Current rise and fall time is corresponding to a small part (voltage plateau) of the Qg versus Vge diagram. Total Vge rise/fall time is considerably higher. You should also consider that the Rg of 10 ohm specified in the table is limiting the gate current to 1.5A peak (for 15V unipolar gate drive). So 9A can't ever happen.
To switch the voltage of the IGBT, you need to deliver the whole Qgc (gate collector) charge, which takes place in the Miller Plateau. From fig 15, it is 100 - 25 (aprox) nC = 75 nC.
The current deliverd from the driver is (15 V - Vplateau) / Rgate = 6/10=0.6 A.
The time which takes to switch the voltage of the Collecter-emitter is the time you take to deliver that charge: Voltage switching time = 75 nC/0.6A = 125 nSeconds.
Apart from the above, you also have the time it takes to reach the threshold voltage and to overdrive the gate which we can omit for the time being.
15 V/10 ohm = 1.5 A is driven just for a very short amount of time when the gate-emitter voltage is 0, so make sure your driver is capable of supplying 1.5 A.
Current rise and fall time is corresponding to a small part (voltage plateau) of the Qg versus Vge diagram. Total Vge rise/fall time is considerably higher. You should also consider that the Rg of 10 ohm specified in the table is limiting the gate current to 1.5A peak (for 15V unipolar gate drive). So 9A can't ever happen.
To switch the voltage of the IGBT, you need to deliver the whole Qgc (gate collector) charge, which takes place in the Miller Plateau. From fig 15, it is 100 - 25 (aprox) nC = 75 nC.
The current deliverd from the driver is (15 V - Vplateau) / Rgate = 6/10=0.6 A.
The time which takes to switch the voltage of the Collecter-emitter is the time you take to deliver that charge: Voltage switching time = 75 nC/0.6A = 125 nSeconds.
Apart from the above, you also have the time it takes to reach the threshold voltage and to overdrive the gate which we can omit for the time being.
15 V/10 ohm = 1.5 A is driven just for a very short amount of time when the gate-emitter voltage is 0, so make sure your driver is capable of supplying 1.5 A.
No. The required current from driver is 1.5 A for a very short amount of time. 0.6 A is taken from the driver just to switch the voltage. You have to search for a driver capable of supplying 1.5 A (if Rg=10 ohms and unipolar drive of 15 V).