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I want to switch power on & off to a module using an MCU output pin. For this I am using a BS170 MOSFET as so:

schematic

simulate this circuit – Schematic created using CircuitLab

This seems to work fine, however I'm wondering if there is any reason to add additional parts for protection, stability or something else? Even the 10k resistor seems to me like something I might be able to drop.

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3 Answers 3

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BS170 is not well specified for 5.0V drive. You would be better off with a 2N7000. There are much, much better parts if you don't mind SMT, which you should definitely explore if the MCU supply voltage is less than 5V.

If the load could be inductive you should have a catch diode across the load.

The 10K resistor will turn the MOSFET off if the MCU port pin goes high-Z. This could be vital if the micro finds itself in a low voltage reset condition- if the load is heavy the MOSFET could be damaged by being partially 'on'.

If your supply to the load is higher than the MCU supply, consider a gate resistor which could protect the MCU if a short occurs across the MOSFET drain to gate pins, say by an errant test probe or a failed MOSFET. It can go between the 10K and the gate something like a few K.

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Depends on the load, it's type and and the current it needs. If it's an inductive load, then a flyback diode is needed to protect the mosfet from the collapsing inductive field creating a voltage spike. If the load is too big for the mosfet, then that's an issue too.

The 10kΩ resistor is not there for protection, per say. The pull-down resistor is there to prevent the mosfet from turning on unexpectedly from a floating pin. When the micro controller is off, or the gpio is in high impedance mode, the gate can fluctuate. If the mosfet is controlling something dangerous, the sure it's for protection. If it's an led, then its more so it doesn't look like an amateur project.

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  • \$\begingroup\$ The load is a module containing several ICs, no inductors whatsoever so I don't a diode in parallel with the load is required. I know the resistor is there in order to make sure the gate it pulled down, however I am wondering if it would be an issue if the gate goes high for a fraction of a sec. before being pulled down by the MCU when it loads. \$\endgroup\$
    – user34920
    Commented Dec 16, 2017 at 8:26
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You need to examine the GROUND currents and GROUND inductances. You can easily create an oscillator if the MCU GND is wrongly wired to the FET GND. Daisy chains are not a good approach to GROUNDING for high current systems.

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