I don't understand the concept of the "virtual neutral" point that's commonly used in BLDC motor controllers. I understand that it's a proxy for the back emf relative to the true neutral point of the motor, but I don't understand why it's accurate or why it works.

Consider the schematic shown below:

enter image description here

Here, the back emf of phase 1 is a trapezoidal function that's generated by sources V8 and V9.

The virtual neutral point is created by the resistor summing network of R2, R3 and R4.

If I perform the standard zero crossing checks of V(p1) > V(V_NEUTRAL_VIRT) and V(p1) < V(V_NEUTRAL_VIRT) then the switching does occur at the correct time for the motor RPM.

My confusion is coming from the fact that V_NEUTRAL_VIRT = V(p1) + V(p2) + V(p3) (or at least some factor of the 3 phase voltages) where V(p1) includes the back emf of the undriven phase. If that's compared to V(p1) then I don't follow how this correctly picks up the zero crossing points.

So my question is what is the justification for building the controller in such a way? I don't dispute that works, I just don't understand why it works.

  • \$\begingroup\$ It's not always virtual. Imagine that the three coils are star-connected. You get three wires out of the motor, and there is the central point inside- you can't access it, but it's there. In a triangle connection it's virtual indeed, but electrically, when you look from outside, you can't tell the difference. \$\endgroup\$
    – TQQQ
    Feb 24 at 21:10

1 Answer 1


In order todo that you need to make some assumption. Give motor phase U,V,W which we give power through U and V as shown. enter image description here

In this case the real netral point will have voltage of \$V_M/2\$ by the voltage divider effect on winding resistant. we can simplify the circuit futhermore by enter image description here

As the motor winding haver very small resitance campare to resistor in virtual ground, so we can ignore them. enter image description here From this simplified model you see the virtual nuetral voltage will equals to actual only if \$V_{EMF} = 0\$ and from voltage divider, $$V_{VN} = \frac{1}{3} V_{EMF} + V_N$$ Event it not equals but it suitable for zero crossing detection because when we measure \$V_{SENSE}\$ respect to \$V_{VN}\$ it will have same sign as \$V_{EMF}\$ and will equals to zero as \$V_{EMF} = 0\$

Note: The property I've used derive from equivalent of resistor circuit which I thought very useful for analyse this kind of problem. enter image description here


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