Pics/Info: Inside the battery pack

[wk057]

One thing I didn't see in any of the pics -- where are the coolant loop connections made between pack and car? You showed the two electrical connectors (HV and LV) on the back edge of the pack (w.r.t how it is mounted under the car). Are the press-fit coolant connections on the front part/edge?

Th coolant loop is interesting to me b/c to support quick battery changes, there is no air-purge step between disconnecting battery #1 and reconnecting battery #2. Loop has to be designed to capture and deal with what little air is introduced during the swap...

The coolant loop quick disconnect is in the front. It has spring loaded caps that close when disconnected so I doubt that much air, if any, would make it in.

Pretty sure its visible in one of my earlier pics. If not I'll post one.

- - - Updated - - -

Yeah there is one. It's captioned "Another shot of the coolant loop connections and front modules."

 

[rabar10]

The coolant loop quick disconnect is in the front. It has spring loaded caps that close when disconnected so I doubt that much air, if any, would make it in.

Pretty sure its visible in one of my earlier pics. If not I'll post one.

- - - Updated - - -

Yeah there is one. It's captioned "Another shot of the coolant loop connections and front modules."

Yep, I missed them. Originally I thought those two black cylinders were coolant reservoirs, but they are clearly sticking up outside of the pack dimensions, and are also visible in the still-assembled pack pics as well. Thanks.

Next question -- did you note the series-connection order of all of the modules? i.e. tracing the series connection path from low-side contactor through the modules/fuse and to the high-side contactor?

My initial guess would have been from the connector/contactor, then up one side, through the stacked modules and fuse, and back down the other side to the other contactor and connector. But your description of the 2/0 cable running the length of the pack mid-line wouldn't make sense in that arrangement...

 

[magnet]

That sucks you had to break it down. It doesn't sound like the pack will be able to be rebuilt by robots easily...

I'm still trying to figure out what is going on with the negative terminal on each 18650. It looks like there is some kind of end cap on each cell. There is also a central oval shape piece on the terminal itself that connects the bond wire. They appear to be made from the same metal by looking at the texture, and appear to be attached to the cell, not part of the cell. It looks like both the endcap and oval are connected to the negative busbar.

On the postive terminal it looks like there is also "two connections" with a bond wire to the main triangle on the the cell and then a U-shaped ring contact. Can you clarify what is connected to what. Thanks

 

[wk057]

Yep, I missed them. Originally I thought those two black cylinders were coolant reservoirs, but they are clearly sticking up outside of the pack dimensions, and are also visible in the still-assembled pack pics as well. Thanks.

Next question -- did you note the series-connection order of all of the modules? i.e. tracing the series connection path from low-side contactor through the modules/fuse and to the high-side contactor?

My initial guess would have been from the connector/contactor, then up one side, through the stacked modules and fuse, and back down the other side to the other contactor and connector. But your description of the 2/0 cable running the length of the pack mid-line wouldn't make sense in that arrangement...

The series connection started at one of the rear modules, then snaked back and forth (side, up, side, up) to the front of the car to the oddball modules, then out of the last one, into the fuse, then back to the contactor in the rear through the 2/0 cable.

- - - Updated - - -

So, Tesla seems to have gone silent on my requests for data on the BMS interface... maybe someone can make use of these high resolution scans of one of the boards from a module and gain some insight. (Click for full res version, ~12MB).

Images I post in this thread and my related commentary are posted and published by me, the original photographer. All copyrights and all other rights reserved. These images may not be copied or otherwise distributed outside of this forum without my express permission.


Front of BMS Module board (one of these on each of the 16 modules)



Back of the BMS Module board.

You can clearly see the bleed resistor setup for balancing now with the board removed (it is hard to see the whole board when installed).

The little halos around some of the components are artifacts of the flatbed scanner passing over the coating that is on the board.

More pics later when I organize them and finish dismantling the pack tonight.

 

[Cottonwood]

Images I post in this thread and my related commentary are posted and published by me, the original photographer. All copyrights and all other rights reserved. These images may not be copied or otherwise distributed outside of this forum without my express permission.


...
You can clearly see the bleed resistor setup for balancing now with the board removed (it is hard to see the whole board when installed).
...

Great pictures!

Below are some enlargements of one of the repeating pattern of 6 bleed circuits, one for each group of paralleled cells; one enlargement for each side of the board. Don't get fooled by the bottom set; all the components are there, just rearranged a little for board layout.

This is a serious multi-layer board; it will be a nice task to recreate the circuit diagram from it, but it is clear that the board contains bleed circuits for balancing. In particular, notice how the four, surface-mount, power resistors with the "1580" label, one of which is labeled "R11", are in parallel for extra power dissipation. My guess is that these resistors are 15.8 Ohms each, and 4 in parallel would be a resistance of 3.95 Ohms. If the full cell voltage is a little over 4 Volts this would mean a power per resistor in parallel of a little over a Watt. With some PWM techniques, this could easily be modulated down to lower power levels.

 

[magnet]

Interesting....they are using a off the shelf TI chip to do monitoring and bleed balancing. What does the host controller board look like?

They are using RF-coupler instead of an opto-coupler for isolation

Are they any other BMS or BMB boards? The more pics the better. Thanks

 

[rlang59]

This is a serious multi-layer board

Looks to me like it is only a 4 layer board.

 

[wk057]

In particular, notice how the four, surface-mount, power resistors with the "1580" label, one of which is labeled "R11", are in parallel for extra power dissipation. My guess is that these resistors are 15.8 Ohms each, and 4 in parallel would be a resistance of 3.95 Ohms. If the full cell voltage is a little over 4 Volts this would mean a power per resistor in parallel of a little over a Watt.

"1580" would be 158 ohms, 1% tolerance. So 39.5 ohms for the parallel set. That'd be 0.44W for the the set (0.11W per resistor) at 4.15V. Each set of 74 cells contains about 885 Wh of energy, so, to drop it 1% would mean about 21 hours of bleeding.

 

[Cottonwood]

Looks to me like it is only a 4 layer board.

Probably correct, but still multilayer, and a pain to trace out the connections...

 

[wk057]

Expanding on that, the BMS could at most draw about 42W from the pack if all 96 bleeders were active at once... which would take 84 days to bring the pack (85kWh) from 100% to 0%, lol.

I think most imbalances would be a few 10s of mV, which should be only a couple hours of bleeding probably.

 

[apacheguy]

This is a serious multi-layer board; it will be a nice task to recreate the circuit diagram from it, but it is clear that the board contains bleed circuits for balancing. In particular, notice how the four, surface-mount, power resistors with the "1580" label, one of which is labeled "R11", are in parallel for extra power dissipation. My guess is that these resistors are 15.8 Ohms each, and 4 in parallel would be a resistance of 3.95 Ohms. If the full cell voltage is a little over 4 Volts this would mean a power per resistor in parallel of a little over a Watt. With some PWM techniques, this could easily be modulated down to lower power levels.

Ok, well that's all well and good, but can you explain this in terms understandable to someone who doesn't hold a degree in electrical engineering? Particularly the bolded phrase. What is the take away that tells us something meaningful about how the pack balances itself?

 

[rlang59]

Ok, well that's all well and good, but can you explain this in terms understandable to someone who doesn't hold a degree in electrical engineering? Particularly the bolded phrase. What is the take away that tells us something meaningful about how the pack balances itself?

The big surface mount resistors are used to bleed down the voltage on cells that measure higher than the others by sinking current though them.

- - - Updated - - -

Probably correct, but still multilayer, and a pain to trace out the connections...

Maybe but since they are using this chip http://www.ti.com/product/bq76pl536A I'd bet looking at the reference design or datasheet would tell you exactly what they are doing.

Here is a link to an app note: http://www.ti.com/lit/an/slaa478/slaa478.pdf

 

[Cottonwood]

"1580" would be 158 ohms, 1% tolerance. So 39.5 ohms for the parallel set. That'd be 0.44W for the the set (0.11W per resistor) at 4.15V. Each set of 74 cells contains about 885 Wh of energy, so, to drop it 1% would mean about 21 hours of bleeding.

Thanks for the correction!

Hmmm.... They seem like big resistors for a 1/10th of a Watt. OTOH, maybe all they need is 0.1% corrections, from time to time.

 

[rlang59]

Thanks for the correction!

Hmmm.... They seem like big resistors for a 1/10th of a Watt. OTOH, maybe all they need is 0.1% corrections, from time to time.

They look like 1206 resistors which are typically rated at 1/8 of a Watt and it's not unusual to de-rate by 20% though for a constant load.

 

[apacheguy]

The big surface mount resistors are used to bleed down the voltage on cells that measure higher than the others by sinking current though them.

Ok, so the excess electricity is wasted as heat, right? Sounds to me that balancing can occur at any SOC regardless of whether or not the car is charging. So does this bust the long standing TMC theory that pack balancing only occurs on a max range charge?

 

[ACDriveMotor]

It does seem to suggest that it could happen at lower SOC. Is there a reason not to balance at, say, a 70% SOC?

 

[wk057]

That sucks you had to break it down. It doesn't sound like the pack will be able to be rebuilt by robots easily...

I'm still trying to figure out what is going on with the negative terminal on each 18650. It looks like there is some kind of end cap on each cell. There is also a central oval shape piece on the terminal itself that connects the bond wire. They appear to be made from the same metal by looking at the texture, and appear to be attached to the cell, not part of the cell. It looks like both the endcap and oval are connected to the negative busbar.

On the postive terminal it looks like there is also "two connections" with a bond wire to the main triangle on the the cell and then a U-shaped ring contact. Can you clarify what is connected to what. Thanks

View attachment 58759


View attachment 58758

Missed this almost...

The 18650 cells are bare, no label or anything. So the entire casing, including the negative end, is negative. So you're looking at the negative end of the cell. The bus plate sits on top of a piece of plastic with cut outs for each individual cell. The cutout in the plastic with it's raised walls around the hole (to protect the small fuse) are what you're seeing on top of the negative end of the cell. The cell level fuse then connects from the bus plate to the cell itself through this opening in the plastic.

The same for the positive side, except the positive end of the cell has a triangle-like positive terminal that is insulated from the negative casing.

Note that the plastic protective cover is in place on the modules in most of my pictures, also.

 

[arg]

It does seem to suggest that it could happen at lower SOC. Is there a reason not to balance at, say, a 70% SOC?

You certainly can perform the balancing at any SoC, but the problem is to be sure of knowing exactly how much balancing is required. All cells equal voltage at 100% SoC is the target, since that's an absolute limit - can't take any cell above the max voltage. Since the voltage/discharge curves might not be identical for all the cells, balancing at 70% might not leave you perfectly balanced when you then charge to 100%.

Of course you could do things like correcting gross imbalance when a 70% charge is requested, and only fine-tuning when doing a 100% charge.

 

[vielster]

Great pictures!

Below are some enlargements of one of the repeating pattern of 6 bleed circuits, one for each group of paralleled cells; one enlargement for each side of the board. Don't get fooled by the bottom set; all the components are there, just rearranged a little for board layout.

This is a serious multi-layer board; it will be a nice task to recreate the circuit diagram from it, but it is clear that the board contains bleed circuits for balancing. In particular, notice how the four, surface-mount, power resistors with the "1580" label, one of which is labeled "R11", are in parallel for extra power dissipation. My guess is that these resistors are 15.8 Ohms each, and 4 in parallel would be a resistance of 3.95 Ohms. If the full cell voltage is a little over 4 Volts this would mean a power per resistor in parallel of a little over a Watt. With some PWM techniques, this could easily be modulated down to lower power levels.

View attachment 58784

View attachment 58783

Great shot of the balancing circuit. You can even see the FET (Q3) which is shunting the current through the resistor bank.

P=V^2/R BTW, so you're talking 4W if those are 15.8ohm in parallel. Not only are the resistors incapable of handling that current, the circuit board doesn't appear to have much in the way of copper pours to dissipate heat in this area so that would suggest the circuit would not be capable of dissipating 4W (or even 1W) continuously without possible damage. For safety you should assume that the balance circuit may get stuck on (say FET fails closed), and this condition should not endanger the battery. I read those as 158x10^0 (so 158ohm resistors), which would be under a half watt.

This passive balancing scheme is fairly common, and it is pretty slow. When batteries are severely out of balance, the charger must be set to a very low charging current towards the end of the charge (once the highest charged cells approach their max voltage). Essentially, the charging current will then max out at the balancing current (in this case about 100mA) until the remaining cells complete their charge. The first charge cycle this trickle charge may take a long time 10s of hours) but once the cells are balanced, they only drift apart slowly over time as they age differently. Balancing these disparities each charge cycle is negligible. Additionally if their BMS does a reasonable job at modeling the cells, they can actually predict the aging and do balancing throughout the entire charge rather than waiting towards the end of the cycle....another reason the firs cycle(s) would be slower since the BMS doesn't have enough data to model the cells in that battery.

My guess is that by the time it enters a car it has already gone through several cycles, thus is balanced and has appropriate info to model the cells to maintain short balancing operations.

 

[scaesare]

I had to first drain the coolant loop. I couldn't find anything that would mate with the quick disconnect easily, so, I rigged up some PC liquid cooling tubing to each outlet, which fit somewhat snug inside them. I then fed air into one side and drained from the other until I didn't get any more coolant. About a gallon of coolant was liberated.

More cool info.

This sounds like it was a gallon or so per module, not the whole pack, right?