P3024 code after reconditioning battery pack

Good work!

Since we know the traction battery ECU measures voltage by module pair, please use the spreadsheet to compute the V load by module pair and then compare the 14 sets of module pairs, to see what you get.

For example, modules 9 and 10 = 15.3V. Compare to modules 23 and 24 = 14.8V. The Delta V between those two module pairs is 0.5V. I believe that is sufficient difference for the traction battery ECU to call a fault.

 

Your module series resistances are a bit high and have too much of a spread. Remember that the battery current can be as high as 100 amps and a resistance difference of 3 miliohms can cause a voltage difference of 0.3 volts which can cause an alarm. You can minimize the problem by matching up the module pairs so that the sum of the two resistances match as well as possible across the 14 pairs. Don't forget to get the voltages lined up a bit better before reassembling your battery.

JeffD

 

That will help along with getting all of the no load voltages lined up.

JeffD

 

You could use the Data Sort function (if using Excel) to sort the table in descending order, using the Internal Resistance (ohm) column. Then, match the highest module with the lowest, the next highest with the next lowest, etc.

However all of this cleverness will be for nothing, if in fact your module measurements are not repeatable which is what you had expressed concern about, in post #21.

 

The mini VCI will give you the internal resistances of all the modules directly. You might want to have someone drive a bit while you watch the numbers update.

A partial printout looks like this:
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With the VCI there is a record facility so that you can record the information while driving. This information is stored in a TSE file and can be played back at any time. Just another very useful function of the VCI. The above print out was a freeze frame from one such recording, although the vehicle was stationary during this recording.

John (Britprius)

 

The use of a "common" wire connection is to connect all of the modules in parallel. This allows the flow of current between the modules so that after a short time all of the modules will be at exactly the same voltage. The other option is to slowly discharge each module while monitoring the voltage to get them all to the same voltage. I prefer using the second method first to get the module voltages close so as to minimize the high current spike as dis-similarly charged modules are put in parallel.

Your method will not equalize the module voltages and accounts for the variation in no-load voltage seen in your measurements,

JeffD

 

You could use a 12 volt to laptop converter "converts from 12 volts to voltage of your choice from 14v to 22v" and will charge the laptop battery on the move. The one I have also has a 5v USB output for phone or tablet.

John (Britprius)

 

It looks correct. After you used the harness were the modules at the same no-load voltage?

JeffD

 

Hi cgates, could you please let us know what Windows version and service pack you end up needing. I'm thinking about trying one of these too and have a couple of old laptops running XP.

 

I just received my mini-VCI and got it working on my old Windows XP Acer Aspire1 netbook (Intel atom processor).

I had to run the TISKEY program from the 7.20 folder before it would work (I had installed the 8.00.34 version).

 

Did you take your no-load (open circuit) voltage measurements before or after the discharge caused by the loading?

There are advantages to using a constant current load as you suggest. This is how automotive (lead acid) battery load testers work. A voltage regulator (e.g. LM1086) may be configured (wired) to provided a constant current rather than its typical use of providing a constant voltage supply. Such a current sink could be used to discharge a battery module and provide a load for measuring internal resistance.

However, your test is not incorrect. Using a fixed-value resistance, you can even do the test without the simultaneous need for a voltmeter and ammeter. Specifically, you can use this eqtn instead:
R_int = ( (V_open_circuit/V_under_load) - 1 ) * R_load
The reason for using the eqtn you list above is that many are using bulbs as the load and bulbs have varying resistance. If it is the current measurement that is making your experiment difficult to repeat, you might consider using the eqtn with R_load rather than I_load.