Series BMS Confusion

Hi Everyone.

I suspect this question should have an easy answer to someone who understands it, and plenty of confusion from people who don't. It involves diodes, bad design, and magic smoke.

The story is that I'm using 3 x 20S BMS instead of 1 x 60S BMS. This is a LTO pack, so 162V max, 144V nominal. Don't judge me. It's hard enough to find any 60S BMS, let alone one that will support LTO.

The MOSFETs are rated for 100v only.

The charger is built from 3 x 48V isolated chargers in series, connected through the BMS.

So far, so good - In theory, I shouldn't need Schottky Diodes to protect the pack, because no BMS will ever see more than the charger voltage.

On the other hand... Regen. On a hard regen stop the system will probably see above the max voltage. Magic smoke time if that causes one of the BMSes to shut down, putting the full regen voltage through the remaining two.

So what are my options?

1. Schottky diodes across each pack? I know this is the recommendation, but no idea WHY, and therefore can't design it. Will a 200V, 70A Schottky be big enough? That's the biggest one I can find. I have no idea how much current I will be making in regen.

2. Set the over-voltage shutoff to be higher than recommended, knowing that LTOs are very tolerant to overcharge?

3. Route around the BMS for discharge, lose all the energy counters, and install my own?

Appreciate any feedback. Thanks.

I am thinking that going with number 3 is a good plan until you know more about what you are running. Routing around the BMS with a metering system that can tell you what you are burning and regen-ing will at least give you better system information then you have now. Once you know what you are up against the you can rewire for known numbers.

The diodes, AFAIK, need to be rated for sufficient current to match any flow the system might have to put thru them, and the highest reverse voltage the system would put across them.




Another potential issue you don't list, assuming you're using the BMS FETs to cut off discharge at LVC, is that when one pack cuts off due to a low cell, it can now have the full voltage of the rest of the pack across it, just like if regen causes an HVC event.

That's probably less of an issue than a regen spike, as teh voltage will be lower, but if it exceeds the FET spec, the same potential for failure exists.

Thanks guys. Appreciate the responses. Heading towards option 3 as e-beach said, because my head is spinning from this. Stay with me through this. Sorry.

When the BMS trips from LVC, that part of the circuit becomes open, right? Doesn't that mean the whole pack now is 0v? How does it get the voltage of the rest of the pack on it?

And if the rest of the pack voltage is 100v or less, does that make the BMS safe? (FETs are rated to 100v) 2/3rds of 162v (fully charged) is 54 volts, though unless something has gone catastrophically wrong, that shouldn't happen... 2/3rds of nominal is 48v, so two remaining packs shouldn't put more than 96v through the BMS.

This is seriously doing my head in! Sorry guys.

Sunder said:

When the BMS trips from LVC, that part of the circuit becomes open, right? Doesn't that mean the whole pack now is 0v? How does it get the voltage of the rest of the pack on it?

Click to expand...

That's an easy one to "see" with a voltmeter. If you have a keyswitch or similar on your bike, measure the voltage across the switch when it is on. You'll see basically zero volts, just like the FETs (also a switch) on the BMS when they are on.

Then measure it when the switch is off, and you'll see the full voltage of the system the switch controls--just like the FETs in the BMS.

The same is true of any switch in series with a voltage in a circuit.

Ok, so a little forward thinking here. So Sunder gives his LTO battery a full charge and allows it to balance. He then disconnects the BMS and takes it for a ride with a proper meter to get real world information on what the pack can do....max amp draw, max regen spike etc. And now, if he needs to add diodes, he can get the ones that are rated higher then his maximum current draw and maximum reverse voltage.

However I am at a bit of a loss at this point. Where does the reverse current hit? What part of the circuit.

@ Amberwolf,

1) would the highest reverse voltage be from the regen spike?
2) would the diodes attach to the B- to stop a wrong way voltage spike to the BMS?

Edit: Put diodes on the P- and B-?

What do you think?

amberwolf said:

The same is true of any switch in series with a voltage in a circuit.

Click to expand...

That's a lot clearer. Thanks so much. The circuit *is* broken, but because it's broken across the FET, the full potential difference of the rest of the pack - through the system (Being presumably a closed circuit), goes through the FET, potentially destroying it - just like having too high a voltage through a physical switch could cause arcing and destruction of the switch.

So the diode, then I presume is to create an alternative path for reverse current flow during charge/regen, but won't protect the BMS in the case of shutdown during discharge?

If I've got that right, I think I've got a fairly good idea what I need to do.

There are two separate settings in this BMS - "When do I start bleeding" and "When do I shut down the charger". Those two values don't have to be the same, and in fact, by default they are separate. I think for Li-Ion, it's 4.28 and 4.35 respectively. I'm pretty sure they're unrelated, so I can set the bleed off to start for LTO at 2.75, and set the cut off VERY high, like 4.35V. That way, there's no way regen or overcharge to cause the BMS to trip.

Likewise for discharge, I'll just turn off LVC or set it as low as I can.

Still feels "fragile", but I can keep an eye on the figures and see what is happening.

e-beach said:

Where does the reverse current hit? What part of the circuit.

Click to expand...

Reverse current (from regen) flows thru the *entire* series circuit, from motor phase wires back thru the controller thru the BMS thru the cells, thru the BMS of the next pack and it's cells, etc, all the way back to the controller....

From charging the reverse current only flows thru the charger, then thru BMS and thru the cells, BMS of next pack and it's cells, etc....

1) would the highest reverse voltage be from the regen spike?
2) would the diodes attach to the B- to stop a wrong way voltage spike to the BMS?

Click to expand...

There is no "wrong way voltage"; the voltage does not reverse (if it did, it would blow up the controller ). The voltage simply *increases* outside the battery during regen (or charging), and causes the *current* to reverse, going into the battery (as the lowest voltage in the system) instead of out of it (when it is the highest voltage in the system).


The diodes *could* be used in series with the BMS input so that current can only flow one way, out of the pack, but that means regen cannot be used at all.

I'm a bit on the worn-out side, but AFAICS at the moment, if the diodes are placed reverse-biased across each pack's output, then when a pack's BMS shuts off the diode provides a path around the turned-off FETs, so the current could still flow thru them.

Ok, then I mis-understood what you meant by this.

amberwolf said:

The diodes, AFAIK, need to be rated for sufficient current to match any flow the system might have to put thru them, and the highest reverse voltage the system would put across them......

Click to expand...

No worries. Hope you get a good nights sleep.

Sunder said:

The circuit *is* broken, but because it's broken across the FET, the full potential difference of the rest of the pack - through the system (Being presumably a closed circuit), goes through the FET, potentially destroying it - just like having too high a voltage through a physical switch could cause arcing and destruction of the switch.

Click to expand...

It might seem pedantic, but it may help understanding:

The *voltage* doesn't actually go "through" the switch or FET, but is simply across it. Whenever the voltage exceeds the breakdown voltage of the FET (max Vds) or the voltage rating of the switch, the possibility greatly increases of it forcing a *current* to flow thru the FET (or switch) anyway, which in the case of the FET causes fast extreme localized heating, and tends to vaporize the silicon, exploding the casing. The vaporized stuff can be very conductive, and plate the whole inside of the FET where the die used to be, leaving the FET shorted in the ON state. (which is why controllers that fail this way often leave the motor hard to turn as the phases are being shorted together by this). It can also destroy the gate driver by passing battery voltage back to this low-voltage circuit not designed to handle that...which is why replacing FETs doesnt' always fix such a problem. :/


In teh case of the switch it tends to arc across the contacts, vaporizing them. The plasma created can allow current to continue to flow and the heat from this can actually cause a fire relatively quickly. (basicaly is arc-welding inside the casing!)

So the diode, then I presume is to create an alternative path for reverse current flow during charge/regen, but won't protect the BMS in the case of shutdown during discharge?

Click to expand...

I...don't know. I'm too tired to think it thru at the moment, but my brain will probalby work it out while I nap.

There are two separate settings in this BMS - "When do I start bleeding" and "When do I shut down the charger". Those two values don't have to be the same, and in fact, by default they are separate. I think for Li-Ion, it's 4.28 and 4.35 respectively. I'm pretty sure they're unrelated, so I can set the bleed off to start for LTO at 2.75, and set the cut off VERY high, like 4.35V. That way, there's no way regen or overcharge to cause the BMS to trip.

Click to expand...

The only potential worry is if the trickle current left between charger and pack is higher than the bleed current, once the pack reaches "full". If that occurs, the cells would continue to charge anyway, and eventually overcharge. As long as the charger voltage is not higher than the pack's full voltage, and the cells do not drop in "full" voltage as they age by much, then it shouldn't cause an issue.

e-beach said:

Ok, then I mis-understood what you meant by this.

amberwolf said:

The diodes, AFAIK, need to be rated for sufficient current to match any flow the system might have to put thru them, and the highest reverse voltage the system would put across them......

Click to expand...

Click to expand...

Yeah, that's because the diodes would be installed reverse-biased (if in parallel with the pack output), so they will always have reverse voltage across them...but the actual system voltages never reverse.

e-beach said:

No worries. Hope you get a good nights sleep.

Click to expand...

That would be nice...but is unlikely right now. :/