3
\$\begingroup\$

I have a circuit where at least some of the capacitance is coming from large, non-radiating, spherical conductors. I would like to model it in SPICE to better understand its operation-- is such a thing possible? The capacitors I have available are all two terminal devices...

I know that the capacitance of my sphere is \$C=4\pi\epsilon_0 r\$, where \$r\$ is the radius of the sphere. I'm trying to figure out how to represent that in a SPICE circuit.


Updated to address points made in comments and answers:

Here is what I originally meant by "not coupled":

The self capacitance of a sphere is \$4\pi\epsilon r\$, where \$r\$ is the radius of the sphere.

If I have two spheres of equal radius the capacitance is:

\$2\pi\epsilon r \sum_{n=1}^\infty \frac{\sinh\left(\ln\left(\frac{d}{2r}+\sqrt{\left(\frac{d}{2r}\right)^2-1}\right)\right)}{\sinh\left(n\cdot\ln\left(\frac{d}{2r}+\sqrt{\left(\frac{d}{2r}\right)^2-1}\right)\right)}\$, where \$d\$ is the distance between sphere centers.

The summation limits to 1 as \$d\$ goes to infinity, and the remaining terms can probably be interpreted as the self capacitance of each sphere in series. So the total capacitance is the self capacitance of each in series, plus, what I will call a "mutual capacitance" caused by interaction of the electric fields and which is a function of distance.

By “not coupled”, I’d meant that this distance dependent mutual capacitance term is arbitrarily small, leaving only the self capacitance. Probably the wrong choice of language. The capacitance value is not dependent upon anything else in the circuit, but obviously Gauss’ Law still holds.

\$\endgroup\$
2
\$\begingroup\$

The capacitor is not coupled to the other nodes.

In this case, the capacitor is not affecting your circuit, so you needn't model it. By Kirchoff's current law, your circuit can't deliver any current onto the spherical capacitor without also being connected to whatever the spherical capacitor is coupled to (which is probably mostly earth ground).

If you want your circuit to be able to deliver charge to the spherical capacitor, you must also connect it to earth (or whatever ground-like object is near the capacitor). Then, of course, earth becomes a node in your circuit and it's no longer true that the capacitor is not connected ot other nodes of your circuit.

Responding to some comments:

Coupled how?

I was just using the word you used. I assumed you meant coupled by electric field lines between the sphere and the things it might be coupled to.

Capacitance is a measure of the energy stored in the electric field, but surely I could raise a sphere far enough from earth that the field strength is negligible on the ground and still store charge (and energy) on the sphere.

You'd have to transport the charge to the sphere from somewhere. That would form a circuit including that other place.

If the charge comes from somewhere in your circuit (elevated in space along with the sphere), then you already had a charge imbalance between the sphere+circuit system and the earth (however far away), and you'd have already had an electric field developed. Moving the charge from the other circuit elements to the sphere wouldn't appreciably change the field between the sphere+circuit system and the earth, so it wouldn't store any additional energy.

Is using KCL circular reasoning: all the current entering a junction must leave the junction therefore there must be a junction?

KCL can also be applied to any closed surface. If you define a closed surface surrounding the spherical capacitor and the rest of your circuit, then the net current across that surface must be zero. Obviously since you've imposed this surface between the sphere and ground (or whatever) you have to include displacement current.

\$\endgroup\$
  • \$\begingroup\$ Many comments concatenated, not meant to sound rapid-fire..: Coupled how? Capacitance is a measure of the energy stored in the electric field, but surely I could raise a sphere far enough from earth that the field strength is negligible on the ground and still store charge (and energy) on the sphere. Is using KCL circular reasoning: all the current entering a junction must leave the junction therefore there must be a junction? KCL is a conservation of charge statement, therefore current entering a node is equal to the current leaving the node plus the current stored at the node. \$\endgroup\$ – Omegaman Jun 25 '15 at 20:36
  • \$\begingroup\$ Thanks for the answers so far. I've updated my question to help get to the last bit I need. \$\endgroup\$ – Omegaman Jun 27 '15 at 19:28
0
\$\begingroup\$

You need to analyse what the capacitance of the sphere is to all relevant points on your circuit - SPICE cannot do this for you - it understands two terminal devices (like a capacitor) but the sphere will have a multitude of capacitors emanating from one common point with each capacitor connecting to relevant nodes in your circuit. You need to do this before looking at the circuit performance implications in SPICE.

Some nodes will be unimportant (like power rails) but high impedance nodes could be very relevant so, calculate the capacitance to each node from the sphere (I'm assuming it is very conductive) and model it in SPICE as a bunch of capacitors connecting each node to a common node. If the sphere is galvanically connected to another node then the common node should also be connected in SPICE.

\$\endgroup\$
  • \$\begingroup\$ I'm looking to model the self capacitance of the sphere itself. There may be some coupling back to parts of the circuit, but I believe they're negligible at best. Mostly there's a whole bunch of charge accumulating on the surface of the sphere. \$\endgroup\$ – Omegaman Jun 25 '15 at 17:34
  • 1
    \$\begingroup\$ Capacitance is between one node and another - if the sphere is just one conducting part then it has no self-capacitance. \$\endgroup\$ – Andy aka Jun 25 '15 at 18:04
  • \$\begingroup\$ I'm pretty sure that's not true. I've added a reference to the question. \$\endgroup\$ – Omegaman Jun 25 '15 at 18:07
  • \$\begingroup\$ Yes I understand but this self capacitance is of no consequence to the other nodes you mention in your question. \$\endgroup\$ – Andy aka Jun 25 '15 at 18:15
  • \$\begingroup\$ I think my comment in response was more clear than the original wording, so I deleted my comment and added the explanation to the question. \$\endgroup\$ – Omegaman Jun 25 '15 at 18:35
0
\$\begingroup\$

You represent a system of N conducting spheres using a circuit comprising N+1 nodes to represent the sphere potentials and 'ground', i.e. the potential of the surroundings. Every node i is connected to every other node j via a capacitance C_ij. That's a total of 0.5 * N * (N+1) capacitors.

A single sphere is represented by a single capacitor (its self-capacitance) with one node grounded and the other at the potential of the sphere. Two spheres require three capacitances. Two between the spheres and ground (their self-capacitances) and one between the two spheres (their mutual capacitance)

The tedious part is obtaining the values needed for these capacitors, which is an exercise in classical electrostatics. For simple systems you can probably use estimates. For more complicated cases you need to use a CAD package.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.