Timeline for Why do I need a ground when simulating a circuit? I thought voltage was relative between two nodes!
Current License: CC BY-SA 3.0
13 events
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| Dec 7, 2017 at 9:06 | comment | added | Curd | @Massimo Ortolano: my point is that not only it would be possible but it would be even easier to set up the equation system without using potentials with respect to a GND reference node. The reason it is not done is because of human convention, not because of mathematics. | |
| Dec 7, 2017 at 8:58 | comment | added | Massimo Ortolano | @Curd I'm quite aware of how Spice works ;-) It is certainly possible to setup a uniquely solvable system of equations in an alternative way, but it would be more complicated than simply fixing the potential of one node. | |
| Dec 7, 2017 at 8:45 | comment | added | Curd | Here is a very short (3 pages) overview explaining how simulations (like SPICE) work. I think it makes some of the issues addressed here clear. You also see that actually it would be more natural to use only voltages across components instead of potantials with respect to a GND node (see all the \$V_i - V_j\$ expressions); but of course it's done the other way because of convention. | |
| Dec 7, 2017 at 8:40 | comment | added | Curd | @Connor Wolf: how does that explain anything as the reference GND potential is very likely NOT the voltage at operation point? (and what if there are more than one operating points? The simulation would still use one reference GND) | |
| Dec 7, 2017 at 8:35 | comment | added | Curd | @Massimo Ortolano: Wrong. It would be just as possible to set up a uniquely solvable equation system representing the circuit without definition of a reference GND node. The voltages to solve for would be voltages across each component (instead of potentials with resepect to a GND node). So the actual reason is just the convention, as James lange said, not a mathematical reason. | |
| Dec 7, 2017 at 2:12 | comment | added | Connor Wolf | Also, even if the system chooses the operating point, floating point number representation has a variable precision based on the absolute offset from 0. | |
| Dec 6, 2017 at 22:34 | comment | added | Massimo Ortolano | @Hamsterrific Briefly, since currents in a circuit depend only on potential differences and not on the potentials alone, if you don't fix the potential of a node, from a circuit you obtain a system of equations that is not uniquely solvable, that is, there is always an infinity of solutions, for which voltages are all differing by a constant. Most methods for the numerical solutions of system of equations work only if systems of equations have a unique solution, otherwise they fail. By fixing the ground, you ensure that the solution is unique (if the circuit is well defined). | |
| Dec 6, 2017 at 21:37 | comment | added | Pedro A | @MassimoOrtolano What would those be? | |
| Dec 6, 2017 at 20:37 | comment | added | Massimo Ortolano | This answer really ignores the inner mathematical mechanisms of simulators. | |
| Dec 6, 2017 at 17:36 | vote | accept | user135172 | ||
| Dec 6, 2017 at 14:21 | comment | added | pipe | I added what I think was a missing point in this excellent answer. Feel free to change it, English isn't my native language. | |
| Dec 6, 2017 at 14:19 | history | edited | pipe | CC BY-SA 3.0 |
Added the missing piece to an otherwise excellent answer.
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| Dec 6, 2017 at 14:10 | history | answered | Solomon Slow | CC BY-SA 3.0 |