Relationship between value of the totem pole output of the gate driver and the speed of turning ON of the MOSFET

Question

We are using LT1243 gate driver for turning ON a MOSFET: if you see the datasheet of the LT1243, then we will know it has a high totem pole output of 1 A. which means that driver will drive a current of 1 A to the gate of the MOSFET to turn it ON.

I was told that the reason for the high current (in terms of Gate driving for MOSFET) 1 A is to charge the parasitic capacitances of the MOSFET (especially Cgd (gate-drain capacitance) though these capacitance are undesirable yet inevitable. As you see, when the high current 1 A charges the parasitic capacitances quickly, the MOSFET also turns ON quickly. And Simultaneously, we have the gate voltage applied by gate driver.

I want to know why the capacitance needs to be charged quickly, so that the gate can be opened by applying the gate voltage. what happens when less current is given, does it affect the speed of the turning ON of the MOSFET. if it affects the turning ON speed of the MOSFET, then How?

Answer 1

I want to know why the capacitance needs to be charged quickly

If you don't charge the gate quickly then the voltage between gate and source (the important voltage for activating a MOSFET) must rise slowly. It comes down to the basic formula of a capacitor: -

$$I = C\dfrac{dv}{dt}$$

If \$I\$ is small then \$\frac{dv}{dt}\$ must also be small i.e. the voltage between gate and source takes more time to reach a value that fully activates the MOSFET channel.

...so that the gate can be opened by applying the gate voltage

We don't say the "gate is opened" because that confuses things. In pneumatics and hydraulics, a valve "opening" causes a flow of fluid but, in EE, a switch "opening" causes a cessation of current flow hence, it's better to use activate or deactivate.