Dear Phantom,
Thank you very much for your help. I promise not to hurt myself. I
will be very careful and keep my distance from the Converter when I set
it on. I plan to record it on film using a camcorder.
I'm beginning to think that what you want is a large current in
cables so that you can measure the magnetic effects of large currents.
If this is so, you should say so because if that's all you want, as
opposed to trying to drive a load with these large currents, it makes
a difference in how I would implement your needs.
I'm also going to assume you don't mind spending some money for what
you're doing, so some of things you need will cost a bit.
I have 3 questions:
1. Can I put an Ammeter or a CLAMP-METER in series connection to
measure the current in the secondary? What king of meter should I use?
I wanted to include in the film the current readings.
A clamp-on ammeter would do the job, provided you can get all 10 of
the 2/0 cables from the transformer through the jaws of the clamp-on.
I see clamp-ons up to 2000 amps with a cursory search of the web:
http://www.stbinc.com/products/fault/ammeter.html
If you can't get all the conductors through the clamp, then you will
need to do what I describe next.
For currents of 3000 amps, you could use a large shunt; for example:
http://www.hotektech.com/TinsleyShunts.htm
http://www.hotektech.com/TinsleyShunt4638.htm
or you could probably find some at surplus outlets. The shunt is
placed in series with the cable carrying the 3000 amps; then a meter
with a 50 millivolt full scale range is connected to the meter
terminals on the shunt.
This arrangement can measure both AC and DC. For AC the
millivoltmeter must respond to AC. You could use an RMS responding
hand-held DVM if, say, 1% accuracy is acceptable. Normally for best
accuracy the shunt and its companion millivoltmeter are calibrated
together.
The typical shunt is manufactured so that when it carries full rated
current, 50 millivolts appears across the meter terminals. Thus your
50 millivolt meter will read full scale when the shunt carries full
rated current. If your meter has a printed scale such that 3000 amps
(or whatever would match the shunt) is full scale, then you can
directly read lower currents. (You might have to get a 4000 amp, or
5000 amp shunt if 3000 amp shunts are not readily available; you just
scale your reading if the millivoltmeter isn't calibrated to the same
maximum value as the shunt).
2. Does the the design you gave me also applies for a DC-DC Step Down
Converter using a Car Battery? If I am not mistaken, don't I need an
interruptor or something or a device that will pulse the DC source to
induce current in the secondary? If I do need an interruptor, can you
recommend one?
The design already given is for AC only. But now that I have come
to believe that you only want to get a high current in a wire for its
magnetic effects, and not to drive a load, I think another approach
would be best for the DC case. And if I am right about your intent,
please let us know.
Assuming I am right about your intent, I think you should proceed as
follows. You should get some 2 volt lead acid cells. such as the gel
cells made by Gates. The thing to do is to get a 12 volt battery of
the cylindrical cells (they can often be found surplus), because you
can remove the connecting straps and rewire the 6 cells (which make up
a 12 volt battery) all in parallel. Use heavy copper conductors for
the rewiring. Copper tubing can be flattened with a hammer and holes
drilled to make the connections. The cells I am thinking of are about
8 inches tall and perhaps 2.5 inches in diameter, and I think they are
rated at about 25 ampere-hours. They can easily supply 125 amps for a
few minutes.
There are several advantages to doing things with such an
arrangement of lead-acid cells instead of a DC-DC converter. You can
wire the 6 cells in series and charge them with a standard 12 volt
charger; then rewire them in parallel for your experiment. This will
be much less expensive than trying to re-configure a DC-DC converter
for such low voltage and high current.
You can use the same shunt and millivoltmeter arrangement I
described above for DC. Just use a 1000 amp shunt (and DC responding
millivoltmeter), and that should cover the 750 amps you want for DC.
And when working with such an arrangement of lead-acid batteries,
you must pay attention to safety. The short circuit current from the
batteries can easily splatter molten copper around, getting in your
eyes and melting holes in your clothes and skin (the same effects can
occur with your AC system also). You must have some disconnect device
in series with the battery (near one terminal of the battery), such as
a DC rated circuit breaker:
http://www.solarseller.com/dc_circuit_breakers_12_volts_to_125_volts.htm
or fuse. If you can't find a circuit breaker rated for 750 amps (such
a breaker will probably cost more than everything else in this
arrangement), you could parallel 3 250 amp DC rated breakers. The
downside of doing this is that the breakers may not share current
equally, and thus the combination might break at a current less than
750 amps. But paralleling like this can't cause the combination to
break at *more* than 750 amps, so the safety aspect is not
compromised. The 2 volt battery could probably supply 2000 amps or
more on dead short, melting copper and lead, but even the 750 amps
which the breakers will pass can do that, so be careful.
In such an arrangement you should mechanically gang the levers of the
breakers together. These sorts of breakers may not be designed to be
switches as well as breakers, but if you don't turn the current on and
off too many times (which I'm guessing you won't do), they will serve
your needs.
Finally, you need some means to control the magnitude of the
current. I would do this by adjusting the length and cross sectional
area of the wires with which you will short circuit the battery (of 2
volt cells in parallel). The short is applied *after* the breaker, of
course. Look up the resistance of standard gauges of copper wire and
use Ohm's law to calculate how long a piece is needed to give your
desired current. For example, 4 gauge wire has a cold resistance of
about 250 microohms per foot. To get 1000 amps from your 2 volt
battery, you would need 2 milliohms of wire resistance. This would be
8 feet of 4 gauge wire. 750 amps would need somewhat more 4 gauge
wire. The resistance of the circuit breakers will need to be
considered; it will probably be high enough that the wire length will
need to be substantially shorter than the theoretical. Start with the
theoretical wire length and shorten it until your ammeter indicates
the desired current. I'm sure you can figure it out.
Be prepared for things to get hot, and of course such large current
draws are hard on the gel cells, so only do it for a few minutes at a
time.
Standard disclaimer: I assmume no responsibility for any injuries
or fires, etc., etc.
3. I tried using the transformer in a soldering gun. I did not get any
low voltage/high current output in the secondary coil... I only got
heat
) What am I doing wrong?
How do you know you didn't get high current? If you got heat, then
you got high current. How did you try to measure it? You can easily
tell when a large AC current is present in a conductor by holding a
rare earth magnet near it. A large (60 Hz) AC current will make the
magnet vibrate like crazy. If you succeed in getting near 3000 amps
AC in a conductor, you will find that the vibrations in a neo magnet
of maybe an inch on a side will be so intense that it will make your
fingers numb if you hold the magnet right up by the conductor.
