# How is high side switching with an N channel mosfet possible in a BLDC motor controller?

I'm trying to understand how this BLDC motor driver circuit works:

Let's say we want to run a current through phases UV. Assuming we want V to be ground, then SW4 would be closed and we would have this circuit:

Where the resistor is the resistance of the phases. Now how is it possible that this circuit works, if we're using the N channel MOSFET as a high side switch? If Vdrive is 30V for example, then assuming a Vgs(th) of 3V we would need the voltage at the gate (PWM1) to be at least 33V. This would break the MOSFET however, because the source is initially 0V and Vgs would then be 33V. However most motor drivers I see don't use MOSFETs with super high max Vgs. How is this switching possible without destroying the MOSFET?

• The gate voltage will be about 10V higher than Vdrive. (Several times the threshold, to turn the FET fully on). It will come from a purpose designed high-side gate driver. Jun 4, 2021 at 19:50