Fusor Forums

Hello, does anyone know whether or not an inner grid made of silver would work? This is for a DD reactor. I can't think of any drawbacks, but if anyone else can, I'd like to know.
Thanks a lot.

It has a low melting temperature. Wouldn't be able to sustain anywhere close to the wattage that a Tungsten grid could without melting.

Ah, so it definitely would not work?
Tyler Christensen wrote:
> It has a low melting temperature. Wouldn't be able to sustain anywhere close to the wattage that a Tungsten grid could without melting.

It probably would melt apart before you could measure any neutrons. There seems to be no rationale (unless you happen to have a spool of it and mere availability is driving the decision) to use it when tungsten wire is easy to obtain, and even stainless steel is serviceable.

-Carl

Wait, you can use stainless steel?! But its conductivity is terrible! And the reason I wanted to use silver was because of its excellent conductivity. I guess I could use stainless steel if it works OK.

The conductivity of the grid is irrelevant so long as it is some conductive metal. It's carrying milliamps, so V=IR will show that the drop is negligible compared to the HV potential.

As mentioned any fair conductivity is adequate. With Ohms Law A=V/R. With house voltage of ~ 110 volts the wire heating due to Ohmic (resistive) heating may be one unit. At a Fusor voltage of 11,000 Volts the heating will be 100 times less at the same current flow. This is why long range electrical power transmission is generally done at high voltages, less energy is lost through heating the wires. The heating of a thin wire at ~ 10,000 volts and several dozen milliamps would be minimal. The wire will not get very hot as even radiation transfer of heat from the wire could probably keep up- no gas convection is needed- thus high vacuums may not change the picture much. What heats the wires is the bombardment by the high speed ions which are an integral part of how Fusors work and cannot be avoided. To control this bombardment heating you use high melting point wire, actively cool the wire, and/or try to magnetically shield the wire. The magnetic shielding could reduce hot ion impacts but would not effect hot neutrals impact, and these hot (high speed)neutrals are unavoidable in normal Fusors. They come mostly (I think) from high speed ions that have recombined with electrons to make a high speed neutral atom. Hot neutrals might also come from up scattered neutrals. What percentage of the neutrals have high velocity is uncertain, though I suspect it is relatively low, but the total neutrals may outnumber the ions by a factor of 10-100 in a normal glow discharge Fusor.

Dan Tibbets

> As mentioned any fair conductivity is adequate. With Ohms Law A=V/R. With house voltage of ~ 110 volts the wire heating due to Ohmic (resistive) heating may be one unit. At a Fusor voltage of 11,000 Volts the heating will be 100 times less at the same current flow.
Wrong! I know you're trying to help, but it does not help if you don't get it yourself, or are too tired to think & write right. The heating will be the same at the same current. It has nothing to do with the voltage between the heated wire and other wires.

> This is why long range electrical power transmission is generally done at high voltages, less energy is lost through heating the wires.
Yes, but. In your previous example, a wire at 11,000 volts (with respect to the return wire) could carry the same POWER as it would at 110 volts,
with 1/100 of the current, and 10,000 times less heating of the wire.

> The heating of a thin wire at ~ 10,000 volts and several dozen milliamps would be minimal. ...
Yes, no different from several dozen mV and several dozen mA. What counts is the voltage drop due to the wire's own resistance.

Rich have you given any thought to the SIZE of the wire in the grid, in terms of Voltage gradient?

George Dowell

George Dowell wrote:
> Rich have you given any thought to the SIZE of the wire in the grid, in terms of Voltage gradient?

Well, in another thread, Chris Trent gave a link to what passes for a design rule document at Shapeways, for their Stainless Steel material. The ordinary minimum section thickness is 3 mm. You can go down to 1.5 mm if it's kept short and closely supported by thicker material.

Putting numbers on the point first made by Tyler in this thread:
A fusor-grid-shaped part with wires that thick, of any plausible metal, would have much less than 1 ohm of resistance between any two points, even at its melting point.
So at fusor currents (order of 10 mA), the voltage gradient will be negligible (as Tyler said) compared to the fusor HV fields.

Dan was right in saying that the ohmic heating would be negligible. I think it would be a challenge to make that term reach 1 mW. (e.g. 100 mA through 100 milliohms, giving a drop of 10 millivolts).

p.s. The Shapeways material data sheet does not give the electrical resistivity, but it does give the Young's modulus (21 MSI), yield strength (66 KSI), and elongation (2.3%).

I admit that my comment about voltage and current is irrelevant as the current is given/ measured. But the resistive heating in the cathode wire is still only a few milliamps and this will not heat a conductor- even a modest conducort very much. Radiation and conduction through the stalk should easily keep up. The current to ground is actually through the plasma. That is why ythe current is zero till a voltage is reached where Pashin dischagre can occur..

In defense, if someone is thinking of 1000 Watts being consumed in a Fusor, they might compare it to a toster oven, running at 110 Volts and this will indeed get hot. with red glowing wires. But, at 11,000 volts the current is 100 times less and little resistive heating of the cathode wire will occur. Again the warmth of the wire cathode and the walls comes from the high voltage- high KE charged particles hitting surfaces. Power is power and heat is the final depository, but the intermediate processes are different.

Another comparison would be a 1.5 volt battery delivering 10 mA through a wire. I don't think anyone would think that the wire would get red hot. With a 10,000 V battery and appropiate resistance in the system, again the same 10 mA would flow through the cathode wire. with the same resistive heating. The expended energy in this case is dominated by voltage effects/ acceleration of charged particles. If this seems obtuse, consider a bullet. At 10 FPS (low voltage) it will not transfer much energy ,. But at 3,000 FPS (high voltage), even with the same bullet weight (Amps) the energy transfer is tremendous.. It is this voltage dependant energy that heats the cathode wire and the wall. Keep in mind that the KE flows both ways. Ions are attracted to and hit the cathode wire and heat it, the electrons are attracted by the grounded or relatively anode walls. The surface area of the wire is much less so it doesn't drain the generated heat as fast as the walls. That is why the cathode wire is red hot, while the walls are only warm..

Dan Tibbets

The power lost due to resistance in the wire of the grid is pretty close to nil. However the fact that silver is a very good reflector means it's a poor absorber and so, in turn, a poor emitter of heat.
You don't want a bright shiny grid: a black one will not get so hot, so you can push it harder before it melts.

All grids get hot.

All grids can melt.

No grid is capable of even getting warm due to electrical internal disappation.

There is no indestructable grid.

If the grid were made of ultra high resistance, pure nichrome, toaster heater wire. It would not heat one iota due to conduction losses in an amateur fusor. Let all discussions related to grid dissapation of electrical energy die here.

Tungsten is the best material that can be easily acquired and is very high resistance electrically, but as above, zero electrical dissapation.

The desired cross section of grid wire is related to its ability to heat due to particle bombardment at a given input power and survive same without melting.

No one has tested a very large cross section pure silver grid as an integral part of a large diameter, cast, pure silver stalk. The heat conductivity of silver is superb and near the top of heat conductive elements. A large 10 ounce grid, stalk and top electrode would take off a lot of heat, but its rather low melting temperature and sputtering might prove to be not worth the effort.

I'll stick with Tungsten.

Richard Hull