Fet Switching times

[jimmythefool]

Hi all,
Hopefully somebody can enlighten me here.
I'm trying to calculate approximate Mosfet gate current.
Using the formula I = Q/t, what time does 't' refer to? Most datasheets include

Turn-On Delay time
Rise Time
Turn-Off Delay Time
Fall Time

Any advice gratefully received

Regards
Jim

 

[Gobbledok]

I could be wrong, but I think that's the time you need to turn the MOSFET on in (i.e. the amount of time you have to overcome the gate capacitance).

Therefore, the faster you need to turn it on (or off), the more current required.

 

[ronsimpson]

If you were charging a capacitor then t= the time to move the voltage on a cap from one voltage to another voltage. (0 to 10V)
A MOSFET is more complex than that.
The turn-On delay is related to G-S capacitance and turn on voltage.
The turn-Off delay is related to G-S capacitance and the on voltage back down to 'gate turn on voltage'.
There is G to D capacitance that you must overcome during rise/fall times.

 

[ronv]

In the data sheet you will find a spec. called Qg (total gate charge). This is the time it will take the FET to switch given 1 amp of gate current. From this you can select a gate resistor or driver to get the speed you want. Lets say Qg is 100. Then it should turn on in 100 ns with 1 amp of gate drive or 1 us with 100 ma of gate drive. I have never gotten them to switch this fast - I think because of other layout parameters - but it gets you close.

 

[crutschow]

If you are concerned about power dissipation during switching (as with a switching regulator or PMW switch for example) you want the transistor to switch as fast as possible. You would then want t to be less then the rise and fall times specified in the data sheet so that the actual rise and fall times are close to the minimum. This can require a high-current push-pull type MOSFET driver to drive the gate.