Dear TI experts.
I have known the topology of working H bridge like under.
First phase, MOSFETs should be turned on to flow the current to the forward or reverse direction.
Second phase, to change the direction of current, All MOSFETs should be turned off during the dead time.
Third phase, Opposing MOSFETs should be turned on again.
I have a question about operating H bridge during dead time.
During first phase, motor (inductive device) can store energy which is the quantity of VM (VM-GND)
And then, during second phase, motor should release the energy which is the quantity of VM.
Because, I think that there should be no difference between the quantity of storing and releasing.
But, TI application note say that the energy of releasing is bigger than the energy of storing under that link.
I don' understand that point why VL = VS + 2VD ( refer to page 5 )
So, could you confirm my question ?
※ Sorry for not attaching image because of the policy of my company.
