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Why does voltage shoot up so quickly when charging?

spendlove

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Oct 22, 2019
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Yesterday I checked our off grid battery bank and it had dropped quite low, under 11 volts. So I cranked up the generator. I came back 3 minutes later to see on the meter the voltage was already up to 12.3. Why does the voltage jump up so quickly when charging with a generator? Seriously....3 minutes is all it took.
 
11V was measured under load / excessive charging current / batteries are degraded.

Pick your poison.

Measuring voltage under load does not give a true indication of the state of the battery. When charging is performed the load will be largely powered by the charger so the battery floats up to its resting voltage and of course is dragged higher by the act of charging.

Excessive charging current will cause the voltage to rise faster than would normally be seen simply because the battery can't cope with the current available, ie chemical reactions can't take place fast enough etc.

If your batteries are aged, sulphated, dry, or any other number of possibilities they will have considerably reduced capacity and will rise up in voltage without taking on much 'charge'.
 
What size is your battery bank? What amperage is your generator charging at? What chemistry is the battery? How old are they?
 
Yesterday I checked our off grid battery bank and it had dropped quite low, under 11 volts. So I cranked up the generator. I came back 3 minutes later to see on the meter the voltage was already up to 12.3. Why does the voltage jump up so quickly when charging with a generator? Seriously....3 minutes is all it took.
Because batteries have an internal resistance, which causes a voltage rise across the battery terminals in response to charging churrent flowing into the battery. The same internal resistance causes a voltage drop across the terminals when current flows out of the battery into a load.

So although your voltage rose quickly in response to the charging from the genny, you would have seen it drop back down quick enough if you shut it off, and only then does the battery voltage give you a true reflection of the extra charge you’ve put into the battery.
 
With the charger on, you read the voltage of the charge source not the battery voltage.
 
The problem I am seeing now, is that there is not enough sun to charge my batteries and use them at the same time, and I am finding I am in the workshop more than I anticipated this winter (blame covid). I have a meter that measures energy in and out not voltage referenced, and I also have an inverter that will flip between mains and battery as per voltage settings, and also contains an 85A battery charger.

So lets say I discharge to 50% (according to the meter), but the voltage is still high because there are only moderate loads - the charger very quickly goes to a float status, even though there are 500AH missing. At float it would take all week to put back in - and these batteries are only a few months old. I simply can't charge the batteries as they wont accept the current.

So following this thread are you saying I need to reduce the charge current to a more acceptable level to get them to accept it.
these are brand new AGM batteries, 1040AH but only ever take 500AH out of them before I ether mains charge or shut shop for them to recover.
It seems the charger sees the voltage and thinks they are full.
 
What voltage is your charger changing to float mode? Absorption for a typical AGM is 14.2 to 14.7V. Once there the charger should not just switch back, it should hold there until the current tails off, for a 1040AH battery call it 30A, or until many hours have passed assuming it has a time limit.

Lifting a half charged 1040AH battery to full charge is going to take many hours with an 85A charger. You'll get to 60% SOC in 2-3 hours or so, from then on it's a long march to full capacity, another 5 hours or perhaps more. This assumes your battery is in good condition, and you took 500AH out of it.
 
What voltage is your charger changing to float mode? Absorption for a typical AGM is 14.2 to 14.7V. Once there the charger should not just switch back, it should hold there until the current tails off, for a 1040AH battery call it 30A, or until many hours have passed assuming it has a time limit.

Lifting a half charged 1040AH battery to full charge is going to take many hours with an 85A charger. You'll get to 60% in 4 hours or so, from then on it's a long march to full capacity, another 5 hours or perhaps more.
I bet that's it then, because it starts at 75A and then drops to about 3 amps after a few hours, even though the battery is hungry for more. Its a 12v system, not the best I know, but when this inverter packs up I will go with higher voltage. It floats at 13.8 and boosts at 14.4 and I disabled equalise. Its all configurable in the menu (within reason).
 
Monitor your battery voltage and current directly, ie use a decent meter. A clamp multimeter etc. Ensure your charger is actually going to absorption/boost voltage. Monitor the amps flowing. If they rapidly fall once you reach absorption voltage either your battery was not really half discharged, or it's crook. The values in this graph do not refer to your specific battery & charger, but they give you an idea of what to expect in terms of percentages.

1604997968239.png
 
If you did stop the charging cycle before measuring, you might not have waited long enough.
There's a "plate effect" (e.g., lots of electrons on the plate that haven't made it into ions yet), typically you have to let them sit 15 minutes or so after charging to get the actual voltage/state-of-charge.
 
All rechargable batteries have two components changing its terminal voltage. First is effective internal resistance of internal connections and electrolyte resistance. Second is kinetic energy to create and move ions.

Lead acid is fairly high on both (compared to LFP). Lead-acid battery has to breakup and recombine a lot of different molecules. This take more overhead energy which manifests itself as added terminal voltage drop during discharge and terminal voltage rise during charging. When battery gets near full charge or near full discharge it is harder to find available molecules needed to convert to meet current demand and kinetic voltage goes up.

There is a third component effecting rechargable batteries during charging. It is called surface charge which is effectively an electrolytic capacitor charge that occurs at layer surfaces. Different chemistry batteries have different amount of surface charge amount and its dissipation rate. Lead acid is moderate, LFP is very low columbic charge amount and dissipates quickly.
 
I bet that's it then, because it starts at 75A and then drops to about 3 amps after a few hours, even though the battery is hungry for more. Its a 12v system, not the best I know, but when this inverter packs up I will go with higher voltage. It floats at 13.8 and boosts at 14.4 and I disabled equalise. Its all configurable in the menu (within reason).

If the battery is discharged to 50% (or about 500AH) then boost charging at 75A should last about 5 hours. An indicator of high cable resistance is if charging current drops to ~3A after a few hours . Charging cable should be at least 1/0. Also, 14.4V boost is quite low.

What brand converter are you using? Which batteries?
 
Hi guys I have sorted it. The problem is that if I turn off and on the charger/inverter whilst charging at high current, it will then think the batteries are fully charged, and go in to float mode. I must have done this whilst working on the system, and never made the connection.

The batteries and inverter are these.



The wire I use is this
 
4 awg cable may or may not be adequate. You want to design to a 1% voltage drop, or cable voltage drop of 0.144V.

At 0.21 m Ohm/ft cable resistance and 75 A charging current, the maximum cable length is:

cable length = 0.144 V / (0.21 mOhm/ft * 75A) = 9 ft (round trip)

The charging source can be max 4.5 ft from battery to maintain 1% voltage drop.
 
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