how do mosfet drivers acheive a "dead time" ?

[Simon]

I'm doing a push pull type power drive so need to drive my gates and effectively from the same output of a 555 but I need to make sure there is no shoot through. I tried the traditional driver with 2 BJT's but that has shoot through. So I will probably use a driver IC on this one but.... how do these IC's that usually have two drivers that are one the negation of the other so that they can drive pairs and bridges with no shoot through, but how do they create the dead time ? the only thing I can think of is a window comparator and maybe a low pass filter on the input to slow the signal down because the high speed switching will actually be produces by the output drivers.

Thoughts ?

[ejeffrey]

If you mean a window comparator on the input, no.  They have a single input threshold and a fixed or programmable dead time, independent of the speed of the rising edge.

I am not sure how the delay is implemented in the real chips, but the way I would do it is to generate the complimentary outputs with standard logic.  Then I would use an RC delay stage with a diode shunting the R.  That would make a small delay for the on->off transition but a larger delay for off->on.  Your window comparator idea would also work.

[Simon]

point is by the time I put a window comparator in I might as well use a pucker driver IC

[SeanB]

Normally they implement the delay internally using a few cascaded weak gates, with a small capacitance on each node that creates the delay. Then it is used with an XOR to gate the drivers such that both are held off on a transition for a small period. Probably would be good to use the pukka driver instead if you are worried. Otherwise you have to implement this on separate pieces of silicon which will be a bit bigger.