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Beginners idiot question
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T3sl4co1l:
The technically correct answer is:
1. The simulation works by iterating matrix inversions.
2. If no absolute reference is defined in the equation, the result is a singular matrix.
3. "Singular" is a synonym for "non-invertible", i.e., there is no solution.

Which is obviously a not-very-helpful answer to an idiot. :P

Expanding on that some more:
- Every component represents an equation, a relationship between variables in the system.  Yes, this is a very mathy thing; simulators lie entirely in the mostly-abstract domain of applied mathematics.
- Every node (set of common connections between components) represents a variable in the system.
- To solve for the voltages and currents in the system, we need to know all of these relations.  Then we put them all together into a matrix.
- A matrix is just a system of equations.  You can solve "ax + by = c; dx + ey = f" by algebra (rearranging and substituting), or you can group the variables in a common order and use an algorithm to solve it.  If we let A = [a, b; d, e] (i.e., a matrix, in order of rows, then columns), and x = [x; y] (a 1-column matrix is also called a vector), and b = [c; f], we have the matrix equation Ax = b, with the solution x = A^-1 b (read: "x equals A inverse times b").  This is practically a trivial solution (it's essentially division by A), so you can see why it's so attractive.
- The hard part is then: writing the matrix A in the first place, and, solving for its inverse.
- A is generated from the schematic, itself not the easiest thing, but just a matter of programming.  It can also be written out in plain text (a netlist), or, for that matter, by inspection (with some experience, of course!).
- The matrix inverse is a PITA to calculate by hand, but it's always done the same way every time -- as long as we can tell a computer how to do this, we're set.

So, "singular matrix" is very much like "division by zero".  It's telling us that there are no unique solutions to this system -- in practice, this is quite understandable, as an isolated circuit can have any voltage at all, with respect to ground.  The difference is, that circuit continues to work regardless of what voltage it's sitting on.  This is partly a consequence of the analysis method we've chosen -- we could simply define a node arbitrarily, or rearrange the equations to remove that ambiguity.  But it's easy enough to stick in a ground, in the designer's preferred location (normally a node with a lot of common connections), so that's what we usually do. :-// :)

Tim
Brumby:
Just noticed this:

--- Quote from: AlfBaz on May 26, 2019, 11:25:14 pm ---Both grounds are at the same potential so with no potential difference there is no current flowing.

--- End quote ---
Sorry, but that is an absolutely incorrect statement.

Current CAN flow through a conductor without any potential difference across it, when it is part of a circuit that has current flowing because of the other circuit elements.  In fact, this is the definition of a conductor (in the normal sense).
DimitriP:
I'd like to swim in this spoonful  of water too but my life preserver doesn't fit in it  :horse:
Zero999:

--- Quote from: Brumby on May 27, 2019, 05:01:52 am ---Just noticed this:

--- Quote from: AlfBaz on May 26, 2019, 11:25:14 pm ---Both grounds are at the same potential so with no potential difference there is no current flowing.

--- End quote ---
Sorry, but that is an absolutely incorrect statement.

Current CAN flow through a conductor without any potential difference across it, when it is part of a circuit that has current flowing because of the other circuit elements.  In fact, this is the definition of a conductor (in the normal sense).

--- End quote ---
Only if it's a superconductor.

Simulators always assume the resistance between ground connections is lower than the wires, which also confusingly have zero resistance. As a thought experiment, imagine the connection you've drawn has a very low resistance, say 1pΩ which is 0.000000000001Ω and the resistance between the ground connections is really 0Ω. All of the current will flow between the ground connections.

Strictly speaking, this isn't what actually happens. SPICE considers the ground nodes first, so if they're connected together with other wires, then the current in those wires is assumed to be zero.
Brumby:

--- Quote from: Zero999 on May 27, 2019, 09:21:16 am ---
--- Quote from: Brumby on May 27, 2019, 05:01:52 am ---In fact, this is the definition of a conductor (in the normal sense).

--- End quote ---
Only if it's a superconductor.

--- End quote ---

I added that highlighted comment specifically to short circuit the superconductor criticism.  Let's not make it any harder for the poor Op with pedantic distraction.
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