Tool/software:
Hi TI Expert,
I'm not familiar with the POB behavior and hope you can help to clarify the following questions I have:
1. According to "7.3.4.9 Powered off Braking (POB)", if the supply voltage is more than VVMOV_POB(60V-69V) and if the voltage of the output nodes rise above the supply voltage by more than VPOB_TH(580mV), the device turns on the two low-side MOSFETs.
So...both of the condition must be triggered otherwise the LS FET will not turn ON?
Which means the OUT1/2 pin voltage during Back EMF will rise up to at least 60.xV?
Also, when it enter into POB protection, how the IC judge when to turn off the LS MOSFET? simply monitor whether the voltage is lower than the VPOB_TH(580mV) again?
2. About the VSD(Body diode forward voltage), DS mark the IO=-4A, may I know the definition of it?
Means the current though body diode or the SOURCE current on OUT1/2 pin?
3. Assume the IC is in low-power sleep mode or when the bridge is disabled (Hi-Z), when we add extra force on the motor then what would be the current waveform during this period before POB protection is triggered?
=> ex: Press the trunk after it's fully open, force the window down when it's already stop, something like that.
Since all of the MOSFET are in Hi-Z mode so pushing the motor means it will create the back EMF.
But there should be no path for current to conduct, so I'm quite confused about this part.
4. Another questions related to OCP during ACTIVE mode, when OCP happen all the MOSFET will enter into HI-Z mode.
Since we turn off all the MOSFET so back EMF will be created and the voltage could be higher than VIN/VM so I think the current will go through body diode like this.
But when the Back EMF voltage(OUT1/2 pin) is lower than VM+VBD, which means there's no other path to discharge the remaining energy of inductor/motor, we can only wait for motor to discharge that or IC has other mechanism to discharge the remaining energy during this moment?
Looking forward to your reply.
Thank you.
