Before I connect these batteries in parallel…

[amberwolf]

There is a 100% chance that you will get more range (more energy out of the 2 packs) if you drain one battery pack, disconnect it, and then connect the second pack and drain that.

In my experiences, the opposite has always been true.

For me, paralleled packs of same cell type and voltage, regardless of differences in age and capacity and cell quality, give greater range (more total Wh) than any of them by themselves.

I suppose if you used packs so large that they were capable of far more current output than the system would ever draw even at peak, and so had no voltage sag and no internal losses to heat, then you would get about the same range / Wh whether paralleled or not.

But I haven't had a case yet where running the packs empty one after the other ever got more range / Wh than running them paralleled. (again, assuming same cell type (chemistry) and pack voltage (number of series cells).

 

[amberwolf]

Paralleled batteries are only as strong as the weakest c-rate cell in the paralleled group.

You do not parallelize a "80A" and a "40A" to get a "120A"...

It is still only a 40A continuous battery stack.

Er....not sure what you're meaning, but if you put two batteries (or cells) in parallel, then the total current you can draw from them does increase to the total of both of the cells. So if you have a cell capable of 2A, and another also capable of 2A, and you put them into a 2p pack, then you do get 4A capability out of that pack, where you would only have 2A if it was a 1p pack.

That's why we parallel cells to get higher current (A) capability from our packs. If that wasn't true, then the only way to get higher current delivery out of a battery pack would be to use cells that were capable of higher current, and you would never put cells in parallel at all (except to increase capacity, Ah).


Now, if you put an 80A and a 40A pack in *series*, then you only get a 40A capable pack...but if you put them in parallel, you do indeed get a 120A capable pack. For instance, if you had a "40A" 14s2p pack that was otherwise the same as an "80A" 14s4p pack, paralleling them gives you a 14s6p pack, capable of 120A.

If that is not true, then you can't have a 14s4p pack that does 80A either, made from the same stuff as the 14s2p pack, and you can't even have a 14s2p pack that is more capable than a 14s1p pack made of the same stuff.


A problematic cell could cause assorted problems with the actual output, depending on what the cell problem was, but with healthy cells there shouldn't be any problem. (if there is, something else, like interconnects or BMS, is causing or contributing to the problem).

 

[DogDipstick]

I was under the impression that C-rate was named that way in relation to Current.

No.

C-Rate = Capacity rate. Is, was, always has been, since 1860. I have some very very old books that do directly refer to it, and the E-Rate also.

E-Rate = ?????? rate.....I bet you can figure that one out pretty easy from the clues..? Eh? Spinny?

"E"bola rate? No.

"E"nterprise warp speed rate? nooo.... How much "E"cstasy is absorbed x Rave / Fire marshals? Noooooo


No one eve mentions the Energy rate, eh? Huh. E-Rate. Very important to know.

E-Rate: Discharge or charge power, in watts, expressed as a multiple of the rated capacity of a cell or battery that is expressed in watt-hours.

C in this context means the battery capacity in Ampere-Hours but divided by hours. "Capacity rate"...

The unit of the C-rate is h−1 ( h= hour) .....equivalent to stating the battery's capacity to store an electrical charge in unit hour times current in the same unit as the charge or discharge current. The C-rate is never negative, so whether it describes a charging or discharging process depends on the context.

Anywhay they are two very intertwined things. No one ever cares bout the E. They just think C.

Look into it. You will find it is and has always been "Capacity " rate. Like I said, I have 1860 books that refer to the calculation.

Go google " Current rate lithium" and see what pops up . Then do " Capacity rate lithium" and see what pops up.

....the...maths dont make sense if it is current. They do when its capacity.

"It is derived from the famous Coulomb’s Law invented by Charles Augustin de Coulomb, a historian, and French physicist. A C-rate can be defined as the measure at which a battery is discharged relative to the maximum capacity of the cell."

Kirchoffs laws, yall. KIRCHOFF KIRCHOFF KIRCHOFF did not lie . All resistors in parallel draw frocking identical current per node. PERIOD.

Hi AMBERWOLFIE NICE TO SEE YOU HERE! MISSED YA!

 

[Comrade]

So if you have a cell capable of 2A, and another also capable of 2A, and you put them into a 2p pack, then you do get 4A capability out of that pack

But not if it's a 3V 2A battery and 2V 2A battery.

It's a 3D problem that you are looking at it from a 2D perspective. I'm not sure if DogDipstick wants to touch this thread again, but he is much better with putting thoughts into words than me it seems. :lol:

 

[amberwolf]

Of course not.

But that's not relevant to the question, because differnet voltage cells can't be paralleled (safely) and neither can different voltage packs (if you do parallel them, then very shortly they'll be at the same voltage, and you'll have wasted energy as heat and potentially damaged things).

My post qualifies this by stating the number of series cells in one example, and simply calling them "the same stuff" in another, even if I did not specify exact voltages.

 

[amberwolf]

Kirchoffs laws, yall. KIRCHOFF KIRCHOFF KIRCHOFF did not lie . All resistors in parallel draw frocking identical current per node. PERIOD.

If they are identical resistors, yes.

If they are different resistances, then no.

If you put a 1ohm and a 2ohm resistor in parallel, then you get twice as much current thru the 1ohm resistor than the 2ohm resistor.

One way to state Kirchoff's first law is "the sum of currents flowing into that node is equal to the sum of currents flowing out of that node". If the paths within that node (like two paralleled resistances) are different resistances, then there are different currents in each of the paths, even though the total of those two currents is the same before it enters or after it exits that node.

In a battery (or set of them), the main power wires leading to and from the set of paralleled cells (or batteries) have the total current, while the current thru each of the parallelled cells (or batteries) is divided up depending on the internal resistance of each of those paralleled items.


As another way to say it, if you have paralleled cells (or batteries) with different internal resistances (whcih you will typically have if they have different current delivery capabilities), then each one will have different current flow thru it than the other, proportionally to that internal resistance.


If all the paralleled cells have identical internal resistance, and so do all interconnects, then every cell will also have the same current flow.

 

[amberwolf]

"Everything" isn't the same. By paralleling, you essentially halved the internal resistance of the overall battery pack.

Well I understand that the packs are under less resistant now but I guess I still don’t understand why there’s more watts.

So at 84v x 80a = 6.720kw and that’s exactly what the display showed before. Now with the second battery added it’s passing that 6.7 kW? I can understand if the last battery by itself failed to achieve the maximum power but it produce exactly 84 V x 80 A.

But anyways I have a little bit extra power definitely a good thing

If there are actually more watts displayed by the power meter under max load, then this is because there is less voltage sag with the paralleled packs because the pack total resistance is lower now. Less voltage drop means more voltage at the controller and motor, so even if the current draw is no greater, the voltage being higher means Amps x Volts = Watts is greater.

If there are no more watts displayed by the power meter during max load but you still get higher torque from the system, then it probably means the reading you see is just an instantaneous max reading that's higher than the actual system loading, and doesn't represent the actual power usage of the system. In this case the system may be dragging the voltage of the single pack down significantly more than the dual pack, while under max load, and so the actual V x A (watts) is significantly lower with that single pack, and you get less torque from the motor.


If you don't have a realtime display but are only seeing a peak reading on the meter, then you can't tell what's going on, and you'd need to use a meter that gives realtime voltage and current (to calculate watts) or realtime watts, to see which one is the case.

 

[999zip999]

Eastwood I see it like this do you have a pond and you have a hose going downhill you open up full blast how much water comes out take that same hoes and you put it beneath the Hoover dam I need to see how much water comes out I think it's the force of the pressure.
What ?

 

[Comrade]

But that's not relevant to the question, because differnet voltage cells can't be paralleled

Take cell A with 4V and 50mOhm resistance. Then take cell B with 4V and 60mOhm internal resistance. Now parallel them. Now connect them to a load. What voltage is cell A putting out? What voltage is cell B putting out? Are they the same or different? I think you just paralleled cells with different voltages.

 

[spinningmagnets]

I will definitely research this. Thank you for the search terms.

 

[Chalo]

Paralleled batteries are only as strong as the weakest c-rate cell in the paralleled group.

You do not parallelize a "80A" and a "40A" to get a "120A"...

It is still only a 40A continuous battery stack.

Incorrect.

All resistors in parallel draw frocking identical current per node.

Incorrect. P=I²R for each resistance. That doesn't change just because you put things in parallel.

I think you may misunderstand the meaning of "node". Each discrete resistance (battery) has its own node. Just because they're in parallel doesn't make them a single node.

frocking Kirchoffs laws. Learn em. I am tired explaining things to people and its not very complicated. Its VERY clear cut.

Evidently it's too complicated for you, because Kirchhoff's laws don't say what you're saying.

Batteries in parallel add their current capacities together, and they move current in proportion to their individual capabilities.

Yelling about it doesn't make you less wrong.

 

[Comrade]

Batteries in parallel add their current capacities together, and they move current in proportion to their individual capabilities.

That's the key. Connect two batteries of the same capacity in parallel that have different internal resistances, and the battery with the lower resistance will be providing more current. So they will be discharging at different rates. But since the batteries are parallel, the battery that is discharged more, will have a lower voltage, meaning the battery with the higher voltage (the one with more charge left) will have to constantly charge the other battery. Taking energy out of a battery is not 100% efficient. Putting energy into a battery is not 100% efficient. Hence less total capacity than if the batteries were not paralleled.

 

[DogDipstick]

I think you may misunderstand the meaning of "node". Each discrete resistance (battery) has its own node. Just because they're in parallel doesn't make them a single node.


Evidently it's too complicated for you, because Kirchhoff's laws don't say what you're saying.

Evidently it's too complicated for you, because Kirchhoff's laws don't say what you're saying.

Lol Chalo. Please dont try to insult my intelligence.

Kirchoffs current law and Kirchoffs voltage law ABSOLUTELY apply to cells in a group and the individual resistances they represent.

100%.

Sum of the current in is the sum of the current out. Period.

Should I find a engineering electrical expert with years of teaching under his belt saying the exact same thing, with references to batteries....? ..... ?

Do you think I pull this stuff out my ass without references standing by? Well accredited references? Since you rarely believe anything I say, friend? Often do I prove you wrong? b ( like the 7005 being stronger than the 6061.. that kind of stuff, friend.,,...)....

Followed by millions of people, electrical designers of many many years, doing the equations node by node, on a whiteboard, for everyone in the world to discuss and ....On camera with a step by step tutorial on how the math works? So we all can, right here, learn about it?

Specifically, the application of Kirchoffs Current law and Voltage law.. to batteries?

Hmm?




CAUSE I CAN .

 

[DogDipstick]

Evidently it's too complicated for you, because Kirchhoff's laws don't say what you're saying.



Yelling about it doesn't make you less wrong.

I got 50 frocking references here waiting. Videos and tutorials. Models. Whiteboard equations. By popular youtubers and academic scholars.

Ya know?

You want them? Chalo? Or is it "too complicated" for me?

Ex:

A state-space model for Li-ion battery packs with parallel connected cells is introduced. The key feature of this model is an explicit solution to Kirchhoff's laws for parallel connected packs, which expresses the branch currents directly in terms of the model's states, applied current and cell resistances. This avoids the need to solve these equations numerically. To illustrate the potential of the proposed model for pack-level control and estimation, a state-estimator is introduced for the nonlinear parallel pack model. By exploiting the system structure seen in the solution to Kirchhoff's laws, algebraic conditions for the observer gains are obtained that guarantee convergence of the estimator's error. Error convergence is demonstrated through an argument based upon Aizerman's conjecture. It is hoped that the insight brought by this model formulation will allow the wealth of results developed for series connected packs to be applied to those with parallel connections.

I did not write that, BTW.

Not that it is too " complicated" for me to understand. Grrrrrrrr BARKBARK! Grrrrrrrr Complicated Grrrrrr.

 

[Eastwood]

If there are actually more watts displayed by the power meter under max load, then this is because there is less voltage sag with the paralleled packs because the pack total resistance is lower now. Less voltage drop means more voltage at the controller and motor, so even if the current draw is no greater, the voltage being higher means Amps x Volts = Watts is greater.

Ahh yeah That makes sense, less voltage sag equals more watts.
Thanks!!

 

[Chalo]

I did not write that, BTW.

And it also doesn't say what you're saying.

You're not making sense. If you put a 6V forklift battery in parallel with a 4 pack of AAA cells at the same voltage, the parallel capacity won't be limited to that of the penlight cells, nor will they move the same amount of current under load as the big battery. But the little cells will contribute according to their capabilities.

 

[Chalo]

Batteries in parallel add their current capacities together, and they move current in proportion to their individual capabilities.

But since the batteries are parallel, the battery that is discharged more, will have a lower voltage, meaning the battery with the higher voltage (the one with more charge left) will have to constantly charge the other battery.

No. If the lesser battery sagged below the voltage of the stronger battery and the load, current flow would stop and the voltage would stop sagging, so there would be no voltage gradient with which to charge the weaker battery. But in reality, the voltage only sags to the point where each parallel pack is picking up the fraction of the load it can manage. The equilibrium voltage is a little higher than what either of the parallel packs would manage on its own. There's no charging going on. This is measurable with inline wattmeters.

 

[Eastwood]

Well while everyone is arguing can someone please try to answer this question? :lol:

So is it safe to parallel the charger ports just like I did the discharge BUT while leaving the discharge connected?? I want to connect them so that I can charge at 8 A so each pack would be getting 4 A each. I mean I’ve charged the first pack at 8 A a few times but I certainly don’t wanna make a habit of putting 8 A through the first pack so it seems like it would be better for the longevity of the battery to parallel the charger ports, so that I can charge at 8 A without all that current going through the first pack.
Is my thinking on the right track?

 

[justlooking808]

i'm not too sure what or how this relates to batteries, but i'm willing to learn?

In short, two dissimilar battery packs connected in parallel to a load do not work as efficiently as a single pack would. They are always interacting with each other, and not just with the load. How inefficiently? It depends on a lot of factors, and the easiest way to measure would be to measure power "in" while charging, adjust for charge-discharge inefficiency of lithium-ion, and measure power "out".

There is a 100% chance that you will get more range (more energy out of the 2 packs) if you drain one battery pack, disconnect it, and then connect the second pack and drain that.

not too sure about this one, might be right, but in my test, this is what i get,

single 10.5 ah battery miles = ~23 - 25 and battery is at 42v = 5volts for every 10 miles
both unmatched battery connected in parallel 30 AH = ~55-60 miles and battery is still at 48v = under 1.5v for every 10 miles
in both senerao's, i'm only using throttle, so idk, maybe exception to the rule?

i think for this one, it's about voltage sag more than anything to help the ebike perform more efficiently as single bat is sagging about 4v+ and dual only sags at most 2v. but idk, according to what i read, i'm doing it wrong.

 

[justlooking808]

Well while everyone is arguing can someone please try to answer this question? :lol:

So is it safe to parallel the charger ports just like I did the discharge BUT while leaving the discharge connected?? I want to connect them so that I can charge at 8 A so each pack would be getting 4 A each. I mean I’ve charged the first pack at 8 A a few times but I certainly don’t wanna make a habit of putting 8 A through the first pack so it seems like it would be better for the longevity of the battery to parallel the charger ports, so that I can charge at 8 A without all that current going through the first pack.
Is my thinking on the right track?

both batteries i have are common port for charge and discharge. so i leave them connected and just use the charge port of one pack to charge both of them after every ride. i ride once a week and at most according to the display voltage, they lose .1v by the time i ride it again, but when i check with a multimeter, it reads 54.8v. so not too sure where that .1v is leaking too. i say you should be ok, but i would check to see if it's common or separate and go from there. maybe someone wiser would be better to answer this question.

 

[justlooking808]

Well while everyone is arguing can someone please try to answer this question? :lol:

So is it safe to parallel the charger ports just like I did the discharge BUT while leaving the discharge connected?? I want to connect them so that I can charge at 8 A so each pack would be getting 4 A each. I mean I’ve charged the first pack at 8 A a few times but I certainly don’t wanna make a habit of putting 8 A through the first pack so it seems like it would be better for the longevity of the battery to parallel the charger ports, so that I can charge at 8 A without all that current going through the first pack.
Is my thinking on the right track?

both batteries i have are common port for charge and discharge. so i leave them connected and just use the charge port of one pack to charge both of them after every ride. i ride once a week and at most according to the display voltage, they lose .1v by the time i ride it again, but when i check with a multimeter, it reads 54.8v. so not too sure where that .1v is leaking too. i say you should be ok, but i would check to see if it's common or separate and go from there. maybe someone wiser would be better to answer this question.

oh, forgot to add, check the BMS if it has one to see what max amps you can charge at as some OEM can only take 5amps max and can prevent charging if it's too high a current.

 

[Eastwood]

oh, forgot to add, check the BMS if it has one to see what max amps you can charge at as some OEM can only take 5amps max and can prevent charging if it's too high a current.

Yeah our batteries are different in the sense mine has separate wires for charging and discharge. So the way I have it now, example my bike is charging at the moment and only one battery is being charged through the BMS while the other pack is being charged through the discharge wires. That’s why I’m thinking it might be safer to parallel connect the charger wire so it does go through bms on both.

But yeah I know one pack can handle 8 amps as I’ve done it about 5 to 10 times while needing to fast charge on some dirtbike tracks. That was when I just had the one single battery. But now I feel like my normal charging should be 8amps since I have two packs and and the current would be divided, but I’m not sure hopefully someone can chime in..

 

[Eastwood]

The charger that comes with the battery only charges at a maximum of 4 A so that will be divided by both packs which equals very slow!
but I have this adjustable charger as well and hoping that I can charge at a higher current while the charger wires are paralleled.

I know I can connect to one battery and charge 8 A and it would charge the other one through the discharge but it seems it’s not so practical for the first battery pack to have all that current running through it so for now I’ll just use the four amp charger and wait all day :lol:

 

[justlooking808]

oh, forgot to add, check the BMS if it has one to see what max amps you can charge at as some OEM can only take 5amps max and can prevent charging if it's too high a current.

Yeah our batteries are different in the sense mine has separate wires for charging and discharge. So the way I have it now, example my bike is charging at the moment and only one battery is being charged through the BMS while the other pack is being charged through the discharge wires. That’s why I’m thinking it might be safer to parallel connect the charger wire so it does go through bms on both.

But yeah I know one pack can handle 8 amps as I’ve done it about 5 to 10 times while needing to fast charge on some dirtbike tracks. That was when I just had the one single battery. But now I feel like my normal charging should be 8amps since I have two packs and and the current would be divided, but I’m not sure hopefully someone can chime in..

did you confirm this by checking the bms?

most times there is a separate charge(dc connection) and discharge port (xt60), but it's soldered on the same line going into the BMS. having separate ports usually cost more and in most cases are only added for custom DIY batteries or custom orders. but only way to know for sure is if you built it your self or you open it up.

for me, i just cut the charge port on the one battery since it was the same and use the other battery's charge port to charge both batteries. i charge a 30ah battery with a 4a charger in 6 hrs. but i think my charger was mislabeled/repurposed when i ordered it and they gave me a higher amp charger and just labeled it as a 4a charger.

 

[Eastwood]

:wink:

oh, forgot to add, check the BMS if it has one to see what max amps you can charge at as some OEM can only take 5amps max and can prevent charging if it's too high a current.

Yeah our batteries are different in the sense mine has separate wires for charging and discharge. So the way I have it now, example my bike is charging at the moment and only one battery is being charged through the BMS while the other pack is being charged through the discharge wires. That’s why I’m thinking it might be safer to parallel connect the charger wire so it does go through bms on both.

But yeah I know one pack can handle 8 amps as I’ve done it about 5 to 10 times while needing to fast charge on some dirtbike tracks. That was when I just had the one single battery. But now I feel like my normal charging should be 8amps since I have two packs and and the current would be divided, but I’m not sure hopefully someone can chime in..

did you confirm this by checking the bms?

most times there is a separate charge(dc connection) and discharge port (xt60), but it's soldered on the same line going into the BMS. having separate ports usually cost more and in most cases are only added for custom DIY batteries or custom orders. but only way to know for sure is if you built it your self or you open it up.

Well when I was setting up my region several months back I talked with the seller and they confirmed that I would have no BMS protection using the region back through the discharge port. So therefore they’re not connected on the same line. When my battery is fully charged I’m just careful to not press the E break :lol:
But let’s be honest our battery is only fully charged until you full throttle for more than just a few seconds especially at 7+ kilowatts