul Jacobs
Hi Dan
I thought that most monorail trains worked on this principle, ie that of a
linear motor.
Good idea, me thinks. Just my pennies worth contribution although I
havent followed this discussion at all.
Regards
Paul
--
Paul Jacobs
pa...@wantree.com.au
Robert Macy
BW>From: in...@intellisys.net (brian whatcott)
BW>I got some serious objections to my post; the most cogent from an ozzie
BW>(which doesn't seem to have made it to the newsgroup yet;) he reminds me
BW>that you need the complex conjugate of the source Z for max power transfer.
Good point. True. forgot that. But sometimes it is difficult to do
that over an infinite spectrum.
BW> It is obvious to most that a low sink impedance prolongs the field.
BW> It is not so obvious that a judiciously chosen C+R should sink the field
BW>energy fastest.
No. The field will sink the fastest with a reverse "infinite" voltage
applied for very short duration. That dumps the field to zero quite
quickly, then remove that voltage.
BW> Evidently, a diode+resistor is a non linear element which is not so easy to
BW>handle analytically (but like so many others, can be handled as an iterated
BW>model in a few lines of Basic code).
Use PSpice demo version. It's free and you can add all the parasitic
elements you want (almost).
BW> I fancy that if a diode must be used, a C+R sink will do the job best.
When you pick the C and the R so that R = sqrt(L/C), you create a
critically damped circuit where there is a single overshoot. But there
is that one overshoot which in the projectile launching realm may
impede. Or not, I don't know anything about the induced fields in the
projectile, nor the kickback fields it generates.
BW> But there's a bright cybernickel just waiting for the person with a
BW>plausible (intuitive) explanation of why an arc discharge dumps the mag fiel
BW>energy slower than either of the two networks mentioned above.
It may be because the arc "fires" at 400+ Volts and sustains the arc
down to 60 (maybe only 90) volts, thus making an arc discharge
suppressor put a low voltage across the coil (which we all know makes
the field come down very slowly).
Does that answer warrant a bright cybernickel? <g>
- Robert -
* OLX 2.1 TD * Elmer Fudd Entepwizes. All wights wesewved.
Terry Harris
>BW> It is obvious to most that a low sink impedance prolongs the field.
>BW> It is not so obvious that a judiciously chosen C+R should sink the field
>BW>energy fastest.
>No. The field will sink the fastest with a reverse "infinite" voltage
>applied for very short duration. That dumps the field to zero quite
>quickly, then remove that voltage.
You open circuit the coil, there is instantly no current flow there is
instantly no field. A perfect coil will generate its own infinite
voltage which is just as impractical as the one you propose applying.
While we are talking impractical consider two identical perfect
capacitors (say 1uF) one discharged and the other with 100v across it.
Short them together with a perfect switch and you get two 1uF
capacitors charged to 50v.
The initial stored energy was 0.5 * 1 * 10^-6 * 100^2 = 5mJ.
The final stored energy is 2 * 0.5 * 1 * 10^-6 * 50^2 = 2.5mJ.
So where did the other 2.5mJ go?
Cheers Terry...
Roy McCammon
no hysterisis, mag field is a linear function of current, no capacitance
across the coil turns. Now, the field will be collapsed at the same moment
that the current goes to zero. To make that happen, let the back emf
go as possible for as long as it takes.
There is usually a practicle constraint as to how high the back emf is
allowed to go. For example, it may destroy the drive circuit. Or there
may be a disruptive arc. Anyway, once you determine the highest allowable
voltage, you want to sustain that voltage until the current goes to zero.
Seems like a zener diode (probably in series with a forward bised diode)
is called for. A resiter allows the peak voltage to reach the max, but
then the voltage decays exponentially, where as a zener would hold it
at the max allowed level.
Opinions expressed herein are my own and may not represent those of my employer.
Larry Deering
<*snip*>
> capacitors (say 1uF) one discharged and the other with 100v across it.
>
> Short them together with a perfect switch and you get two 1uF
> capacitors charged to 50v.
>
> The initial stored energy was 0.5 * 1 * 10^-6 * 100^2 = 5mJ.
>
> The final stored energy is 2 * 0.5 * 1 * 10^-6 * 50^2 = 2.5mJ.
>
> So where did the other 2.5mJ go?
>
> Cheers Terry...
This is almost as good as alien electronics! OK, Terry, I *know* that you
know what the answer is, but for the on-line headscratchers I have a
thought experiment. DO NOT TRY AT HOME! (or work for that matter)
Take one very large Capacitor, say a 33,000 ufd charged up to say 35
Volts. Calculate the energy, then short across the terminals with a big
screwdriver :( (if you must do this, wear eye protection, no sense
being blind as well as an idiot!). Now calculate the remaining energy,
which is zero.
Can anyone figure where all of that energy went?
Happy July 4th!
Larry
Tom Bruhns
: But there's a bright cybernickel just waiting for the person with a
: plausible (intuitive) explanation of why an arc discharge dumps the mag field
: energy slower than either of the two networks mentioned above.
Ooooh, an easy one. An R or RC can allow kilovolts of back EMF, which
discharges the inductor quickly; once an arc forms, the voltage drop is
in the 100 volt region (depending slightly on path length and more so on
what's in the plasma).
--
Cheers,
Tom
to...@lsid.hp.com
k...@together.net
;>
;>It turns out that the power transfer theorem covers this case adequately:
;>maximum power is transferred when the load and the source impedences are
;>The source is the coil, the load is the diode resistor combination.
Don't forget the reactive portion of this equation. I do believe that there
should be a capacitor in there somewhere too. Look at an automobile
ignition for a good example.
/----------------------------------------------------------
/ Keith R. Williams
/ k...@together.net
/ Burlington Vermont
asta...@vitgcab5.telecom.com.au
Can anyone suggest how I could organise for my PC to act as both a fax and
answering machine. Functionaly, it must possess the following attributes:
* Must be able to distinguish between types of incomming calls.
* Must treat as a fax if it is an incomming fax call.
* Must treat as answering machine if it is a voice call.
* Must send a usual message - eg. "Sorry I'm not in right now..." to
the caller and must be able to recieve and record what the caller says
(possibly as a *.wav file"
I know that this must be handled through the sound blaster somehow.
What concerns me is how to correctly integrate the SB with the (modem,
Telephone lines).
What's involved ?
Claude Frantz
>There is usually a practicle constraint as to how high the back emf is
>allowed to go. For example, it may destroy the drive circuit.
Have a look to ftp://bauv106.bauv.unibw-muenchen.de/pub/claude/p3896367/
where you can find a self-protecting circuit for driving inductive loads.
--
Claude
(cla...@bauv106.bauv.unibw-muenchen.de)
The opinions expressed above represent those of the writer
and not necessarily those of her employer.
Malcolm Blake
More interestingly from a theoretical viewpoint, consider two capacitors
(of any value) connected through a resistor and a switch, charged to
any initial (different) voltage. If the power loss is calculated (after
the switch is closed) as the integral of i^2.R between 0 and infinity
this results in the following equation:
Power loss = (0.5) * (C1*C2)/(C1 + C2) * (Vc1 - Vc2)^2
Which confirms the original result. Significantly the R has dropped out
of the equation. Unfortunately the mathematics do not prdict the level
of blindness or degree of burn liable to be suffered as a result of this
process.
Hope that helps.
---------------------------------------------------------
DR.M.O.Blake e-mail: cex...@coventry.ac.uk
School of Engineering tel: (0044) 01203 838808 (direct)
Coventry University fax: (0044) 01203 553007
Priory Street room: T209
COVENTRY CV1 5FB
U.K. Dept. tel:(0044) 01203 838366
----------------------------------------------------------
Winfield Hill
>Take one very large Capacitor, say a 33,000 ufd charged up to say 35
>Volts. Calculate the energy, then short across the terminals with a big
>screwdriver :( (if you must do this, wear eye protection, no sense
>being blind as well as an idiot!). Now calculate the remaining energy,
>which is zero.
>
>Can anyone figure where all of that energy went?
O.K., I'll take my 462,000uF 200 volt capacitor bank (220 caps at 75 cents
each from ELI on Hampshire Street). The bank is bolted up with 1/4 x 1/2
inch copper bars in a custom-made box. Measured esr is under 1 milli-ohm.
Inductance is in the nano-Henry territory.
OK, now I'll charge it up to 230 volts (over-voltage is allowed for a few
minutes). This takes several minutes.
OK, now I'll get in the next room, and send a signal to my bank of four 750A
SCR pucks (rated 20,000A peak each = 80,000A peak total) which are wired
across the 3/8-inch bolt outputs with four parallel size 2/0 cables. These
SCR gates take 150mA each to trigger, so I'll use 0.6 Amps for that.
OK, ready?
What happens next?
Hello? ANYbody?
--
Winfield Hill hi...@rowland.org
Rowland Institute
Cambridge, MA 02142
Terry Harris
>Terry Harris wrote:
><*snip*>
>> While we are talking impractical consider two identical perfect
>> capacitors (say 1uF) one discharged and the other with 100v across it.
>>
>> Short them together with a perfect switch and you get two 1uF
>> capacitors charged to 50v.
>>
>> The initial stored energy was 0.5 * 1 * 10^-6 * 100^2 = 5mJ.
>>
>> The final stored energy is 2 * 0.5 * 1 * 10^-6 * 50^2 = 2.5mJ.
>>
>> So where did the other 2.5mJ go?
>>
>> Cheers Terry...
>
>This is almost as good as alien electronics! OK, Terry, I *know* that you
>know what the answer is,
No if were are talking perfect components then I don't know what the
answer is.
If the switch isn't perfect but has some resistance then what ever
current it takes to waste 2.5mJ goes through the resistor, and it
doesn't matter what value resistor is used you still end up with 50v
on each capacitor.
Well it doesn't matter till the resistance goes down to zero, in that
case I guess the infinitely small inductance of the perfect switch and
the capacitors would ring at an infinite frequency for ever. Except
it would be radiating so the 2.5mJ would go as electromagnetic waves.
Or you could argue that you get infinite current through 0 ohms
flowing for zero time, the dissipated energy being infinity squared
times 0 squared which is around a sensible value
Or a combination of the two - beats me!
Cheers Terry...
Terry Harris
>O.K., I'll take my 462,000uF 200 volt capacitor bank (220 caps at 75 cents
>each from ELI on Hampshire Street). The bank is bolted up with 1/4 x 1/2
>inch copper bars in a custom-made box. Measured esr is under 1 milli-ohm.
>Inductance is in the nano-Henry territory.
>
>OK, now I'll charge it up to 230 volts (over-voltage is allowed for a few
>minutes). This takes several minutes.
>
>OK, now I'll get in the next room, and send a signal to my bank of four 750A
>SCR pucks (rated 20,000A peak each = 80,000A peak total) which are wired
>across the 3/8-inch bolt outputs with four parallel size 2/0 cables. These
>SCR gates take 150mA each to trigger, so I'll use 0.6 Amps for that.
>
>OK, ready?
>
>What happens next?
>
>Hello? ANYbody?
You will hear a bit of a thud and dissipate about as much energy as
held in a single alkaline AA cell - scary stuff.
Cheers Terry...
Bob Wilson
: Can anyone suggest how I could organise for my PC to act as both a fax and
: What's involved ?
Considering that you will have to have the PC running continuously, at
over 250 Watts power consumption (including monitor), and considering the
cost to repair it if it eventually dies of continuous running (MTBF of
the fan motor in the power supply isn't that good, for example, and the
MTBF of a Pentium chip cooling fan is even worse), doesn't it make rather
more economic sense to spend $400 or $500 and buy a proper Fax/Answering
machine? Bringing a $2000 (or more) PC to an early death to save a few
hundred bucks doesn't sound right somehow. In fact, it will cost you
something like $200 per year (at $0.10 per KW Hour) just to run the
thing.
Bob.
Wilson Lee
:
: Can anyone suggest how I could organise for my PC to act as both a fax and
: answering machine. Functionaly, it must possess the following attributes:
: * Must be able to distinguish between types of incomming calls.
: * Must treat as a fax if it is an incomming fax call.
: * Must treat as answering machine if it is a voice call.
: * Must send a usual message - eg. "Sorry I'm not in right now..." to
: the caller and must be able to recieve and record what the caller says
: (possibly as a *.wav file"
:
: I know that this must be handled through the sound blaster somehow.
: What concerns me is how to correctly integrate the SB with the (modem,
: Telephone lines).
:
: What's involved ?
Buy a fax/voice modem. i use MaxTech 14.4k. Forgot the model number but it
is a real steal. it does not even need a sound card. works fine for me with
the bundled software and i think it mees all your requirements above.
wilson
Tom Bruhns
: Lets assume an ideal inductor: inductance is not a function of current,
: no hysterisis, mag field is a linear function of current, no capacitance
: across the coil turns. Now, the field will be collapsed at the same moment
: that the current goes to zero. To make that happen, let the back emf
: go as possible for as long as it takes.
: There is usually a practicle constraint as to how high the back emf is
: allowed to go. For example, it may destroy the drive circuit. Or there
: may be a disruptive arc. Anyway, once you determine the highest allowable
: voltage, you want to sustain that voltage until the current goes to zero.
: Seems like a zener diode (probably in series with a forward bised diode)
: is called for. A resiter allows the peak voltage to reach the max, but
: then the voltage decays exponentially, where as a zener would hold it
: at the max allowed level.
Yes, a zener diode would do the trick, but it would dissipate the energy
that was in the magnetic field. You can just as well use a diode off to
a high-voltage rail and pump that energy back into a power supply so you
can use it for something else. If you are talking serious energy, this
could be a significant design issue.
--
Cheers,
Tom
to...@lsid.hp.com
Tom Bruhns
: In <4ki73m$d...@zoom2.telepath.com>, in...@intellisys.net (brian whatcott) writes:
: ;>
: ;>It turns out that the power transfer theorem covers this case adequately:
: ;>maximum power is transferred when the load and the source impedences are
: ;>equal. That's how a given amount of energy is dissipated fastest.
: ;>The source is the coil, the load is the diode resistor combination.
: Don't forget the reactive portion of this equation. I do believe that there
: should be a capacitor in there somewhere too. Look at an automobile
: ignition for a good example.
Since the coil is Rl+jwL, the complex conjugate would be Rl-jwL, or a
negative inductance. With a capacitor, the reactance goes down as the
frequency increases, but with an inductor, the reactance goes up. A
"negative inductance" isn't a component you can go out and buy. You can
synthesize one with amplifiers and feedback, but it's not worth the
effort! You can get a match at one frequency with a capacitor, but not
over a wide range. On the other hand, if you choose to match at a very
high frequency, the discharge will be very fast -- and at a very high
voltage. In practice, the max voltage you can hope to achieve is
limited by the resonance of the inductance with stray capacitance (in
the coil and in the drive circuit). And the maximum safe voltage, as
others have pointed out, is limited by the drive electronics and perhaps
by discharge paths that could break down in a spark. (After all, that's
how old-fashioned auto ignition systems work: build a magnetic field in
a coil and turn off the current rather suddenly; the back EMF breaks
down the gap in a spark plug.)
--
Cheers,
Tom
to...@lsid.hp.com
brian whatcott
says...
///
>BW> But there's a bright cybernickel just waiting for the person with a
>BW>plausible (intuitive) explanation of why an arc discharge dumps the mag
>BW>field energy slower than either of the two networks mentioned above.
>
>It may be because the arc "fires" at 400+ Volts and sustains the arc
>down to 60 (maybe only 90) volts, thus making an arc discharge
>suppressor put a low voltage across the coil (which we all know makes
>the field come down very slowly).
>
>Does that answer warrant a bright cybernickel? <g>
>
> - Robert -
close enough to join the other big winners of the week:
****Tom Bruhns and Harry Conover.******
But intuition has its limits....I KNOW that the stored energy of the field
HAS to be dissipated - arc, heat, motion, e.m radiation; whatever.
And it seems reasonable that an arc provides a rather low impedance path to
a (presumably) low impedance power source, hence holds the current longer
than necessary. Or if there's no arc, the ringing self oscillation of the
coil with its self-capacitance has to decay via the internal coil resistance.
I think I'm going to take up that sensible suggestion to look at the snubbers
in the SCR/Thyristor manuals before venturing further.... :)
Brian
dla...@mainelink.net
>Larry Deering <ldee...@haven.ios.com> said...
>>Take one very large Capacitor, say a 33,000 ufd charged up to say 35
>>Volts. Calculate the energy, then short across the terminals with a big
>>screwdriver :( (if you must do this, wear eye protection, no sense
>>being blind as well as an idiot!). Now calculate the remaining energy,
>>which is zero.
>>
>>Can anyone figure where all of that energy went?
>O.K., I'll take my 462,000uF 200 volt capacitor bank (220 caps at 75 cents
>each from ELI on Hampshire Street). The bank is bolted up with 1/4 x 1/2
>inch copper bars in a custom-made box. Measured esr is under 1 milli-ohm.
>Inductance is in the nano-Henry territory.
>OK, now I'll charge it up to 230 volts (over-voltage is allowed for a few
>minutes). This takes several minutes.
>OK, now I'll get in the next room, and send a signal to my bank of four 750A
>SCR pucks (rated 20,000A peak each = 80,000A peak total) which are wired
>across the 3/8-inch bolt outputs with four parallel size 2/0 cables. These
>SCR gates take 150mA each to trigger, so I'll use 0.6 Amps for that.
>OK, ready?
>What happens next?
>Hello? ANYbody?
>--
>Winfield Hill hi...@rowland.org
>Rowland Institute
>Cambridge, MA 02142
you place a call to the cambridge fire dept.
Bob Duckworth
>Roy McCammon (rbmcc...@mmm.com) wrote:
>: Lets assume an ideal inductor: inductance is not a function of current,
>that was in the magnetic field. You can just as well use a diode off to
>a high-voltage rail and pump that energy back into a power supply so you
>can use it for something else. If you are talking serious energy, this
>could be a significant design issue.
>
You want to pump down the rail(s).
I thought that was how these things worked?
-bob
--
Bob Duckworth Consulting, 960 Ralph McGill Blvd. Atlanta GA 30306-4447
bobs' address is r...@ka4ybr.netmha.com 404-888-0389(V) 892-2301(FAX)
Buy Sell Trade Surplus Computer Electronics Datacom Telecom since 1981.
Fax or email your list for a fast cash offer. Watch for listserv catalog.
Ron Wolenty
terry....@dial.pipex.com (Terry Harris) wrote:
> Larry Deering <ldee...@haven.ios.com> wrote:
>
> >Terry Harris wrote:
> ><*snip*>
> >> While we are talking impractical consider two identical perfect
> >> capacitors (say 1uF) one discharged and the other with 100v across it.
> >>
> >> Short them together with a perfect switch and you get two 1uF
> >> capacitors charged to 50v.
> >>
> >> The initial stored energy was 0.5 * 1 * 10^-6 * 100^2 = 5mJ.
> >>
> >> The final stored energy is 2 * 0.5 * 1 * 10^-6 * 50^2 = 2.5mJ.
> >>
> >> So where did the other 2.5mJ go?
> >>
> >> Cheers Terry...
> >
> >This is almost as good as alien electronics! OK, Terry, I *know* that you
> >know what the answer is,
>
> No if were are talking perfect components then I don't know what the
> answer is.
>
Even given perfect (lossless) components some energy has to be expendened
to "push" those electrons from one capacitor to another.
Ron
Mustafa Soysal MS57
brian whatcott <in...@intellisys.net> wrote:
>In article <4kbsk9$6...@news.hal.com>, el...@hal.COM says...
>>
>>In article <Dp9so...@news2.new-york.net>, fin...@where.com writes:
>>|> adu...@rhea.trl.OZ.AU (Allan Duncan) wrote:
>>|> just use a diode across the coils ...this would also protect any solid
>>|> state devices switching on the coils ..
>////
>> Allen is incorrect. Using a diode will not quickly
>>dump the current in the coil. The rate that the energy
>>is removed from the coil is related to the resistance that
>>the coil feeds the current into. The diode will look like a
>>low ohms resistance to the energy in the coil and have a long
>>time constant. The bigger the resistance the quicker the coil
>>will dump its current. Using a diode with a series resistor
>>will have a better effect. The value of the resistor should
>>be chosen to keep the voltage flyback of the coil below the
>>breakdown of the switching device.
>///
>
>Hmmm...Duncan advocates a diode
>and Elvey advocates a diode and as high a resistor value as will not pop the
>coil driver.
>
>It turns out that the power transfer theorem covers this case adequately:
>maximum power is transferred when the load and the source impedences are
>equal. That's how a given amount of energy is dissipated fastest.
>The source is the coil, the load is the diode resistor combination.
Yet, to calculate the coil resistance, you have to take into account the
magnetic material and its resistance to changes in magnetic field as
well. (which does change over time significantly....)
>
>Brian
>
Dwight Elvey
|>
Hi Brian
The need is for quick dumping not maximum power transfer. If
it was maximum power, one would match the resistance of the coil.
In this case one doesn't really care where the power goes but
it is desireable to stear it out of the coil because the coil
may be hard to cool. The quickest time constant is with the largest
value resistance. A infinite resistor would remove the energy
of the coil instantly. Of course the voltage across the coil
would be infinitely large also. With the large value of resistor,
most of the energy would be in the resistor and not in the coil.
Dwight
Dwight Elvey
|> In <4ki73m$d...@zoom2.telepath.com>, in...@intellisys.net (brian whatcott) writes:
|> ;>
|> ;>It turns out that the power transfer theorem covers this case adequately:
|> ;>maximum power is transferred when the load and the source impedences are
|> ;>equal. That's how a given amount of energy is dissipated fastest.
|> ;>The source is the coil, the load is the diode resistor combination.
|>
|> should be a capacitor in there somewhere too. Look at an automobile
|> ignition for a good example.
|>
|>
|> / Keith R. Williams
|> / k...@together.net
|> / Burlington Vermont
|>
Hi
The capacitor in a car's system is there to slow the initial
rise of the coils voltage so that the points can open far
enough to keep them from arcing. The capacitors size is
choosen to be a compimise that causes even ware on both
of the points contact. If the capacitor is too large on contact
will ware out first because of the charging current when
the points close ( if it is way to large, it will soak up
so much energy, that the coil won't develop the high voltage.)
If the value is too small the points will arc when opening
causing the other contact of the points to ware. In each case
the result will also reduce the amount of energy in the
spark. In one case the capacitor steals too much from the
coil and the other the arc in the point takes too much.
Dwight
ia...@gabhsann.demon.co.uk
> Use PSpice demo version. It's free and you can add all the parasitic
Is Pspice (or any spice type analysi prog) downloadable from anywhere?
Maybe I'm not looking in the right places but a search failed to turn
anything up.
--
Cheers,
Iain.
Mustafa Soysal MS57
Daniel P. Engel III <den...@hom.net> wrote:
>d...@pyramid.com (Don Abernathy) wrote:
>
>
>>Huh? Wouldn't you want to make the slug out of a ferrous metal?
>
>I don't know who asked the original question about a rail gun, but
>would it be possible to build one that pushed, rather than pulled the
>mass? I've never studied them, so maybe it's a dumb idea, but if a
>copper "ring" was used as the payload, then wouldn't a rising magnetic
Sure, at least it could be mounted on the payload and activated just
before the 'ignition'....Superconducting materials would be ideal for
that, yet you have to do more to give them mechanical stability to deal
with the forces and stay superconducting.
You still have to deal with "building" or reeasing the magnetic field
quickly to get a high d'phi'/dt.
>field induce a current that would cause an opposing magnetic field.
>Then, you could turn the coils ON as the payload passed, and not worry
>about collapsing the field--only about establishing it at the proper
a push-pull combination switched along the way of the payload with
pumping the energy into the next push-pull cascade would be also
interesting.
Jeffery C. Condit
Bob Ludlum
Adrian
ia...@gabhsann.demon.co.uk writes:
> Is Pspice (or any spice type analysi prog) downloadable from anywhere?
> Maybe I'm not looking in the right places but a search failed to turn
> anything up.
>
>
Try www.intusoft.com
David K. Bryant
>>
>>I don't know who asked the original question about a rail gun, but
>>would it be possible to build one that pushed, rather than pulled the
>>mass? I've never studied them, so maybe it's a dumb idea, but if a
>>copper "ring" was used as the payload, then wouldn't a rising magnetic
>Sure, at least it could be mounted on the payload and activated just
>before the 'ignition'....Superconducting materials would be ideal for
>that, yet you have to do more to give them mechanical stability to deal
>with the forces and stay superconducting.
I have a NASA Tech Brief that describes a superconducting rail gun.
I can send you a copy.
Russell Eberhardt
>
>In article: <96041415...@engineers.com> rober...@engineers.com
>
>> Use PSpice demo version. It's free and you can add all the parasitic
>
>
>
>Maybe I'm not looking in the right places but a search failed to turn
>anything up.
>
http://www.paranoia.com/~filipg/HTML/FAQ/BODY/F_Free_Spice.html
I recommend using spice32/Nutmeg32 if you're using Windows95.
Happy simulating!
--
Russell Eberhardt | email: r...@telecam.cityscape.co.uk
Managing Director | tel: +44 1223 578055
Telecam Electronics Limited | fax: +44 1223 578054
ia...@gabhsann.demon.co.uk
Thanks to all who replied. I've now got myself a copy.
Now all I've got to do is remember how to use the thing... :-)
--
Cheers,
Iain.
Jerrold Foutz
> > >> While we are talking impractical consider two identical perfect
> > >> capacitors (say 1uF) one discharged and the other with 100v across it.
> > >>
> > >> Short them together with a perfect switch and you get two 1uF
> > >> capacitors charged to 50v.
> > >>
> > >> The initial stored energy was 0.5 * 1 * 10^-6 * 100^2 = 5mJ.
> > >>
> > >> The final stored energy is 2 * 0.5 * 1 * 10^-6 * 50^2 = 2.5mJ.
> > >>
> > >> So where did the other 2.5mJ go?
The energy goes into heat, including contact arcing if a relay is used,
or into radiation (EMI), usually undesirable. You can easily see the
heat part by adding a small resistor in the path and calculating the
energy loss in the resistor. It is the same no matter how large or small
you make it and equals the energy lost from the capacitors.
Check my Web page in about two weeks for a full explanation
including references to some papers discussing the subject.
--
Jerrold Foutz
SMPS Technology
Web Personal Page: http://www.cyberg8t.com/smpstech/
Email: smps...@cyberg8t.com
Charles Wenzel
I don't buy the assumption that the final energy will be 2.5 mJ. If the
components are loss-free and radiation is somehow prevented then the energy
will bounce back and forth at a rate dependent on the physical dimensions of
the capacitors and interconnecting wires. The connecting wires become the
inductor in a tank circuit which rings "forever" (assuming ideal components).
The energy will remain 5.0 mJ. Obviously, in real life, electromagnetic energy
is radiated (RF, heat, light). It is an interesting exercise to prove that
exactly one half of the energy is lost under all real-world circumstances.
bryden Qiurk
In article <dbryantD...@netcom.com>, dbr...@netcom.com says...
Tom Faloon
The total charge remains constant, and is equally divided .
The initial charge is 100 uC, on a 1 uF capacitor. The energy is 5 mJ
Two 1 uF capacitors charged with 50 uC each, can, only contain 2.5mJ
of energy, so the additional 2.5 mJ MUST go.
If we assume ideal capacitors, zero resistance leads, and no
inductance, then we can not have oscillations. So where will the
energy go. There is nothing left but radiation. Or is there?
If we could prevent energy from being radiated, it would STILL have to
go somewhere, or dissobey the laws of physics, as we know them.
WHERE would it go?
There are many other examples of this lost energy .
e.g Imagine two identical reservoirs full of water, one full, one
empty. Now connect a pipe between them, at the base. Water will flow
until the level in each is equal. Once again, when the water is at
rest, the total potential energy is half of what it was to begin with.
I assume the water gets a little warmer!
Tom Faloon
Brian Denheyer
In article <83192808...@tfaloon.demon.co.uk>,
t...@tfaloon.demon.co.uk (Tom Faloon) writes:
>
>There are many other examples of this lost energy .
>e.g Imagine two identical reservoirs full of water, one full, one
>empty. Now connect a pipe between them, at the base. Water will flow
>until the level in each is equal. Once again, when the water is at
>rest, the total potential energy is half of what it was to begin with.
>I assume the water gets a little warmer!
>
Doesn't each reservoir have 1/2 the initial energy and therefore 2 x
1/2 is 1, which is what your started with...
--
Brian Denheyer
bri...@iccom.com
Roy McCammon
Think of the water in the top half of the initially full resevoir being
transfered to the bottom half of the empty reservoir. The potential energy is
clearly less.
Opinions expressed herein are my own and may not represent those of my employer.
Roy McCammon
t...@tfaloon.demon.co.uk (Tom Faloon) wrote:
>If we assume ideal capacitors, zero resistance leads, and no
>inductance, then we can not have oscillations. So where will the
>energy go. There is nothing left but radiation. Or is there?
>
>If we could prevent energy from being radiated, it would STILL have to
>go somewhere, or dissobey the laws of physics, as we know them.
>WHERE would it go?
If you could transfer the electrons from one cap to the other thru zero
resistence and zero inductance and you could keep them from radiating
and the capacitor plates were zero resistence, then the missing energy
would be found in the kinetic energy (KE) of the electrons.
Of course, if the plates of the capacitor were not infinite, then the paths
of the electons are going to have to deviate from a straight line. That means
acceleration which means radiation. Or, if the resistence of the plates
were not zero, then the KE would turn into heat.
And heat is where most of the missing energy is going to wind up in almost
any realizable experiment. The heat will either wind up in the conductor
between the caps, of in the cap that receives the charge carriers (probably
electrons).
Any conduction mechanism that transports the electrons from one cap to the
other will either have to allow the conversion of potential energy to
KE and thus deliver electrons with KE equal to the missing energy, or else
extract energy during transport in some form such as heat, or radition,
or even useful work.
Tom Faloon
bri...@iccom.com (Brian Denheyer) wrote:
>Doesn't each reservoir have 1/2 the initial energy and therefore 2 x
>1/2 is 1, which is what your started with...
No Brian,
Each reservoir finishes with a quarter of the initial energy, just
like the capacitors.
Let's say they each has a rectangular floor, and vertical walls, and
their floors are at the same level.
When all the water is in one, the potential energy (relative to the
floor) is given by mgh (mass X gravity X perpindicular height to
centre of mass)
When the water is equally divided between the two reservoirs, then
both the mass and the height to the centre of g is halved, so each
contains only a quarter of the initial energy.
Tom
>--
>Brian Denheyer
>bri...@iccom.com
Tom Faloon
rbmcc...@mmm.com (Roy McCammon) wrote:
>If you could transfer the electrons from one cap to the other thru zero
>resistence and zero inductance and you could keep them from radiating
>and the capacitor plates were zero resistence, then the missing energy
>would be found in the kinetic energy (KE) of the electrons.
>Of course, if the plates of the capacitor were not infinite, then the paths
>of the electons are going to have to deviate from a straight line. That means
>acceleration which means radiation. Or, if the resistence of the plates
>were not zero, then the KE would turn into heat.
>And heat is where most of the missing energy is going to wind up in almost
>any realizable experiment. The heat will either wind up in the conductor
>between the caps, of in the cap that receives the charge carriers (probably
>electrons).
>Any conduction mechanism that transports the electrons from one cap to the
>other will either have to allow the conversion of potential energy to
>KE and thus deliver electrons with KE equal to the missing energy, or else
>extract energy during transport in some form such as heat, or radition,
>or even useful work.
>Opinions expressed herein are my own and may not represent those of my employer.
Well put Roy.
But if we could achieve the (purely hypothetical) situation where
electrons have KE equal to the missing energy, and nothing was lost by
radiation ot heat, then the capacitors will contain twice as much
energy as expected. Is E=1/2 C V squared no, longer true or could we
measure a higher V than we expect.
I know this is all getting silly, and I don't expect a reply. On the
other hand if ....
Best wishes,
Tom Faloon
Roy McCammon
t...@tfaloon.demon.co.uk (Tom Faloon) wrote:
>
>But if we could achieve the (purely hypothetical) situation where
>electrons have KE equal to the missing energy, and nothing was lost by
>radiation ot heat, then the capacitors will contain twice as much
>energy as expected. Is E=1/2 C V squared no, longer true or could we
>measure a higher V than we expect.
>
>I know this is all getting silly, and I don't expect a reply. On the
>other hand if ....
>
>Best wishes,
At least this news group will let you be silly without setting you on fire.
Sometimes, when we get silly, we learn something: usually that one of
our assupmtions was incorrect.
It is a principal of logic, that if you allow a contradiction in your set
of assumptions that you can then prove just about anything, so we find ourselves
asking "if the impossible were possible ...". So, suppose we had these
impossible capacitors ...
Well lets think about two identicle resovours of water, one full and the other
empty. Usually if I drain one into the other by gravity, the potential energy
after draining is less than before, but if there were no viscosity, and no
other friction, etc, then we would find that the water would have kinetic energy
in addition to potential energy. The water would swirl forever and more energy
would be available than the formula for potential energy would imply. And in
fact, we could extract some of that kinetic energy in addition to the potential
energy.
Back to capaciters. Yes, if you could keep those electrons in motion, the
capacitor would have more energy than it should based on the voltage. I think
that we will have to accept that the formula C*V*V/2 is the formula for
a portion of the energy of a cap.
The voltage of a cap, is not fundamentally a measure of the energy in the cap
but rather a meausre of how much energy it takes to transport an electron
from ground to the capacitor plate ( E = Vq). Lets take two identicle caps with
identicle charges. On one the electrons are at rest, on the other,
the electrons pair off and travel in small circles. One member of the pair
travels clockwise and the other travels counter clockwise. Remember, we assumed
these guys don't radiate. Since the electrons in each pair are going opposite
directions, there is no magnetic field. The second cap has a lot more total
energy, due to the motion of the electrons, but the voltage on both caps is the
same. The electrons on each cap repell an incoming electron with the same force
whether they are at rest or moving (the circles are realy small).
So, C*V*V/2 doesn't measure the total energy in a cap. No surprise, after all
a cap also has chemical energy (we could burn it), nuclear energy ( mc^2),
gravational potential energy (it has weight, we can drop it to the floor),
thermal eregy (its not at absolute zero). It also might (for a breif time)
hold some energy as movement in its charge carriers.
brian whatcott
In article <4no5om$r...@dawn.mmm.com>, rbmcc...@mmm.com says...
>
///
> It also might (for a breif time)
>hold some energy as movement in its charge carriers.
>
It turns out that for any reasonable electric field gradient,
the electrons in a conductor move (drift) rather slowly...<1 mm/sec
Brian
Don Young
See other post with same title SCOPE PATTERNS and more details.
Your help is appreciated. Thanks.......
I did not post the initial article posing the question on collapsing a
magnetic field but have an application where I want to do just that.
After reading (and printing and rereading) the 30 + responses I still
do not know what the quickest way is to collapse a magnetic field. My
application is much simpler than the rail gun proposed in the initial
article. It involves only one coil, 200 micro Hy inductance, 0.3 ohm ,
attached to the collector of a power transistor and to the positive
(+12v) rail of the circuit. It is pulsed by a 555 timer, 11 times a
second, positive (on) duration is about 300 micro seconds. Info
I have says " At switch-off the magnetic energy of the transmitter
loop is transferred to and stored in a capicitor. This gives a
switch-off time of approximately 50 micro seconds." Being a novice I
do not understand how to do this nor do I know if there is a faster
way to collapse the field in this coil. Can one of you, most talking
over my head, tell me how this is done or something better.
I appreciate any help I can get. Suggestions welcome.
Thanks, Don Young , Dallas, Texas
E-mail dcyoung@airmail net or post to this news group
Don Young
I want to collapse the fieldin a coil of 200uh. It is connected from
collector of 2N3055 to positive 12V rail of the circuit.
See other post with same topic for more details.
Any help is greatly appreciated....... Thanks.......
When a square pulse is itroduced into base of 2N3055 coil the results
on my scope looks like the coil pulse on the left below. What I want
is something that looks like the pulse on the right. TO CHOP OFF the
edge of the PRIMARY NEGATIVE spike as soon as possible by
collapsing the field.. I don't care about the positive spike. It is
OKAY the way it is. The reverse of the spikes are induced in a
secondary coil and I want the energy of the negative spike from the
primary coil to go away as soon as possible after it induces the
positive spike in the secondary coil without any interference with the
secondary coil.
Sure hope this diagram makes it without getting scrambled.
What I HAVE
-------- 12V Same square wave in
square l l The results I
wave _l l___l WANT is this
. ||
coil A . A V
pulse l l . l l V
/ \ . / \
/ \ . / \
_/ \ . / \ l
\ . / \ l
\ . / \ l
\ / :-((( :-)) \ l
l l A l lA
V | V |
NOT this This
Thanks, Don......Dallas, Texas
dcy...@airmail.net
Bill Sloman
In article <4t3k2r$n...@library.airnews.net>, dcy...@airmail.net (Don Young) says:
>
>
>See other post with same title SCOPE PATTERNS and more details.
> Your help is appreciated. Thanks.......
>
>attached to the collector of a power transistor and to the positive
>(+12v) rail of the circuit. It is pulsed by a 555 timer, 11 times a
>second, positive (on) duration is about 300 micro seconds. Info
>I have says " At switch-off the magnetic energy of the transmitter
>loop is transferred to and stored in a capicitor. This gives a
>switch-off time of approximately 50 micro seconds."
>
useful to concentrate on the current through the inductor.
When your power transistor turns on, the current through your coil ramps
up at about 60,000amps/sec (12V divided by 200uH). After 300usec this
would be 18A, if the power transistor doesn't run out of base drive, and
neglecting resistive losses in the circuit - in fact 18A over your 0.3ohm
is only 5.4V. In fact, what you should see would be an exponential
approach to 40A (12V divided by 0.3 ohm), which gives 14.49A after 300usec.
At this point, the inductor is storing an energy of 0.021 joules
(E=0.5 L I**2). You now turn the power transistor off. The 14.5A current
continues to circulate through the coil, but now it charges up the
capacitances within the circuit.
These invariably include the interwinding capacitance of the coil, which
can run from 1pF (for a single layer coil) to nFs (for a normal multilayer
coil) and the collector capacitance of your transistor (of the order of
a nanofarad?).
If this roughly one nanofarad of capacitance was the only destination
available for the energy stored in the coil, it would charge up to +6.5kV
(E= 0.5 C V**2) in 0.7usec (f= 1/(2 pi root L C), then ring from +6.5kV
to -6.5kV at 360kHz for a couple of milliseconds.
In reality, your transistor breaks down sometime before the voltage hits
6.5kV, which is why real circuits have someplace else to dump the current.
This can be a bigger capacitance; if you want to limit the initial swing
+50V or less, you need 30uF or more, and the voltage will peak after
about 120usec, or more. If you put a (15A+) diode in series with the
capacitor, so that the coil can only charge it, you have killed your
magnetic field in 120usec. You then need an 470R 0.5W resistor in parallel
with your capacitor to discharge it in the 90msec you've got between pulses.
You can replace the capacitor and resistor with a 38V zener diode returned
to the +12V rail (still through the diode). The current through the coil
then decays at 190,000A/sec, so your 14.5A decays to zero in 76usec -
somewhat faster. You need a tough zener/transient suppressor diode such
as the Philips BZW03C27 to survive the peak current.
If you want to kill the magnetic field faster, you have to be able to
sustain a higher voltage across coil; swap your drive transistor for one
that will take more than 50V, and start paralleling your BZW03C?? with
a resistor in series with each one to distribute the current more or
less evenly over the bunch, or look for something really big from
Motorola or International Rectifier.
I wonder what the answer to the actual problem is?
Bill Sloman
Earl Kiosterud
In article <4t3k2r$n...@library.airnews.net>,
dcy...@airmail.net (Don Young) wrote:
> (how to make a coil's field collapse quickly)
Don,
Short of running current in the reverse direction for a period of time, the
fastest collapse will be to let it collapse. This means no clamping diode
across the coil. This also means a very high voltage spike as the field
collapses. You might want to consider a high-voltage silicon diode in series
with the coil, maybe more that one diode. You can get silicon rectifier
diodes in the kilovolts range. Don't know if they're fast-switching enough
for your application, but probably are.
Earl K. Virginia Beach, VA ea...@livenet.net