I Potential difference between a battery's terminal and Earth ground

[sophiecentaur]

With reference to the simulation in post#22, from my previous work, LTspice uses a Thevenin voltage source with a series resistor of 2 mOhm (or 500 Siemens Norton conductance) to drive the node referred in the .IC V(nxx)=<value> global directive.

In this case the node Vp is driven with a Thevenin voltage source of 24V with a series resistor of 2 mOhm. With the .tran simulation, LTspice evaluates first the Initial Transient solution (aka initial DC solution) opening the capacitor and shorting the inductors. Therefore in our case (before the grounding switch is turned on) the Initial DC solution for node Vn is 24-12=12V w.r.t. ground. It is also the state initial condition for the numeric integration of the C2 capacitor voltage.

Said that I believe, as @Dale pointed out before, that even the battery EMF is responsabile for the charges stored on the battery's terminal/ground capacitors (i.e. C1 and C2 in the model). Consider the following: suppose no external charge is placed on the battery terminals and the inter-terminal capacitance is negligible. The PDs between each terminal and ground cannot be both zero. Hence the battery EMF must charge someway the battery terminal plates.

My point was that people do not read the small print. There is a much clearer way to understanding by using basic circuit theory because there is only BIG PRINT.

 

[Baluncore]

My point was that people do not read the small print. There is a much clearer way to understanding by using basic circuit theory because there is only BIG PRINT.

You have not defined what you mean by small print or large print. I use LTspice to communicate the schematic under discussion, not for fundamental understanding of the functionality.

It is necessary to translate fundamental understanding of the physics into SPICE without error. As an example, the nodes in SPICE are equipotentials where currents flow. Electric charge is NOT stored on those nodes. Charge is stored on the plates of the discrete capacitor components connected to those nodes.

 

[sophiecentaur]

I use LTspice to communicate the schematic under discussion, not for fundamental understanding of the functionality.

You are possibly being disingenuous here. You are an experienced and capable Engineer whotook on board all the "small print" years ago. I was, of course, referring to the uninitiated who assumes the simulation is reality.

It is necessary to translate fundamental understanding of the physics into SPICE without error.

Absolutely; well said.

 

[Baluncore]

Sure. The small capacitance between the terminals and ground is different from the internal capacitance.

When I simulate the terminal capacitance, with an ideal voltage source as the battery, the simulation behaves reasonably.

When I replace the ideal voltage source with the 80kF and 4kF capacitor battery model, I get digital noise. That is because the terminal capacitance has then become insignificant. 80kF to 1pF gives sixteen orders of magnitude, that cannot be simulated with the 16 digits of double precision floating point.

When I reduce the battery model from kilofarads to millifarads, the simulation works well again, as well as it does with the simple voltage source. The ratio is then 80mF to 1pF, ten orders of magnitude, well within the dynamic range of double precision.
Battery voltage is then stable at 12.000 volts.
I(C3) is between 1 fA and zero, that is one LSB of double precision.

Notes:
1. I have staggered the values of C1 and C2, to separate the lines on the plot.
2. The schematic of the voltmeter switch looks real, but the virtual switching is actually done by the .IC initial conditions, applied at t=0. The switch is always closed by the shared node "n" label.

 

[Dale]

When I simulate the terminal capacitance, with an ideal voltage source as the battery, the simulation behaves reasonably.

Excellent. That is the one really crucial test. Of course, in different situations different models will be reasonable. So you should use the appropriate model for the circumstances and not over or under complicate the model

 

[cianfa72]

When I reduce the battery model from kilofarads to millifarads, the simulation works well again, as well as it does with the simple voltage source. The ratio is then 80mF to 1pF, ten orders of magnitude, well within the dynamic range of double precision.
Battery voltage is then stable at 12.000 volts.
I(C3) is between 1 fA and zero, that is one LSB of double precision.

I double checked your simulation using LTspice XVII: indeed the battery voltage is stable and I(C3) current is under 1 fA (LSB stands for Least Significant Bit) ?

Btw, why did you choose to model the "battery internals" that way ?

Ps. how did you draw the slanted line for the switch in LTspice schematic ?

 

[Baluncore]

(LSB stands for Least Significant Bit) ?

Yes. One least significant bit of double precision is 10^-16 ≈ 0.1 fA.

Btw, why did you choose to model the "battery internals" that way ?

Dale gave a link to it in post #42.
"Temperature-Dependent Battery Models for High-Power Lithium-lon Batteries"

Ps. how did you draw the slanted line for the switch in LTspice schematic ?

Draw a short orthogonal offset wire. Select the free end, then "drag" it to a diagonal line. You can "move" that line segment if you need to.
You may need to change the drafting options.

 

[cianfa72]

Dale gave a link to it in post #42.
"Temperature-Dependent Battery Models for High-Power Lithium-lon Batteries"

Ah ok. With this battery model the initial conditions V(p) and V(n) are actually two folded. They implicitly assign "external electrostatic charges" on terminal/ground capacitor C1 and C2 and at the same time "charge" the battery (i.e. charges the internal capacitors Cc and Cb and the inter-terminal capacitance C3). The .tran simulation starts from these initial conditions: the ground switch via the 10meg voltmeter discharges the external electrostatic charges from C1 and C2, meanwhile the PD at battery terminals stays put and the inter-terminal capacitance C3 isn't discharged at all.

 

[Baluncore]

Correct.
That is why we can ignore C3 in the "flying battery" circuit model.
C3 is dominated by the fixed battery voltage; C⋅dv = dq = zero.