Hi
I'm currently trying to the operation for Qmax, i have followed the calibration process and now the Qmax learning procedure. below is one of the few i have found on this forum
1. Send an IT Enable (21) command to turn ON FETs. Send a Reset command (41) to disable Ra-table updates.
2. Discharge to empty
3. Wait 5 hours
4. Send IT Enable command to set a DOD0 reference point.
5. Charge to full
6. Wait 2 hours Qmax will be learned at this step. You should see VOK turn green after Qmax has learned. You should see Update Status change to a 05.
7. Discharge at typical low rate of target application (usually C/5 for notebooks) down to Term Voltage. Using low rate will allow to update resistance points at deep states of discharge. Rate of discharge must be higher than C/10 rate, otherwise resistance will not update.
8. Wait 5 hours for relaxation. You should see Update Status change to a 06 during the rest period.
9. Repeat steps 5-8 while running EV Software log to verify accuracy.
Questions
1
all is going well until step 6. when i turn on the Qmax learning the VOK bit stays low. my question is what are the factors that must be correct for VOK to go high ? does the discharge have to be of a certain rate ? also i let the battery relax for at least 2 days while i was still connected to the bq34z100 (not in sleep mode).
sorry, i have a few more questions whose answers will help clarify things for me.
2.
if i need to leave Qmax learning mode, how do i do that if there is only a command to enter it (21) ?
3
I have a 4 serial x 2 parallel battery configuration which gives me a nominal voltage of 14.8V at 4.5A, its max charging voltage is 16.8V.my charger is set for 16.8 volts, which value do i use for the highest expected battery voltage for the 'voltage divider' entry (step 3 in design steps on page 29 of BQ34z100 datasheet)
4.
on the bq gas gauge evaluation software 'DataRAM' screen, should the 'Full charge capacity' show the total capacity of one cell of the battery pack or of the whole pack ?
5.
is the Qmax Cell 0 value the value of one cell of the battery pack or of the whole pack (4500mA) ?
thanks for any help given. I know some of these questions may seem straightforward but i just want to clarify i have certain things correct.
many thanks
Alan