For my application I need a torque of \$\tau=0.87 \, Nm \$ and a speed of \$ w=140 \, RPM \approx 14.65 \, rad/s \$. This gives me a mechanical power \$ P_m=12.75 \, W\$. $$ P_m=\tau w $$ I have selected this stepper motor, which gives me the desired speed torque characteristics. enter image description here Now, I want to double check that the motor will generate enough electrical power to match or exceed the mechanical power requirements (\$P_m<=P_e\$). $$P_e=UI$$ From the motor spec \$I=6.37\$, \$R=0.3\$. As recommeded here, I obtain \$ U=IR\$; this gives me a value \$U=1.91 \, V\$.

Now this gives me an electrical power of \$P_e \approx 12 \, W\$, which is lower than my \$ P_m\$. Looking at the graph I would expect a \$P_e\$ considerably higher than my \$P_m\$ requirement. Question: why is \$P_e\$ not higher?

  • \$\begingroup\$ 12w is the electrical power consumed by one coil when operating as a space heater - what you might call the "parasitic resistance loss". \$\endgroup\$ – Chris Stratton Nov 18 '14 at 13:49

To much text for a comment.

  • Your link leads to 6 differnt moters, none of them is listed with 0.3 Ω
  • You can not use the DC resistance to make power calculations
  • The inductance is not stable, it changes with speed and load
  • Stepper motors are supplied with a pulsed DC voltage/current, that makes power calculations a lot more difficult than with DC or AC currents
  • Stepper motors have a bad efficiency, if the diagram above has no bugs, it shows a situation with Pm = 46.2 W and Pe = 305 W (Eventually this is peak and not average power)

I don't think that i'm wrong when i concluded that you will need a lot of profound electrotechnical and mathematical knowledge the make reliable calculations with stepper motors.

| improve this answer | |
  • \$\begingroup\$ All I need is a reasonable ballpark estimate to: (i) know that the motor will have the desired mechanical performance (I guess the torque -- speed characteristic graph is sufficient for that?) (ii) choose the correct power supply. \$\endgroup\$ – A.L. Verminburger Nov 18 '14 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.