Batteries, continually improving, but keep going flat

[Moonshed]

Batteries today are in a different league compared to those of just 20 years ago. The ones used in mobile phones and tablets are a good example.

However, they all suffer from the same problem, which is after being fully charged, and even if left completely unused, the battery will, sooner or later, go flat.

Tbis is my question, daft though it probably is. Where does all that energy go that was put into the battery to charge it up?

That’s it.

Thanks 

Keith

[saac]

9 minutes ago, Moonshed said:

Batteries today are in a different league compared to those of just 20 years ago. The ones used in mobile phones and tablets are a good example.

However, they all suffer from the same problem, which is after being fully charged, and even if left completely unused, the battery will, sooner or later, go flat.

Tbis is my question, daft though it probably is. Where does all that energy go that was put into the battery to charge it up?

That’s it.

Thanks 

Keith

Ultimately it dissipates as heat following conversion from electrical potential to chemical energy. 

Jim 

Edited October 6, 2023 by saac

[Moonshed]

1 minute ago, saac said:

Ultimately it dissipates as heat following conversion from chemical energy. 

Jim 

Really, it’s that simple, the battery dissipates the stored energy as heat? That being the case is that why batteries kept in  the cold go flat sooner than those kept in the warm because they are able to radiate heat faster? Or is that not true about batteries in the cold going flat quicker?

I can’t get over there being such a simple explanation. I expected genius types (no disrespect to yourself of course I’m sure you are also super clever) with Masters degrees in physics and university lecturers and theoretical particle physicists and scientists at CERN, all to get into terrible arguments about it. But instead you come along and say the energy is dissipated as heat! I’m sorry but that’s just not good enough! Dear oh dear!  Please come up with some super clever and confounding theory that sounds a lot cleverer.

Thank you

Keith

[globular]

Fairies pinch the power to charge their 90s game boy consoles.
In cold weather the fairies also use the power to heat their wing warmers.

[Moonshed]

6 hours ago, globular said:

Fairies pinch the power to charge their 90s game boy consoles.
In cold weather the fairies also use the power to heat their wing warmers.

Now that’s more like it!

[George Jones]

15 hours ago, Moonshed said:

However, they all suffer from the same problem, which is after being fully charged, and even if left completely unused, the battery will, sooner or later, go flat.

These days, can't Auto-Tune be used to fix this?

Edited October 7, 2023 by George Jones

[DaveL59]

The problem with a lot of modern electronics gear is there is no true off switch, it's done instead by a momentary action switch that triggers a component to energise the main board. So there's likely some leakage there and if a BMS is fitted it may be adding a small drain also.

On some cars that have a bulb monitoring system you can also find that they will add drain on your car battery if left parked for long periods, more so if you switch from filament to LED bulbs that are CANbus types. The Alfa 156 IIRC could suffer from this which was likely part of the issue I had with it, battery would be down to 50% in 2 weeks on the v6 so I kept it hooked to a smart charger when not in use.

Other than that, as Jim said, batteries will chemically return to a neutral state over time, the decay lost slowly as heat. Downside can be the degradation of the electrodes/electrolytics meaning less total capacity over time.

[saac]

19 hours ago, Moonshed said:

Really, it’s that simple, the battery dissipates the stored energy as heat? That being the case is that why batteries kept in  the cold go flat sooner than those kept in the warm because they are able to radiate heat faster? Or is that not true about batteries in the cold going flat quicker?

I can’t get over there being such a simple explanation. I expected genius types (no disrespect to yourself of course I’m sure you are also super clever) with Masters degrees in physics and university lecturers and theoretical particle physicists and scientists at CERN, all to get into terrible arguments about it. But instead you come along and say the energy is dissipated as heat! I’m sorry but that’s just not good enough! Dear oh dear!  Please come up with some super clever and confounding theory that sounds a lot cleverer.

Thank you

Keith

Yep, it really is that simple, unless you want to look at the specific chemistry changes for a particular battery.  Energy moves from a high state to a low state always, if you wish to put a fancy scientific/engineering tag on it then call it the first law of thermodynamics and the march of entropy    When the battery was charged for every 1V of potential difference across the terminal 1 J/ coulomb of electrical energy was being  invested in the charge carriers changing the constituent chemistry of the battery.  From a thermodynamic perspective this raises the internal energy of the system , bound up in the chemical state of the battery.  If the chemistry within the battery was static, immune to change then that charge would remain, theoretically indefinitely as an energy store. However the chemistry is not static and the constituents of the battery will continue to react (degrade), and as the first law of thermodynamics demands energy will move from a high energy state to a low energy state. Low grade heat tends to be the ultimate dissipation of energy in the universe.  Low temperatures are just facilitating the degradation of the chemical constituents but in many cases may actually reduce the rate of discharge by slowing the chemical reaction.   Now if you had charged a capacitor instead of a battery and assuming there was no leakage then the capacitor could in theory hold the charge indefinitely.  Isolating the capacitor from the rest of the universe to prevent leakage of charge is the trick. 

As you ponder this, if you still feel  cheated that the answer is not tied up in quantum superposition and the influences of distant black holes merging, pour yourself a hot cup of coffee.  Set it aside and marvel at the loss of heat after a few minutes. If you find yourself ever doing that and the temperature of the coffee increases, go back to bed, you have woken up in the wrong universe  

Jim 

Edited October 7, 2023 by saac

[saac]

Here, this article provides an explanation which explores the chemistry changes taking place.  It may satisfy your search for a complex explanation.  

Jim 

https://sciencing.com/calculate-e-cell-2671.html

[Mandy D]

20 minutes ago, saac said:

When the battery was charged for every 1V of potential difference across the terminal 1 coulomb of electrical energy was being  invested in the charge carriers changing the constituent chemistry of the battery.

Say what?

I cannot follow your logic.

"1 coulomb of electrical energy"? Energy is not measured in coulombs, it's unit is the joule. Charge is measured in coulombs, where the charge on a single electron is 1.6 x 10^-19 J. The ampere is a rate of flow of charge of 1 coulomb per second, hence given the charge on the electron we can calculate how many flow past a given pont per second for a given current.

So, if you want the energy transferred into the battery (or from) it is the product of volts x current x time = volts x charge.

Hopefully, that makes sense.

[saac]

10 minutes ago, Mandy D said:

Say what?

I cannot follow your logic.

"1 coulomb of electrical energy"? Energy is not measured in coulombs, it's unit is the joule. Charge is measured in coulombs, where the charge on a single electron is 1.6 x 10^-19 J. The ampere is a rate of flow of charge of 1 coulomb per second, hence given the charge on the electron we can calculate how many flow past a given pont per second for a given current.

So, if you want the energy transferred into the battery (or from) it is the product of volts x current x time = volts x charge.

Hopefully, that makes sense.

It's the definition of the Volt.  1 volt is defined as 1 Joule per coulomb of charge (1V = 1 J/C) .  The emf of a cell therefore is the amount of energy given to each coulomb of charge that passes through the cell.  

Jim 

Edited October 7, 2023 by saac

[Mandy D]

14 minutes ago, saac said:

It's the definition of the Volt.  1 volt is defined as 1 Joule per coulomb of charge (1V = 1 J/C) .  The emf of a cell therefore is the amount of energy given to each coulomb of charge that passes through the cell.  

Jim 

Yes, that is correct and makes sense now you have reworded it, thank you. But, the coulomb is not the unit of energy, which was my main point.

[saac]

9 minutes ago, Mandy D said:

Yes, that is correct and makes sense now you have reworded it, thank you. But, the coulomb is not the unit of energy, which was my main point.

Correct the coulomb is not the unit of energy it is the unit of charge.  The unit of energy is the Joule (J) and the volt is defined as J/C.  

The effect of temperature on battery discharge is quite complex and I guess must be influenced a great deal by the chemistry of the battery itself. Internal resistance increases as the battery depletes and eventually a point is reached where, while still showing a sizeable emf across the terminals,  the terminal potential difference (tpd) is insufficient to push current through the internal resistance of the cell.  This can be the source of confusion with car drivers in winter wondering why their battery, showing an emf of 13V, won't start the car.  The condition of the battery really needs to be taken under load (the tpd) .

I wonder if we will get to a stage where super capacitors  replace many battery applications.

Jim 

Edited October 7, 2023 by saac

[Moonshed]

1 hour ago, saac said:

Yep, it really is that simple, unless you want to look at the specific chemistry changes for a particular battery.  Energy moves from a high state to a low state always, if you wish to put a fancy scientific/engineering tag on it then call it the first law of thermodynamics and the march of entropy    When the battery was charged for every 1V of potential difference across the terminal 1 J/ coulomb of electrical energy was being  invested in the charge carriers changing the constituent chemistry of the battery.  From a thermodynamic perspective this raises the internal energy of the system , bound up in the chemical state of the battery.  If the chemistry within the battery was static, immune to change then that charge would remain, theoretically indefinitely as an energy store. However the chemistry is not static and the constituents of the battery will continue to react (degrade), and as the first law of thermodynamics demands energy will move from a high energy state to a low energy state. Low grade heat tends to be the ultimate dissipation of energy in the universe.  Low temperatures are just facilitating the degradation of the chemical constituents but in many cases may actually reduce the rate of discharge by slowing the chemical reaction.   Now if you had charged a capacitor instead of a battery and assuming there was no leakage then the capacitor could in theory hold the charge indefinitely.  Isolating the capacitor from the rest of the universe to prevent leakage of charge is the trick. 

As you ponder this, if you still feel  cheated that the answer is not tied up in quantum superposition and the influences of distant black holes merging, pour yourself a hot cup of coffee.  Set it aside and marvel at the loss of heat after a few minutes. If you find yourself ever doing that and the temperature of the coffee increases, go back to bed, you have woken up in the wrong universe  

Jim 

Thank you @saac that is an excellent explanation and one that I can understand. It makes sense now.

Cheers

Keith

[Mandy D]

23 minutes ago, saac said:

Correct the coulomb is not the unit of energy it is the unit of charge.  The unit of energy is the Joule (J) and the volt is defined as J/C.  

The effect of temperature on battery discharge is quite complex and I guess must be influenced a great deal by the chemistry of the battery itself. Internal resistance increases as the battery depletes and eventually a point is reached where, while still showing a sizeable emf across the terminals,  the terminal potential difference (tpd) is insufficient to push current through the internal resistance of the cell.  This can be the source of confusion with car drivers in winter wondering why their battery, showing an emf of 13V, won't start the car.  The condition of the battery really needs to be taken under load (the tpd) .

I wonder if we will get to a stage where super capacitors  replace many battery applications.

Jim 

When the engine is cranking over at normal speeds and failing to start, the number of people who suggest the battery is "flat" never ceases to amaze me!

Failing availability of a proper battery tester, the simplest thing to do is stick the headlights on and watch the battery terminal voltage rapidly fall and the headlights dim.

The Russians use super-capacitors (are they monolithic carbon, I forget?) in parallel with truck starting batteries in Siberia as they perform far better in sub-arctic temperatures and will deliver the current that the battery is incapable of, yet the battery can slowly charge them for a second and subsequent attempt, despite appearing to be discharged. As you said earlier, the battery voltage might be sitting at 13 volts, but that is with no load. There are too many confusing specifications on vehicle batteries for the average bod to understand, but maybe the most important one is CCA (Cold Cranking Amps), which gets abused by battery manufacturers who choose their own conditions rather than SAE or DIN standards.

* I see your recent correction that now clarifies the point we have been discussing.

Edited October 7, 2023 by Mandy D
Footnote added

[saac]

We have some 10F and I think 20F supercapacitors at work which are fun to play around with. There are a few youtube videos out here where folk have rigged up capacitor banks using super capacitors and placed them in cars.  

Jim 

[saac]

7 minutes ago, Moonshed said:

Thank you @saac that is an excellent explanation and one that I can understand. It makes sense now.

Cheers

Keith

Somebody will hopefully come along and explain it from a chemistry point of view   

Jim 

[Mr Spock]

The loss of charge is caused by a microscopic singularity passing through the battery. Somehow, the energy emitted by the singularity shifts the chroniton particles in the battery into a high state of temporal polarisation thereby dissipating its energy flux. 

Note: reversing the polarity doesn't reverse the process.

[Mandy D]

10 minutes ago, saac said:

We have some 10F and I think 20F supercapacitors at work which are fun to play around with. There are a few youtube videos out here where folk have rigged up capacitor banks using super capacitors and placed them in cars.  

Jim 

Try this one: 130F @ 62 V. By my reckoning that is 250 kJ.

https://uk.rs-online.com/web/p/supercapacitors/1797440?gb=s

I had some incidental involvement in a project where they were dumping capacitor stored energy into lasers at power levels exceeding the power available from the entire UK grid!!!

Edited October 7, 2023 by Mandy D
Correction

[saac]

12 minutes ago, Mandy D said:

Try this one: 130F @ 62 V. By my reckoning that is 250 kJ.

https://uk.rs-online.com/web/p/supercapacitors/1797440?gb=s

I had some incidental involvement in a project where they were dumping capacitor stored energy into lasers at power levels exceeding the power available from the entire UK grid!!!

Yep nominal 250 kJ, that is pretty useful  

Jim 

[Zermelo]

1 hour ago, Mandy D said:

Try this one: 130F @ 62 V. By my reckoning that is 250 kJ.

https://uk.rs-online.com/web/p/supercapacitors/1797440?gb=s

I had some incidental involvement in a project where they were dumping capacitor stored energy into lasers at power levels exceeding the power available from the entire UK grid!!!

That's very impressive (and expensive).

When I was at school, the Farad seemed to be a hopelessly optimistic unit. 100μF was a big capacitor! Most were measured in nF, or even pF.

[DaveL59]

On 07/10/2023 at 21:47, Zermelo said:

That's very impressive (and expensive).

When I was at school, the Farad seemed to be a hopelessly optimistic unit. 100μF was a big capacitor! Most were measured in nF, or even pF.

so true, tho some of those large paper capacitors were quite impressive when given high voltage (mains) as confetti generators 😉 

[Carbon Brush]

A point not yet raised is cell (battery) construction. Think about NiCd or NiMH packs in power tools - yes 10 years back. They are lithium now.

When constructing these cells, the maker has to juggle the capacity (amps x hours) against their ability to retain charge when left.
The internal construction is a roll of metal foil with blotting paper on top soaked in clever chemical, then another metal foil.
Fasten a wire onto each metal foil. Roll it up tight put it into the can. You have a cell.
OK there is more to the chemistry and construction but the simple explanation works for me.

If the blotting paper is thin, you get a cell that will deliver a high current for your electric drill - it has low internal resistance.
It has high capacity (amps x hours) for a given volume so you can drill lots of holes between charges.
Great so far. Unfortunately this construction leaks charge away quickly.
After a year or two of use, you find the battery goes flat in a few days of storage.
If you had cells with thicker blotting paper, the drill wouldn't perform as well. But it would be charged when removed from the tool cupboard.

Another useless snippet of information.😁