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Original question:

4 phase control

DRV8881E

ken dias
Intellectual 795 points
Other Parts Discussed in Thread: DRV8881

I have the 8881EEVM and am using it with the MC33033 TO CONTROL A 4-PHASE motor. Problem is the EEVM comes with R3, R4 Resistor banks which have to be disconnected. I don't have the skills or equip to do this, it is SMT Is there any solution i.e. can I get the EEVM without R3,R4 or can I get the DRV8881E on a board with pinout connectors?

  • 0
    TI__Guru** 114905 points
    Hi Ken,

    The R3,R4 resistor banks are SMT. The resistors are included on the EVM.

    This is the only option available. If you do not have the proper equip, there is still an option available.

    An small Xacto knife can be used to cut the traces. Score the traces perpendicular to the trace path between the resistor banks and the header. Then using the tip of the knife pry up the trace on a corner.

    The cut should not be deep and the width of the cut can be small (1.25mm). After each cut, use an ohmmeter to verify the cut.

    If you do not feel comfortable, there are a few traces to experiment on. The two traces to the unused J3 RETRY connector are a good example.

    If you want to reconnect, scrape some of the solder mask off the trace and solder a surface mount resistor or small piece of wire across the cut.

    In the worst case the GUI and firmware will no longer work. If successful in reconnecting the cut traces, the GUI and firmware will work again.
  • 0 in reply to Rick Duncan
    Intellectual 795 points

    Yes,scraped out the solder traces and checked resist. Before it showed zero ohms, now it shows a high resistance (not infinity) across the broken traces. Is there some parallel path and is this OK?

  • 0 in reply to ken dias
    TI__Guru** 114905 points
    Hi Ken,

    High resistance (>100kOhm) should be OK. The problem with on board measurements is there could be some sneak paths through diodes.
  • 0 in reply to Rick Duncan
    Intellectual 795 points
    Hi Rick
    Ref your email of Oct 15:
    I am using 20VDC for both the MC33033 and the DRV8881E. So, will connect a resistor of 4.7 kohms from Bt and Ct to this 20VDC. Then will use a volt divider at Bb and Cb outputs i.e. two 4.7 kohms in series from the 20VDC to gnd and connect Bb and Cb to the connection point of these two 4.7 k resistors. Just want to be sure I understood your email correctly?
    Thanks
    Ken
  • 0 in reply to ken dias
    TI__Guru** 114905 points
    Hi Ken,

    I don't think you understood it correctly. Look for a reply shortly.
  • 0 in reply to Rick Duncan
    Intellectual 795 points

    Thanks.

    Your reply makes it clearer. Only problem now is I don't understand the 2nd para of the reply:

    ' Assuming Vcc = 20V, it is recommended to reduce voltage to 5V by creating a V divider. For the resistor divider, you can use 14.1 kohm from Bb to the centre point and a 4.7 k from centre point to GND. The centre point can now be connected to CD4070.'

    Intuitively I would connect Bb to a R divider with 14.1 k to Vcc AND 4.7 k to GND. Similarly for Cb. however, want to be sure as it does not match up with the advice offered.

    Any thoughts would be welcome.

  • 0 in reply to ken dias
    TI__Guru** 114905 points
    Hi Ken,

    I replied in a private message. The Xb outputs (Bb and Cb) are used to provide inputs to the CD4070 which provides the correct signals to the DRV8881. The Xb outputs can be as high as the MC33033 VCC supply voltage which needs to be reduced.

    The resistor divider reduces the output of the Xb outputs to 25% of the original voltage, making it safe for the CD4070.