When the frequency is increased whilst keeping the voltage constant, the magnetic field strength is going to weaken... I don't understand why and what actually happens that causes the field strength to drop.

If anybody has an information on this I'd appreciate some clarity on the matter.

  • 1
    \$\begingroup\$ A hint: consider how that affects the current, given that motors are inductive \$\endgroup\$ – Hearth Oct 14 '19 at 16:13
  • \$\begingroup\$ Moving rotor generates almost the same voltage as driver (BEMF) and thus with no load current is only 5%~10% of load current. Surge current at stall is V+/ DCR of coil can be 5~10x due to DCR/Zcoil(f) at full loaded RPM. Thus to flatten torque = current vs RPM, V is increased with f. \$\endgroup\$ – Tony Stewart EE75 Oct 14 '19 at 16:38
  • \$\begingroup\$ current thru an inductor, which your coils are, is I = 1/L * integral(V*dT). This means faster frequency gives a smaller dT, and the peak current is reduced. \$\endgroup\$ – analogsystemsrf Oct 14 '19 at 17:10

Consider the equivalent circuit for one phase of an induction motor shown below. The current, \$Im\$ in the magnetizing inductance \$Xm\$ determines the stator magnetic field strength. For a simple approximation, we can assume that \$Xs = 0\$ and everything else in the circuit except \$Xm\$ is infinite. We can then say that \$Im = \frac{V}{Xm}\$

Since \$Xm = 2*Pi*f*Lm\$, we can say that $$Im = \frac{V}{f}*\frac{1}{(2*Pi*Lm)}$$.

If we include all of the other components in the analysis, we will see that they have some influence, but the simple approximation shows pretty well what is happening.

enter image description here Diagram adapted from Malcom Barnes "Practical Variable Speed Drives and Power Electronics"


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