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TimWescott
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Efficiency vs. Current:

Define an ideal motor as one that is 100% efficient. Then a real motor is an ideal motor wrapped in problems, like armature resistance, core losses, windage, friction, and other mechanical losses.

If \$k_m(i) = \frac{T_{out}}{i}\$ is the ideal motor's torque "constant" at any given current (I put "constant" in quotes, because here it depends on current), then by conservation of energy, it simply must be the reciprocal of the ideal motor's speed constant: \$\omega = \frac{V_a}{k_m(i)}\$. You can even, after an adventure in SI units, show that \$ \mathrm{\frac{N \cdot m}{A}} \$ has the same units as \$\mathrm{\frac{V \cdot s}{radians}}\$.

So it's certainly mathematically possible for the efficiency to be high at high currents, as long as the copper losses are low enough. If that's really true, I'd expect that a motor excited at a constant voltage would actually increase in speed at high torque. I find this highly counter-intuitive, but -- that's what drops out of the data sheet's claims.

Losses vs. Harmonics

I did not see anything in the data sheet that goes into this, although I may have overlooked this.

Yes, harmonics could be a problem, because some types of core losses increase with increasing frequency. You may need to just experiment with this -- if you have a high-enough power square wave generator, you could just hit the thing with the harmonics you expect and measure the power consumed (or measure the current and calculate the power).

Define an ideal motor as one that is 100% efficient. Then a real motor is an ideal motor wrapped in problems, like armature resistance, core losses, windage, friction, and other mechanical losses.

If \$k_m(i) = \frac{T_{out}}{i}\$ is the ideal motor's torque "constant" at any given current (I put "constant" in quotes, because here it depends on current), then by conservation of energy, it simply must be the reciprocal of the ideal motor's speed constant: \$\omega = \frac{V_a}{k_m(i)}\$. You can even, after an adventure in SI units, show that \$ \mathrm{\frac{N \cdot m}{A}} \$ has the same units as \$\mathrm{\frac{V \cdot s}{radians}}\$.

So it's certainly mathematically possible for the efficiency to be high at high currents, as long as the copper losses are low enough. If that's really true, I'd expect that a motor excited at a constant voltage would actually increase in speed at high torque.

Efficiency vs. Current:

Define an ideal motor as one that is 100% efficient. Then a real motor is an ideal motor wrapped in problems, like armature resistance, core losses, windage, friction, and other mechanical losses.

If \$k_m(i) = \frac{T_{out}}{i}\$ is the ideal motor's torque "constant" at any given current (I put "constant" in quotes, because here it depends on current), then by conservation of energy, it simply must be the reciprocal of the ideal motor's speed constant: \$\omega = \frac{V_a}{k_m(i)}\$. You can even, after an adventure in SI units, show that \$ \mathrm{\frac{N \cdot m}{A}} \$ has the same units as \$\mathrm{\frac{V \cdot s}{radians}}\$.

So it's certainly mathematically possible for the efficiency to be high at high currents, as long as the copper losses are low enough. If that's really true, I'd expect that a motor excited at a constant voltage would actually increase in speed at high torque. I find this highly counter-intuitive, but -- that's what drops out of the data sheet's claims.

Losses vs. Harmonics

I did not see anything in the data sheet that goes into this, although I may have overlooked this.

Yes, harmonics could be a problem, because some types of core losses increase with increasing frequency. You may need to just experiment with this -- if you have a high-enough power square wave generator, you could just hit the thing with the harmonics you expect and measure the power consumed (or measure the current and calculate the power).

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TimWescott
  • 47.1k
  • 1
  • 45
  • 108

Define an ideal motor as one that is 100% efficient. Then a real motor is an ideal motor wrapped in problems, like armature resistance, core losses, windage, friction, and other mechanical losses.

If \$k_m(i) = \frac{T_{out}}{i}\$ is the ideal motor's torque "constant" at any given current (I put "constant" in quotes, because here it depends on current), then by conservation of energy, it simply must be the reciprocal of the ideal motor's speed constant: \$\omega = \frac{V_a}{k_m(i)}\$. You can even, after an adventure in SI units, show that \$ \mathrm{\frac{N \cdot m}{A}} \$ has the same units as \$\mathrm{\frac{V \cdot s}{radians}}\$.

So it's certainly mathematically possible for the efficiency to be high at high currents, as long as the copper losses are low enough. If that's really true, I'd expect that a motor excited at a constant voltage would actually increase in speed at high torque.