BLDC: Coil Windings and their effect on maximum RPM 3

[hyboxis79]

I understand that there's a direct connection between the number of windings on each pole and the maximum RPM the BLDC can achieve. I understand that torque is proportional to the current in the windings and since less windings usually enables enlarging the winding wire thus enabling more current can enter the motor. Consequently enabling the motor to reach higher acceleration. But what about the correlation between the number of windings and the created magneto motive force F_mmf=N i
is that parameter only pertinent to the induced emf or is it just that only after certain point the number winding of provide better RPM - so called the turning point

Appreciate all who can help

 

[sreid]

The basic force and Voltage equations apply

F = B x l x I

V = B x l x v

If you reduce the length of the wire (number of turns), the force per unit current goes down as does the voltage per unit velocity.

Electrically, the maximum motor speed is determined when the motor BEMF equals the supply voltage.

 

[waross]

If a motor is originally well designed it will be running close to magnetic saturation. Trying to increase torque by changing the windings will push it into magnetic saturation and you will develop increased heat instead of increased torque.
By changing the number of turns you will be changing the induction of the windings. With AC and pulsed DC the inductive reactance is more important than resistance in limiting the current.

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

 

[hyboxis79]

thanks sreid and Bill.

two more related questions


sreid

Isn't the figure for maximum acheivable torque also dictating some speed limits?


Bill

about magnetic saturation
Why and how does magnetic saturation happens?

thanx

 

[waross]

Put simply, the maximum magnetic field strength that a material is able to develop is close to the saturation point. I say close because saturation is not an exact value but describes the point on a curve where the curve changes slope.
Saturation limits the maximum lines of force in a magnetic material.


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

 

[cswilson]

If you look in almost any brushless servo motor catalog, you will see a variety of motors listed on the same page that have the same "iron" but different windings.

From the first relevant page in a Kollmorgen catalog I yanked off my shelf:

Winding A has a resistance of 1.12 ohms and an inductance of 1.7 mH. This provides a torque constant Kt of 0.157 N-m/amp and a back EMF constant Ke of 16.4 V/krpm. Continuous current is rated at 6.14 amps.

Winding B has a resistance of 7.06 ohms and an inductance of 10 mH. This provides a torque constant Kt of 0.391 N-m/amp and a back EMF constant Ke of 41.0 V/krpm. Continuous current is rated at 2.46 amps.

What's the difference? Winding B has more turns (2.5x) of narrower wire with the same "fill". Resistance and inductance increase with the square of the number of turns, so are over 6x greater (but note that the L/R time constant remains constant). Kt and Ke both increase linearly with the number of turns. Of course, with a higher Ke and a given supply voltage, your top speed is lower. The net effect compared to winding A is essentially the same as adding a mechanical gear reducer of factor 2.5 to the output shaft.

Of course, if you kept increasing the number of turns, you could drive the iron into magnetic saturation (which occurs when virtually all of the micro magnetic domains in the iron are already aligned, and adding extra external field from more current has little to no effect), but most motor designs of this type are capable of taking a variety of useful windings before getting significant magnetic saturation.

Curt Wilson
Delta Tau Data Systems

 

[sreid]

The continuous torque a motor can achieve is limited by motor heating (temperature). Some motors can achieve this torque through out their speed range. Some motors have derated torque at higher RPM due to additional heating from Iron Losses (histerysis and eddy current losses).

 

[hyboxis79]

thanks a lot everybody

 

[hyboxis79]

Still on the matter of windings and so

I have some additional questions and I would appreciate your help very much.

In this discussion we talked about the effects of adding windings and the pros and the cons of it. If extra windings hurdle the maximum achieved speed why not automatically stick to as least windings as possible. I know that at this situation the speed achieved per volt will be the maximum. Other than that I know that the wire will be thicker thus I will be able to drive more current into the motor which will help me with more torque. Wouldn't it be best speed wise and torque wise. Or will the torque be hurdled and for better torque more windings is preferable

 

[cswilson]

hyboxis:

Note that the Ke back-EMF constant and the Kt torque constant vary together. If you cut one in half, then you cut the other in half as well. In consistent units (SI, with Ke in V/(rad/sec) and Kt in N-m/A), they are the same numerically.

This is a consequence of the law of conservation of energy, or colloquially, "there's no such thing as a free lunch".

The motor torque equation is: T = Kt * I

The motor generator equation is E = Ke * w

By conservation of energy, electrical power must equal mechanical power: E * I = T * w

Substituting: Ke * w * I = Kt * I * w

Therefore: Ke = Kt

So if you reduced the number of wirings, you would reduce the torque per unit current. So you need more current to get the same torque. Because heating is proportional to the square of current, even with reduced resistance, you will not be able to get more torque out.

Curt Wilson
Delta Tau Data Systems

 

[hyboxis79]

thanks for the answer Curt.
and another one
The numerical equality between Ke and Kt is based on making no references to mechanical or electrical losses because usually there's some difference between them right?
And for the sake of semantics do most catalogs offer Ke and Kt with the field flux factor already included
(which in brushed dc engine is usually in the stator)

 

[cswilson]

Yes, the match between Ke and Kt is based on the reversible (lossless) transfer of power between the electrical and mechanical realms. On both sides of this transfer, there are losses.

On the electrical side, the terminal voltage (let's use the simple case of a DC motor) can be expressed as:

V = Ke*w + R*I + L*di/dt

Of course, the R*I term represents a loss.

On the mechanical side, the shaft output torque is the generated torque minus losses (e.g. windage, friction).

So if you were just to measure electrical input and mechanical output and calculate as if the whole system were lossless, you would not derive identical values for Ke and Kt.

The values of Ke and Kt are dependent on the magnitude of the field as well -- in the absence of magnetic saturation, directly proportional to this field strength.

Curt Wilson
Delta Tau Data Systems

 

[hyboxis79]

thanks a lot Curt....

 

[hyboxis79]

On the matter of added current conduction....
Looking on a BLDC motor catalog for motors with the same iron I've noticed different motor windings but similar power ratings.
Since you have a thicker wire more current will be drawn from the power supply. Although the current goes up the voltage which is dictated by the power source stays the same. I understand that in such a case it necessarily means that I have to deliberately restrict the voltage to prevent over heating.
If my system can't reach the nominal power output in anycase how do I know if its problem isn't with the voltage? Meaning that having the ability to draw more current can help me.

 

[abudabit]

I want to increase N-m/A by 3.44 times and inversely decrease RPM/V but I don't want to decrease the volume / length of the wire by so much. I will of course add more powerful permanent magnents to compensate.

Do I increase the wire length 3.44 times or will there be more to it since I won't be changing the wire diameter like in the example with Hyboxis? Is the change in inductance the only significant factor in determing the N-m/A to RPM/V ratio during a rewind (assuming no magnetic oversaturation)?

 

[ShamilaWije]

abudabit,

The problem with trying to increase the wire length by 3.44 times without changing the wire gauge is you'll run out of space in the slots.

If you do increase the length Nm/A should increase by about the same factor assuming no magnetic saturation.

 

[abudabit]

Thank you very much Shamila.

 

[abudabit]

Now I'm worried about efficiency.....

If I increase wire length 3.44 times, decrease current draw by only 50%, keep wire diameter the same.... does that mean it dissipates(3.44 / 2)^2 (2.96)times more heat from the current through windings?

What effect does increasing the diameter of the rotor have on N-m/A vs RPM/V? Is it linear with diameter, or perhaps linear with circumfrence?

 

[waross]

If I increase wire length 3.44 times, decrease current draw by only 50%, keep wire diameter the same.... does that mean it dissipates(3.44 / 2)^2 (2.96)times more heat from the current through windings?

decrease current draw by only 50%

At 50% current the I^R losses will be 50%^2 or 25%.

If I increase wire length 3.44 times

Increasing the wire length and holding the current constant will increase the losses proportionally. A 3.44 increase in length will cause a 3.44 increase in losses at the same current.
50% current and 3.44 times more wire length will result in 3.44/4 or 86% of the base losses.

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

 

[abudabit]

Thank you very much, that makes a lot more sense. Glad I won't have to redesign the rotor and magnets much since I'm utterly in the dark with those.