The development of Aether continues on schedule for release in September. Capabilities for time-domain simulations are complete are we are creating several examples for the tutorial manual. In this example, the full three-dimensional capabilities of the code are used to characterize a pulsed power system. The goal is to design a dummy load to test a low-impedance generator that drives a long parallel-plate transmission line. The design of the load resistor is a rather poor one to illustrate how the code can highlight problems. The example highlights two Aether techniques: 1) modeling a portion of a long periodic system using symmetry boundaries and 2) generation of a pulse with desired properties in a transmission line.
The first figure below shows the geometry. The transmission line has a long length along. The load consists of a series of identical resistors. The length along x of each cell is b = 8.0″. The spacing between the inner and outer conductors of the transmission line is a = 2.00″. The resistive solution inside the alumina housing has radius r = 0.875″ and length L = 4.0″. Most of the assembly volume is filled with purified water. A voltage pulse of magnitude V0 = 1.2 MV and risetime tr = 3.0 ns is incident from the right-hand side.
Although only a short length (D = 6.00″) of the transmission line is included in the model, we would like the line to behave as though it extended an infinite distance past the right-hand border. The technique is to include a termination layer on the boundary and to excite the pulse with an internal current source. Any reflected waves pass through the source and are absorbed by the layer. The second figure, a two-dimensional projection in the plane x = 0.0″, illustrates the computational mesh and the regions of the calculation. An element size of 0.10″ gives a good representation of the curved surfaces. The mesh contains about 600,000 elements. The outer wall, center conductor, metal cap and connecting rod are represented as metals in the calculation. The purified water and resistive solution have εr = 81.0, while alumina has a relative dielectric constant εr = 7.8. We need to determine two quantities: 1) the current density of the source layer to produce a 1.2 MV pulse and 2) the conductivity of the resistive solution for a matched termination.
The impedance of purified water is η = 377.3/√81 = 41.922 Ω. The characteristic impedance of a section of the dual-sided parallel-plate transmission line of length b is
Z0 = η (a/2b) = 5.2403 Ω.
For a 1.2 MV pulse, the source should supply a total current 2V0/Z0, where half the current travels down the line and half is lost to the absorbing layer. Inserting values, the peak total current is 458.0 kA. Therefore, the current in the top and bottom source regions should have peak magnitude I = 229.0 kA. The source has dimensions 0.1″‘ by 8.0″‘, so the cross-section area is 5.161E-4 m2. The peak current density magnitude is therefore jy = 4.437E8 A/m2. The cylindrical load has resistance
R = L/(σπr^2).
Taking R = 5.2403 Ω and inserting the dimensions, we find that the matched conductivity is σ = 12.49 S/m.
Here are links to the input files:
http://www.fieldp.com/myblog/examples/chargeline.min
http://www.fieldp.com/myblog/examples/chargeline.ain
Some entries in the Aether script are of interest:
- The simulation time is 30.1 ns, long compared to the transit time of 4.57 ns along the 6.0″‘ length of transmission line.
- The SMod command invokes the standard normalized Step function starting at ts = 0.0 ns with a risetime tr = 3.0 ns.
- The upper and lower boundaries along x are set as symmetry planes. The assignment is valid because 1) pulse propagation is parallel to the planes and 2) the magnetic field is symmetric about the planes.
- The two source regions have the dielectric constant of water and peak current densities jy = ±4.437E8 A/m2.
- Snapshots of the field distribution are recorded at 5.0 ns intervals, and probes are located at the midpoints of the transmission line and resistor.
The Aether run takes about 5 minutes. The final figure shows the electric field distribution in the plane x = 0.0″‘ at early (5.0 ns) and late (20.0 ns) times. The view at the top shows the initial plane pulse moving down the line. The field magnitude agrees with the theoretical prediction, |Ey| = V0/a = 2.362E7 V/m. The bottom view shows the distribution at t = 20.0 ns. The reflected pulse in the transmission line indicates a poor impedance match. The source of the problem is the high inductance of the small-radius connecting rod and resistive solution. Another concern is the strong concentration of electric field on the resistor end cap and across the insulator housing.
Please use this link for more information on Aether: http://www.fieldp.com/aether.html.
