MOSFET pull down resistor placement

Hi,

I'm planning on using a (logic level) MOSFET to drive a DC motor. After reading up on MOSFETs I understand I need to use 2 resistors : a small one (200-300 Ohm) just after the Arduino pin and a large one (10k Ohm) between Gate and Source. The small one is needed because "the gate is highly capacitive and can draw a big instantaneous current when you try to turn it on". The high one is to ensure the MOSFET stays 'off' when the Arduino pin is set to LOW and to protect against noise which might be above the MOSFET's threshold voltage. It's the latter bit I'm unclear on.

If I undertstand things right, if I have the Source connected to Ground, then to keep the MOSFET OFF then I need to have the Gate set to 0V, which I would do by setting the pin to LOW.

Where is this noise generated? At the output pin? As in it's meant to be 0V but it might not quite be? If so, shouldn't the large resistor be between the output pin and the Gate? All the schematics I've seen put the 10k resistor between the Gate and the Source.

I'm probably missing something really basic here but google hasn't yielded an answer yet!

The high one is to ensure the MOSFET stays 'off' when the Arduino pin is set to LOW and to protect against noise which might be above the MOSFET's threshold voltage. It's the latter bit I'm unclear on.

The 10K is there to turn off the MOS fet in event the o/p from the Arduino ever goes to Hi impedance (i.e. at power up time).

Ah yes, what I wrote is wrong. I was thinking of this (from another site) : “The resistor holds the gate low when the arduino does not send a high signal. This is here incase the arduino comes loose, or the wiring is bad it will default to off. You don’t want this pin to ever be floating as it will trigger on and off.”

In any case, would this scenario (loose wire, arduino startup, etc) be equivalent to having some random voltage being applied? Is it anything like the attached diagram?

PullDownQuestion.png

Yes when a FET input is not connected to anything it is said to float. In this state it picks up interference and can be any voltage so the FET can be on, or off or only partially turned on.

The 10K (or 20K, or 50K...whatever..) "pull down" would be used if leakage from the MCU or EMI sufficient enough to charge the gate capacitance is a problem at any point during the operation cycle of the final design. Such leakage/interference could conceivably turn the MOSFET on and cause problems. Using such a resistor is a 'good practice' but is not always necessary. Being an incredibly cheap solution of a potential problem, it might not be a bad idea to include it if you think the motor 'twitching' at the wrong time might be undesirable.

Thanks for the replies. I suppose my question then is why is it that when everything is working correctly (i.e. everything connected, etc) the 10k doesn't affect the 5V HIGH, but when it floats and we assume some random voltage, the 10k kicks in and keeps the mosfet off?

(apologies I really think I'm missing something basic here! :/ )

oh, and yes I will be using one regardless, I'm just trying to understand how things work exactly! :)

The short answer is, it does affect the 5V, but only very slightly as the 200 ohm and 10K resistors form a voltage divider. So you'd loose about .01 V to the arrangement on a 5V signal coming form the MCU.

However, EMI or leakage represent very tiny currents that are easily drained away through the pull down resistor.

When the uC resets, the IO pins become inputs, and the MOSFET gate can float until such time as the sketch kicks in and drives the pin high or low.
When high, the uC can easily drive a 10K resistor with 0.5mA needed to drive the pin high. When not driving (in reset) and pre-sketch startup), the 10K is sufficent to keep the gate low.

Right, I think it’s starting to sink in. Is the following explanation correct then :

When we supply 5V at the pin it charges the gate and turns the MOSFET ON. Sure there is that resistor there connecting the gate to GND but because we made it quite high not a lot of current goes through it. So it’s as if it not really there.

On the other hand, when the gate floats it may pick up a charge. If we don’t give it a pathway to dissipate (connect to GND) then it might turn the MOSFET ON. Connecting the Gate to GND even with a high resistor is enough to dissipate that charge.

Yes, no, maybe?

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Yes, exactly.

Yeah! :D

8) Like that!

+1