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Pics/Info: Inside the battery pack

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I'm not sure how out of balance the cells get from one another... but even with all 6 sets of bleed resistors enabled on a module through an 8 hour charge we're talkkng about 20Wh dissipated which is pretty negligible.

For simplicity it probably only does balancing at high states of charge so the differences are more accurate.

There is almost certainly no circuitry for inter-module charge shuffling here considering the presence of the isolation chip.
 
If you charged at the same rate as balancing, you could effectively bypass a module. However this would be only be available at very low charge currents (like at the end of a charge cycle.)

This might be why Tesla use specific 158 ohm, 1% resistors, as knowing cell voltage, they would know cell balance current accurately.
 
Inter-module active balancing is done by shuttling during charging but this shuttling is not done in the traditional dangerous way. Balancing is done by increasing the voltage not only bleeding. Bleeding is not an efficient way of balancing as it just creates heat. Virtual shuttling is done by continuously changing the target balance voltage which will in effect shift charge through the whole pack

Magnet, I've asked earlier for some source for earlier claims, but didn't see a reply.

Thus far, it doesn't appear that anything in wk057's documented tear down support these ideas. Is this conjecture on your part? If not, what is the source for your assertions?

Thanks.
 
Magnet, I've asked earlier for some source for earlier claims, but didn't see a reply.

Thus far, it doesn't appear that anything in wk057's documented tear down support these ideas. Is this conjecture on your part? If not, what is the source for your assertions?

Thanks.

I've fixed my own BMB (a few times) in my Roadster, Model S pictures posted here just seems to refine the existing tech, Magnet posts regarding "active balance" and "active impedance" etc.. seem like conjecture to me.
 
I am trying to understand what I am seeing here. Are the individual module BMS boards entirely capsulated within the module? Meaning they operate entirely one their own within the module requiring no communication or connection to the other modules or the rest of the car for that matter. Thus each module is self balancing within itself? Is this correct or am I not seeing the interconnection between the other modules to the main BMS board?
 
Do you have any pictures of your BMB? It'd be interesting to compare it.
Here is a fuzzy pic from an old email, the better ones at home. Has same fuses (mine are on the back of the PCB), similar bleed resistor logic, CAN bus transceiver isolator. Though looks like they removed the on-board dc-dc converter, and removed two of the temperature sensors.

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Are the individual module BMS boards entirely capsulated within the module?
What your seeing is BMB boards, they talk to the BMS.

Meaning they operate entirely one their own within the module requiring no communication or connection to the other modules or the rest of the car for that matter.
They operate on their own, but talk to each other over CAN bus and to the the BMS.

Thus each module is self balancing within itself?
If it's anything like the Roaster, it will only self balance when the BMS tells the BMB it to bleed off certain bricks down to lower the other cells to balance it. Normally the BMB just reports back voltages to the BMS.

Is this correct or am I not seeing the interconnection between the other modules to the main BMS board?
Your not seeing the interconnections.
 
The particular IC on the Model S BMB is autonomous so to speak. The BMS will not command it to sink currently directly, but the BMS will set a target voltage and it will bleed off to reach that target.

Yes. Each BMB on each module has it's own target balance voltage (TVB) for each string in the module, but each BMB also sends it's highest TBV value to the BMS which then resets the global highest TBV of all the modules. Shuttling is a bad way to describe what is going on. You could call it "lossless balancing" or "lossless equalization" or "distribution balancing"

It is not totally lossless of course as the BMB is powered by the module. Now what happens if you "charge/distribute" through the balance wires and that "charger/distribute-r" is powered by the battery itself. The total module voltage would be decreased but you would redistribute the charge to balance each string.
 
The particular IC on the Model S BMB is autonomous so to speak. The BMS will not command it to sink currently directly, but the BMS will set a target voltage and it will bleed off to reach that target.

Below is a diagnostic screen available to Tesla service people that shows the Voltage and Temperature measurements from the BMBs. Notice high and low Voltages and Temperatures being highlighted. Also notice how tight the range is with a Voltage spread of 0.013 Volts and a Temperature spread of 0.6˚ C. The car had been charging at 19 kW for at least 20 minutes at this point after the tech fixed my HPWC.

BMS.JPG
 
Magnet, I've asked earlier for some source for earlier claims, but didn't see a reply.

Thus far, it doesn't appear that anything in wk057's documented tear down support these ideas. Is this conjecture on your part? If not, what is the source for your assertions?

At least part of what he's saying makes sense, given the circuitry on the per-module management boards:

Inter-module active balancing is done by shuttling during charging but this shuttling is not done in the traditional dangerous way. Balancing is done by increasing the voltage not only bleeding. Bleeding is not an efficient way of balancing as it just creates heat. Virtual shuttling is done by continuously changing the target balance voltage which will in effect shift charge through the whole pack

Think of it this way.. From the negative terminal of the entire pack to the positive terminal, there are sixteen modules, and let's say 400V for a nice round number. From the per-module BMB's perspective, there are negative (0v) and positive (~25v) ends. If you count from the pack negative, through module A, then module B, and so on until you reach the positive terminal, module A's negative is the pack negative, and mod A's positive is mod B's negative, and so on, in series. Clear?

Sorry, I know this is basic and well-understood, but bear with me. It's critical that everyone is up to speed on this much to make sense of what's next.

So, each BMB (per module) can bleed a group (a group of parallel cells) to equalize it with its (series) neighbors. It does this by shorting out that group's positive and negative ends (a difference of ~4v) through the bleeder resistor for that group. When you do this, you have at most 158R / 4 (paralleled resistance) = 39.5R draining the group's voltage down to where the BMB wants it to be. Still clear?

OK, now think about what happens when the pack is charging. If the bleeder resistors are turned on by the FET, instead of bleeding power out of the group, you are effectively limiting the ability for that group to charge. Not completely, the cells will still receive some of the charge current, but not as much as if the cells in the group were the only electrical path through. Remember, the charge is current-limited, so by shunting around a group, you take the available current and divide it through two paths -- one through the group, one around it. This limits the voltage available to that specific group while the others still receive their full "dose". This isn't "shuttling" of charge from one group or module to another, but it is a way to control how much each group/module charges.

Furthermore, the FET can either be turned on hard, or activated in its linear region, depending on whether the BMB's signal to the FET's gate is digital logic or analog. What this means is that the FET itself can be anything between a dead short (or close to it -- the Rds(on) spec), and open circuit. With a DAC feeding the gate, you could variably control the impedance through that bleeder circuit (fixed resistors in series with the variable FET resistance), for any one cell group, and ultimately for the entire module. This enables active control over the amount of charge of any cell group in the entire pack. Ergo, that "active impedance" thing someone spoke of earlier. Pretty clever.
 
Yes. Each BMB on each module has it's own target balance voltage (TVB) for each string in the module, but each BMB also sends it's highest TBV value to the BMS which then resets the global highest TBV of all the modules. Shuttling is a bad way to describe what is going on. You could call it "lossless balancing" or "lossless equalization" or "distribution balancing"

It is not totally lossless of course as the BMB is powered by the module. Now what happens if you "charge/distribute" through the balance wires and that "charger/distribute-r" is powered by the battery itself. The total module voltage would be decreased but you would redistribute the charge to balance each string.
This is still passive balancing, and is not loss-less -- the extra energy is lost via heat in those shunting resistors.

It IS intelligent balancing, i.e. per-paralleled-cell-group inside each module, and then among all modules in the pack. But the system is not actively removing energy from one module/cell in the pack and placing that energy into another cell/module.

When any of those shunt FETs are on, a current will flow proportional to that paralleled-cell-group's voltage. That current will either come from the cells themselves (when not being charged), or from externally-supplied current i.e. if the pack is being charged.

Tesla has chosen to carefully match and balance cells/modules before they are integrated into the pack. That way the circuitry required to keep those cells/modules in balance, over the life of the pack, is minimized.

The up-sides of avoiding active balancing are both decreased component cost and also fewer failure-modes -- fewer places where a component failure can cause bad things to happen (like internal shorts or larger continuous drains on a cell group).

The down-side is that the entire pack is always limited by its weakest paralleled cell group capacity. There's no way to prop up one weak group (via active balancing) to access any additional Ahrs available in the stronger groups. Not an issue if your QC is good and your faith in all the cells to perform/degrade similarly over time is high.

Active cooling also helps a lot here -- since these cells' performance/degradation is so temperature-dependent, ensuring that everything is as near the same temperature as possible helps a lot.
 
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And remember the resistive bleed circuits are pretty delicate in their load. The 4 158 Ohm resistors in parallel are 39.5 Ohms with the cell Voltage about 4 Volts, that means the bleed current is about 0.1 Amp, dissipating about 0.4 Watts. The Amp-hour capacity of these packs is about 85 kWh / 370 Volts or about 230 Amp-hours. That means that it would take 2.3 hours to bleed 0.1% off of a set of paralleled cells. That's a very delicate and gentle balance that only works in a reasonable time, if the cells are almost matched to start with.
 
I drew up this quick schematic to get an idea of what the balancing circuit looks like. There's a diode as well, but it appears to only be used for reverse protection/transients. (Might be a zener.) It's obvious that conducted noise was a concern to Tesla - I suppose when you're pulling 1000A from a battery at maybe 30kHz, there's going to be a little noise! So they appear to have a network of filters, starting with L7 and C24, and the network of 11.3k + C26/C27, and 10k + C56/C57.

Also of note is that Tesla follow good engineering guidelines for ceramic capacitors in "mission critical" systems. A short of any of the capacitors attached to the battery cell would lead to a pack contactor abort, so they put two capacitors in series. One capacitor is probably sufficient to achieve the desired attenuation at the frequency of operation so the battery pack can operate with one capacitor bad in each series group just fine. They also fuse both the positive and negative of the pack despite this not being strictly necessary (you would only need to fuse the positive.)
 

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At least part of what he's saying makes sense, given the circuitry on the per-module management boards:



Think of it this way.. From the negative terminal of the entire pack to the positive terminal, there are sixteen modules, and let's say 400V for a nice round number. From the per-module BMB's perspective, there are negative (0v) and positive (~25v) ends. If you count from the pack negative, through module A, then module B, and so on until you reach the positive terminal, module A's negative is the pack negative, and mod A's positive is mod B's negative, and so on, in series. Clear?

Sorry, I know this is basic and well-understood, but bear with me. It's critical that everyone is up to speed on this much to make sense of what's next.

So, each BMB (per module) can bleed a group (a group of parallel cells) to equalize it with its (series) neighbors. It does this by shorting out that group's positive and negative ends (a difference of ~4v) through the bleeder resistor for that group. When you do this, you have at most 158R / 4 (paralleled resistance) = 39.5R draining the group's voltage down to where the BMB wants it to be. Still clear?

OK, now think about what happens when the pack is charging. If the bleeder resistors are turned on by the FET, instead of bleeding power out of the group, you are effectively limiting the ability for that group to charge. Not completely, the cells will still receive some of the charge current, but not as much as if the cells in the group were the only electrical path through. Remember, the charge is current-limited, so by shunting around a group, you take the available current and divide it through two paths -- one through the group, one around it. This limits the voltage available to that specific group while the others still receive their full "dose". This isn't "shuttling" of charge from one group or module to another, but it is a way to control how much each group/module charges.

Furthermore, the FET can either be turned on hard, or activated in its linear region, depending on whether the BMB's signal to the FET's gate is digital logic or analog. What this means is that the FET itself can be anything between a dead short (or close to it -- the Rds(on) spec), and open circuit. With a DAC feeding the gate, you could variably control the impedance through that bleeder circuit (fixed resistors in series with the variable FET resistance), for any one cell group, and ultimately for the entire module. This enables active control over the amount of charge of any cell group in the entire pack. Ergo, that "active impedance" thing someone spoke of earlier. Pretty clever.

Yes! You get it! This the "active impedance" control is only enabled towards end of charge when the charge is slowing the the string voltage is more stable. Is it totally lossless? No of course not. But is is safer than true shuttling and it is more efficient than simple bleed balancing.