The quest for inexpensive resistors that can handle high voltages and energies leads to what are known as “water resistors”.A water resistor is essentially a tube filled with a salt solution and an electrode at each end. You can control the concentration of the salt to control the resistance, for a given size tube.
There are three basic types of resistor required in a high voltage system. They differ primarily in their required resistance: tens/hundreds of ohms, 10K-20K, and multi megohms. The very high resistance, high voltage resistors are used in voltage dividers to measure multi hundred kilovolt signals without destroying your test gear. The really low resistance units (around a hundred ohms) are used as load resistors for an impulse generator, or to control the circuit Q. Finally, the middle range, 10K-20K are used as charging resistors to limit the current. The last two applications require handling a fair amount of energy, often with very high peak powers.
Copper Sulfate is popular as the salt, since it doesn't corrode copper (or copper alloy, i.e. brass) electrodes. I started using PVC water pipe and some surplus copper bolts, but found that you always wind up with air bubbles and sealing problems. As the temperature changes, the air in the bubble expands and contracts creating leaks. Also, the resistance is orientation sensitive, as the bubble moves around changing the conductive path (in fact, they use this effect in sensitive electronic levels).
My latest scheme is flexible tubing (e.g. vinyl (Tygon) or polyethylene) and brass machine screws. I've also used brass hose barbs, but they have a hole in the middle which you have to seal, although in some cases this might actually be an advantage. You stick the tube into a bucket of salt solution, fill it up and, keeping it submerged, shove/screw in the bolts. Don't use steel or cad plated hardware or the copper sulfate will react, plating out on the steel, and leaving sludge in your resistor. I haven't looked into aluminum electrodes and soluble aluminum salts, but I suppose it would work.
Why not use brass or copper rod as the electrode? There aren't those handy barbs or threads to aid in sealing the bolt to the tubing. For better sealing, you could use crimp-on brass ferrules as used in air hoses and the like. You could use a screw hose clamp like they use on radiator hoses, but the edges on the clamp will doom you to corona problems, unless you are going to pot the whole assembly, which raises all sorts of other problems.
You don't need the purest reagent grade copper sulfate for this either. Regular old technical or purified grade is sufficient. It runs about five or six dollars per pound. The kind with lumps is harder to dissolve, and costs more to boot (they use it for killing tree roots).
When designing your water resistor, you generally choose the diameter and length so that it can absorb the required energy (or dissipate the required average power) and hold off the required voltage (length). Then, you adjust the salt concentration to get the value you need. As a practical starting point for your calculations: A 1 meter length of ¼” id tubing filled with 0.1 Molar Copper Sulfate (1.6 g in 100 cc water) has a measured resistance of 39.8 Kohms.
You want to keep the diameter up fairly large (> .25 inch) to reduce corona problems. A more critical issue for high energy (e.g. Marx Generator and capacitor dump) applications is making sure you won't melt the hose with the heat dissipated in the resistor. You need to calculate how much volume you have, its thermal capacity, the thermal resistance of the plastic, etc.
Sample calculation:
10 kJ pulses every minute.
Handy conversions: 1 kJ = 240 calories (approx)
Vinyl tubing is good to around 90 degrees C, so our maximum temperature rise is 40 degrees above an assumed ambient of 40 degrees.
1 cm ID (3/8”) tubing, 50 cm long. Approximately 35 cc of salt solution, or 35 g, assuming specific gravity of 1. We'll also assume a specific heat of 1 degree/cal (= 4.184 Joule/K). The delta T for this resistor is 7 degrees C per kilojoule. If you are dissipating 10 kJ per pulse, you are looking at a 70 degree rise, which will probably melt the tubing. Even if you use a meter of tubing, doubling the volume, you are still looking at a 30 degree rise. A 200 cm length gives you a 16 degree rise.
Try 1.9 cm (ID) diameter tubing. A 50 cm length is 153 cc, or about 1.6 degree per kilojoule, which is much better. Our design 10 kJ pulse now results in a delta T of 16 degrees.
How fast will this heat be carried away? The primary limiting thing is the thermal resistance of the tubing wall, nominally .125 inches (.3175 cm) thick. You need to figure with the mean logarithmic area which is about 350 sq cm. The thermal conductivity is about .003 W/cmK. So:
Watts = .003 * 350 / .3175 * 16 = 52.5 watts at 16 degrees above ambient.
Log mean diameter = (od - id) / ln(od/id)
= (2.535-1.9)/ln(2.535/1.9)
= 2.2 cm
Log mean area = pi * log mean diam * length
= 3.14159 * 2.2 * 50
= 350 ( approx)
At this rate it will take some 200+ seconds for the water to cool back down from a 10 kJ pulse.
Going back to another case, the 1cm ID (3/8 nominal) with 1/16” walls, a 200 cm long piece with a 16 degree rise dissipates about 250 watts, for a time constant around a minute, which is getting reasonable.
You can continue this process and figure out the most cost effective approach, given that bigger diameter terminal bolts cost more. Tubing runs anywhere from US$0.10/ft to US$1.00/ft, depending on the diameter. 3/8” id runs around US$0.33 /ft, so the 2 meter piece is about $2.00, add another $2 for the two terminal bolts, and you have a resistor for around $4, plus a bunch of labor. Even if takes you an hour to make a resistor, compared to the $100 for the commercial item, this isn't a bad deal. Not like making capacitors, where you spend a lot more time for an inferior product. (Prices as of 1997)
A water resistor can also be used as a high power load resistor. For instance, when testing my Marx generator ( 40 kJ at 600 kV), I need a repeatable load that can handle the energy without a lot of noise and fuss. For this application, you are looking at a total resistance of a few hundred ohms, and a lot more volume to absorb the energy without appreciably heating. I started with a provisional design of a 2 meter length of 3 inch diameter PVC pipe. The thermal calculation runs as:
Volume of liquid: 9.1 liters
Thermal transmission area: 5000 sq cm
Temp rise: .026 deg/kJ
Thermal Resistance: 47.5 Watts/degree
Time constant: 810 seconds
So, for my 40 kJ pulses, I expect the liquid to gain about a degree. If I were to hit the load with a pulse every 5 minutes (i.e. avg power of about 130 Watts), the system would reach thermal equilibrium at a temperature rise of about 3 degrees, certainly reasonable for the plastic.
There are some practical details with electrodes on something of this scale, making sure you don't have a current concentration that causes localized heating at a rate faster than convection can distribute it through the electrolyte.
For purposes of measuring high voltages from low current sources (like a Van deGraaf generator), it is nice to have a resistor of thousands of megohms which can stand a megavolt or so. For this, you need a very small diameter tube, and lots of it. You can also use almost pure distilled water, since you probably want something like 1 Megohm cm resistivity. For instance, 10 meters (1000 cm) of 1 sqcm tubing filled with 1 Mohm cm water would be a 1000 Meg resistor. The problems that will creep in here is the gradual change in the conductivity of the water due to ions from the electrodes. One way to solve this is to circulate the water through the system and continuously deionize it with a cartridge deionizer. Actually, if you are building a voltage divider, you probably don't care what the actual resistivity is, just that it be uniform. A pump to circulate the water would fix this. With high resistance values, you have to be concerned about dirt on the outside of the tube getting damp from atmospheric moisture and the leakage path being lower resistance than your resistor.
Another problem with high resistance water resistors is that the resistivity of water, itself is zero for DC, that is, it gradually polarizes and electrolyzes. For AC signals, this may not be a significant problem. Distilled water has been used in pulse forming capacitors and transmission lines that only remain charged for a microsecond or so quite successfully. The high dielectric constant of water (85) makes it attractive in these applications.
Copyright 1998, Jim Lux/ resww.htm / Back to Resistor page / Back to High Voltage Main Page / Back to home page / Mail to Jim (jimlux@earthlink.net)