Best way to connect a switching fet to another chip

[Moriambar]

Hi there,
I'm wondering: when connecting a fet to an MCU/cmos logic chip/other IC, which configuration is better and why?
I put just simple resistors as a load example, but it can be something more complex, but let's assume that it doesn't require a "special" treatment (eg: fast switching or fast pwm), just an on-off for a load.

I can think of these three circuits, I usually hook up #2, but I'm wondering if it can be done better. For example I see some circuits without resistors between mcu pins and fets…
Can you help please?
Cheers

[globoy]

They all have their uses.  Which one (or combo) that you select depends on other aspects of your circuit.

Circuit 1 protects the MCU for the case that the FET fails with a D-G short and VCC > MCU operating voltage.   The series resistor limits the current being sunk by the IO port ESD protection diode array.

Circuit 2 ensures the FET is turned off in the case the MCU IO is an input (which often occurs during and after reset before the MCU has a chance to initialize the port as an output and set an initial low level).

Circuit 3 can be used if VCC <= MCU operating voltage and the load resistance will sufficiently limit current in the condition of a shorted FET (or you need to save every penny and you don't care what happens if the FET shorts).

[Moriambar]

They all have their uses.  Which one (or combo) that you select depends on other aspects of your circuit.

Circuit 1 protects the MCU for the case that the FET fails with a D-G short and VCC > MCU operating voltage.   The series resistor limits the current being sunk by the IO port ESD protection diode array.

Circuit 2 ensures the FET is turned off in the case the MCU IO is an input (which often occurs during and after reset before the MCU has a chance to initialize the port as an output and set an initial low level).

Circuit 3 can be used if VCC <= MCU operating voltage and the load resistance will sufficiently limit current in the condition of a shorted FET (or you need to save every penny and you don't care what happens if the FET shorts).

thanks, very well explained.

[Manul]

It is good practice to always add series resistor, like in circuit 1. The only exception is if you drive very small fets with very low gate capacitance. Calculate series resistor based on maximum permissible pin current. For example, if voltage is 5V and maximum pin current is 20mA, use at least 250 Ohms series resistor.

For logic ICs, MCUs and other fast switching ICs, capacitive load presents almost a short circuit to ground during fast transitions. So for a brief moment current might hugely exceed maximum limits. Although it's very short and usually does not present significantly increased thermal dissipation, high current peaks might cause momentary sag of supply voltage, either for the whole IC or some section of the IC. Especially if IC does not have very good bypassing. And in general stresses the ouput of IC without reason. If you want to drive a mosfet gate fast, you use some sort of dedicated driver.

FYI, power mosfets typically have 1-10nF gate capacitance, that is significant.

The need of gate pull-down resistor depends on the application. It is good practice to add it, except if you are sure that driving pin can never be high-z, aka floating.