Question about parallel packs with mismatched voltage charging each other...

Klauts

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Nov 10, 2022
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Ok so... My question, if you had a 48v and a 52v li-ion pack paralleled but which direction would the lower voltage pack get charged via? I.E. would the current flow from the higher voltage pack into the negative terminal on the lower voltage pack or would it flow through the positive terminal?

The reason I ask is I've got a a 48V and a 52v battery that are hooked in parallel using a 40A (My controller is rated for a peak of 25A) Schottky Diode. The diode is working perfectly and there isn't and flow between batteries but I want to implement a secondary failsafe...

If the current is flowing into the negative terminal on the 48v I was thinking I could add a 24v 2A in-line thermal breaker just before the negative terminal wire ....

If it's flowing the other way then I'm lost....

Please assist me in not dying 🤣😅
 
Would never have battery packs of different voltages paralleled, no buts. Not safe.
Packs should be brought to same voltage before connecting them together.
Or maybe I don't understand.
 
ZeroEm said:
Would never have battery packs of different voltages paralleled, no buts. Not safe.
Packs should be brought to same voltage before connecting them together.
Or maybe I don't understand.

A schottky diode... It only allows current to flow in one direction, they're used regularly for parallelling different chemistries as well as different voltages so long as the nominal voltage difference isn't too great.

I've seen many bikes sold with 48v/52v combos with a "battery blender" which is just a schottky diode...

I'm trying to add an extra layer of protection.

A thermal fuse rated for very low amps seems like the perfect solution?
 
I should add that I am planning to keep the larger pack switched off until it hits 54.6, the diode is in the event the smaller pack experiences excess drain and triggers a charge loops.

If the diode fails I want a backup.

The backup would also serve to protect me if the second pack were to get flipped on accidentally at the same time the diode decided to fail.

Does my logic track? I'm not an electrician by any means...
 
Klauts said:
if you had a 48v and a 52v li-ion pack paralleled but which direction would the lower voltage pack get charged via? I.E. would the current flow from the higher voltage pack into the negative terminal on the lower voltage pack or would it flow through the positive terminal?
Both. Current flows in a circuit, requiring a "circle", or "circuit", so it goes thru both. If there was only one connection, current couldn't flow (which is what your fuse would do, it would blow if current exceeded it's limit by enough amps for a long enough time*** and break the circuit, stopping the current flow).

It doesn't matter which wire it is in, + or -. Breaking either one breaks the entire circuit, as long as it is breaking the wire at the battery itself, before that wire connects to any other wire outside the battery. The only way it matters is if you have other things connected to the battery so that a current path still exists somewhere.

But I don't think a fuse that's rated to handle the current your system requires (at least 25A) is going to blow under the conditions of your diode failing shorted and allowing the 52v pack to charge the 48v pack. The worst case condition would be a full 52v at presumably ~58v and a full 48v pack at ~52v, or a 6v difference. Let's assume your packs are low resistance, say 100milliohm each. Let's also assume the diode has failed as a 100milliohm short (would probably be higher resistance than that, but failure modes vary so can't be sure of what it would be until it happens). Let's say all connectors/wiring/etc total another 100milliohms.

That's 400milliohms, or 0.4ohms. Current is V / R, so 6v / 0.4ohms is 15A. I would bet that the total resistance is higher than that, but even if it's lower it's not a huge current--not enough to blow a fuse that can handle the current the system must be able to pull thru it in order to operate.



If the current is flowing into the negative terminal on the 48v I was thinking I could add a 24v 2A in-line thermal breaker just before the negative terminal wire ...
.

With a 2A fuse in series with the battery, it can then only supply about 2A before the fuse will blow. Presuming that sometimes the 48v pack is going to supply the entire current, you're going to need a fuse that can handle that. I recommend a 30A (or whatever will work for your system) ANL inline bolt-on fuse, so that you do not have to worry about the fuseholder causing you grief (a common problem with any of the cheap ones, or even the good ones if environmental conditions degrade them.


Note that If the fuse is only rated for 24v, it may not be able to break the current if the voltage between the two packs is greater. Arcing can occur in that situation, creating a risk of fire from the arc heat (just like a welder), if the current is great enough. As long as the votlage difference between the packs does not exceed the fuse rating, it is guaranteed to break the current flow when it blows.

Real Eaton Bussman ANLs are ok up to 80v. Dunno about others; you'd have to check their spec sheets from their actual manufacturer.


***see the manufacturer spec sheet for the specific fuse you are using for the current vs time curve chart, to find out how long it will actually take to blow (or if it will at all) at any particular current, and it's real max voltage, etc.
 
Klauts said:
I should add that I am planning to keep the larger pack switched off until it hits 54.6, the diode is in the event the smaller pack experiences excess drain and triggers a charge loops.
By larger, do you mean the 52v pack? Because leaving that off means it will never drop in voltage, and only the 48v pack will be providing any current.


If both packs are on and connected thru the diode, the higher voltage pack will provide all the system current until it drops in voltage to where the other pack equals it, at which point they will begin to share current. Once the 52v pack drops to it's LVC point and it's BMS shuts it's output off, then the 48v pack will supply all the current.


If you really want a backup to the diode, put a battery cutoff switch in series with each battery, so that you can turn off the one you are not using. Even one of the crappy ones HarborFright sells will do the job, if you are switching them when there is no load on the system, even though they are not rated for the system voltage you have. (they might not break the circuit under load, but they'll keep it open if switched when parked).
https://www.google.com/search?q=battery+cutoff+switch&tbm=isch
 
amberwolf said:
Klauts said:
I should add that I am planning to keep the larger pack switched off until it hits 54.6, the diode is in the event the smaller pack experiences excess drain and triggers a charge loops.
By larger, do you mean the 52v pack? Because leaving that off means it will never drop in voltage, and only the 48v pack will be providing any current.


If both packs are on and connected thru the diode, the higher voltage pack will provide all the system current until it drops in voltage to where the other pack equals it, at which point they will begin to share current. Once the 52v pack drops to it's LVC point and it's BMS shuts it's output off, then the 48v pack will supply all the current.



If you really want a backup to the diode, put a battery cutoff switch in series with each battery, so that you can turn off the one you are not using. Even one of the crappy ones HarborFright sells will do the job, if you are switching them when there is no load on the system, even though they are not rated for the system voltage you have. (they might not break the circuit under load, but they'll keep it open if switched when parked).
https://www.google.com/search?q=battery+cutoff+switch&tbm=isch


Thank you for such a detailed and helpful response! I had it backwards there l, meant to say I'd keep the small pack turned off until the 52v was run down.

My thinking with the 20amp breaker was that because if things are running properly on my set-up the 48v battery should never draw more than 12.5amps at any time.

If there's a sudden failure of the diode then it'd be allowed to draw the full 25amps the system will allow which should trip the breaker?

I have this breaker here...

H6OCkG6.jpg


It's rated for 24v 20amps but the seller said it'd trigger on a 48v system ?

I'd install it inline directly out of the batteries positive terminal before it reached the diode?

I can test whether it would trip properly by running the 48v on its own and asking for 25a from the motor...
 
Klauts said:
My thinking with the 20amp breaker was that because if things are running properly on my set-up the 48v battery should never draw more than 12.5amps at any time.

It will draw more and more as the 52v battery gets closer and closer to empty, and when the 52v battery is empty and it's BMS shuts off the 48v battery will supply all the current.

Exactly how much current it will draw will depend on the actual load on the controller at any time.



If there's a sudden failure of the diode then it'd be allowed to draw the full 25amps the system will allow which should trip the breaker?

Which purpose is the diode and fuse/breaker there for?

You'll need to decide and state exactly which things you want to happen, under what conditions, so we can determine what hardware is needed to accomplish those things. If the hardware you have has limits you haven't stated (like if the 48v pack can't supply more than x number of amps before failing, etc), those would be helpful to know, too.

I may have misunderstood, but based on the start of this thread, I thought you were intending to use them to prevent the 52v pack from overcharging the 48v pack?

If so, then a failure of the diode, if it fails short circuit, and depending on where it is in the circuit, could allow current to flow from the 52v pack "down" into the 48v pack, during any portion of the 52v pack's state of charge that is higher voltage than the 48v packs' SoC at that same moment.

If the diode fails open, then it will be just like a fuse in that it will prevent any current flow thru it in either direction, so no matter where it is in the circuit it will prevent overcharge of 48v by 52v.

If the fuse / breaker is intended to prevent the overcharge by blowing when the diode fails short circuit and would allow the 52v pack's current to flow back into the 48v pack, I don't think it could work without also preventing the 48v pack from being able to run (or help run) the system in any useful way.


If instead the diode / breaker is supposed to prevent the 48v pack from supplying more than 12.5A of current (why?) then it won't do that. The diode, failed or working, wouldn't stop that at all (unless it fails open circuit and is in the 48v's current path). The breaker or fuse would stop current flow once it exceeds it's limit for long enough (less time the farther it exceeds the limit).


I have this breaker here...
H6OCkG6.jpg


It's rated for 24v 20amps but the seller said it'd trigger on a 48v system ?
If by "trigger" you mean it will still "trip" (break the circuit) when current exceeds it's limit by enough amps for a long enough time, then yes, it would do that at any voltage. Overcurrent is what activates a breaker (or fuse).

However, if a fuse or breaker is designed for a lower voltage than that which will end up across it's terminals when the circuit is broken, the current may never stop flowing because the voltage can cause an arc (like a welder) across the opening contacts (and this can cause a fire if it goes on long enough to heat things around it that much).

It almost certainly won't trip at exactly 20A. That *should* be the current that it can sustain indefinitely, above which it will trip after some amount of time (from milliseconds to minutes or more, depending on how far above that current it is).

Most of these things are really meant to protect against hard short circuits, where you could have hundreds to thousands of amps, enough to cause a fire quickly, if the circuit isn't broken rapidly. So they will do that, at those currents. But when near their sustained max limits, it takes longer for the heat buildup in them to happen to the point they will trip or blow, and the closer the current is to the max the longer that will take; the higher it is above that limit the shorter time it will take.

To know what this breaker can do, you must look at the spec sheet from it's manufacturer (not the seller), and use the chart it contains. Here is an example for an Eaton Bussman ANL fuse:
https://www.eaton.com/content/dam/eaton/products/electrical-circuit-protection/fuses/data-sheets/bus-ele-ds-2024-anl.pdf

If it doesn't have a manufacturer name and model on it that will get you to a website that has a datasheet like that, then you cannot know how it will actually perform, except in the exact situation that you test. In any other situation it will perform differently, but you wont' know what that difference is without testing it.

If you need to know exactly what will happen in "any" situation, you'd want to get a breaker (or fuse) that has a datasheet you can use to determine that...or you'd need to test every situation that you need it to protect things in...or else just "trust" that someone did design and manufacture it correctly .


I'd install it inline directly out of the batteries positive terminal before it reached the diode?
It can be installed anywhere in the circuit it is supposed to break.




For further info, if you want to spend a while reading / learning how these thing work / are used, this document is for house/etc AC stuff, but the principles in it apply to any type of "protective device" in a circuit
https://www.eaton.com/content/dam/eaton/products/electrical-circuit-protection/fuses/technical-literature/bus-ele-br-3002-spd-2017.pdf
 
If the fuse / breaker is intended to prevent the overcharge by blowing when the diode fails short circuit and would allow the 52v pack's current to flow back into the 48v pack, I don't think it could work without also preventing the 48v pack from being able to run (or help run) the system in any useful way.

This right here is my intended use case, I also won't ever be allowing the 48v pack to run on its own. It would be only run in parallel. I still have 45% charge in the 52v after my 75Km commute it appears so I can't imagine ever running down to the point of relying solely on the 48v.

My 48V pack has an absolute peak of 30Amps buts it seems to get out of balance at anything over 20...

I'm Ok with the breaker tripping meaning that the 48V pack is unusable in the system until it's flipped and the diode is replaced. I've got the wiring setup so I can easily bypass the diode and plug either battery directly to the controller in the event of a failure.

I feel like the breaker I've got probably won't do though... Any tips on where I could find any appropriately rated breaker? I guess somewhere around 48v 15Amps would be suitable?

I'd prefer to use a break over a fuse simply to make it easy to bypass the diode and plug into the controller...
 
I don't know of any that will do what you want, because as per the guesstimate math I did before, the current is so low that if it trips at all, it will probably take too long to do much protecting, unless you size the breaker so low that the 48v pack can't provide any useful current.

A 15A breaker will trip somewhere above 15A, once the overload has gone on long enough...but that might not be soon enough to stop the charging event from doing whatever damage it is going to do.

You would have to look at the datasheets for the breakers you might want to try, to see what overload actually trips them, and how long that overload has to last before it trips.

Or test it under the specific (range of) conditions you need to protect against.

Same for fuses.

They just aren't designed to do what you want them to do.

If you had a common-port BMS on the 48v pack, then assuming the charge current from the 52v pack wasn't too high, the BMS itself would prevent overcharge by shutting off that charge port whenever any cell reaches HVC. The only issue with this is it's just like the diode--it coudl have a failure mode where the FETs are shorted, stuck on, and can't actually stop the charging current.

In addition, the discharge port would still be active, most likely, so the 48v pack could still supply current if the 52v pack was disconnected or ran low enough voltage.


You could build something with hardware or an MCU-based approach (arduino, etc) that detects current flow direction in the 48v pack connection, and either alerts you so you can physically discconnect it if it is ever charging during a ride, or has a contactor to electrically disconnect it automatically.
 
Do you know where I might start reading about programming an Arduino for that purpose? I happen to have about a dozen Arduino Nanos lying around... And a couple rpi4s
 
23a43e381a9a193ba7c3a769329874765b0a08d9.jpeg


It turns out I've got more than just a Schottky Diode at play... I contacted the vendor I purchased it from and was sent that schematic of the internals as well as this description of what it's doing

2f4a391196f8c5b51cf19c56c42923516e98292e.jpeg



This makes me think that I might already be safe?
 
It also has a com port I can apparently use to monitor voltage of each battery and adjust the LVC...
 
Well, they provide incomplete and conflicting information in all of that, including at least one impossible statement given the hardware involved. Not surprising, as that's the norm for information supplied by sellers (when they supply any at all). :(

I myself wouldn't feel safe at all with that device, and because the info provided doesn't even match itself, wouldn't trust that it does any of the functions it claims to, including even the diode protection, without reverse-engineering it completely (given it claims an MCU in there, if there is one, I wouldn't be able to really tell for sure how it's software worked and so couldn't trust it's functions to operate as designed).

For instance, the last picture, the schematic, must be incomplete (I hope). If it's not incomplete, then it doesn't work as described, and doesn't protect the way you probably want it to.

It shows only one schottky, oriented so that it always allows power from battery B to flow into battery A. (BTA, BTB).

However, it would always prevent power from BTA from flowing into BTB.

The MOS (FET) switch also can only cut off BTA from the output, it cannot cutoff BTB.

So BTB would, despite being named B, be the primary battery of the system, since it cannot be disconnected no matter what.

The switch diagram above the schematic shows that it disconnects or connects both the + and - of one of the batteries to the output, which is not how the schematic shows things working. A switch designed the way the diagram shows will also only connect one battery at a time, and never connect both. It only switches between them. If they are just using their own schematic methods instead of standard ones, then the switch could do anything at all and we have no way of knowing what that is. It could even connect just the + of one pack and hte - of the other, which would not provide any power to the output at all. Or it could connect the plus and minus of the same pack together and cause a fire. Etc. Literally no way to know what they intend by that diagram if it is not a standard method of showing it...and if it is the standard....it doesnt' do what the other stuff says it does).

So it must be (hopefully) for a different module than the schematic is. They can't both be for the same device. And the description below that doesnt' match how either of the diagrams work.

The first statement in paragraph 2 conflicts with the switching diagram; it can't make both of them discharge together.

The statement "and adjust the current output for each battery" is not possible without much more circuitry than this module could hold--it would need to have a complete independent DC-DC converter, that has a current-limiting output, adjustable by the MCU, for each battery, to actually adjust the current output.

The last statement in the same paragraph is sort of true--if each battery is identical in characteristics including complete internal resistance, then yes, at the same voltage the current would be divided equally between the two...but that's almost never the case, and is not the case for two entirely different batteries such as you have.

The last paragraph, 3, means that if you set it up to stop discharging to protect the 52v pack, you will not be able to use much of the 48v pack. If you set it up to protect the 48v pack, then the 52v pack will be completely drained first, until it's own BMS LVC kicks in and turns it's output off. However, if the schematic is correct, it can never disconnect BTB regardless of how low it's voltage gets.


Can they also send you the program to read and write settings and data to the unit? And tell you what kind of cable it takes to do this (do they sell the cable?) If not, that part of the unit is useless to you, unless you can "sniff" the data stream (assuming it just always outputs this and doesn't have to be commanded to do so) and reverse engineer what the data means, then write a program and build hardware to display the voltages (or write a program to send the data over BT to a phone, etc, and an app for the phone to display them). Still wouldn't let you program it, but you'd have made a complex dual multimeter out of it. ;)


Regarding using Nanos...I have as yet learned very little in my projects, don't yet know enough to program one. I recommend http://arduino.cc as a good source of info and other people's projects to dive into, however, if you already know how to program them. You could find pieces of others' projects that do things you want to do, and copy those code bits into your code to save time writing it, for instance.
 
Assuming I was using two similar packs (same voltage range), and required some form of protection against one feeding the other, and didn't mind the wasted power in the diode, I would personally feel safer buying a plain old TO-format dual schottky diode, bolting it to a sufficient heatsink for the usage and environmental conditions, (with fan if necessary to keep the heatsink small), wiring connectors to it, and plugging the batteries into that.

For your case, I'd just use a switch to go from the primary pack to the secondary, so there wouldn't be any risk of higher charging lower, if I didn't just unplug one and plug the other in.

One possible search that finds such dual-packs. It's filtered for over 40A current and over 70v, either common anode or common cathode or an unconnected pair.
https://www.mouser.com/c/semiconductors/discrete-semiconductors/diodes-rectifiers/schottky-diodes-rectifiers/?configuration=Dual~~Dual%20Cathode%20Common%20Anode&if%20-%20forward%20current=40%20A~~40%20A%20%282%20x%2020%20A%29%7C~44%20A~~50%20A%20%282%20x%2025%20A%29%7C~52%20A%7C~55%20A~~60%20A%20%282%20x%2030%20A%29%7C~65%20A%7C~70%20A%7C~78%20A~~80%20A%20%282%20x%2040%20A%29%7C~88%20A%7C~109%20A%7C~120%20A%20%282%20x%2060%20A%29%7C~160%20A%20%282%20x%2080%20A%29%7C~200%20A%7C~240%20A&mounting%20style=Through%20Hole&product=Schottky%20Diodes~~Schottky%20Silicon%20Carbide%20Rectifiers&vrrm%20-%20repetitive%20reverse%20voltage=80%20V~~90%20V%7C~100%20V&rp=semiconductors%2Fdiscrete-semiconductors%2Fdiodes-rectifiers%2Fschottky-diodes-rectifiers%7C~Product%7C~Configuration%7C~If%20-%20Forward%20Current%7C~Vrrm%20-%20Repetitive%20Reverse%20Voltage

One example, pin 1 could be BTA+, 3 BTB+, 2 would be controller+.
https://www.mouser.com/ProductDetail/Vishay-General-Semiconductor/VX80M100PW-M3-P?qs=7MVldsJ5UaxLd2iLl%2FcyBQ%3D%3D
https://www.mouser.com/datasheet/2/427/vx80m100pw-2897285.pdf
 
Sarcasm is not intended but why make mounting two batteries so darn complex? Use a double pole double throw DC switch. Dpdt Toggle Switch. No programming no need for matching packs. flip the switch and done. Nearly every day I read a forum post from someone stressing over parallel packs.
 

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Klauts said:
a 48v and a 52v li-ion pack paralleled
Please assist me in not dying 🤣😅

Hi Sir, You have to do some math.
48V battery is a 13S (48V/3,7V = 13) and a 52battery is a 14S (52V/3,7V = 14)

case battery 48V:
min 2,7V x 13S = 35V and max 4,2V x 13S = 54,6V
case battery 52V:
min 2,7V x 14S= 37,8V and max 4,2V x 14S = 58,8V
35V {37,8V;54,6V} 58,8V
common interval match between elements (how do U call this at math? :flame: )

soo only range between 37,8V and 54,6V will be parallel supply.. :bolt:
from 58,8V down to 54,6V only one battery in action, when reaches 54,6V second battery will turn on, soo parallel.
down to 37,8V, under that only one battery supply until cut at 35V

Imagine one battery is at 54,6V and other at 37,8V and than you connect those together, soo there is a 16,8V difference . if one battery has 100mohm IR and other has 80mohm IR, so total IR will be 180mOhm , using math at ohm law I = U/R soo 16,8V/180mOhm equals 93Amperes draw from one battery to other
:bolt: WORST CASE: if you connect one battery at 58,8V to other at 35V is 23,8V difference, so ohms law 23,8/0,18 equals 132Ampers charge :warn: *edited*
allways connect parallel batteries at same level voltage or near.

hope U did understand tha math :bigthumb:
 
MAKE SURE both 48V and 52V batteries have BMS charge/discharge common port, if not one battery may overcharge other!! Be careful :flame:
 
amberwolf said:
Well, they provide incomplete and conflicting information in all of that, including at least one impossible statement given the hardware involved. Not surprising, as that's the norm for information supplied by sellers (when they supply any at all). :(

I myself wouldn't feel safe at all with that device, and because the info provided doesn't even match itself, wouldn't trust that it does any of the functions it claims to, including even the diode protection, without reverse-engineering it completely (given it claims an MCU in there, if there is one, I wouldn't be able to really tell for sure how it's software worked and so couldn't trust it's functions to operate as designed).

For instance, the last picture, the schematic, must be incomplete (I hope). If it's not incomplete, then it doesn't work as described, and doesn't protect the way you probably want it to.

It shows only one schottky, oriented so that it always allows power from battery B to flow into battery A. (BTA, BTB).

However, it would always prevent power from BTA from flowing into BTB.

The MOS (FET) switch also can only cut off BTA from the output, it cannot cutoff BTB.

So BTB would, despite being named B, be the primary battery of the system, since it cannot be disconnected no matter what.

The switch diagram above the schematic shows that it disconnects or connects both the + and - of one of the batteries to the output, which is not how the schematic shows things working. A switch designed the way the diagram shows will also only connect one battery at a time, and never connect both. It only switches between them. If they are just using their own schematic methods instead of standard ones, then the switch could do anything at all and we have no way of knowing what that is. It could even connect just the + of one pack and hte - of the other, which would not provide any power to the output at all. Or it could connect the plus and minus of the same pack together and cause a fire. Etc. Literally no way to know what they intend by that diagram if it is not a standard method of showing it...and if it is the standard....it doesnt' do what the other stuff says it does).

So it must be (hopefully) for a different module than the schematic is. They can't both be for the same device. And the description below that doesnt' match how either of the diagrams work.

The first statement in paragraph 2 conflicts with the switching diagram; it can't make both of them discharge together.

The statement "and adjust the current output for each battery" is not possible without much more circuitry than this module could hold--it would need to have a complete independent DC-DC converter, that has a current-limiting output, adjustable by the MCU, for each battery, to actually adjust the current output.

The last statement in the same paragraph is sort of true--if each battery is identical in characteristics including complete internal resistance, then yes, at the same voltage the current would be divided equally between the two...but that's almost never the case, and is not the case for two entirely different batteries such as you have.

The last paragraph, 3, means that if you set it up to stop discharging to protect the 52v pack, you will not be able to use much of the 48v pack. If you set it up to protect the 48v pack, then the 52v pack will be completely drained first, until it's own BMS LVC kicks in and turns it's output off. However, if the schematic is correct, it can never disconnect BTB regardless of how low it's voltage gets.


Can they also send you the program to read and write settings and data to the unit? And tell you what kind of cable it takes to do this (do they sell the cable?) If not, that part of the unit is useless to you, unless you can "sniff" the data stream (assuming it just always outputs this and doesn't have to be commanded to do so) and reverse engineer what the data means, then write a program and build hardware to display the voltages (or write a program to send the data over BT to a phone, etc, and an app for the phone to display them). Still wouldn't let you program it, but you'd have made a complex dual multimeter out of it. ;)


Regarding using Nanos...I have as yet learned very little in my projects, don't yet know enough to program one. I recommend http://arduino.cc as a good source of info and other people's projects to dive into, however, if you already know how to program them. You could find pieces of others' projects that do things you want to do, and copy those code bits into your code to save time writing it, for instance.


Since I've been gone I took the time to learn how read schematics and understand the basics of MOSFETs, diodes DC/DC boost and FETS... It's been a wild ride haha

A lot more to it than it appears at first glance and definitely a lot of red flags with that first adapter. It turned out the schematic was for a different product, a battery switcher they also sell. They weren't able to provide the schematic for the adapter I had only to say it used P and N MOSFETs and and MCU...

So I did some more searching for an adapter equipped with a bit more knowledge as to what I should be looking for.

I ended up finding this product:

100A Double Battery Discharge Converter for E-Bike XT90 Two Batteries Parallel Equalization Module Increase Battery Capacity 24-72V DC https://a.co/d/19uRH7c

They don't have a schematic available on the advert but when I contacted the seller they were very knowledgeable about the product itself.

It contains a P and N MOSFETs, dual diodes on each battery positive input, a DC/DC boost regulator and an MCU controlling things.

The seller is going to try and get an actual schematic for me from the manufacturer but this seems to me like a solid product... Id be tempted to trust it more than the DATEx2...

Thoughts?
 
Ok I finally got the schematic from the manufacturer and a more detailed explanation:

AkYsuPs.jpg


Working Process
1. MCU monitor voltage of both batteries in real time and send the volatage value by UART if UART interface is available.
2. Auto Battery on-line detecting. Switch to the battery and make power output if only one battery was connected with this module.
3. If the voltage of two batteries is not same, one is lower and other is higher , module will use the battery with higher voltage to discharge first till two batteries voltage approaching similar.
4. Two batteries will discharge together when two batteries voltage is similar.
48V and 52V batteries can be mixed, our module is only responsible for the maximum and minimum voltage, the system automatically to manage the identification, priority to use the highest voltage.
5. Proprietary Intelligent short circuit, overvoltage, over temperature and over current protections.

The schematic appears to be incomplete still but I'm feeling pretty comfortable with it....
 
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