GUIDE: Modifying the DeWalt A123 charger to not be useless

[OneEye]

To pull power from the inner terminal is going to require wiring up the battery as shown in figure 1 of the following document

http://www.neodymics.com/Images/V24Prot ... 70818E.pdf

There should also be a lead from + to pin 9 (enable) to power the logic of the BMS (as shown in figure 2 of the .PDF)

Then you can draw through the inner spade terminal and test whether the BMS does anything useful during discharge and what the current limits are for this wiring configuration (up to 22A?)

WARNING: Doing any of the above is almost certain to result in releasing the magic smoke in the battery

[Doctorbass]

Ok ok guy.. I understod...


I will do this famous test for you all.

I have 7 or 8 BMS that i have to blow for you

For sure.. i will not blow all of them just for fun...!

I have a very big non inductive dummy load in my lab (a 3kW 0 to 10 ohm).. and....... a 0.25ohm 1.2kW..(60C at 36V ha haaa!)

I will try to enable it and see what i can get on the controlled output at different current.. and will finally TORTURE one just for you.

I know teh circuit board have a 0.002ohm shunt to monitor the current... maeby i will bypass it even if it cut due to protection circuit Let's get REAL results!
What are your guess? will it blow at 20, 25, 30... 45??.... 50A??
let me know! i will try to give you the answer after the tests tonight!

Doc

[Doctorbass]

For additional info, you can see the pic of the pic of a connector that come with the 36V drill:



Also, about the flashlight, i know the circuit is similar, but it have an inductor on it.

Finally, already found an interesting disscussion on RCgroup forum about the BMS pins. here it is:

I have noticed that the top and bottom row of the 14 muti-pin connector seem to be the same top and bottom, so really only 7 connections besides the 3 main terminals. I can get the FET to turn on by pulling the center multi-pin to the positive side of the battery, but only with no load across the positive and FET output terminal. With a 470 ohm resistor across the positive to FET output, I get a short series of pulses across the load and then nothing untill I reset the process. I'm wondering if the input requires some type of pulse width control? Or it could just need a combination of control pins connected. I suppose the wise move would be to get a charger and drill, and then scope the logic on the multi-pin while in all modes of operation.

Kevin


and about the drill:
The circuit board in the impact wrench shorts two pins together, supplying power to the battery module. It also includes a resistor divider to apply voltage to a different pin, presumably indicating the presence of a tool (and perhaps even identifying the tool as well). If the signal connector pins are labeled with pin 1 next to the positive spade connector, then pins 3 and 4 are shorted and 2/3 of that voltage is applied to pin 5.

The board otherwise contains a freewheeling diode to protect the battery from polarity reversal (during motor direction changes) and a pair of transient voltage suppressors on the motor leads. The common terminal for the TVS's goes to pin 1 on the connector. Data logging, maybe? There is a 40 amp automotive-style blade fuse on the power input as well.

Bryan

Doc

[OneEye]

Nobody on this end is going to try to force you into destroying your toys; I'm just encouraging you to play with them some more

All in the interest of SCIENCE!



I'm guessing the BMS shuts off at ~25-30A.

I am still skeptical about whether it provides a low voltage cutoff to protect the cells... I'd love to be proven wrong.

At least you know your little mad-scientist experiment won't harm the cells, you've proven them under far worse conditions.

[OneEye]

I'm guessing the inductor on the flashlight is to clamp/spread out current spikes like the neodymics folks have on their e-bike system.

The drill is a somewhat inductive load anyway, so an additional inductor isn't needed.

[Doctorbass]

I just built the circuit to activate the BMS with potentiometer instead of resistor differents value.

IT WORK!

I will post results tonight as promissed.

Here some pics of my circuit
(i use 3 x 10k pot + 1x 250uhenry inductor and a simple switch)

Doc

[Doctorbass]

I started to test with a combinaison of load that give 1.5ohms load and I just got only 10volts when using the R valur of de desing in the PDF that are:
7.7K for the R2
8.5K for R3
and 1K for R4

I heard a low hf noise and the coil (a 250uH) created oscillation and the BMS cut after 5sec... I decided to bypass it and that solved the problem

I began to change value of these resistor that i replaced by 3 pot and got better results.
I got 29.37V at 19.7A that give 578W.

This time i heard no noise indicating me that probably the PWM reached the full duty cycle.

here are my resistor value for full PWM:

7.26K for the R2
7.28K for R3
and 4K for R4

Note that changing value of R4 did'nt change voltage at output.

I suspect that normally, the r valur that control the duty cycle should be a simple pot that replace the two resistor R2 and R3 that are in serie an dthat the pin 5 should be the middle connection of the voltage divider between pin 1 and 7... I supose...

Doc

[OneEye]

Beautiful, pulling nearly 20A through the circuit without a blowup.

Any luck the BMS shutting down the pack on a low voltage condition?

[CGameProgrammer]

It would be really great if you could monitor the voltage for each cell while discharging to see what happens. At 20A it'll take you 7 minutes to drain the battery, so you can discharge and watch each cell, turning it off when any cell voltage gets too low if the BMS doesn't do it itself. Then we'll know exactly what effect the BMS has on discharge.

[fechter]

That sure is a mighty big resistor you got there Doc

I guess if you were running a brushed motor, you could use the PWM as your controller.

[Doctorbass]

+++++UPDATE+++++


Officially, THE LOW VOLTAGE CUT OUT WORK WELL !!!

I just connected the same circuit as shown above and i watched the voltage drop on the main terminal by monitoring the output POS and NEG(I monitored the neg from the from the 15A fuse that is not passing thru the mosfet)

The other NEG was for sure.. connected to the resistor!

I got:

24.2V cut out Voltage and the current faded to zero from 26.5V to 26.0V at 19A.

At 7A i got a cut out voltage of 26.5V

At 24V, the current decreased completly and the voltage rise a bit after that up to 26.97V

That make sense since the dewalt charger detect bad pack at 25V(when the 3 led flash together)


NOW YOU CAN OFFICIALLY:

=== USE THE BMS AS LOW VOLTAGE CUTOFF===


I posted a video here:
http://www.youtube.com/watch?v=vituFUmEnU8




Doc

[CGameProgrammer]

That makes sense; the increase in voltage is simply because it's no longer under load. Anyway good test and informative video; it's nice that the BMS smoothly lowers current for a few volts. Also nice to know the "official" low voltage -- 26V no-load.

Today I used my eight DeWalt battery packs for the first time, using stock packs but with bypassed BMS, in a 2s4p configuration (20s4p in terms of cells). Initial voltage for each pack was between 34.8 and 35.6 (it varied). After using up 5.2 Ah (out of 9.2), I tested the voltage of each battery pack:

32.7
32.8
32.8
32.8
32.8
32.8
32.8
32.8

And that was without the BMS. So these things really do self-balance impressively; I think Doctorbass is right, it's safe to hook individual cells up in parallel and charge them as one. Of course I didn't measure each cell, only each pack, but given the above results, I think the odds are high that the cells are in balance as well...?

By the way, with each pack pushing 10 amps (40 total), the voltage under load was 31V per pack initially, and was 30.5V per pack at the end. So minimal sag, but the true voltage is much lower than the stated 36V.

[OneEye]

Thank you. A beautiful test! Applause to Doctorbass for jumping in and running a solid test for us.

I like your final conclusion the BMS will step in and reduce power and eventually cut off the pack as it nears a low voltage condition. It's also nice to see there isn't a 15A limit on the pack in its pristine condition: you can avoid the 15A fuse on the outer spade terminal by correctly hooking up the other pins of the BMS, raising the output to almost 20A.

To clarify for me... does the BMS shut down the pack temporarily if the load attempts to draw too much current? or does it provide a current limiting mechanism instead?

[Jozzer]

Yes, good work Doc, without a decent load to use on the workbench I never would have got to the bottom of it. As it was I had a full evening tryng to balance a bad pack before it went out.
With the cutoff so low, it must be very easy for the system to flatten the lowest cell completly unless it IS monitoring each cell too. When an imbalanced pack gets run down on a 1A load just to 29v on a poor pack with no bms, usually 9 cells are over 3.2v, and 1 cell is less than 1v (unless watced). When put back in the charger the dewalt charger refuses to charge, despite overall pack voltage to be above what you mentioned was the cutoff for the charger. Did you by any cance try it with a pack with 1 cell not fully charged?

[rf]

Has anyone looked into cranking up the current on the Dewalt charger to charge a couple packs simultaneously at a faster rate?


Richard

[Doctorbass]

Has anyone looked into cranking up the current on the Dewalt charger to charge a couple packs simultaneously at a faster rate?


Richard

That would be difficult. that cahrger pulse the current to the batt and the circuitboard is not built to hold more current than 3A. You can cahrge many pack with one and having only one of the pack with the BMS, but the time to cahrge them will be proportional to the number of pack.

Also, the internal BMS of the pack is not built to be able to balance and protect the cells with higher current.

By my experience, balancing cells always give better results when charging at low rate.
Doc

[fechter]

Do you think you could parallel more than one charger? Certainly it would work if you had parallel packs and split them for charging. I'm just wondering what would happen if you kept the cells in parallel and paralleled some chargers. This tends to work OK for lead-acid chargers.

[Doctorbass]

Like i said, this charger pulse the current to the battery pack fonction as the level of charge is reached. The dead time between the pulse is used to measure the cells voltages and it re-adjust the cahrge rate if it is needed. This cahrger have more than two pin to connect to the battery. it have 8 active pins for some digital/analog feedback between the BMS and the charger. Because of that it would not be suitable to join two cahrger together and partalleling these pins. (for exemple, you would not try to plug two parallel port from a PC with an Y to use two printer in the same time with one computer!... digital data comunication can't be paralleled.

You can use only ONE BMS on a dewalt charger. I already found on the internet a guy that tried to parallel two BMS together on the input pins... he blowed those.... that was on the RC universe or RCgroup forum if i remember...

By paralleling ONLY the cells on a BMS out pins would simply share the current thru the cells and avoid problem. I thing it have a constant current sens so the load on the balancer pins and the out of the BMS to the battery pack are not overloaded.

If your goal is to cahrge faster than 45min ot 1h for one pack, then use a RC charger with balancer with at least 10s or buy a switching powersupply like Condor built (they are excellent) and chose the optioin with V and I linit adjustment... then buy an external balancer with 10s.

Doc

[CGameProgrammer]

How about amplifying current between the BMS and the battery pack? So the normal amount of current goes through the charger and the BMS, but it gets amplified before being sent to the cells themselves. If that could be done, it would be a great way of taking advantage of the cells' ability to be quickly charged.

[GGoodrum]

So, I've now read this thread three times, and I think I know what to do to in order to enable the LVC function, and allowing current loads of 19A, but I have a few questions.

• First, if a pot is used, connected between pins 1 and 7, with the center connected to pin 5, what is the value this pot needs to be?
• The inductor does not need to be used, right?
• Once the cutoff occurs, will it reset automatically if the voltage rises back up above some value, or does the circuit need to reset in some fashion?

I want to add this circuit to this mount:



One thing I'm also wondering about is what will happen if the BMS is used in this fashion with packs in parallel? Should the Schottky diodes be used to isolate these, or does it matter?

Great stuff, really. I've been wondering what the heck the BMS does in these packs. All I've done for the last year is cut these off, and strip out the cells.

-- Gary

[GGoodrum]

I have another question. I assume you can use two BMS-enabled packs in series, as each LVC function in each BMS will work independently to protect each pack. What about if these series connected BMS-enabled DeWalt pairs are paralleled, in 2p, and/or 4p configurations? Would it matter? If it is a problem, would using the Schottky diode isolators help?

Normally, when you parallel two packs, the voltage ends up being the average of the two. I always assumed this would be roughly analogous to having the level of water in two containers equalizing, if they are connected. To make that happen, some water has to flow from the higher level container to the lower level one. I would think the same sort of thing happens with two paralleled batteries. Is this the case, what happens if you parallel two BMS-enabled DeWalt packs? I guess another way of asking the question, can you recharge the cells through the switched connections? I would think not, but what happens if you try?

Interesting.

-- Gary

[OneEye]

I thought it was more because the current in each parallel system would be in some way proportional to its ability to supply that much current at that voltage, so a 'weaker' battery that would normally experience a high voltage drop at moderate loading will not carry as much current, it will only contribute current at the rate available for the shared voltage. In the same way, the stronger cell will be contributing a larger share of the current as it can provide this current without as much voltage drop.

The neodymics schematic is a good jumping off point for a serial/parallel 72V system. You should be able to add more parallel branches based with the same basic philosophy.

[GGoodrum]

The neodymics schematic is a good jumping off point for a serial/parallel 72V system. You should be able to add more parallel branches based with the same basic philosophy.

Yep, that obviously works, so adding the diodes is probably a good thing. I'm just curious if it is really necessary.

[rf]

The neodymics schematic is a good jumping off point for a serial/parallel 72V system. You should be able to add more parallel branches based with the same basic philosophy.

Yep, that obviously works, so adding the diodes is probably a good thing. I'm just curious if it is really necessary.

The resistor network is the more important part in my opinion. It allows you to use the BMS to shut off battery output when voltage of any cell drops below minimum -- thus keeping the battery from damaging itself.

The diodes protect you from dumb mistakes, like not charging all the batteries evenly.

Richard

[GGoodrum]

I'm still confused about a couple of things with the BMS. First of all, where is the fuse on the unswitched negative terminal? I took the BMS out of one of the DeWalt case carcusses I have laying around (about 15-20, leftover from the RC and eBike packs I've made...), and there is nothing visible. It must be internal. If so, what's the point of having a fuse if the whole BMS is trashed if it blows?

I'm also confused about whether or not the values for the resistors actually control the current limitation function. If the values for R2 and R3 used in the Neodymics PDF are used (7.7k and 8.5k, respectively...), I assume the PWM circuit will reach 100%, but will the current be something less than 20A that Doctorbass was able to achieve, or do you really need to use the 7.26k/7.28k values?

I was going to try a 10k pot in series with a 5k resistor, but what I'd really like to do is just use two fixed resistors, but this will be hard to do if values HAVE to be exactly what Doctorbass came up with in his test.

Finally, which negative blade terminal does the charger use to supply the main charging current, the switched one, or the unswitched/fused one?

-- Gary