200 Amp Load

There must be something in the air. This is the fifth or sixth GA I have voted today.

But it is easy to do so when a response is spot on. Well done.

Thanks! I am just doing my part to combat the ever increasing forces of ignorance and unnecessary/pointless over engineering that tends to go with it.

Right, rheostat bank with cooling arrangement do the job,

Inductive?? Are the cables and wires formed into current-carrying coils?

While tcmtech's reply was spot on, in that you don't need much more than some sort of resistance for the load, be aware that 200A @ 24V is not far off 5kW, so there'll be lots of heat.

Way back, I built a load for testing a PSU in that power range using elements from radiant electric heaters. When it was near the top end of power x time (and things started smoking), we put it in a big bucket of water, and left it with cold water flowing in, and an overflow to the drain.

Provided the output's isolated from the mains supply (and it'd be a deadly welder if it wasn't), water's no problem, and very good for getting rid of unwanted heat.

load bank or sea water

KISS principle. Load banks with resistors or individually switched lamps. Volt and Ammeter. Cooling. Thats it.

You don't need to make anything inductive, just use the original cables of the welder, which will test them as well, just as in actual welding......in fact making something inductive may cause you problems. All the inductance in a welder, IS IN THE WELDER!!!

I would find a source of say 110 volt heater bars, say 1 Kw each (more would be better say 10KW bars!). Each one will about 12.1 ohms. To get 0.12 Ohms (for 200 amps), you need 100 of these in parallel......or 10 of the 10 KW bars!!) Put them in clean fresh water to keep them cool. Distilled would be even better.

I would not use bulbs as they changer their resistance depending upon how hot they get, which depends on the voltage/current etc...

A simpler/cheaper way might be to take two thick disks of metal and solder thick copper wire between the disks till you get the right resistance, keep under water when using.....t6hats the cheapo method.....

Never done this before though, but thats approximately the way to go.....I hope I remembered my Ohm's law correctly!!!

Assuming the 1Kwatts bars=

Watts divided by voltage = amps. 1000watt element/110 volts = 9.09 Amps

Volts divided by Amps = Ohms. 110/9.09 = 12.1 Ohms.

Resistance = Volts/Current. 24/200 = 0.12 Ohms

100 1Kwatts bars in parallel = 0.12 Ohms.

Buy a few water heaters, tanks and all.

The more practical rout is to simply make your own banks of resistor elements by buying the right amount of properly sized Nichrome wire online. Its cheap and designed to do what it does best which is turn electricity into heat.

Add a pair of digital amp and volt meters, some good adequately rated switches and a good fan and you have yourself a reliable adjustable load bank system.

solenteng; we built a 250 amp load bank using 1/2" dia. carbon rods 12" long with 5 100a aircraft toggle switches for a 200 amp D.C. welder with no load voltage 75 volts in a portable 24x24 x36 package with metering, with the idea of over loading 25%, with the 5th switch,this was air cooled. perry

Many thanks for all the realy helpful reposponces.

I noted your comments about being overkill.

I was using inductances as I wanted a small mobile unit which was both portable and did not generate huge amounts of heat, and would provide accuracy of +/- 5 A.

My original design was indeed a huge resistor bank with fans, but I needed a very small bench unit, also I could not dynamically load the units as the welder design incorporated a pulsed design which reacted to short, open, pinch mode etc.

The load resisitor method is great for static load but I could dynamic test the different modes of a weld, this includes simulating a short and then specific current steps which are controlled by a CPU.

Are there any ways of using resisitors to provide dynamic loads as ultimately they are only wire wound and so are inductive by nature when operating in the KHZ range.

Thanks for the great help and nay further information.

Forget about trying to use an inductive load. If you really want to simulate the cable inductance, the cables between the welder and the load will do just as well as in a welding operation.

" ...and did not generate huge amounts of heat...": facts is facts: power = volts x amps. You have to get rid of nearly 5kW, without melting a hole in the bench.

For your ±5A accuracy, it will be a lot easier and cheaper to use a calibrated ammeter, and switch the load (to give approximate values) rather than try to control the load accurately. Alternatives will be very expensive.

Thanks very much for the information really appreciated.

I was using inductances, as using the standard Ohms law of 2 volts across the load circuit at 300 A is only 600 Watts as the volatge delay is used to accuratly produce the load current level of 300 A with only 2 to 3 volts drop across the whole load circuit thats a load smaller than 5KW.

I used a similar device for 100 A loads @ 24V by using mosfets, as this produced a small and vey compact device witout the heat. 3 Volts @ 100 A using only 300 Watts, this was mainly due to the cabling between the devices and line.

Although this did have other effects such as phase lag etc, which i could have corrected with PFC circuit.

Many thanks will build the switched resistor version to do the inital testing ,as the experiences indicate great forum for info thanks

In your original post, you specified 24V @ 200A. This seems to have morphed into 2V @ 300A.

Also, 24V @ 100A may have been the rating of your similar device, but if you had actually used those values (rather than the 3V @ 100A) you would have had 2.4kW of heating to deal with.

Please clarify.