Battery Discharged - But Potential difference exists without any current flow

[FreshmanNewbie]

My below question is regarding batteries.

I have a USB charger and a mobile battery (Li-ion battery).

So, my mobile battery percentage has dropped to 0%.

So, what I do now - Take the USB Charger. Plug it in the wall socket and measured the output voltage of the USB Charged. I measured it to be 5V.

But once I connected it to my mobile battery, I measured the voltage in the USB Line and the current through it. (Ignored the D+ and D- lines)

I measured the voltage to be around 4.6V and current to be 0.92A.

My questions :

1. When my mobile battery voltage is showing 0%, but the voltage between my mobile battery terminals in 4.6V. I have read that it stops around that voltage because some chemical reactions cannot occur. But My fundamental question is - Even while having 4.6V, why cannot my battery provide current to the mobile and keep it working? Like, the battery has potential difference but is not providing any current? Why is that?

(This is not specific to the only Li-ion batteries. I have seen 12V Lead acid batteries also not providing current even though the voltage across the terminals is around 11V? Why is this happening even though they have the potential difference? Like potential difference should cause current flow right? If potential difference is present, why is the current not flowing from the battery)

1. My USB charger was showing 5V before connecting to the mobile battery. But once I connect the charger to the mobile battery, the voltage on the line is showing 4.6V and the current is around 0.92A? The output of the USB charger is supposed to be constant regardless of the load current, right? If not, then it is not a good regulator or charger, right? My question here - Why is the voltage dropping to 4.6V instead of being 5V, when connected to the mobile battery and what determines the current value to be 0.92A?

My USB Battery charger rating is 5V & 1A. So, shouldn't it output 5V & 1A instead of 4.6V and 0.92A?

Please help me to clear my misunderstanding the about two concepts of charging batteries

 

[d123]

Hi,

Not sure but it might be related to battery impedance.

Vout = Vin x 1/(1 + (Rsource/Rload))
Zin = Vin/Iin
Zout = (Vin - Vout)/ Iout

Zout << Zin = Rload >> Rsource

If battery impedance is lower than source impedance (the charger), it's like a voltage divider so the 5V is reduced to 4.6V.

4.6V/0.92A = 5 Ohms

Also, I have a cheap lab supply whose 12Vout is never above 11.76V for very light loads, and the 5V setting isn't much better... Your charger may be struggling with the load or just have loose regulation.

I think, I might be wrong, battery impedance is dynamic, so the more charged it is, the higher the impedance/resistance and so the charging voltage rises as it is providing less current. More current, less voltage and vice versa.

Hope I haven't said anything incorrect.

 

[betwixt]

I think you confuse the terminal voltage with the battery capacity. Most batteries will still show a voltage across their terminals when discharged but it drops away as soon as you try to draw current from it. A charged battery will sustain the voltage under load. if you want a better representation of how much charge there is, you have to measure the voltage while at the same time drawing current.

In layman's terms it is like d123 explains, the 'effective series resistance' increases rather than the voltage dropping. It certainly isn't the technical reality but you can think of the battery always providing full voltage but a restriction being placed between the voltage source and the terminals. As the restriction increases (the battery discharges) it doesn't change the voltage but it does change how much of it gets past the restriction to where you need it.

Brian.

 

[FvM]

the battery has potential difference but is not providing any current? Why is that?

Most likely because a LiIon battery includes a protection circuit cutting discharge current below a certain voltage. Without protection, the battery might catch fire on deep discharge.

My USB Battery charger rating is 5V & 1A. So, shouldn't it output 5V & 1A instead of 4.6V and 0.92A?

A battery charger implements a current limiting function. Presumed you measured correctly, the current limit seems to be set a bit below 1A. Not unusual.

 

[FreshmanNewbie]

Most likely because a LiIon battery includes a protection circuit cutting discharge current below a certain voltage. Without protection, the battery might catch fire on deep discharge.


A battery charger implements a current limiting function. Presumed you measured correctly, the current limit seems to be set a bit below 1A. Not unusual.

When you say, "a protection circuit cutting discharge current below a certain voltage might be included" - does this include the 12V lead acid batteries? Could you please explain what happens in the case of the 12V lead acid batteries a little briefly

 

[FreshmanNewbie]

I think you confuse the terminal voltage with the battery capacity. Most batteries will still show a voltage across their terminals when discharged but it drops away as soon as you try to draw current from it. A charged battery will sustain the voltage under load. if you want a better representation of how much charge there is, you have to measure the voltage while at the same time drawing current.

In layman's terms it is like d123 explains, the 'effective series resistance' increases rather than the voltage dropping. It certainly isn't the technical reality but you can think of the battery always providing full voltage but a restriction being placed between the voltage source and the terminals. As the restriction increases (the battery discharges) it doesn't change the voltage but it does change how much of it gets past the restriction to where you need it.

Brian.

Thank you very much. But I am little confused with regard to the battery discharge. Could you please explain this with regard to battery chemistry on how the battery stops discharge when it is less than its nominal voltage by some amount and how the internal resistance comes into play and how the internal battery chemistry is stopped in the case of 12V Lead acid battery?

 

[KlausST]

Hi,

Not the battery nor the chemistry stops discharge ... it's the protection circuit that stops the discharge.

You should read some articles on batteryuniversity.com.

Klaus

 

[FreshmanNewbie]

Hi,

Not the battery nor the chemistry stops discharge ... it's the protection circuit that stops the discharge.

You should read some articles on batteryuniversity.com.

Klaus

But a 12V lead acid battery does not have a protection circuit. Could you please tell if where it would be for a 12V lead acid battery?

 

[d123]

Hi,

You can't compare lead acid behaviour to li-ion, or to Ni-Cad, or to NiMH... An hour of Googling and reading about it yourself saves a week of questions.

An admittedly very questionable/dubious analogy might be of an elevator/lift where the emergency lighting works but the main electricity supply is down - it won't go anywhere. Or an upward pointing hose with a weak water source, the water hasn't got the strength to reach the top as it is only a weak trickle, not a gushing flow. My apologies if either example is nonsense.

Lead acid is 'very hard' to destroy so it has no protection circuitry except in the more sophisticated charging devices, old ones are pretty much just a transformer. Li-ion is easy to overheat, has a limited charging temperature range and can so cause a fire so protection is often included in the battery pack. Lead acid isn't prone to thermal runaway, Li-ion is. Lead acid battery in cars in winter or middle of summer, a robust technology.

Some Li-ion protection may have an under-voltage cut-off circuit to prevent excessive discharge making the battery/cell unusable, besides over-temperature, over-voltage and over-current protection circuitry. Maybe some have a Coulomb counter to cut-out when cell charge is down to 10% or 20%.

If you compare a few lead acid charger circuits to professionally manufactured Li-ion charger ICs and their associated circuits and look at the inner workings in datasheets that show them, it's really interesting.

 

[Robert L. Stripling]

yes you are right. i have been also facing this problem while using my welding helmet. i literally could not understand the performance of even high end batteries in high end welding helmets.

 

[KlausST]

Hi,

i literally could not understand the performance

What do you mean?

A battery provides electrical energy .... but the energy is limited...

Klaus

 

[Pjdd]

It's not good for the life of a rechargeable battery to extract the last ounce of energy from it. When your phone tells you that it has zero charge left, it does not actually mean that. When the charge has dropped to a safe minimum level, the phone cuts it off to protect the battery. Also, the phone may not work correctly with a very low battery and that's another reason to cut it off at some level.

Another factor is that, as the charge in a battery is used up, the internal resistance increases. This reduces its ability to provide current to a load. Even when the voltage is still fairly high without a load, the voltage will drop as soon as a significant load is connected.

Regarding the 12V lead-acid battery, you didn't specify the type of load connected to it, or if the 11V reading was taken with or without a load. Maybe the load has a cut-off mechanism; maybe 11V was the reading without a load and the voltage dropped severely when the load is connected. Otherwise, if 11V is the voltage with a load, then it should provide current flow.

 

[Audioguru]

Look up Lithium-ion battery at batteryuniversity.com. It says one cell is 4.2V when fully charged and its charging current has reduced by the battery to a low amount, and its minimum voltage is 3.0V to 3.2V. it becomes a fire hazard if it is discharged below its minimum allowed voltage.

The battery in a cell phone has a protection circuit in it that disconnects it if its voltage becomes low and the phone has a charging circuit in it that is powered from 5V.

 

[c_mitra]

Let me try to explain the basic electrochemistry. Batteries are electrochemical devices. They develop a potential and also supply current. When they are supplying current and also maintains a potential, it is delivering energy. This much is simple.

First thing first. What is the cause of the potential? The potential is produced at the electrode liquid interface and there are two electrodes and hence two interfaces. The battery potential is the algebraic sum of these two potentials. This is very important to understand.

Now the things start getting complicated. The potential of an electrode is determined by the charge transfer process at the electrode. So the potential of an electrode depends only on the material of the electrode and the contents of the solution (and the nature of the charge transfer process). It does not matter whether the electrode is small or big or the solution is just a drop or a bucketful. The concentration of the reactive species does matter and that is the reason that battery voltage drops by a small amount when it gets discharged. It is appearing complicated because I am trying to compress too much into too little space.

Now about current. As soon as you try to extract some current (you are taking away free energy from the battery) from the cell, chemical reactions will take place. When the battery is resting, there will be no current and no chemical reaction (ignore leakage current). This chemical reaction cannot go on indefinitely because the chemicals will get over sooner or later. You can have 12V lead acid battery in a small size or a large size. As far as chemistry and physics are concerned, they are identical except of the mass. The larger cell has more material and can supply current for a longer time.

When a battery is down, it just means that the chemicals are over and it cannot supply energy. The voltage has reduced but not become zero. The current is also reduced (become very very small) but is still there. It is not sufficient to run the phone.

The chemicals that are consumed during the discharge process can be regenerated by charging the battery. The process is not 100% efficient but let us not bother too much at the details.

 

[Robert L. Stripling]

Hi,


What do you mean?

A battery provides electrical energy .... but the energy is limited...

Klaus

yes, you are right but it must be good enough that you can perform your welding work for a while that you are doing.

 

[KlausST]

Hi,

If the battery (capacity) is correctly dimensionated then the battery will provide enough energy for the expected worktime...without getting in self destructive "under discharge voltage" region.

The "healthy battery voltage" region is specified .... if the welding helmet does not work the expected time, then the calculation was wrong. Blame it on the designer, not on the battery .... Unless the battery is old and needs to be replaced.

Klaus

 

[barry]

Your basic problem is that you are equating voltage and power. Power, as I hope you know, is voltage TIMES current. Just because your battery has 12V (or 100V or 1000V) at its terminals doesn't mean it can supply ANY current.

 

[dick_freebird]

Batteries have decreasing charge and increasing
series resistance as they discharge. At deep discharge
there ain't no more current to give, whatever the
voltage, so Rs appears high and there you are. You
may measure a sane appearing voltage as long
as you don't want anything from it.

The question of "battery gauge accuracy" in consumer
products is a whole 'nother topic.