BLDC Speed and torque 1

[ehat001]

I am having trouble wrapping my head around this. Everyone says BLDC motors are speed controlled via voltage BUT the force (torque) produce by a coil or wire in a magnetic field is equal to F=BlI (Force(torque) = B,flux density * l,length of wire * I,current in wire). So the more current through a wire produces more force accelerating the wire through the magnetic field faster. So when you adjust the voltage across the winding it adjusts the current, increasing or decreasing both the torque and speed.
Do I have all of that correct?

In my application I am using a 3 phase Y connected BLDC with 3 half bridges run with 6 step commutation and PWM. It is speed controlled.

Is the speed of the motor adjusted by the PWM on time or how fast you run through the 6 commutation steps?
To me its a combination of both.. to go faster you need more current (voltage) through the windings (higher PWM percentage (more on)) to accelerate the winding through the magnetic field faster in addition to going through the commutation steps faster. Is that correct?

As an example (the numbers are not right)..
12V or 1A (1ohm winding) produces 1Nm of force at 10,000RPM...
The 1Nm is my torque constant = 1Nm/Amp
and 10KRPM/Nm (Speed/Torque gradient)
and speed constant would be 10000rpm/12V = 833.3RPM/V

Now my application is speed controlled and I want to run at 10,000rpm but my load torque is only 0.5Nm. Where does the rest of that torque go???
Is it that you are still going through the 6 commutation steps fast to produce the 10,000RPM but you PWM slower to adjust the current through the windings which produces less torque? and the commutation steps is what is controlling the speed?
If this is all true then the Speed Constant, Torque Constant, and Speed/Torque gradient are irrelevant as long as the motor can produce the maximum speed and torque without over heating. Right?
I hope that is clear and makes sense.
Thanks in advance.
Erik

 

[waross]

A BLDC motor is an AC motor and the speed is controlled by the frequency, ie. the speed with which you go through the commutation steps. More voltage and current and more torque means less angular displacement of the rotor position for a given load.
At higher speeds the induction of the coils prevents the current of each pulse from reaching the maximum steady state value and so current and torque are inversely proportional below the optimum Volts/Hertz ratio. The maximum voltage is fixed by the power supply voltage. As the speed and frequency rise the maximum torque will drop.
As the load is increased the speed will remain constant until maximum torque is exceeded and the rotor will be pulled out of step with the rotating magnetic field. Depending on the load, the motor may run slow or stall completely.
Others will add more and restate this in different ways.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ehat001]

Bill, thanks for the explanation. So the speed of the motor is controlled by the speed in which you run through the 6 commutation steps. Then what or why PWM during the commutation blocks? Is PWM to adjust the torque?
Thanks Again

 

[waross]

To hold the effective Volts/Hertz ratio below optimum at slower speeds. Exceeding the optimum V/Hz ratio increases current and heating while producing no benefit.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[benta]

The easiest way to understand a BLDC motor is to picture it as a normal brushed DC motor turned "inside out".
The windings are on the stator instead of the anchor/rotor, and the magnets are in the anchor/rotor.
The electrical characteristics are the same as for a PMDC motor, which means that back EMF is the controlling factor for speed.

The big difference is the commutation, which in the BLDC is electronic instead of mechanical.

It is important to understand that for both motors, commutation is controlled purely by the position of the anchor/rotor. For PMDC mechanically, for BLDC usually using Hall sensors (sensorless control is possible, but I'll ignore it here for simplicity).

Now, PWM control: again no difference, if you keep in mind that the PWM frequency has NOTHING to do with the commutation frequency. Picture a three-phase bridge doing the commutation based on the feedback signals from the Hall sensors. Now PWM modulate the supply voltage to the bridge at a completely different, usually much higher frequency. Again, no real difference to the PMDC motor.

waross' explanation is a very special use of a BLDC motor. Using a volt/hertz control scheme is possible, but would normally not be called BLDC, but PMSM (permanent magnet synchronous motor). The motor is basically the same, but the control circuit completely different.

Hope this helps,

Benta.

 

[ehat001]

I thought the BEMF was used to "look" at the speed... the voltage of the BEMF is directly proportional to the speed (generator) from that speed (voltage) you adjust accordingly to the required speed by way of commutation...

 

[benta]

You can phrase it in a lot of ways.
Basically, when you apply a voltage to a DC motor, it will accelerate until the back EMF is equal to the applied voltage (minus losses, of course).
This is independent of whether it's PMDC or BLDC motors.

Benta.

 

[cswilson]

The phase of a brushless motor aligned to produce torque when current is pushed into it will have a back EMF voltage on it that is proportional to the motor velocity. Only the difference between this back EMF and the supply voltage (the "headroom") is available to push current into the phase. So if the back EMF is 10V and the supply is 12V, you only have 2V available to overcome the IR drop. In your example of R = 1 ohm, you can only push 2A in at this speed, but if the back EMF were 8V, you would have 4V headroom and could push in 4A. So as your speed increases, you have less capability to produce torque, and you can get to the "no load" speed where you have no headroom to push in current and produce torque.

If your PWM modulated the supply voltage to be on 11/12 of the time, and so produce a time-averaged 11V, this would produce 1A if the back EMF were 10V, but 3A if the back EMF were 8V.

As the motor turns, the next phase will come into alignment so you will want to push current into that phase. The higher your speed, the sooner this will happen. This commutation frequency, which is speed dependent, is independent of your much higher PWM frequency.

Curt Wilson
Delta Tau Data Systems

 

[mikekilroy]

perhaps reducing a DC brush, DC brushless, or AC brushless motor its 2.5 governing equations would simplify understanding?

Vm=Ke*N+IR
T=Kt*Im
& N=120*Freq/poles (just like in ac induction motor)

where
..Vm is voltage across motor winding
..Ke is back emf constant, typical unit can be volts/1000rpm
...N=speed say in rpm
..Im=amps into motor winding
..R resistance of motor winding
..Kt is same as Ke but in different units - typically like nm/amp

On DC brush motor, the commutator keeps the freq of switching mechanically correct, so last equation is not required for understanding.

On brushless motor, switching freq must match exactly the speed - this is done with commutation and can be ignored in trying to understand the workings of V & I since it is assumed you would have some sort of drive that controls it properly for you according to the third equation.

Next it is best to look at current since it is a direct relationship to output torque. so if Kt=1nm/a, 1 amp WILL produce 1nm torque, 10 amp WILL produce 10nm torque. No if,ands,or buts about it. linear within motor rating.

motor voltage then is used to counter the internal motor back emf AND make a forcing function to make current in the motor resistance; it is NOT linear. here is how current and voltage interact. if one of these motors is rotating at speed N, then it will be generating voltage Ke/N; example: 1000rpm on motor with 100v/krpm means it will be generating 100v.

so if you put 100 volts across the motor it will do nothing for you! You will get no torque since T= Ke/I and I=0. now put in 110v and you get current flow per equation above: 100v dissapears against the back emf leaving you 10 volts TO MAKE TORQUE IN THE IR TERM. So say motor resistance is 10 ohm - you will have 1 amp flowing into the motor so due to T=Ke*I for the Ke of 1nm/amp, you will produce 1 nm of torque.

playing this these 2 basic equations should make it clear as mud after a while

Then you can go on to see why the motor speeds up or slows down with mechanics equation T=jw/t+Tfric.... in summary on it, if load requires less torque than you produce above, the motor will speed up until it gets limited by volts equation or the torque needed matches produced. Ditto the other way.

 

[mikekilroy]

darn missing edit button....

I showed "/" instead of "*" 2 places. fix:

then it will be generating voltage Ke*N; example: 1000rpm on motor with 100v/krpm means it will be generating 100v.

so if you put 100 volts across the motor it will do nothing for you! You will get no torque since T= Ke*I

 

[ehat001]

Thank you all. Much clearer now!