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how the battery behaves in this diagram

how a battery behaves in parallel to the charger with a shared load but with a voltage difference

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    \$\begingroup\$ Because the battery and charger are in parallel, they will be at the same voltage. You need to give more information about the type of battery and what the charger is intended to charge to determine how they get from their nominal voltages to their actual voltages. \$\endgroup\$
    – Neil_UK
    May 24 '19 at 5:46
  • \$\begingroup\$ How are you wanting them to behave? \$\endgroup\$
    – HandyHowie
    May 24 '19 at 6:50
  • \$\begingroup\$ Depends on the battery, the charger, and the load.It might work, it might shorten the battery life, or you might need to keep the fire brigade on speed dial. \$\endgroup\$ May 24 '19 at 7:14
  • \$\begingroup\$ You got that right. @BrianDrummond #speeddial \$\endgroup\$ May 24 '19 at 15:00
  • \$\begingroup\$ @Jerson. Your edit needs to be a totally new question which obsoletes comments and answers. This will require a BMS in addition to the MPPT charger to track battery current vs load current and charge voltage so when the battery has reached a certain SoC e.g.90~95% the voltage is reduced so the charger current is effectively directly going to the load and the battery is at the float level of whatever 3.7~3.9V/ cell. This will prevent Dendrites (nano shorts) from growing if the battery stays at 4.2 supplying the load > x% of CC for cutoff current in CV mode \$\endgroup\$ May 24 '19 at 15:09

My Xtal ball says you are thinking of this;

51V (58.8V) 5A Li-ion / LiFePO4 Charger Works with 14S Li-ion batteries and 16S LiFePO4 batteries.

When you put a load on the battery while charging it defeats the cut-off detected by the current in CV mode dropping below 5% ( or so) . Failing to do this can significantly reduce battery life by the amount of time spent in CV mode.

Otherwise, in CC mode, no problem since your charger will share CC (5A) between battery and load current, so it takes longer. ( just like your mobile)

But if in CV mode, you must limit the time spent charging at 58.8V if the system does not monitor Battery current separate from the load. ( read up at Battery University site)

  • \$\begingroup\$ "beware your load is now operating at 58.8V" - Just as it is when the battery is 100% full before the charger is disconnected and the load is turned on. The load will be able to handle it. \$\endgroup\$
    – JimmyB
    May 24 '19 at 11:19
  • \$\begingroup\$ @JimmyB A 48V motor with a 5 Ω load for equiv impedance at rated current may have a DCR of about 10% +/-2 of this or 0.5 Ω . If the charger is on at 58.8, it may have start-stop surge powers of 6.9 kW vs 4.6kW at 48V \$\endgroup\$ May 24 '19 at 12:40
  • \$\begingroup\$ Maybe. But the point is: A 14S Lithium at 100% SoC is at 58.8V too, even without the charger connected; so if the load can operate with a fully charged 14S Lithium, it can also operate when a 58.8V charger is connected in addition/instead. \$\endgroup\$
    – JimmyB
    May 24 '19 at 12:52
  • \$\begingroup\$ Over time, the open circuit voltage will settle to between 3.70V and 3.90V/cell. This is 100% SoC not 4.2 \$\endgroup\$ May 24 '19 at 12:58
  • \$\begingroup\$ My question is this. I'm using this ic mppt SM72445 along with a 250 watt solar panel, mppt output configured to 50.4 or 51 volts, 12-cell lithium-ion battery pack and bms, if the load is less than the output of mppt the battery will be charged, but what happens when the load is greater than the mppt output and there is a voltage difference between the mppt output and the battery? example ; mppt output = 50.4V 5A / battery pack = 46.8V 3.9Vcell * 12 / load = 10A my doubts are for the difference in voltage between the two sources mppt and battery in parallel \$\endgroup\$ May 24 '19 at 14:53

The charger seems to be limited to 5A, so if the load+battery tries to draw more than that, the charger will reduce its output voltage to a lower value until no more than 5A are drawn from the charger.

As long as the charger's voltage is higher than the battery's voltage, current will flow into the battery and charge it. If the load draws 5A or more, the charger's voltage will drop and finally reach the battery's voltage and any extra current draw beyond the 5A will be sourced from the battery.

If you're wondering if the setup will work:

Yes, it can work as desired, if the charger is not too smart and can handle extended periods of maximum output current.

A "smart" charger may detect that something seems to be wrong with the battery because it seems to draw much more current/accept much more mAh than it should. The most simple safety mechanism that could get into your way is just a safety timer in the charger set to turn off after a certain time after which the battery should definitely be fully charged. Notice that in your circuit the battery will never be charged but in fact be discharged although the charger is connected, which may trigger a couple of different safety mechanisms in the charger.

  • \$\begingroup\$ It must sense battery current separate from load current as in laptops but not in the case of external chargers. Where sensing is external. So it may not be protected for this.. \$\endgroup\$ May 24 '19 at 12:49
  • \$\begingroup\$ Li-ion does not need to be fully charged as is the case with lead acid, nor is it desirable to do so. In fact, it is better not to fully charge because a high voltage stresses the battery. Over time, the open circuit voltage will settle to between 3.70V and 3.90V/cell. THis is the correct 100% SoC not 4.2V/cell \$\endgroup\$ May 24 '19 at 12:58
  • \$\begingroup\$ @SunnyskyguyEE75 That's right, the "charger" may be only a 'dumb' power supply. \$\endgroup\$
    – JimmyB
    May 24 '19 at 12:59
  • \$\begingroup\$ No it is not a dumb supply , but if it is external it cannot distinguish between battery and load current. \$\endgroup\$ May 24 '19 at 13:00
  • \$\begingroup\$ @SunnyskyguyEE75 "3.70V and 3.90V/cell. This is the correct 100% SoC not 4.2V/cell" - I have never seen a BMS considering 3.7V as "100%". \$\endgroup\$
    – JimmyB
    May 24 '19 at 13:00

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