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

[cianfa72]

If you try to electrostatically charge one terminal of the battery only, that charge will appear on both terminals in proportion to their effective capacitance.

You mean the terminal own capacitance (just to be clear their self capacitance).

The difference in charge density between the terminals, will be related to the battery voltage, not which terminal you charged.

Ok, so the two battery's terminals will have a different charge on it (the total electrostatic charge will be redistributed according to the terminal self-capacitance and battery EMF).

The external insulated case of the battery will also hold some residual surface charge that failed to leak across the surface to a terminal.

Ok, this makes sense.

 

[Dale]

Since there is not a closed circuit involving the battery and the Earth

You can always make a closed circuit by including the capacitance with the ground.

 

[cianfa72]

You can always make a closed circuit by including the capacitance with the ground.

Ah ok, you mean the capacitors C1 and C2 in post#22 schematic when the battery is flying. Therefore, as you pointed out, the effective charge on battery's terminals depends on battery EMF and the "history" as it flies around the space.

As I said before, I'd add a capacitor C3 between the nodes Vn and Vp to model the "internal/intrinsic capacitance between terminals".

 

[Baluncore]

As I said before, I'd add a capacitor C3 between the nodes Vn and Vp to model the "internal/intrinsic capacitance between terminals".

That is pointless.
A voltage source has zero impedance. The Vp and Vn terminals are voltage clamped internally. Since the differential voltage cannot change, C3 will have a constant charge; C = Q / V = dq / dv .
Delete the idea of C3.

 

[Baluncore]

Let us assume that during the connection and grounding process, the excess electrostatic charge on the surface of the battery terminals is fixed.

As the flying battery falls towards the ground, the area, a, of the terminals remains constant, while the separation distance, d, between the terminals and ground, is being progressively reduced. The terminal to ground capacitance is proportional to, a / d, so the capacitance is increasing.

The definition of capacitance; c = q / v ;
v = Q / c .
Since the charge is fixed, and the value of capacitance is increasing, the battery terminal voltage will be falling relative to ground. At the instant of grounding, the voltage will reach zero, and the transported excess charge will be dumped, from the battery terminals to the ground.

 

[cianfa72]

Delete the idea of C3.

From a circuit perspective, sure, adding the capacitor C3 is pointless. However my point was to model how things actually work: the battery EMF will "move" electrons away from the terminal that will be the positive to the other terminal that will be the negative one. This process will keep going until the battery inter-terminal intrinsic capacitor will be fully charged at a voltage/PD equals to the battery EMF.

Let us assume that during the connection and grounding process, the excess electrostatic charge on the surface of the battery terminals is fixed.

Ok, as you pointed out before, the total electrostatic charge will be distributed on the surface of battery's terminals according to the relative effective terminal capacitances and battery EMF -- see Capacitance matrix

As the flying battery falls towards the ground, the area, a, of the terminals remains constant, while the separation distance, d, between the terminals and ground, is being progressively reduced. The terminal to ground capacitance is proportional to, a / d, so the capacitance is increasing.

The definition of capacitance; c = q / v ;
v = Q / c .
Since the charge is fixed, and the value of capacitance is increasing, the battery terminal voltage will be falling relative to ground.

Sorry, by the battery falling towards the ground, you actually mean the process by which the wire attached to the negative terminal is progressively brought closer and closer to the ground until the instant of grounding.

At the instant of grounding, the voltage will reach zero, and the transported excess charge will be dumped, from the battery terminals to the ground.

Ok, yes.

 

[Dale]

Delete the idea of C3

I am not so sure about that. Batteries have rate constants which could be meaningfully represented by an internal resistance and a capacitance. I am not sure if those would best be in series or parallel with the EMF.

 

[Baluncore]

I am not so sure about that. Batteries have rate constants which could be meaningfully represented by an internal resistance and a capacitance.

One or two picofarads of inter-terminal capacitance, is quite irrelevant to the model of a battery of chemical cells.

If you replace the battery with a model voltage source, any parallel capacitance becomes meaningless, as the capacitor voltage will never change.

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

 

[cianfa72]

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

What about an ideal voltage source V with a series resistor between the node Vn and Vp of post#22 schematic with a parallel C3 inter-terminal capacitance ?

 

[Baluncore]

What about an ideal voltage source V with a series resistor between the node Vn and Vp of post#22 schematic with a parallel C3 inter-terminal capacitance ?

As a battery, the series resistance would need to be less than 1Ω, or it would not be used. The time constant of the series resistance, with a 1pF terminal capacitance, would be of the order; 1Ω * 1pF = 1 picosecond. That is shorter than any simulation step.

For a 12 volt battery, the charge on the 1pF terminal capacitance would be;
q = c * v = 12 pC.
A 1 A⋅hr battery will store and deliver, 3600 C. The 12 pC terminal charge, will be lost in the round off of the 15th digit, during a double precision simulation.

 

[cianfa72]

Btw, I believe a sensitive approach, at least in principle, boils down to setting up a (quasi) Dirichlet problem for the Laplace equation. The boundary condition is the PD between battery's terminals (including the zero potential for the "Earth conductor"). As the battery flies through space, the system capacitance matrix changes accordingly, and battery EMF pushes/moves charges (i.e. electrons) on its terminals to keep fixed their PD voltage (in order to cope with the varying terminals capacitance matrix coefficients w.r.t. Earth).

In this model the charges on the 3 conductors involved (the two battery's terminals + Earth) no longer have to be equal even in magnitude.

 

[Dale]

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

I am not convinced that it will be 1-2 pF. For the internal capacitance the important structure would be the electrodes, which are usually much larger than the terminals.

One or two picofarads of inter-terminal capacitance, is quite irrelevant to the model of a battery of chemical cells.

You have staked your position on a number that I don’t accept.

The time constant of the series resistance, with a 1pF terminal capacitance, would be of the order; 1Ω * 1pF = 1 picosecond. That is shorter than any simulation step

Real batteries have measurable time constants that are longer than typical simulation steps. So one or both of these estimates is wrong.

Here is a 12 V Li battery model using two internal capacitances: https://www.nrel.gov/docs/fy01osti/28716.pdf I am not intending that as a definitive general model, just a single experimentally validated battery model that indeed shows capacitance.

 

[Baluncore]

Here is a 12 V Li battery model using two internal capacitances:

That models internal resistance as about 2 milliohm.
The short term chemical storage is modelled as 80 kilofarad.
Any Vp to Vn terminal capacitance, is swamped by the internal virtual voltage source.
Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

 

[Dale]

That models internal resistance as about 2 milliohm.
The short term chemical storage is modelled as 80 kilofarad.
Any Vp to Vn terminal capacitance, is swamped by the internal virtual voltage source.
Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

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

 

[cianfa72]

Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

This because we have a loop consisting of voltage source and two capacitors (the capacitances between each terminal and ground). Technically the charges/voltages on such capacitors are not both state variables since there is a KVL constraint for the loop. Therefore the total charge is redistributed between the two terminal capacitance accordingly.

Indeed the above loop doesn't have a unique solution: one has to fix the charge on a capacitor (or their sum) in order to get an unique solution.

However, as I said in #41, I believe the above is just a rough model. This because there is no reason why the charges on the "Earth's plates" should be equal in magnitude to the charges on the respective battery terminal "plates".

 

[Baluncore]

Indeed the above loop doesn't have a unique solution: one has to fix the charge on a capacitor (or their sum) in order to get an unique solution.

The state variables are:
Q, excess charge on the flying battery.
C1, varying capacitance between ground and terminal p.
C2, varying capacitance between ground and terminal n.
Vp, voltage on p, relative to ground.
Vn, voltage on n, relative to ground.

Vb, the battery voltage, is a parameter. Vb = Vp - Vn .

Solve for Vn ;
Vp = Vn + Vb .
Q = ( Vp * C1 ) + ( Vn * C2 ) .
Vn = ( Q - Vb * C1 ) / ( C1 + C2 ) .

So the excess charge, Q, is shared between the terminal capacitors. That charge flows through the low-impedance voltage source during the balancing process.

Any inter-terminal capacitance, C3, between n and p is irrelevant, as the voltage is fixed by the low-impedance voltage source. V( C3 ) = Vb.

 

[cianfa72]

Solve for Vn ;
Vp = Vn + Vb .
Q = ( Vp * C1 ) + ( Vn * C2 ) .
Vn = ( Q - Vb * C1 ) / ( C1 + C2 ) .

 Ok. The excess charge you were talking about is just the total "external" electrostatic charge placed there or it also includes charges stored there by battery EMF ?

So the excess charge, Q, is shared between the terminal capacitors. That charge flows through the low-impedance voltage source during the balancing process.

By balancing process you mean the process that "redistribute" the excess total charge Q on battery's terminals/ground capacitor "plates".

 

[Baluncore]

The excess charge you were talking about is just the total "external" electrostatic charge placed there or it also includes charges stored there by battery EMF ?

The charges on the surface of the battery terminals, created by the internal chemical EMF, will appear between the terminals of the virtual C3 in this discussion. C3 is a single capacitor, so the charges on the terminals of C3 must be equal and opposite. The excess external charge, stored on C1 and C2, is independent of the charge on C3.

 

[sophiecentaur]

Ah ok, you did a simulation using LTspice. C1 and C2 are both discharged through R1 since the current through R1 is always the sum of I(C1) and I(C2).

Simulation programs can be a snare and delusion because they tend to make assumptions about those components in a circuit that you don't specify (or even notice). You should hesitate to conclude that the software 'got it wrong'; it just tried to help you by filling in the gaps in your data with its own estimates.

It's incredibly easy to find a disagreement between what your basic theory and a simulation tells you. So beware and choose the direction of what the basic theory tells you. When there's a problem then start looking for something that your basic approach has missed or failed to specify. EMFs can be difficult to quantify.

 

[cianfa72]

Simulation programs can be a snare and delusion because they tend to make assumptions about those components in a circuit that you don't specify (or even notice). You should hesitate to conclude that the software 'got it wrong'; it just tried to help you by filling in the gaps in your data with its own estimates.

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 referenced 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 .tran simulations, 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.

 

[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.