Voltage Breakdown over two parallel rods

[WalrusFuzz]

Hello everyone.

I'm building a Jacob's Ladder just for the hell of it, and am currently trying to figure out how far apart to place the two rods so that an arc will pass over them.

Paschen's law seems to vary depending on the shape of the electrodes; the equations that google brings up are probably for parallel plates, as I tried plugging in the numbers just for the hell of it, however not surprisingly, this didn't work because my resulting answer didn't make any sense.

Is there any way to calculate the distance between the two rods, or is the measurement entirely empirical? Wasting boards isn't exactly a preferred method.

 

[Okefenokee]

It's not as complicated as you might think. Voltage is simply an electric field times a distance. Here's a simplified version that will work for the ladder. We can use this formula to calculate the field intensity E:

$$V = E \cdot x$$

The not so simple version is this:

$$V = \oint{\vec(E) \cdot \vec{ds}}$$

Don't worry about it though. For the ladder, we'll use the simple one (because the electric field is approximately constant between the rods).

Insulating materials have something called a dielectric breakdown. This is the point where the electric field E becomes so intense that the insulator starts to conduct. For air, it's 3MV/m. That's 3 million volts per meter.

So, here are two examples. Let's say you have a circuit that outputs a 30 kV pulse. You could find the necessary distance like this:

$$E = 3MV/m = 30kV / x$$

Solve for x to get 10 cm.

Now, let's say you have a ladder with a gap of 1 cm. You need to know how much voltage you have to generate. Solve the following:

$$E = 3MV/m = v / 1cm$$

The voltage required is 3kV.

The dielectric breakdown of air can vary. That's why the distance between is varied.

 

[BertHickman]

Hi,

You want to make sure that the gap at the bottom of the ladder will spark over so that the arc can start. Unfortunately, the above analysis has a math error that makes it off by a factor of 10. The DC breakdown voltage for a 1 cm sphere gap is actually closer to 30 kV (actually an electrical field of about 30 kV/cm) for spherical electrodes with a diameter that is much larger (>4X) than the gap length. The breakdown voltage for a rod-rod gap (the case you have) may be significantly less. If you are using smaller diameter electrodes, the peak electrical field at the electrode surface is intensified to V/r where V is the applied peak voltage (in volts versus ground) and r is the electrode radius (in cm). The air next to a 0.25 cm diameter electrode will begin to break down, forming corona at the surface of the electrodes, at a much lower applied voltage.

Once corona forms, free charges (called space charge) are injected into the gap, which then distorts the shape of the overall E-field within the gap. If these changes enhance the E-field within the gap, as it does especially with AC or RF voltages, the chances for a complete breakdown increase dramatically. A rod-rod gap will break over at a lower voltage when the applied voltage is 50/60 Hz AC, and at a considerably lower voltage for high frequency AC. However, estimating the actual breakdown voltage of the gap under these conditions can be quite difficult.

Since you'll be using relatively small diameter wires, the breakdown voltage at the bottom of your ladder will be lower than 30 kV since the electrical field will be enhanced by corona and space charges. And, since you'll probably be driving the ladder from a high voltage transformer powered from a sinusoidal AC source, the peak output voltage will be 1.414 times the RMS value. So, a 15,000 volt neon sign transformer (NST) will actually have a peak voltage of about 21,200 volts.

From a practical standpoint, a climbing arc setup (a "Jacob's Ladder") that has electrodes that are 1/4" in diameter or smaller can easily be powered from a 15 kV NST as long as it has an initial gap of 1 - 1.5 cm at the bottom.

 

[WalrusFuzz]

So if I have a flyback that has ~20kV, is set in a circuit like "www.angelfire.com/80s/sixmhz/flyback.html"[/URL], and both rods are 1/8'', I should be able to place the rods about 3 1/3 cm apart?

 

[BertHickman]

You might be able to do 2 cm. Make your ladder adjustable and lock it at the point where you get the best results. Because the circuit you referenced is low power, you may not be able to generate much heat in the arc, so you might not be able to separate the rods very much at the top...