Gate Driver Output Wiring (TC4431)

Question

I don't know how to how to properly wire the two outputs of the driver IC TC4431 to the MOSFET gate.

In the datasheet, there's a test circuit diagram (@page7, FIG4-1), should I connect the outputs together to the gate pin? Since resistors might be needed to fine tune Turn-On and Turn-Off timings on the actual PCB (as mentioned in page 6), I need to consider placing spaceholder pinouts (If not needed I'll populate with 0 Ω resistors). Which is the right wiring or should the resistors be pull-up pull-down as it's common on MOSFET gates (and left unpopulated if not needed)?

^{simulate this circuit – Schematic created using CircuitLab}

Datasheet: TC4431

Answer 1

The application diagram suggests the pins can be shorted together if need be.

Diodes are certainly not required.

Separate resistors may be desired for changing rise/fall times independently, or perhaps for reducing shoot-through current.[1]

If equal resistors would be used, the outputs can be shorted together and a single resistor used instead.

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[1] Shoot-through is not documented on this type. At best, Fig.2-1 hints at it, but the no-load consumption could be due to internal paths as well as output shoot-through per se. And though 4.7µF is indicated (a rather large bypass capacitor, especially in relation to the ~nF typical loads shown), it might simply have been chosen out of habit rather than need. More characterization is needed to determine if shoot-through can be managed this way.

Whatever the case, shoot-through (both outputs being active, pulling up and down, simultaneously) is only a concern with respect to supply current and device dissipation, not basic device operation and function. Maybe EMI. The possibility would simply be this: even if one intends to use equal up and down resistors, supply current may be slightly lower with independent resistors, rather than shorting the pins together and using one resistor.