COMSOL 6.3 - Frequency Domain Modeling of a Capacitor

Frequency Domain Modeling of a Capacitor

A capacitor with an applied sinusoidally time-varying voltage difference is modeled. A wide frequency range is considered and the impedance of the device is computed. Solver accuracy is addressed. The relationship between the frequency domain impedance and the steady-state capacitance and resistance of the device is discussed.

Figure 1: A simple capacitor composed of a disk of dielectric with metal plates on either side, and lead wires. An AC signal is applied.

Model Definition

The capacitor being modeled is shown in Figure 1, where two metal disks with leads are separated by a disk of quartz glass with relative permittivity εr = 4.2 and a small electric conductivity σ = 10-14 S/m. A region of surrounding air (εr = 1.0, σ = 0 S/m) is also modeled to account for the fringing fields. A sinusoidally time-varying 10 kV source is connected to the leads and the resulting electric potential distribution is computed, along with the device admittance.

It is assumed that the capacitor plates are made of a highly conductive material, so that their total effective resistivity is much lower than that of the quartz. It is further assumed that the absolute magnitude of the current flow is small, which implies negligible magnetic fields and magnetically induced currents. Under these assumptions, the terminals can be assumed to be equipotential at any instant in time.

A separate electrostatic analysis can be used to compute the capacitance of the device, C. Similarly, a separate stationary Electric Currents analysis can be used to compute the conductance G of the cylinder of quartz. Under the previous assumptions, and for a device with no frequency-dependent material properties, the admittance can be computed from

where ω is the angular frequency of the excitation. It can be observed from this equation that as the frequency increases, the imaginary component of the admittance, the capacitive contribution, becomes many orders of magnitude larger than the real component, which is the conductive contribution.

In such cases, where the model is dominated by either conductive or capacitive effects, it can become necessary to change the default solver settings. The default solver for 3D electromagnetic problems uses an iterative approach, which means that the solution is found within a relative tolerance. The default relative tolerance for the iterative solver is 10-3, but can be tightened up to 10-6 if needed. Beyond a relative tolerance of 10-6 it may become difficult for the iterative solver to converge. If even higher accuracy is needed, it is also possible to use a direct solver, although this does take more time and memory to solve. Both iterative and direct solvers are used and compared in this example.

Results and Discussion

The electric potential is plotted in Figure 2. The admittance is computed and the real and imaginary components of the admittance are shown to be related to the steady state conductance and capacitance. When using the iterative solver, and as the capacitive effects begin to dominate the admittance, the conductive component of the admittance becomes less accurate. Once the real component of the admittance is less than six orders of magnitude smaller than the imaginary component, the results are below the tolerance threshold of the iterative solver, and the real component of the admittance is no longer accurate. Using the direct solver can improve this accuracy, if desired.

Irrespective of the solver used, and the tolerance specified, the dominant component of the admittance is computed correctly. This device is primarily capacitive, and such capacitive behavior is captured correctly, even with the default solver settings. The total resistive losses within the dielectric are also computed correctly, regardless of solver method.

Figure 2: The electric potential and resistive losses in the dielectric and in the air domain surrounding the capacitor.

ACDC_Module/Introductory_Electric_Currents/capacitor_ac

Modeling Instructions

From the menu, choose .

New

In the window, click

Model Wizard

In the window, click

In the tree, select > > .

Click .

Click

In the tree, select > .

Click

Geometry 1

In the window, under click .

In the window for , locate the section.

From the list, choose .

First, create a cylinder for the model domain.

Cylinder 1 (cyl1)

In the toolbar, click

In the window for , locate the section.

In the text field, type 20.

Click

Choose wireframe rendering to get a better view of the interior parts.

Click the

button in the toolbar.

Then, add a cylinder for the disc of dielectric with the two metal plates.

Cylinder 2 (cyl2)

In the toolbar, click

In the window for , locate the section.

In the text field, type 10.

In the text field, type 4.

Locate the section. In the text field, type 8.

Click to expand the section. In the table, enter the following settings:


Layer name	Thickness (cm)
Layer 1	5[mm]

Clear the checkbox.

Select the checkbox.

Finish the geometry by adding two cylinders for the leads.

Cylinder 3 (cyl3)

In the toolbar, click

In the window for , locate the section.

In the text field, type 0.75.

In the text field, type 8.

Click

Cylinder 4 (cyl4)

Right-click and choose .

In the window for , locate the section.

In the text field, type 12.

Click

Electric Currents (ec)

The model is composed of a disc of dielectric material with metal plates on either side and two lead wires. To get a better view, hide some of the boundaries. Begin by selecting the interface, then add a node.

Definitions

Hide for Physics 1

In the window, expand the > node.

Right-click and choose .

In the window for , locate the section.

From the list, choose .

Select Boundaries 1, 4, and 23 only.

Even though not strictly necessary for a closed electrode, the domains inside the ground surfaces are disabled with the next instructions. This is the suggested model setup whenever the electrode is considered to be equipotential as this is excluding any possible numerical issue due to the effect of the large contrast in material properties.

Electric Currents (ec)

In the window, under click .

Select Domains 1, 3, 4, and 6 only.

Terminal 1

In the toolbar, click

and choose .

Select Domains 4 and 6 only.

In the window for , locate the section.

From the list, choose .

In the V0 text field, type 10[kV].

Definitions

Explicit 1

In the toolbar, click

Select Domains 2 and 5 only.

In the window for , locate the section.

From the list, choose .

Electric Currents (ec)

Ground 1

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Next, assign material properties to the model. Begin by specifying for all domains.

Add Material

In the toolbar, click

to open the window.

Go to the window.

In the tree, select > .

Click the button in the window toolbar.

Override the dielectric disc with glass (quartz).

In the tree, select > .

Click the button in the window toolbar.

In the toolbar, click

to close the window.

Materials

Glass (quartz) (mat2)

Select Domain 3 only.

Mesh 1

In the window, under right-click and choose .

Study 1

In the window, click .

In the window for , locate the section.

Clear the checkbox.

Step 1: Frequency Domain

In the window, under click .

In the window for , locate the section.

In the text field, type 1 10 100 1000 10000 100000.

From the list, choose .

In the text field, type 1e-6.

In the toolbar, click

Add a to the first solution. This selection will only include the air and the glass. Domains that are excluded from the selection will be hidden in the corresponding plots.

Results

In the window, expand the node.

Study 1/Solution 1 (sol1)

In the window, expand the > node, then click .

Selection

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Select Domains 1 and 3 only.

Add a duplicate solution and set a selection for the metal parts.

Study 1/Solution 1 (2) (sol1)

In the toolbar, click

and choose .

Selection

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Select Domains 2 and 4–6 only.

3D Plot Group 1

In the toolbar, click

The metal parts can be visualized by choosing for the coloring type. In this case the purpose of the plot is not to show a quantity, but to show a shape. The variable that the plot is based on is of no importance. Create a surface plot for this purpose.

Surface 1

Right-click and choose .

In the window for , locate the section.

From the list, choose .

Click to expand the section. From the list, choose .

Locate the section. From the list, choose .

From the list, choose .

Add slice plots for the norm of the electric potential and resistive losses.

Slice 1

In the window, right-click and choose .

In the window for , locate the section.

In the text field, type 1.

Locate the section. From the list, choose .

Slice 2

Right-click and choose .

In the window for , click in the upper-right corner of the section. From the menu, choose > > > .

Locate the section. From the list, choose .

In the text field, type 1.

Locate the section. From the list, choose .

In the toolbar, click

Compare the resulting plot with that in Figure 2.

Global Evaluation 1

In the toolbar, click

In the window for , locate the section.

In the table, enter the following settings:


Expression	Unit	Description
ec.Y11	S	Admittance, 11 component

Click

Global Evaluation 2

In the toolbar, click

In the window for , locate the section.

In the table, enter the following settings:


Expression	Unit	Description
imag(ec.Y11)/ec.omega	F

Click

Volume Integration 1

In the toolbar, click

and choose > .

Select Domain 3 only.

In the window for , locate the section.

In the table, enter the following settings:


Expression	Unit	Description
ec.Qrh	W	Volumetric loss density, electric

Click

Study 1

Next, use the direct solver to improve the accuracy of the conductive component of the admittance.

Solution 1 (sol1)

In the window, expand the > node.

In the window, expand the node.

In the window, expand the > > > node.

Right-click and choose .

In the toolbar, click

Results

Global Evaluation 1

In the window, under > click .

In the window for , click

Global Evaluation 2

In the window, click .

In the window for , click

Volume Integration 1

In the window, click .

In the window for , click

3D Plot Group 1

Now, compare the values in . Set the table to using the buttons in the tables toolbar. The real part of the admittance should be much more consistent at higher frequencies when using the direct solver. For the capacitance and the resistive losses, the difference is less prominent (see and ).