0
\$\begingroup\$

It seems standard for a lithium-ion charger to cut off the applied voltage when the CV-mode current draw dips below 0.1C (or thereabouts). Why is this necessary? Why can’t the charger continue to apply 4.2V indefinitely?

\$\endgroup\$
  • \$\begingroup\$ According to Battery University: Li-ion cannot absorb overcharge. When fully charged, the charge current must be cut off. A continuous trickle charge would cause plating of metallic lithium and compromise safety. To minimize stress, keep the lithium-ion battery at the peak cut-off as short as possible. See batteryuniversity.com/learn/article/… \$\endgroup\$ – Bimpelrekkie Jul 5 '17 at 12:41
  • \$\begingroup\$ Thanks. So I'm under the impression that in a series pack configuration the BMS would not be able to detect this sort of overcharging (since the cell voltages each stay at 4.2V). It's up to the charger to cutoff the applied voltage. Is that correct? \$\endgroup\$ – scttnlsn Jul 5 '17 at 12:54
  • \$\begingroup\$ Also, not sure how you detect the cutoff current if there's also a load present. Is it possible to charge li-ion batteries with a load connected? \$\endgroup\$ – scttnlsn Jul 5 '17 at 12:56
  • \$\begingroup\$ For cells in series the proper way is to detect the voltage across all individual (sets of) cells. So that if one cell is not full yet (say at 4.0 V) but the others are (at 4.2 V) that this must be detected and charging must stop. This is what is done in any properly designed laptop battery for example. Relying on all cells being equal and only using the sum of all voltages is not a good idea. \$\endgroup\$ – Bimpelrekkie Jul 5 '17 at 12:58
  • \$\begingroup\$ A load connected makes no difference, part of the charge current will be used by the load (assuming Iload < Icharge), the batteries will simply be charged more slowly. In the case Iload > Icharge then the cells will slowly discharge. \$\endgroup\$ – Bimpelrekkie Jul 5 '17 at 12:59
2
\$\begingroup\$

Applying 4.2V indefinitely to a LiIon cell will damage or destroy it.
Worst case metallic Lithium may "plate out" and the classic "vent with flame" failure mode may occur.

A LiIon cell is mechanically stressed by charge/discharge cycles due to the transport of Lithium ions in and out of the cell structure. [LiFePO4 cells almost eliminate this affect by providing a mechanical olivine "cage" into which the Li ions are "intercalated".

A look at the charge/voltage curve for a charging LiIon cell shows that towards the end of charge the voltage rises at a rate which is increasingly greater than the rate of charge absorption. This reflects an increasingly inefficient charging process as the last available areas are used to 'store' ionic Lithium.

Bythe time Ich under CV mode has dropped to 10% of Imax the cell is almost completely charged to capacity - "road warrior" mode where maximum possible energy is stored at the expense of lifetime. Cells charge terminates at 0.1 x Icv will have a short cycle life. Terminating at 0.5 x Icv still achieves > 90% of maximum capacity and stopping at 0.25 x Icv is probably preferable to 0.1 x Icv in most cases.

Proper design of charging circuitry makes it easy to detect current into and out of the battery while load varies from 0 to 100%. The battery and charging circuitry are fed from the charger input point as is the load. The load current may affect the amount of current available for charging (depending on charger capacity) but battery current is monitored regardless of direction or magnitude.

When charging multiple cells in series a modern BMS monitors the voltages of ALL cells individually. When one or more cells in a string reach their final state of charge ahead of others balancing circuitry 'shunts' current around the cell so that the cell is effectively separated from the charging string.

\$\endgroup\$
  • \$\begingroup\$ Thanks. Very helpful info. So with LiFePO4 cells you can hold them at 3.6V w/o risking safety or cell longevity? \$\endgroup\$ – scttnlsn Jul 5 '17 at 16:06
  • \$\begingroup\$ electronicsweekly.com/market-sectors/power/… \$\endgroup\$ – Bruce Abbott Jul 5 '17 at 16:47
  • \$\begingroup\$ @scttnlsn I'd not guarantee 100% that floating LiFePO4 at 3.6V would not to some extent reduce lifetime BUT it is probably tolerable. LiFePO4 have a very unusual charging behaviour under CC charge. They approach 3.6V or so at a reasonably linear rate of voltage increase then voltage rises very rapidly towards 4.2V (and beyond.). They seem to tolerate this final sudden increase well. I do not know the mechanism but suspect it is caused by Lithium ions absorption without being "intercalated" into the olivine matrix - ie you have a very small LiIon cell appended to the LiFePO4 main cell. ... \$\endgroup\$ – Russell McMahon Jul 7 '17 at 2:27
  • \$\begingroup\$ ... Some charger ICs (several from TI) make use of this characteristic and use only CC charging with cutoff at about 4.2V and no CV phase. \$\endgroup\$ – Russell McMahon Jul 7 '17 at 2:27
  • \$\begingroup\$ @BruceAbbott That article looks useful but I'm uncomfortable with the accuracy of some of the general assertions. But overall it seems useful. \$\endgroup\$ – Russell McMahon Jul 7 '17 at 2:28
0
\$\begingroup\$

All lithium-ion cells will continue increasing in voltage until the charge current is cut. So unless you have a damaged cell, the cell's voltage will pass 4.2 and can go as high as 4.5 or 4.6 V. So it is possible that with an ideal CV supply, to keep it at 4.2 V, and let the cell naturally cut off the current as its voltage rises. A cell receiving above 4.2 V will be incrementally damaged.

The problem arises when the CV supply is not perfect. I.e. as the current diminishes, its voltage would rise and exceed 4.2 V. Or a malfunction with the charger will make it supply 4.6 V instead of 4.2 V, at very low currents. So lithium-ion chargers in order to account for a malfunction of the supply, would want to cut the current at some stage. Some chargers have a timer, and all chargers will cut the current if it drops below a threshold (0.1C or even lower).

Then there is the problem of damaged cells that refuse to go above 4.2 V. In this case there should be the timer or undercurrent detector.

Also note that balancer-protection boards usually cut off above 4.2 V. In a series configuration, the balancer-protection board will cut off when any one of the cells exceeds the cut off voltage. It can be as high as 4.35 V given the tolerances, and this can damage the cell. That is why you should not rely on the balancer-protection to keep the cell voltage below 4.2. In a nS series connection, I recommend the charger to cut off at n * 4.2 V, and not rely on the balancer protection to kick in. Note that not all balancers have protection circuitry and thus may not cut off the current. Also note that balancers shunt the cell with only a minor current, as little as 50 mA. Thus at high-C charging, balancing may not happen if the cells in series are significantly different. I also believe that after cut off, whether by the charger or the balancer, balancing will continue until each individual cell drops below the balancing voltage (which could be as high as 4.35 V).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.