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?

  • \$\begingroup\$ 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. \$\endgroup\$
    – jpa
    Commented Nov 16, 2022 at 7:11
  • \$\begingroup\$ @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 \$\endgroup\$
    – Wiers
    Commented Nov 17, 2022 at 18:23

1 Answer 1


As fast as possible

Gate driver currents are impedance limited, not heated limited as far as I know. Therefore, leave room for a gate resistor and jumper with zero Ohms. If you scope excessive ringing between the MOSFET gate-source then start increasing the gate resistance.

If parallel MOSFETs you need one gate resistor per MOSFET right next to the gate. No sharing because this will not stop certain forms of ringing (such as parallel MOSFETs ringing between each other).

  • \$\begingroup\$ +1- Another reason you might want to increase your gate resistor value even if you think the ringing is acceptable is if your circuit is causing excessive EMI. \$\endgroup\$
    – John D
    Commented Nov 15, 2022 at 22:38
  • \$\begingroup\$ Isn't a 0 Ohm resistor bad for parasitic inductance aswell? So if I build a prototype and it works with 0 Ohms I should try to remove it for the production version? Or is the effect a 0 Ohm resistor has neglible? \$\endgroup\$
    – Wiers
    Commented Nov 17, 2022 at 18:21
  • \$\begingroup\$ @Wiers Up to you but it is probably negligible. There are also actual jumpers which are copper and not zero Ohm chip resistors. Ease of production and part BOM count would matter much more than inductance. Ferrite is also on the table for that footprint if you know what you are doing to avoid resonant peaking. You can also short across the footprint and cut the trace to use the footprint. \$\endgroup\$
    – DKNguyen
    Commented Nov 17, 2022 at 18:27
  • \$\begingroup\$ @DKNguyen As my FET is switching 20 A it is probably important to avoid ringing somehow but it is also important to keep the switching time low, otherwise it will generate a lot of heat. As I know the switching frequency (100 kHz), should I try to pick a ferrite that has a low resistance at 100 kHz and then goes as high up as possible beyond the 100 kHz? \$\endgroup\$
    – Wiers
    Commented Nov 17, 2022 at 18:40
  • \$\begingroup\$ @Wiers Ferrites can cause more problems than they fix if you pick the wrong one. Ferrites also saturate at significantly below their heat limited current. you want to suppress. You don't actually know what frequency your ringing is so you can't pick the ferrite until you measure. It's the ringing frequency that you want the ferrite to be lossy at. Just make a footprint that you can find ferrites, resistors, and jumpers for and start testing. \$\endgroup\$
    – DKNguyen
    Commented Nov 17, 2022 at 18:52

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