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DRV8889-Q1: How to flow the current in the H-bridge during the dead time

Jeongho Cho
Prodigy 40 points
Part Number: DRV8889-Q1

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 ?

www.ti.com/.../slva321a.pdf

※ Sorry for not attaching image because of the policy of my company.

  • 0
    TI__Prodigy 251 points

    Hello Jeongho,

    Dead time (where all of the FET switches are off) is added in order to allow time for the energizing FET switches to switch to their OFF state and for inductive loads to carry existing current through body diodes. The equation VL = VS + 2VD works as, based on figure 2-1 (specifically the one for Dead Time) and on figure 2-2, the voltage applied to the inductive load is inclusive of the voltages of the two body diodes and the source voltage.

    In terms of your question about quantity of stored and released energy, can you share a link to where you read this information from to provide some more context?

    Regards,

    Jaya Hari

  • 0 in reply to Jaya Hari
    Prodigy 40 points

    Based on the conservation of energy, it is the ideal that the quantity of storing energy and the quantity of releasing energy should be no difference.

    As you know, It is the general theory for all electrical engineering.

    The energy equation of inductor is that (w = 1/2 * L * I^2).

    On the upper equation, L is the motor coil and a constant. I is the current and might be a constant. 

    So, there should be no change for inductor energy.

    And the inductor equation is that (V = L * di/dt) .

    I think that di/dt might be constant. 

    So, the voltage across the inductor might be constant.

    However, as I mentioned before, VL = VS (VS - GND) in first phase. VL = VS + 2VD in dead time phase.

    I do not understand how the voltage across the inductor is increased at the dead time. 

    Is di/dt in dead time greater than di/dt in fisrt phase ? 

     

  • 0 in reply to Jeongho Cho
    TI__Prodigy 251 points

    Hello Jeongho,

    I will make sure to get back to you within 24 hours from this response!

    Regards,

    Jaya Hari

  • +1 in reply to Jeongho Cho
    TI__Prodigy 251 points

    Hello Jeongho,

    During Dead Time, the energy stored in the inductive loads will flow back to the supply through the body diodes.

    This would mean that the voltage across the inductive load includes the two forward voltages of the diodes, causing an increase in the voltage across the inductor in dead time.

    Regards,

    Jaya Hari

  • +1 in reply to Jaya Hari
    Prodigy 40 points

    Hello Jaya,

    I understand your reply for my question.

    I would add my comment for just sharing my opinion about this

    Hope it useful for you. 

    The quantity of storing energy to inductive load and releasing energy from inductive load should be no different by the equation.

    So, the inductance and the current would be constant before and after.

    The reason why the voltage across the inductive load is increased before and after is the two diode forward voltage as you mentioned.

    I think It might be affected by di/dt of inductive load. 

    Maybe, the value of di/dt during dead time might be bigger than the value of it in forward/backward cases.

    Because, as i mentioned, the current is constant but the time of dead time and time of forward/backward cases might be different.

    The time of dead time is shorter than the time of forward/backward cases.