Can a DC brushless motor produce the same torque at different power levels?Brushless motor specs to maximize...

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Can a DC brushless motor produce the same torque at different power levels?


Brushless motor specs to maximize stall torqueHow are current and voltage related to torque and speed of a brushless motor?Comparison of the efficiency of DC motor current limiting / control methods?How can brushless DC motor has constant torque?How can I calculate the inductance of brushless DC motor?Overloading BLDC Motor (Between nominal and stall torque)What is the state space model for a BLDC motorVoltage Regulation to Brushless DC MotorDoes increasing voltage increase the torque of a brushless motor?What happens to the torque/speed curve of a brushless motor when I increase the voltage?






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$begingroup$


I am looking these Neo Brushless motors for a project. Here are the links to the data sheet and also to stall testing from the company:



http://www.revrobotics.com/neo-brushless-motor-locked-rotor-testing/
http://www.revrobotics.com/content/docs/REV-21-1650-DS.pdf



So in the stall torque testing, on the last test with 80A limiting, they were able to achieve a torque of about 2.3 N-m.



enter image description here



You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.



Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W. How is this same motor producing the same torque at wildly different power values? I assume this isn't gearing related since these are motor tests to figure out the limits of the motor itself. Does it have to do with current limiting?



enter image description here










share|improve this question







New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$














  • $begingroup$
    The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
    $endgroup$
    – Kevin White
    8 hours ago










  • $begingroup$
    When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
    $endgroup$
    – Marko Buršič
    7 hours ago






  • 1




    $begingroup$
    Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
    $endgroup$
    – Marko Buršič
    7 hours ago




















1












$begingroup$


I am looking these Neo Brushless motors for a project. Here are the links to the data sheet and also to stall testing from the company:



http://www.revrobotics.com/neo-brushless-motor-locked-rotor-testing/
http://www.revrobotics.com/content/docs/REV-21-1650-DS.pdf



So in the stall torque testing, on the last test with 80A limiting, they were able to achieve a torque of about 2.3 N-m.



enter image description here



You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.



Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W. How is this same motor producing the same torque at wildly different power values? I assume this isn't gearing related since these are motor tests to figure out the limits of the motor itself. Does it have to do with current limiting?



enter image description here










share|improve this question







New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$














  • $begingroup$
    The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
    $endgroup$
    – Kevin White
    8 hours ago










  • $begingroup$
    When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
    $endgroup$
    – Marko Buršič
    7 hours ago






  • 1




    $begingroup$
    Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
    $endgroup$
    – Marko Buršič
    7 hours ago
















1












1








1





$begingroup$


I am looking these Neo Brushless motors for a project. Here are the links to the data sheet and also to stall testing from the company:



http://www.revrobotics.com/neo-brushless-motor-locked-rotor-testing/
http://www.revrobotics.com/content/docs/REV-21-1650-DS.pdf



So in the stall torque testing, on the last test with 80A limiting, they were able to achieve a torque of about 2.3 N-m.



enter image description here



You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.



Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W. How is this same motor producing the same torque at wildly different power values? I assume this isn't gearing related since these are motor tests to figure out the limits of the motor itself. Does it have to do with current limiting?



enter image description here










share|improve this question







New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




I am looking these Neo Brushless motors for a project. Here are the links to the data sheet and also to stall testing from the company:



http://www.revrobotics.com/neo-brushless-motor-locked-rotor-testing/
http://www.revrobotics.com/content/docs/REV-21-1650-DS.pdf



So in the stall torque testing, on the last test with 80A limiting, they were able to achieve a torque of about 2.3 N-m.



enter image description here



You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.



Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W. How is this same motor producing the same torque at wildly different power values? I assume this isn't gearing related since these are motor tests to figure out the limits of the motor itself. Does it have to do with current limiting?



enter image description here







voltage current brushless-dc-motor current-limiting torque






share|improve this question







New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question







New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question






New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked 9 hours ago









Ryan CRyan C

173 bronze badges




173 bronze badges




New contributor



Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




Ryan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • $begingroup$
    The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
    $endgroup$
    – Kevin White
    8 hours ago










  • $begingroup$
    When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
    $endgroup$
    – Marko Buršič
    7 hours ago






  • 1




    $begingroup$
    Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
    $endgroup$
    – Marko Buršič
    7 hours ago




















  • $begingroup$
    The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
    $endgroup$
    – Kevin White
    8 hours ago










  • $begingroup$
    When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
    $endgroup$
    – Marko Buršič
    7 hours ago






  • 1




    $begingroup$
    Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
    $endgroup$
    – Marko Buršič
    7 hours ago


















$begingroup$
The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
$endgroup$
– Kevin White
8 hours ago




$begingroup$
The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM.
$endgroup$
– Kevin White
8 hours ago












$begingroup$
When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
$endgroup$
– Marko Buršič
7 hours ago




$begingroup$
When stalled, motor outputs exactly 0 watts. $P=TcdotOmega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss.
$endgroup$
– Marko Buršič
7 hours ago




1




1




$begingroup$
Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
$endgroup$
– Marko Buršič
7 hours ago






$begingroup$
Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different?
$endgroup$
– Marko Buršič
7 hours ago












4 Answers
4






active

oldest

votes


















3














$begingroup$

The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention



As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.



This is spelled out in the OP's linked page describing the testing -




Please take the following into consideration when interpreting the data below:



Average motor phase current (or winding current) is different than the average input current to the motor controller.



Average Input Current = Average Phase Current x Duty Cycle



Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.




2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.






share|improve this answer









$endgroup$















  • $begingroup$
    +1 for deciphering what really begs the OP.
    $endgroup$
    – Marko Buršič
    6 hours ago



















2














$begingroup$

Torque is proportional to current. Power out = torque x rpm.



At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).



If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.



This is how the same motor can produce the same torque at 'wildly different power values'.



However,




You can see the current is about 25A and the voltage is about 10V.
This gives a power of 250W at stall.



Now, looking at the data sheet, they list a stall torque of 2.6 N-m.
They also give a voltage of 12V and current of 105A. This equates to
power draw of 1260W.




When a controller is used to limit current the motor current may be higher than the power supply current.



The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.



This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.



But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.






share|improve this answer









$endgroup$















  • $begingroup$
    This also equally answered my question.
    $endgroup$
    – Ryan C
    1 hour ago



















1














$begingroup$

A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.



enter image description here



EDIT:



I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $V=Icdot R$ and not the DC link voltage, the driver does PWM.






share|improve this answer











$endgroup$























    0














    $begingroup$


    How is this same motor producing the same torque at wildly different power values?




    By generating mechanical power as well as dissipating power as heat.



    If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).






    share|improve this answer









    $endgroup$


















      Your Answer






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      $begingroup$

      The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention



      As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.



      This is spelled out in the OP's linked page describing the testing -




      Please take the following into consideration when interpreting the data below:



      Average motor phase current (or winding current) is different than the average input current to the motor controller.



      Average Input Current = Average Phase Current x Duty Cycle



      Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.




      2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.






      share|improve this answer









      $endgroup$















      • $begingroup$
        +1 for deciphering what really begs the OP.
        $endgroup$
        – Marko Buršič
        6 hours ago
















      3














      $begingroup$

      The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention



      As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.



      This is spelled out in the OP's linked page describing the testing -




      Please take the following into consideration when interpreting the data below:



      Average motor phase current (or winding current) is different than the average input current to the motor controller.



      Average Input Current = Average Phase Current x Duty Cycle



      Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.




      2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.






      share|improve this answer









      $endgroup$















      • $begingroup$
        +1 for deciphering what really begs the OP.
        $endgroup$
        – Marko Buršič
        6 hours ago














      3














      3










      3







      $begingroup$

      The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention



      As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.



      This is spelled out in the OP's linked page describing the testing -




      Please take the following into consideration when interpreting the data below:



      Average motor phase current (or winding current) is different than the average input current to the motor controller.



      Average Input Current = Average Phase Current x Duty Cycle



      Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.




      2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.






      share|improve this answer









      $endgroup$



      The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention



      As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.



      This is spelled out in the OP's linked page describing the testing -




      Please take the following into consideration when interpreting the data below:



      Average motor phase current (or winding current) is different than the average input current to the motor controller.



      Average Input Current = Average Phase Current x Duty Cycle



      Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.




      2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 7 hours ago









      Phil GPhil G

      4,3611 gold badge7 silver badges15 bronze badges




      4,3611 gold badge7 silver badges15 bronze badges















      • $begingroup$
        +1 for deciphering what really begs the OP.
        $endgroup$
        – Marko Buršič
        6 hours ago


















      • $begingroup$
        +1 for deciphering what really begs the OP.
        $endgroup$
        – Marko Buršič
        6 hours ago
















      $begingroup$
      +1 for deciphering what really begs the OP.
      $endgroup$
      – Marko Buršič
      6 hours ago




      $begingroup$
      +1 for deciphering what really begs the OP.
      $endgroup$
      – Marko Buršič
      6 hours ago













      2














      $begingroup$

      Torque is proportional to current. Power out = torque x rpm.



      At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).



      If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.



      This is how the same motor can produce the same torque at 'wildly different power values'.



      However,




      You can see the current is about 25A and the voltage is about 10V.
      This gives a power of 250W at stall.



      Now, looking at the data sheet, they list a stall torque of 2.6 N-m.
      They also give a voltage of 12V and current of 105A. This equates to
      power draw of 1260W.




      When a controller is used to limit current the motor current may be higher than the power supply current.



      The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.



      This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.



      But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.






      share|improve this answer









      $endgroup$















      • $begingroup$
        This also equally answered my question.
        $endgroup$
        – Ryan C
        1 hour ago
















      2














      $begingroup$

      Torque is proportional to current. Power out = torque x rpm.



      At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).



      If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.



      This is how the same motor can produce the same torque at 'wildly different power values'.



      However,




      You can see the current is about 25A and the voltage is about 10V.
      This gives a power of 250W at stall.



      Now, looking at the data sheet, they list a stall torque of 2.6 N-m.
      They also give a voltage of 12V and current of 105A. This equates to
      power draw of 1260W.




      When a controller is used to limit current the motor current may be higher than the power supply current.



      The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.



      This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.



      But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.






      share|improve this answer









      $endgroup$















      • $begingroup$
        This also equally answered my question.
        $endgroup$
        – Ryan C
        1 hour ago














      2














      2










      2







      $begingroup$

      Torque is proportional to current. Power out = torque x rpm.



      At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).



      If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.



      This is how the same motor can produce the same torque at 'wildly different power values'.



      However,




      You can see the current is about 25A and the voltage is about 10V.
      This gives a power of 250W at stall.



      Now, looking at the data sheet, they list a stall torque of 2.6 N-m.
      They also give a voltage of 12V and current of 105A. This equates to
      power draw of 1260W.




      When a controller is used to limit current the motor current may be higher than the power supply current.



      The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.



      This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.



      But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.






      share|improve this answer









      $endgroup$



      Torque is proportional to current. Power out = torque x rpm.



      At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).



      If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.



      This is how the same motor can produce the same torque at 'wildly different power values'.



      However,




      You can see the current is about 25A and the voltage is about 10V.
      This gives a power of 250W at stall.



      Now, looking at the data sheet, they list a stall torque of 2.6 N-m.
      They also give a voltage of 12V and current of 105A. This equates to
      power draw of 1260W.




      When a controller is used to limit current the motor current may be higher than the power supply current.



      The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.



      This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.



      But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 6 hours ago









      Bruce AbbottBruce Abbott

      29.4k1 gold badge24 silver badges40 bronze badges




      29.4k1 gold badge24 silver badges40 bronze badges















      • $begingroup$
        This also equally answered my question.
        $endgroup$
        – Ryan C
        1 hour ago


















      • $begingroup$
        This also equally answered my question.
        $endgroup$
        – Ryan C
        1 hour ago
















      $begingroup$
      This also equally answered my question.
      $endgroup$
      – Ryan C
      1 hour ago




      $begingroup$
      This also equally answered my question.
      $endgroup$
      – Ryan C
      1 hour ago











      1














      $begingroup$

      A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.



      enter image description here



      EDIT:



      I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $V=Icdot R$ and not the DC link voltage, the driver does PWM.






      share|improve this answer











      $endgroup$




















        1














        $begingroup$

        A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.



        enter image description here



        EDIT:



        I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $V=Icdot R$ and not the DC link voltage, the driver does PWM.






        share|improve this answer











        $endgroup$


















          1














          1










          1







          $begingroup$

          A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.



          enter image description here



          EDIT:



          I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $V=Icdot R$ and not the DC link voltage, the driver does PWM.






          share|improve this answer











          $endgroup$



          A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.



          enter image description here



          EDIT:



          I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $V=Icdot R$ and not the DC link voltage, the driver does PWM.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Marko BuršičMarko Buršič

          12.2k2 gold badges9 silver badges13 bronze badges




          12.2k2 gold badges9 silver badges13 bronze badges


























              0














              $begingroup$


              How is this same motor producing the same torque at wildly different power values?




              By generating mechanical power as well as dissipating power as heat.



              If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).






              share|improve this answer









              $endgroup$




















                0














                $begingroup$


                How is this same motor producing the same torque at wildly different power values?




                By generating mechanical power as well as dissipating power as heat.



                If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).






                share|improve this answer









                $endgroup$


















                  0














                  0










                  0







                  $begingroup$


                  How is this same motor producing the same torque at wildly different power values?




                  By generating mechanical power as well as dissipating power as heat.



                  If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).






                  share|improve this answer









                  $endgroup$




                  How is this same motor producing the same torque at wildly different power values?




                  By generating mechanical power as well as dissipating power as heat.



                  If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 8 hours ago









                  TimWescottTimWescott

                  14.8k1 gold badge13 silver badges29 bronze badges




                  14.8k1 gold badge13 silver badges29 bronze badges


























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