Hi,

i know that when both MOSFET turn on it will cause a low resistance path which leads to shoot through current.

can someone tell/explain to me how/why it happens???

thanks guys

 

It's quite simple. If you apply a simultaneous drive to turn on one transistor and turn off the other, the normal transistor delays, which are not generally the same for turn on and turn off, means that the two transistors will be momentarily off during the transition.
This generates a large shoot-through current.

To prevent that the drive circuit must be designed with a small delay between turn off and turn on, so that both transistors on same side are never ON simultaneously.

 

If driving an H-Bridge from a microcontroller, it is much better to use four port pins; one per gate. Then the delays between top and bottom turn-on of each side of the bridge can be handled in software. Turn off, wait, before turning on other.

 

the two transistors will be momentarily off during the transition.

I think crutschow meant "momentarily on" .

 

thanks guys.

but why when 2 MOSFETs turn on at the same time it will form a low resistance path???

 

I think crutschow meant "momentarily on" .

Certainly. Don't know how I managed to say "off".

thanks guys.

but why when 2 MOSFETs turn on at the same time it will form a low resistance path???

Do you know what a MOSFET does when it turns "ON"?
It acts like a low value resistor from drain to source, generally being much less than an ohm.
So two of them ON in series will form a low resistance path.