4

See the video in the following link: https://www.youtube.com/watch?v=olnXy-J48SY

"Variable field magnet motor"

How does this work and is it actually efficient?

FatalSleep
  • 145
  • 1
  • 6

2 Answers2

3

That's clever! I have had it happen by accident occasionally, but never thought to use it for good effect.

It works by moving the rotor out of line with the stator so the effective magnet strength is lower. This increases Kv (velocity constant) because the motor has to run faster to produce enough back-emf to match the supply voltage. It is the permanent magnet equivalent of varying field current in a shunt-wound motor.

Motor speed could be raised by simply increasing voltage, but this also increases 'iron' losses due to hysteresis and eddy currents. Field weakening reduces the magnetic field strength as it raises rpm, reducing the loss at light loading. However it also reduces torque and output power, so it is more suited to applications which require high efficiency at both low rpm / high torque and high rpm / low torque.

Bruce Abbott
  • 55,540
  • 1
  • 47
  • 89
  • 1
    OP asked about efficiency. I'm not going to provide a competing answer just to say that his intuition that it's not efficient is partially correct. It's not as good as a motor custom designed for the new speed, but it is better than simply increasing the voltage. – Neil_UK Mar 18 '20 at 04:44
  • To be clear, reducing (or increasing) magnetic field strength does not increase efficiency, it changes the point of peak efficiency. At the extremes you can have a motor so weak that it can hardly overcome brush friction, or so strong that it needs extreme cooling to avoid burning up. I have installed weak ferrite motors in my N scale model trains that only draw a few milliamps, but are 10 times more efficient in that application than same size neodym motors used in drones etc. – Bruce Abbott Mar 18 '20 at 05:13
  • Worth mentioning that field weakening used to be a standard method for speed control . in the days of wound field motors (solenoid instead of permanent magnet to produce the field) all you had to do was reduce the field current. Efficiency is OK under light loads, but nosedives if the motor has to produce any torque. –  Mar 18 '20 at 10:43
  • 4
    In a brushless motor, the only real advantage of field weakening is that it reduces the conversion range of the drive circuitry. Drive circuitry has to be sized for peak current, but many applications (e.g. traction drives) are closer to constant power than constant torque output. By using field weakening you can run drive circuitry within a smaller current range. Also, this whole design is a bit pointless, as you can do the same thing electronically by shifting the phase. – Jon Sep 08 '20 at 08:54
1

As this pictire form hyperphysic

Pic

\$\tau = ILWBsin(\alpha)\$

And

enter image description here

\$E = \dot\theta WBL cos(\alpha)\$

When we apply supply voltage actoss motor's terminal and current flow through armature coil as (supply voltage)/(armature resistant), the motor will produce torque and start spinning. As it spin, it create Back-EMF voltage proportional to its speed which counter supply voltage and reduce current flow as (supply voltage - bemf voltage)/(armature resistant). At equilibrium point or steady state (Assumed no-loss) maximum speed (aka. no-load speed) equalto supply voltage divied by BEMF constant.

When moving rotor in and out, will change the length, \$L\$ which proportional to torque constant and BEMF constant.

As BEMF constant getting smaller, the motor will spin faster but the torque constant has reduced at the same ratio.

From motor power deliver equation

enter image description here

You see that

\$P = \frac{K_\tau v^2 (a-a^2)}{R K_B} \$

As length parameter, L change both \$ K_\tau \$ and \$ K_B \$ with same number so it cancel out it self so power transfer equation still the same but output speed and torque will change equivalent to gear reduction.

From mathematical proove above, the motor still transfer same amount of power theorically.

For efficiency if you meant "Power effiency", it hard to tell in reallity which loss came from damping friction and ironloss If we consider as "Power transmition efficiency" as how much power we can take form motor, you see it very efficient because it can deliver same amount power as rotor mving plus if we can match bemf constant to maximum power deliver it will deliver power more than normal motor. So I called it more effiecient than motor without this system.

Note: I capture image from my own video but unfortunately, I record it in Thai language.

M lab
  • 1,191
  • 6
  • 21