Gate drive Requirements in Active free-wheeling BLDC motor drive

Thread Starter

gonespa

Joined Sep 23, 2010
3
Hi,

Background: Trying to understand how a BLDC motor controller works. The system would comprise a microcontroller generating the PWM signals (6 of them, one for each MOSFET), which would be fed to three IR2110 MOSFET drivers using a synchronous rectification (active freewheeling) drive schema. (the microcontrolles handles the deadtime problem). It would be using high and low N-Channel MOSFET.

I have been struggling to understand how is it possible, for a MOSFET driver IC (IR2110 for instance) to effectively activate a low-side N-Channel MOSFET whenever it is handling the freewheeling current from the motor. As I understand, MOSFETs can conduct from Drain to Source as well as from Source to Drain provided the device is "on", ok, but... in this particular moment the source of the transistor would be higher than ground so, in order for the IR2110 to provide enough gate voltage, it would have to generate voltages greater that the Vcc (Low side supply), isn´t it?. I can´t see how can that happen, what am I missing?.
 

Thread Starter

gonespa

Joined Sep 23, 2010
3
See http://tahmidmc.blogspot.ca/2013/01/using-high-low-side-driver-ir2110-with.html
You turn on the low side mosfet in order for the high side charge pump to operate.
Max.
I understand the need to recharge the charge pump capacitor in order to be able to switch on the High side MOSFET. What I do not understand is, when driving inductive loads as a BLDC motor, how is it possible to switch on the Low side one (in the same Phase which is being driven by the controller at a particular time) to conduct the freewheeling current that the inductor is generating.

At that moment the high side mosfet is switched off, So the normal path for the current to go is interrupted. The inductor reverses the Voltage at its terminals, and to allow the current to go somewhere we turn on the Low side MOSFET (the one located in the same phase), so it can free-wheel through it. But in that situation the Source of the MOSFET is above the Drain (otherwise the current would not flow...). Since the Gate has to be above the Source in order to open de MOSFET, and the IC is apparently unable to do so (not in the Low side), How can this work?
 

MaxHeadRoom

Joined Jul 18, 2013
28,619
Not quite clear as to what you are asking, but in the case of a DC brushed or BLDC motor while the motor is in motion the generated field is in the same polarity as the applied voltage and opposes it.
There is also normally reverse connected Schottky diodes placed across each Mosfet.
Max.
 
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