Efficiency of charging new and old Li-ion batteries in electric cars

[Frodo]

TL;DR

How does the efficiency of charging a Li-ion battery vary as the battery gets older?

An electric car has, say, a 50 kWhr battery.

1. How much electricity, in kWhr, is needed to add 1 kWhr of stored electricity to the battery?

2. After several years, the battery capacity has, say, fallen to 25 kWhr. How much electricity, in kWhr, is now needed to add 1 kWhr of stored electricity to the battery?

I am trying to understand how the efficiency varies with battery age. Small electric cars average about 5 miles per kWhr of battery charge but I am wondering what mileage they get per kWhr of mains electricity.

 

[anorlunda]

A good source for battery technical issues is
https://batteryuniversity.com/

Ultra-fast charging, or boost charging, must be done under the right conditions. Charge efficiency varies according to SoC and battery temperature. As a battery ages, the internal resistance and the cell balance worsen and the charge rate must be slowed accordingly.

That suggests that you could keep charging efficiency higher if you slow the charging rate. So it comes back to you. Do you mean change in efficiency with age at the same charging rate, or when charge rate decreases with age?

 

[Frodo]

Thanks.

I am trying to answer the question:

How many miles do I get per kWhr of mains electricity

when the battery is new?​

when the battery is old?​

 

[Frodo]

Battery University is interesting. It says "Charge efficiency is about 99 percent and the cell remains cool during charge" so, at least for a newer battery, the charge efficiency is excellent.

It also says
"EV makers must further account for capacity fade in a clever and non-alarming way to the motorist. This is solved by over sizing the battery and only showing the driving range. A new battery is typically charged to 80 percent and discharged to 30 percent. As the battery fades, the bandwidth may expand to keep the same driving range. Once the full capacity range is needed, the entire cycle is applied. This will cause stress to the aging battery and shorten the driving ranges visibly."

 

[anorlunda]

How many miles do I get per kWhr of mains electricity

I hope you understand that the answer depends on how literally we read that question because it depends as much on the charger as it does on the battery. Smart/dumb/expensive/cheap/thermal-sensing all kinds of chargers have different efficiencies. Old fashioned voltage regulators in cars were sometimes only 50% efficient.

But if you mean to ask about the battery only, and not the battery plus charger combination, this figure from battery university should help.

Power loss during charging is $I^2R$. If we hold $I$ constant (which most chargers do not do) and hold temperature constant, then increasing R by the ratio 90/38 will increase losses by about 2.6. If the new battery was 99% efficient, then there was 1% losses. When the battery is old, the losses would increase to 2.6% making the charging 97.4% efficient. Is that what you are looking for?

But battery temperature is often more important than R, as the curve below shows. That is why they make the chargers smart. A smart charger that senses temperature automatically adjusts for the changed characteristics of the battery when it gets old and adjusts for ambient temperature. In that case, you can't answer the question about mains charging efficiency without detailed knowledge of power losses inside the charger.

Edit: By the way, my new PC has three settings for charging. It can charge up to 100% or 80% or 60% of rated capacity. Stopping at lower levels makes the battery age slower and extends the lifetime.

 

[Frodo]

Thanks. That is the answer I was looking for - basically, there isn't much difference in charging efficiency between old and new batteries so my "miles per kWhr of mains electricity" won't change much as the battery ages.

I think Battery University says you should always power off your PC before charging it and not keep the charger plugged in as chargers have problems when there are parasitic loads across the battery.

 

[TonyStewartEE75]

Be careful on your assumptions. If you use an ultra-rapid charge-current the energy dissipated in the ESR of the loop dictates that all circuit series resistance of I^2*ESR *t is lost energy. So the ratio of ESR/ Ah capacitance must be extremely low such that I\$^2\$*ESR*t <1% of 1/2CV\$^2\$ of the storage capacity. In other words ... I'll let you compute the ESR/ Ah ratio to satisfy that 1% by converting Wh to J * 3600.

Although this you would never do this. Connect a full discharged battery to a fully charged battery. The result is the voltage decays to the midpoint but the energy stored is not the same as before. e.g. connect two caps with 1uOhm ESR with one at 10V , the other at 0V. The result after is each cap is charged at half the voltage but only 1/4 of the energy * 2 so half the energy was lost in the ESR. But for batteries the range in voltage would be about 15%.

On the laptop, if the charger cannot supply max load and CC rate, it is a poor design .

 

[NTL2009]

...

Although this you would never do this. Connect a full discharged battery to a fully charged battery. The result is the voltage decays to the midpoint but the energy stored is not the same as before. e.g. connect two caps with 1uOhm ESR with one at 10V , the other at 0V. The result after is each cap is charged at half the voltage but only 1/4 of the energy * 2 so half the energy was lost in the ESR. But for batteries the range in voltage would be about 15%.

I think there is a flaw in your example. It didn't make intuitive sense to me. Assume the ESR was zero, why should we lose any energy at all with purely reactive components?

I plugged some numbers into this calculator:

https://www.omnicalculator.com/physics/capacitor-energy

and it became obvious to me when a certain 'magic number' appeared, and I was questioning the 5 V assumption because discharge is exponential. A 1F cap at 10V has 50 Joules energy. If we split that energy across two caps, the 'magic number' 7.071 V appeared. So the voltage will drop to V/(2^.5), V * 0.7071, not V/2.

On the laptop, if the charger cannot supply max load and CC rate, it is a poor design .

That's a separate issue. The problem described above is that it is challenging to know when the batteries are at full charge if there is also some current being bled off to the laptop. The charger can't 'know' if that is battery current, or laptop current.

 

[DaveE]

Assume the ESR was zero, why should we lose any energy at all with purely reactive components?

This has been discussed in a few other threads. This one, for example:
https://www.physicsforums.com/threa...-gained-by-the-capacitor.1006597/post-6534477

Also this: https://www.hep.princeton.edu//~mcdonald/examples/twocaps.pdf

 

[NTL2009]

This has been discussed in a few other threads. This one, for example:
https://www.physicsforums.com/threa...-gained-by-the-capacitor.1006597/post-6534477

Also this: https://www.hep.princeton.edu//~mcdonald/examples/twocaps.pdf

Interesting - I see it now. My thinking was that a low ESR should mean low loss, but... a low ESR means high instantaneous current, and the full voltage delta across that low ESR at t=0. So large current at X volts means a lot of watts! Example, 10V and one milliohm is 10,000 amps and 100,000 watts (for an instant).

 

[dlgoff]

This may be helpful:
From: https://batteryuniversity.com/

https://batteryuniversity.com/article/understanding-lithium-ion
https://batteryuniversity.com/article/bu-409-charging-lithium-ion