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I need a resistor of low ohmic value (about 10 to 20 Ohms), non inductive, capable of withstanding 100kV and more. I want to use this as a wavefront damping resistor, for a marx bank.
I was wondering, which would be the best way to make such a resistor? Wirewound? Plastic tube coated with some conductive paint? Tube filled with a salt solution??
Account deactivated by user request on 6/11/2009. Registered Member #1071
Joined: Fri Oct 19 2007, 02:13AM
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Posts: 44
The easiest and most robust would be a copper sulphate or potassium bromide resistor. They are just tubes filled with a solution that gives you the resistance you want. They can be made to whatever resistance you want and can be made as large as you want if you are putting a lot of energy into them. People use them all the time in resistive dividers for looking at voltages over 500 kV, it is easy to make a resistive divider with them because you can make them several feet long if you want to.
A CuSO4 resistor was what i had in mind as well. The needed resistance value is a bit low, so I might need a dense solution. I've read that 1mol/kg CuSO4 has a conductivity of ~4 S/m (CuSO4 solubility in water is ~1.5mol/lt). A 0.5m length x 0.1m diameter tube will marginally reach that resistance value.
Account deactivated by user request on 6/11/2009. Registered Member #1071
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That is kinda long for that resistance, if you dont care to make it some exact value you can put a lot in these things and they still seem to pretty much work. We made some that were about 7 inches long and 2 in diameter and they were 1 ohm.
I just did an initial solution with sea salt (1kg NaCl salt in a total 5liters of volume) and the reading that i get is a little strange. The resistance does not seem to be quite stable, it increases for as long as i hold the multimeter on the water load terminals, and it also changes by reversing the position of + and - terminals of the multimeter.
Does this have anything to do with the electrodes/salt/bubbles and any electrolyzation that takes place?? Electrodes are circular brass plates of 5cm diameter. Load is 60cm long and 10cm in diameter.
What should i take as the correct resistance value of such a load?
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Joined: Tue May 08 2007, 03:47AM
Location: New Jersey, USA
Posts: 616
With brass plates as electrodes there will no doubt be some electrochemistry going on, in the form of a voltage appearing across the resistor. This is probably throwing off your multimeter. To correct this you would need platinum or graphite electrodes.
I would recommend putting it in series with a voltage source and measuring the current, then just use ohms law.
Alternatively, you could make it part of a resistive divider with a known value resistor. Use a much higher voltage (say 24V). This will make the small voltage produced by the cell negligible and allow you to measure the new voltage across it, which along with the supply voltage and known resistance will allow you to calculate the resistance.
You should be able to keep the brass terminals because even if there's a voltage across it, it will still "look like" a regular resistor to such a high voltage and current as a Marx generator produces.
Account deactivated by user request on 6/11/2009. Registered Member #1071
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Oh, I forgot about that. I use a nice resistance meter that can make an AC measurement at 100kHz. I dont think you can make a DC measurement of them. There is probably some way to make a homemade AC resistance meter to get an accurate measurement. Another option would be to use this in series with another known resistor and send an AC signal across it and measure the voltage so you could infer the resistance of your unknown resistor in a simple divider.
The voltage source/current amount seems a better idea for the water load test. I did a NaCl solution just for getting an initial idea of the amount of salt needed. I will test again with CuSO4 though, probably this will work better with the brass plates.
I guess a drive with a square waveform will reveal any self inductance of this load, but i reckon this will not be much (200nH maybe?).
I made a box out of acrylic glass, measuring inner dimensions 10cm*8cm*48cm. I filled the box with saturated CuSO4 solution (more than 1kg of salt in 4lt of water - so that no more could be dissolved) and put brass terminals of 4cm in diameter. The solution kept measuring a nonconstant resistance with the multimeter (~20Ohm with one polarity and ~35Ohm with reverse polarity) with a minor tendency to increase with time.
So I measured the AC frequency responce of the load, by driving it with a signal generator (10Hz to 15MHz) and checking the amplitude with an oscilloscope.
The load actually presents an "impedance" of 17Ohm up to 1MHz, and after that has a rapid increase in the impedance ("a lot bigger than 50Ohms") just at 10MHz. These values were extracted by using the load line of the (50Ohm) sweep generator.
Conclusion? It is difficult to get really low resistance values with CuSO4, unless you make a really fat and short resistor (i.e. getting resistor length over area ratio really small).
But i guess, for starters, my construction will do fine.
EDIT: Actually I calculate it as ~[15Ohm + 1.4uH].
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DIY high voltage resistor for a marx bank...
