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

enter image description here

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:

enter image description here

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?

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  • \$\begingroup\$ 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. \$\endgroup\$ – user_1818839 Jun 4 at 19:50
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Easy, they either use isolated supply (very popular in China with cheap transformers) or boost capacitors.

The capacitors are charged through a diode when the low side is on, then keep the voltage for the gate driver of the high side while it's on. The drawback - your pwm duty cycle is limited to about 95-98%. Usually it doesn't matter.

One thing to remember with boost capacitors-charge currents are high, their frequency is very high. So layout should be considerate of that.

And taking a step back, why would they even use NMOS? Because PMOS of the same size and cost will hold only about 30% of the current. That's why in power electronics you almost don't see PMOS

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  • \$\begingroup\$ *bootstrap capacitors \$\endgroup\$ – winny Jun 4 at 17:11
  • \$\begingroup\$ @winny depends on what side do you break eggs. \$\endgroup\$ – Gregory Kornblum Jun 4 at 22:20

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