HELP With Step-Down Transformer and Converter

I have two Questions:

  1. Can you please help me design a Step-Down Transformer AC-AC Input : Electrical Outlet 15A/220 Volts Output: 3000 Amps/1.10 Volts

  1. Can you please help me design a Step-Down Converter DC-DC Input : Car BAttery 75A/12 Volts Output: 750 Amps/1.20 Volts

If there is an existing equiptment that approximate the above specs, I would rather buy than build.

Thank you very much in advance.

Gene

Reply to
gene
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I forgot to mention, my goal is to have an output voltage in the range of 0.90 to 1.5 Volts and the maximum possible current output.

Thanks again.

Gene

Reply to
gene

I read in sci.electronics.design that gene wrote (in ) about 'HELP With Step-Down Transformer and Converter', on Fri, 16 Sep 2005:

Both of your requirements are impossible to satisfy, because they require 100% efficiency, and that cannot be achieved for anything except a heater.

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
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Reply to
John Woodgate

Anything close to what you ask for will be a custom made device. You can make your own by passing a big copper pipe through a 20 amp Variac that you have removed the wiper structure from.

I'm talking about something like this:

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These come up on Ebay quite often.

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Reply to
John Popelish

Dear John,

My immediate need is to use a DC Car Battery as my main source and step-down the voltage and Step-up the Current. Voltage should be between 0.90 to 1.50 volts and Current should be as high as possible.

I know I can't have 100% efficiency but I dont mind. I do not know how to set it up properly and what materials to use. I also don't know what to use as an interruptor to induce a lower voltage/high current.

Could you help?

Gene

Reply to
gene

Dear John,

Thanks for the reply. Actually, 100% efficiency is not required. Even 50% to 80% efficiency would suffice.

I need some advice on how to design a step down DC-DC converter using a car battery as source, with an output of 0.90 to 1.50 Volts and current output of as high as I can get.

I know I can hook-up two copper coils but I have have no idea as to the size of the coils and what interruptor to use to induce high current on the 2nd coil.

Can you help me? If a similar device already exist that uses a car battery (DC) as source, I would rather buy. If not, then I have to build one.

Gene

Reply to
gene

Dear Phil,

Thanks for the tip on the Spot Welding... I will try it. I tried using the coils from a soldering gun but I don't seem to generate a high current... just heat.

I still need the DC-DC converter. I know I can't have 100% efficiency ... 50% to 80% is ok. I tried making a primary coil (lots of windings and finer wire) and a secondary coil (fewer windings and thick copper wire) but I can't seem to induce a high current/low voltage on the secondary wire. I know I'm missing something ... like an interruptor perhaps?

Can you help me with the DC-DC Step down converter.

Thanks.

Gene

Reply to
gene

"gene"

** That is an electrical device similar to a spot welding transformer.

You will have to reveal the application for any sensible advice.

** If AC output is OK then you do not need DC in this case.

You will have to reveal the application for any sensible advice.

........... Phil

Reply to
Phil Allison

As Phil said, if you will tell us why you want to do this, it would help us give advice. What are you trying to do with 3000 amps?

Reply to
The Phantom

Dear Phil & Phantom,

There is no particular application. I'm trying to test Faraday's Induction Laws. I want to built a similar device but also test the limit of how much maximum current can be generated in a DC-DC Step-Down Converter.

Thanks.

Gene

Reply to
gene

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 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.

  1. 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?

  2. 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 :o) What am I doing wrong?

I appreciate your help.

Gene

The Phantom wrote:

Reply to
gene

OK, Gene. I don't understand why you have to test at such high currents, but here goes. The 60 Hz transformer will be much easier. You will need a transformer capable of handling about 3000 watts, with a bobbin. You will need a square stack of what are known as EI250 laminations. You will need to wind *about* 200 turns of wire of appropriate gauge (fill about half the bobbin) for a primary. Then the secondary will need to consist of copper conductor of about 1 square inch cross section (I calculate that you will need to use about

10 lengths of 2/0 cable in parallel). Thread the 10 pieces of 2/0 wire in parallel through the space remaining on the bobbin (that's one turn). This secondary will be able to handle about 3000 amps continuously (I say continuously with tongue in cheek. You could be dissipating upwards of 3000 watts. Things are going to get hot quickly). You will only have about a volt, so you can't get the 3000 amps through much of anything other than a dead short of the secondary. The leakage inductance of such a transformer may prevent getting the full 3000 amps, but it will give you a start.

You must put a current limiting resistor in series with the primary if you are going to run with a shorted secondary; perhaps a big light bulb. Put a 500 watt incandescent bulb in series with the primary; if that doesn't give you enough current (put an ammeter in series with the primary), put another bulb in parallel with the first, adding more as needed. Be aware that finding 240 volt 500 watt bulbs may be difficult. You could perhaps use 300 watt, 120 volt bulbs, which are fairly common at regular stores, with two of them in series to stand off 240 volts. Put multiple two-bulb combinations in parallel to get the needed current in the 240 volt primary.

Try not to hurt yourself, or burn down the house!

Reply to
The Phantom

Keep in mind that large enough DC currents will generate a force due to the earth"s magnetic field. If you hook up two car batteries with jumper cables, and go +>- and ->+ the jumper cables will go flying through the air.

Tam

Reply to
Tam/WB2TT

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.

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:

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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:

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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).

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:

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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.

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.

Reply to
The Phantom

dubious . They may fly apart, dont thinks its the earth that is causing it

martin

Reply to
martin griffith

Take it from someone who tried to build a battery tab spot welder. It's a LOT harder than you think.

Do the math. If you have 1V and 1 milliohm of resistance, you ain't gonna get more than 1000 amps no matter what you do.

Now, think about how much resistance is in each connection to the secondary and your load. It gets past a milliohm instantly. You can't use wire and interconnect schemes you're used to. Big wire ain't necessarily the answer, 'cause you can only connect to the surface.

Save yourself a lot of grief and go buy a harbor freight 220V spot welder. They go on sale frequently for $100-$150.

and this:

I want to built a similar device but also test the limit of how much maximum current can be generated in a DC-DC Step-Down Converter.

is nonsense.

There's no way you're gonna test the limits of ANYTHING with anything you can build in your garage and with the experience you appear to have. You will, however test YOUR limits. That can be a fun project in itself.

The sensible thing is to find some other way to waste your money. Paypal to me would be an excellent choice ;-)

mike

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FS 512MB 45X SD Flash memory.
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Reply to
mike

Dear Phantom;

I can't thank you enough for your help. Don't worry, I will be careful. I will stand in another room holding the switch, wear goggles, leather gloves, and all the necessary precautions. I know it is dangerous to handle such high voltage. I have my camcorder setup to record evrything that happens so I really don't need to be close.

Your assumption is correct. I only want to get a high current in a wire for its magnetic effects, and not to drive a load, and YES, I would not mind spending some money on it.

In the Dc-Dc setup using gel cells, do I need to have an interruptor? I understand that if I use AC, I do not need an interruptor since AC is already turning ON and OFF by itself, but in a DC setup, shouldn't I have an interruptor to induce a high current on the secondary?

To Answer your question on the soldering gun: How do you know you didn't get high current? How did you try to measure it?

If you got heat, then you got high current.

sensitive ampmeter you suggested. I was just wondering, if the soldering gun produces a high current at it's tip, would it not short circuit the CHIP that one is soldering or would it not electricute the user holding the lead?

What is a rare earth magnet? Where can I buy one?

Lastly, Mike suggested that I should just buy a portable harbor freight

220V spot welder... Does this portable welder use DC as source? Is MIke correct in saying that I would achieve and be able to measure the same high current induction?

My deepest gratitude to your efforts in helping me.

Gene

Reply to
gene

Actually, it's high *current*, not high voltage you're dealing with in the DC setup, and in the AC setup the high voltage (240 volts) is only on the primary side of the transformer. But really high currents pose a danger of a different sort. You won't get electrocuted or even shocked by the high current circuits you propose to create. The danger is molten metal splattering around.

Your are th "You must have some disconnect device in series with the battery (near one terminal of the battery), such as a DC rated circuit breaker:

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or fuse."

That's a type of "interrupter", but it's not to get "transformer action". It's more commonly called a "disconnect". The idea is that if anything goes wrong while the experiment is running and some of the wiring becomes red hot (this has happened to me as a teenager. We all learn some of these lessons the hard way), when you panic and try to undo the bolts connecting the wiring to the battery, you discover that red hot wires burn you. You need a disconnect so that all you have to do is flip the switch and everything is turned off. The disconnect should be the first thing in the circuit connected to one of the battery terminals where no other wire can bypass around it. What I am recommending is that you use the big DC rated circuit breakers as switches to turn the current on and off, and since they are breakers too, they will automatically disconnect if the current gets too high. If you chose to use a fuse for the current limiting function, you must still have a switch there for the safety disconnect. That's the advantage of breakers; they are switches too. (Such high current fuses are expensive, by the way.)

In your high current experiments the danger of molten metal comes from making connections when the circuit is energized. When you touch two pieces of metal which will carry 3000 (or even 750) amps when the connection is made, the sparking at the point of contact can splatter molten metal. You should turn off the breaker or switch while making the connections to the ammeter and whatever else is in the high current circuit. This is just common safe practice when working with electricity. Don't work on energized circuits. Hook everything up the way you want it, *then* turn on the switch or breaker. If you made a mistake and too much current passes, the breaker will open and avoid serious damage. Then you have to troubleshoot your setup and figure out what you did wrong.

For your AC setup, the house breaker in the 240 volt circuit will provide the safety disconnect in the event of a short circuit. If you build the transformer I described, you should mount it in an appropriate electrical enclosure (box) with a switch on the primary (240 volt) side so you can turn it on and off while you're working on the high current side without having to run to the house breaker box. You don't want to have to provide a switch on the 3000 amp secondary side; it would be very big and expensive and it's not necessary. Just turn off the 240 volts to the primary of the transformer when you're working on the high current side. The breakers in the house service entrance are often not designed to be used as switches for many on-off cycles anyway, so the extra switch takes the wear and tear off the house breaker.

The DC setup I described doesn't need an interruptor. It doesn't have a transformer and so there isn't a secondary; there's just the battery. The lead acid cells can put out a very large current on their own. Have you ever shorted a car battery accidentally when connecting jumper cables? When you start your car, the battery is supplying hundreds of amps to the starter motor. Lead acid batteries are dangerous because of the high currents they can supply. Most modern cars have a plastic sheath that covers the terminal(s) of the battery, because if you are working on your car and accidentally drop a wrench across the terminals of the battery without that sheath, you get a most impressive fireworks display!

ampmeter you suggested.

The high current is only in the tip; it doesn't get into the thing you're soldering.

electricute the user holding the lead?

The soldering gun has essentially the circuit I described for your AC setup. It's just a transformer with a single shorted turn of very heavy conductor for a secondary. If you remove the actual tip that does the soldering and measure the voltage at the bolt-on connections, it's probably quite low. My own Weller soldering gun has .35 volts open circuit at the bolt-ons, and .24 volts when the tip is in place. That voltage isn't enough to hurt anything. The current in the tip of a 100 watt gun must therefore be about

100/.24, or about 400 amps. But the reason that can't hurt you is that the 400 amps can only flow in the copper tip, not in your body nor in the thing you're soldering. It takes much more *voltage* to cause that much current in a human body (thousands or volts, although it doesn't take 400 amps to electrocute a person so a much lower voltage than thousands can electrocute, but .35 volts is completely harmless through skin). Only a virtual short circuit, such as provided by the heavy copper tip on the soldering gun can cause that much current when only .24 volts is available to "push" the current.

A rare earth magnet is a magnet made from mixture of substances which include what are called the "rare earths". The elements samarium and neodymium are the two most common rare earth elements used in modern magnets. These magnets are *very* strong, and the big ones can hurt you by pinching you really hard when a couple of them smack together on your fingers. You can buy them at:

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Get a couple at least of number 0013 and several number 0016 and whatever else strikes your fancy. They are fun to play with! Don't let them get close to your watch, your TV or your computer monitor or any floppy disks; and especially don't let them get close to your credit cards. They will erase the magnetic strip on the card; don't even hold a magnet in your hand and drop your hands down to your waist near a pocket with your wallet.

The designation 220V means that it gets its power from the 220 volt AC coming into your house. I wouldn't know what it does after that without seeing a spec sheet, but Murphy's law says that it won't do what you want.

Reply to
The Phantom

STOP RIGHT NOW!!!!!!!!!! Stop before you burn your house down, and probably kill someone.

Go directly to the public library, do not pass go, do not collect $200.00 and learn the difference between Voltage and Current.

THEN, go read some basic electronics tutorials and lurk in news:sci.electronics.basics for awhile.

You have no business going anywhere near anything that uses mains power.

Sorry, no insult intended, but just from what you're posting I can tell that you have no idea what your doing, ergo, it will be dangerous, and very probably lethal.

Good Luck, Rich

Reply to
Rich Grise

Dear Phantom, I have broken-down what I need as follows: A. 2 Volt Lead Acid Gel Cells, RAted 25 ampere-hours (6 pcs) B. 12 Volt Cylindrical Battery (1 pc), looks like a big AAA Batery right? C. Ammeter, DC responding (1 pc) D. Switch for the safety disconnect E. 5000 amp shunt (1 pc) F. millivoltmeter F. Circuit Breaker 12 Volt DC-Rated, 750 Amp (1 pc) G. 12 Volt Charger (AC to DC)

Series Connection When Charging: AC Outlet >>>>> Charger >>>>> +Battery #1- >>>>> +Battery#2+

Parallel Connection when Generating High Current _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ | | | | | | | Battery#1 Battery#2 Battery#3.....Battery#6 Cir.Breaker====| Ammeter |_ _ _ _ _ |_ _ _ _ _|_ _ _ _ _ _ _ _|_ _ _ __| |_ _ _|

  1. Where do I put the Switch, Shunt and Millivoltmeter in the diagram above?
  2. Do you mean that I should dissect the 12 Volt Cylindrical battery and use the wires inside it as connectors to the Lead Acid Batteries?
  3. Although I can get very high current in the above setup, my voltage would exceed the 0.90v to 1.50v range that I wanted, isn't it? Basically, high current is generated by shorting the battery, is this right?
  4. If I decide to have a high DC Current and still maintain the DC Volts at 0.90v to 1.50v range (to test faraday's induction law), I can still use the design for the primary and secondary coil you provided, right? Then I would need an interruptor to induce high current, right? What should I use for an interruptor or where can I buy one? Is it the same as pulsed DC?

I tried using a 12Volt Cylindrical BAttery some time ago and had a smaller model with a thin Primary and a thick Secondary but I was not able to induce a higher current in the secondary. My research tells me that what I need is an interruptor that will correctly turn the primary circuit ON/OFF at precise time. I used an ordinary ON/OFF switch as my interruptor but nothing happens.

Gene

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Reply to
gene

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