Again, that's if you need the fastest switching speed.
If it's just for an on/off function, than obtaining the fastest switching speed just adds unnecessary complexity to the circuit.
Turning the motor rapidly on and off is only needed if you are doing something like PWM switching to control the motor speed.

My design goal is always to make the circuit as simple as possible to do the required task.

Thank you so much ?

Regards,

 

@iimagine

In simulation, D1 in that circuit does not seem to affect output. However, since both Microchip and XBee promote or use that circuit for level shifting, it is the circuit I decided to post. Of course without it, the MCU driving the load needs to sink the entire current, which is less than a mA in this simulation. I modified the circuit by adding an addition diode drop to get well above Vgs(th) (shown below). All simulations here are with the modified circuit.

With D1&D3 disconnected:
View attachment 183278


Voltage at "output:"
View attachment 183277

Note that the "low" voltage is below 1.2V (i.e., lowest Vgs(th) of the original mosfet).

Schematic simulated with D1&D3 disconnected:

View attachment 183276

Source and sink series resistance for outputs is not provided in the datasheets for the PIC's I checked. Since outputs (source and sink) are limited to 25 mA at 3.3V, I assigned 132 Ω. That value doesn't make a whole lot of difference. The 91k was added to "discharge" the mosfet gate but is also probably not required. It does drop Vgs (low/off) a little.

This was just a thought at the time. I have never tested it for this purpose and would have chosen a more appropriate mosfet to begin with to avoid all of these gate drive complications.

Could you explain a bit how this circuit works.

1. High from MCU :
Diode anode at 5V and cathode at 3.3V, diode conducts and Vo = Vf+Vf =2Vf

2. Low from MCU
Diode anode at 5V and cathode 0V and now diode conducts and Vo = Vf+Vf =2Vf

Correct me where i am wrong.
Can this voltage be increased to more than 5V

Regards,

 

@sbkenn
What's your question?

 

This assumes that the uC output would have been high enough to turn the MosFET fully in in the first place. If that doesn't apply, then the driver would also provide the voltage to do that.

That does not make sense.
The driver always provides the needed gate voltage, as well as the current drive, to rapidly charge/discharge the gate capacitance and turn the MOSFET on or off.
All the mircro voltage does is drive the driver logic input signal.

 

What's your MCU?
Just look for a logic level MOSFET
Also look in any of the MCU forums (Arduino etc.) and you will find loads of information on driving MOSFETs
Here's a popular one
https://www.sparkfun.com/products/10213
Get a few and try them