# Gate resistor necessary and value calculation

I want to drive the SiHD240N60E with the UCC27511A and I want it to switch as fast as possible. The PCB traces are very short so the resistance is just 1 mΩ. The voltage is 12 V. Now when I check the datasheet of the SiHD240N60E I see a "Gate input resistance" which is between 0.8 and 3 Ω, 1.5 Ω typical. So I calculate 12 V / 0.8 Ω = 15 A which is beyond the 4A/8A the driver can supply. However, the 3 Ω would mean exactly 4 A so that would be perfectly fine. The driver also says the 4 A/8 A are "Ensured by Design" in a footnote, does that mean it limits the current automatically and I don't have to worry about it? In that case I don't need a resistor at all?

Those are my thoughts but I still haven't really reached a conclusion on what resistor would be the right one, or if it is even necessary to put a resistor in there at all. The gate driver has some resistance in it aswell, the capacitor that bypasses the driver also has some small resistance and there is also the inductance of the PCB that would kinda limit the inrush current a little so it's not a complete short at the beginning when the highest current flows. Is there even a proper calculation method or is it necessary to actually try out the real design and somehow measure the current that's flowing during switching? I assume when I go over those limits the driver won't immediately fail so I can't just assume that if it works I'm running it within the specs?

• One thing of note: pay attention to the return path from MOSFET source to the gate driver ground pin and supply capacitor. If you happen to have e.g. current sense resistor in the path, decoupling with a ceramic capacitor for shorter path may be necessary.
– jpa
Commented Nov 16, 2022 at 7:11
• @jpa Thanks, that's not the case here though. The FET source is connected using short traces to the driver, if I understood it correctly it is necessary for best EMI performance to keep that loop as small as possible Commented Nov 17, 2022 at 18:23