4S Li-ion pack charger and protection complete circuit

[VEGETA]

I found this very cheap IC that has everything: CN3795. What do you think?

This charger IC is made for buck topology. You will need a bit higher supply then 5V to use it with 4S Li-ion pack.

well, then I use the 3608 boost IC before it. easy.

However, I hope you could help me to answer previous questions about how to determine discharge current since there is nothing in the datasheet (that i know of) that says that. I couldn't figure out how to choose and configure Rsense and so on.

I mean many BMS boards using the same 8254A IC but with different maximum current, how?

[mvs]

I mean many BMS boards using the same 8254A IC but with different maximum current, how?

This boards may have different versions of 8254A, different Rsense and different MOSFETs. If you have some photos of this boards you might see used components. Rsense would be usually a large 2512 package or even 2 in parallel with clearly readable value on it (like R010 for 0.01 Ohm). 

[VEGETA]

I mean many BMS boards using the same 8254A IC but with different maximum current, how?

This boards may have different versions of 8254A, different Rsense and different MOSFETs. If you have some photos of this boards you might see used components. Rsense would be usually a large 2512 package or even 2 in parallel with clearly readable value on it (like R010 for 0.01 Ohm).

I wanted to know from the datasheet. I don't think mosfets are of any relation except to be powerful enough to handle the current but not to determine the maximum current itself.

See an example board here (10 amps): https://cbu01.alicdn.com/img/ibank/2017/294/031/4501130492_1344841943.jpg    -> it has 2 of R010 in parallel
another one (6 amps): https://my-live-01.slatic.net/p/af3bb4846bb0ecbb822f3087715dca0f.jpg        -> it has 2 of R020 in parallel

This one is 10 amps: https://img.joomcdn.net/5e2c6ead056818092e50dc2e871701ff131835f9_1024_1024.jpeg   --> only one R010

this one is very clear: https://images-na.ssl-images-amazon.com/images/I/71QhMB--BGL._SL1004_.jpg

it says 6 amps and has 2xR020 in parallel with 51R0 which is 51 ohms recommended value of Rvss.

thus maybe we could say this: I choose 100x2 in parallel = 50 ohms as Rvss and choose something like 0.01R as Rsense to allow 6 amps or 0.005R to allow 10 amps. both are more than enough but maybe i can go with 0.005R since it is used in the sepic pre-regulator.

^ How about this suggestion?

I will modify the circuit to add CN3795 with H3608.

[VEGETA]

I have updated the schematic and added charging IC. Charging current is maximum of 1.2 amps which is good.

Notes:

1- R14 = 100R instead of 120R, C14+C15 = 200nF instead of 220nF. I guess this is ok right?

2- L1 = 10uH to ensure it works.

3- R27 = 0.005 as agreed previously.

4- No LEDs for CN3765 IC, no need for them.

What do you think now?

[fcb]

I would fit the LED's for debugging purposes and delete for MFR.  R11 not necessary if you don't have LED's.

Any voltage drop across Q1/Q2 will affect the charging circuit - also, what happens if you disconnect (i.e. open Q1/Q2) the battery??

R14/C14|C15 compensation network changes - perhaps this will be fine, I wouldn't deviate from manufacturers recomendations until you can test it.

Your potential divider on U1 is still a bit mad. Bear in mind that 1% means +/-1% so your exact resistance is likely to be frustrated by tolerancing - are you doing somesort of select on test, and anyway the tolerance stack-up of other items makes it a bit pointless - not to mention that cell terminal voltage is affected by temperature and there is no temperture compensation in the your design.

Connection to your lower cell and/or shunt resistor still missing.

Boosting to then buck down seems wasteful. But it will probably work.  Go and prototype it.

[VEGETA]

I would fit the LED's for debugging purposes and delete for MFR.  R11 not necessary if you don't have LED's.

yes I will probably do it. I added R11 to lower the current a bit, but it seems not so important.

Any voltage drop across Q1/Q2 will affect the charging circuit - also, what happens if you disconnect (i.e. open Q1/Q2) the battery??

The drop voltage will be so little, they are necessary anyway. Perhaps you meant making the charging voltage more than 16.79v to compensate for this?

S-8254A should connect and disconnect the battery pack automatically AFAIK. I will dig into this more. However, the charger IC CN3765 can power the device without the battery as said in datasheet. I don't know if it can do it while battery is charging.

R14/C14|C15 compensation network changes - perhaps this will be fine, I wouldn't deviate from manufacturers recomendations until you can test it.

Mine makes nearly the same time constant so I guess it is ok. if anything happened i can just de-solder these and solder recommended ones to test.

Your potential divider on U1 is still a bit mad.

I know it is but... consolidation is a beast. This whole circuit is just a part of another bigger project, hoping I could design all of them properly.

Bear in mind that 1% means +/-1% so your exact resistance is likely to be frustrated by tolerancing - are you doing somesort of select on test, and anyway the tolerance stack-up of other items makes it a bit pointless

I am open to your suggestions, I did all this to consolidation. No other reason. I didn't think that tolerances will be so hugely important here, nice that you taught me that.

not to mention that cell terminal voltage is affected by temperature and there is no temperture compensation in the your design.

I don't know what you mean by this. So you kind lecture is awaited xD.

Connection to your lower cell and/or shunt resistor still missing.

Yes, I will just put a GND (and B-) after Rsense right?

Boosting to then buck down seems wasteful. But it will probably work.  Go and prototype it.

Absolutely, but still no other choice. Assuming you use 5V 2.5A USB chargers which are cheap and available these days... then I guess it will work. Even when using 12v supply works. Efficiency is not an issue here since no solution can be done.

[VEGETA]

BTW, I just noticed CTL pin of S-8254A IC. I tied it to LOW since this is what I understood from the datasheet. It is not tied to anything in the typical application circuit.

what do you think?

[fcb]

You aren't using the CTL pin to control charge/discharge?  Just relying on the cell voltage state machine?

Where are you connecting the negative side of the battery to? I can guess, but it's not on drawing?

Why not use the sense resistor 'Rsense' (which is 100||100 - to high?) to do your current sensing for charging.

I'm lost why you have a constant voltage boost (HX3608) setup for charging. Especially with that mad 7 resistor potential divider. It'll just sit there and hiccup. Have you read how to charge Lithium cells?

Going to ask where do V+ and V- (for the opamp) come from before I declare that won't work.

What does "charger(-16.8v)" mean? Why negative?

[VEGETA]

You aren't using the CTL pin to control charge/discharge?  Just relying on the cell voltage state machine?

Where are you connecting the negative side of the battery to? I can guess, but it's not on drawing?

Why not use the sense resistor 'Rsense' (which is 100||100 - to high?) to do your current sensing for charging.

I'm lost why you have a constant voltage boost (HX3608) setup for charging. Especially with that mad 7 resistor potential divider. It'll just sit there and hiccup. Have you read how to charge Lithium cells?

Going to ask where do V+ and V- (for the opamp) come from before I declare that won't work.

What does "charger(-16.8v)" mean? Why negative?

I didn't quite understand how to use it. I want this circuit to be autonomous without intervention from MCU or anything.

[VEGETA]

From the datasheet, putting that pin to ground makes it "normal operation" so this is what we want right?

[mvs]

Boosting to then buck down seems wasteful. But it will probably work.  Go and prototype it.

Absolutely, but still no other choice. Assuming you use 5V 2.5A USB chargers which are cheap and available these days... then I guess it will work. Even when using 12v supply works. Efficiency is not an issue here since no solution can be done.

16.8V@1.2A is around 20W of power, with efficiency about 70% you will need >30W PSU.
5V@2.5A USB charger has only 12.5W, and it may not be designed for continuous operation at this power level.
Look better for 19 - 20V laptop PSU.

[VEGETA]

Boosting to then buck down seems wasteful. But it will probably work.  Go and prototype it.

Absolutely, but still no other choice. Assuming you use 5V 2.5A USB chargers which are cheap and available these days... then I guess it will work. Even when using 12v supply works. Efficiency is not an issue here since no solution can be done.

16.8V@1.2A is around 20W of power, with efficiency about 70% you will need >30W PSU.
5V@2.5A USB charger has only 12.5W, and it may not be designed for continuous operation at this power level.
Look better for 19 - 20V laptop PSU.

Hmm you are right, but 1.2A is like the maximum so even at say 0.5A it should work properly right?

Another method I could do is to put another power input which is a DC jack to get 12v but I fear someone will plug 24v instead. Also, what if they plug 2 power sources at the same time? maybe a shotcky diode in series is a good method to make only the largest one (12v) to be the default right?

If these 2 methods are available, one could either plug a powerful 12v supply or just stick to USB. what do you think?


Also, looks like you didn't check the datasheet for the CTL pin.  I checked it and it seems that "normal operation" is when CTL pin connected to ground or "low" like what I did in schematic.. so it seems ok right?

[mvs]

Another method I could do is to put another power input which is a DC jack to get 12v but I fear someone will plug 24v instead. Also, what if they plug 2 power sources at the same time? maybe a shotcky diode in series is a good method to make only the largest one (12v) to be the default right?

And if somebody plugs it into USB port of a PC, laptop or car radio with 0.5A limit?
Schottky diode will cause a voltage drop and will reduce efficiency further. On 5V rail with 0.5V drop you will lose another 10%.

If these 2 methods are available, one could either plug a powerful 12v supply or just stick to USB. what do you think?

You need to be able to switch between different charge current limits. Otherwise it is pointless in my eyes.

And if you plan to use 12V supply, why not to use 20V laptop brick? You would not need boost circuit and you would have higher charging current, if needed.

[VEGETA]

And if somebody plugs it into USB port of a PC, laptop or car radio with 0.5A limit?

then he knows his limits. If he works with low currents (the final device is gonna be linear lab psu with switching sepic pre-regulator from batteries), then say 100mA load current... then even if charge current is as less as 200mA it would be enough. even if he takes more, he won't drain the battery as hard.

Schottky diode will cause a voltage drop and will reduce efficiency further. On 5V rail with 0.5V drop you will lose another 10%.

having the option of 2 charging ratings is the best here.

You need to be able to switch between different charge current limits. Otherwise it is pointless in my eyes.

why pointless? say we assigned 1.2A as maximum... then if user has 5v @ 0.5A - diode drop - boost drop - etc... = 0.3A total. then the circuit will provide 0.3A of current. If he connected a good 12v-24v supply, then it will give the maximum of 1.2A. However, it is gonna protect and charge the batteries in bot conditions.

Now, the performance depends on your choice that you understand.

And if you plan to use 12V supply, why not to use 20V laptop brick? You would not need boost circuit and you would have higher charging current, if needed.

Most laptop or chinese supplies will give 12-24v DC @ 5A or so which is nice. I will modify the circuit to make the boost chip provide up to 25v so it can take all types of inputs (5v, 12, 18, 20, 24) since if I assigned it at 20v and suddenly connected 24v... then what happens?

my main point was to allow people to connect it to USB (assuming they understand the limitations) + high power DC.

let's try to solve the CTL pin now. shall we?

always thanks to your cooperation.

[VEGETA]

A quick thinking for isolating the 2 power sources came to these 2 solutions:

1- using a 24v relay: https://lcsc.com/product-detail/Relays_SRA-24VDC-CL_C127865.html

I can tie its coil to the 24v laptop charger as well as tying its NO to it, while the NC is for 5v USB. This way 5v USB is by default unless you plug the 24v dc supply. this can go for around 0.3$ which is kinda ok. contact resistance is about 100 mOhms which is not perfect but can be enough (wasting 0.05 drop voltage for 5v supply).

2- using a SPDT switch, panel mount like this: https://lcsc.com/product-detail/Relays_SRA-24VDC-CL_C127865.html

here the user has to pick manually.

[mvs]

why pointless? say we assigned 1.2A as maximum... then if user has 5v @ 0.5A - diode drop - boost drop - etc... = 0.3A total. then the circuit will provide 0.3A of current. If he connected a good 12v-24v supply, then it will give the maximum of 1.2A. However, it is gonna protect and charge the batteries in bot conditions.

Because it is not that easy. You need to limit charge current not only for battery, but also to match your power source.
1.2A after boost converter is >4A before boost converter, if you convert 5V to 18.6V. This means that 5V 500mA USB port fuse will trip or USB port burns. 5V 2.5A USB charger will likely go in overload/short circuit mode...

[VEGETA]

why pointless? say we assigned 1.2A as maximum... then if user has 5v @ 0.5A - diode drop - boost drop - etc... = 0.3A total. then the circuit will provide 0.3A of current. If he connected a good 12v-24v supply, then it will give the maximum of 1.2A. However, it is gonna protect and charge the batteries in bot conditions.

Because it is not that easy. You need to limit charge current not only for battery, but also to match your power source.
1.2A after boost converter is >4A before boost converter, if you convert 5V to 18.6V. This means that 5V 500mA USB port fuse will trip or USB port burns. 5V 2.5A USB charger will likely go in overload/short circuit mode...

My calculations were like this:

5v 500mA USB = 2.5W -----> 16.8v @ 2.5W = 148mA of charge current maximum. Thus it won't reach the 1.2A maximum limit. That is unless you mean the charging circuit will force 1.2A which will cause voltage drop at 5v USB.

5v 2.5A USB = 12.5W ------> 16.8v @ 12.5W = 744mA of charge current.

Kindly explain to me if I am wrong. I assumed the charging IC will not exceed the current limit and won't force constant current if it cannot reach it.

So far we have no problems using the DC jack input and we have a problem with USB input. what solution do you suggest we work on?

[mvs]

My calculations were like this:

5v 500mA USB = 2.5W -----> 16.8v @ 2.5W = 148mA of charge current maximum. Thus it won't reach the 1.2A maximum limit. That is unless you mean the charging circuit will force 1.2A which will cause voltage drop at 5v USB.

To not exeed 500mA at 5V rail, you will need to enforce 148mA limit on your charge IC. Just forget 1.2A, in this case it should be 148mA ( - some margin for losses).

If you connect 12V 2A supply, you may reconfigure your charging IC for full 1.2A charging current.

[VEGETA]

To not exeed 500mA at 5V rail, you will need to enforce 148mA limit on your charge IC. Just forget 1.2A, in this case it should be 148mA ( - some margin for losses).

If you connect 12V 2A supply, you may reconfigure your charging IC for full 1.2A charging current.

Well it is easy then.

Charging current = 0.120 / Rsense

Option 1 (USB 5v): Rsense = 1 ohm -> charging current = 120mA.
Option 2 (12-24v): Rsense = 0.1 ohm -> charging current = 1.2A   ----> this could also be configured to have more than 1.2A if needed like ~ 1.5A or something.

I will modify the circuit to show this assuming the use of SPDT relay, until we reach a better solution. Also, we need to determine the CTL pin although I am kinda sure it is connected properly.

[VEGETA]

Updated schematic to version Alpha 3.

Updates in brief:

1- Added a SPDT relay to choose which power supply (USB or DC jack). I will make a custom model and footprint for it later.
2- Modified boost to give 25.2v to always ensure working with up to 24v of input.
3- Now has 2 shunt resistors for charge current set: 1.2A and 120mA for DC and USB respectively.
4- Shunt resistors are automatically chosen depending on active source... via N-channel MOSFETs.
5- Added a 5.1 zener diode to DC jack input to act as enable voltage for 1.2A charge current.

I barely made it fit into one page, I still don't know how to make another page in KiCAD (but within same schematic sheet).

I remind you that CTL is "low" which means "normal operation" according to datasheet.

how about it now?

[fcb]

Looking at your schematic.

1. You've definitely not read/understood how the buck converter/charge IC (CN3765) works.  L2 is the buck inductor and is wired all wrong.
2. Is D2 necessary?
3. Risk of pulling DC jack out, relay de-energising, C5 dumping straight into USB+, you could possibly argue that the relay will take longer to reclose than the boost converter will take to empty C5, but still not nice. You could just put a 2nd HX3608 in to deal with the 2nd voltage source, probably cost less than the relay.
4. My head hurts looking at the multiple current shunts (x3) and I'm not going to start looking up the Vgs of Q4/5 to work out what is going on.

My advice would be to start again, simplify your specification (avoid mission creep) and prototype something.

[mvs]

2- Modified boost to give 25.2v to always ensure working with up to 24v of input.

You do not need to do this. Feedback set to 19V is absolutely fine. Higher input voltage will find its way to the output (throught inductor and boost diode) in any case.

3- Now has 2 shunt resistors for charge current set: 1.2A and 120mA for DC and USB respectively.
4- Shunt resistors are automatically chosen depending on active source... via N-channel MOSFETs.

120mA shunt can stay always connected, no need to switch it.
You have battery voltage (up to 16.8V) on source of the MOSFET. It would not open with just 5V on the
gate.
Note that RDSon of the MOSFET and ressistance of the tracks on PCB will affect results of curent measurement. Kelvin connection (Fig. 3 in the CN3765 datasheet) is not possible.
 

I remind you that CTL is "low" which means "normal operation" according to datasheet.

Connect CTL to VSS, not to the minus pole of the battery.

[VEGETA]

You do not need to do this. Feedback set to 19V is absolutely fine. Higher input voltage will find its way to the output (throught inductor and boost diode) in any case.

Ok understood, I will modify it.

120mA shunt can stay always connected, no need to switch it.

It can affect the high current but not by much so yeah you are right. It would be around 1.32A instead of 1.2A so it is ok. but still need to switch the 0.1 shunt.

You have battery voltage (up to 16.8V) on source of the MOSFET. It would not open with just 5V on the
gate.

I was misguided.. logic level mosfet need Vgs to be 5v or more but this one has Vgs in the minus due to source voltage as you said. Putting DC jack voltage won't do a thing too. What do you suggest?

Note that RDSon of the MOSFET and ressistance of the tracks on PCB will affect results of curent measurement. Kelvin connection (Fig. 3 in the CN3765 datasheet) is not possible.

Rdson is low here and PCB tracks will be as short as possible.

why kelvin connection is not possible? Also, I was going to take the connection from the pads of the PCB directly, I don't have 4 terminal resistors.

Connect CTL to VSS, not to the minus pole of the battery.

will do.

You've definitely not read/understood how the buck converter/charge IC (CN3765) works.  L2 is the buck inductor and is wired all wrong.

I was wiring it correctly but when I tried to fit everything into the small space I moved it like that but forgot to re-draw it properly. sorry for this.

Is D2 necessary?

datasheet doesn't say about it but rather it says we can eliminate D1. Not an issue right now.

Risk of pulling DC jack out, relay de-energising, C5 dumping straight into USB+, you could possibly argue that the relay will take longer to reclose than the boost converter will take to empty C5, but still not nice. You could just put a 2nd HX3608 in to deal with the 2nd voltage source, probably cost less than the relay.

it definitely costs less (assuming inductor + others are cheap) and will try to do it.

still the output of the 2 boost converters will be connected together. Yes the shotcky diodes will do some protection but still...

if both sources got connected... 5v on one of them while +12 on the other... could the diodes deal with isolation and protection? I guess we will be forced to put the feedback resistor divider before the diodes to make sure nothing gets triggered right?

My head hurts looking at the multiple current shunts (x3)

I will try to modify them to make it better... as consolidated as possible.

and I'm not going to start looking up the Vgs of Q4/5 to work out what is going on.

mvs spoke about Vgs of the mosfets and I took the note... working on it.

My advice would be to start again, simplify your specification (avoid mission creep) and prototype something.

I will work again to do all of these modifications. I don't have any parts to prototype but rather will finish the design and order the pcbs with materials.

always thanks for both of you...!!

[fcb]

If you put a 2nd boost converter in: just set the  voltage from the EXT.DC powered converter a little higher than the USB one.

[VEGETA]

Updated to version Alpha 4.


Changes in brief:

1- Added another HX3608 boost, now dedicated boost for each supply.
2- Added an NPN transistor (Q4) to enable the DC shunt resistor which allows +1.3A while the 120mA one is always connected as we agreed.
3- Re-arranged feedback resistors as agreed.

I hope this solves the problems. The only thing I am not sure of yet is what happens if both supplies are connected at the same time since this will make both boosts on. USB boost is set to 18.6v while DC one is 24.6v. Both connected to the same node... so what is the situation?