Re: Stepper Motor Power Calculations
Posted by
Mariss Freimanis
on 2006-01-23 19:20:05 UTC
There are two causes of heat dissipation in a motor and inductance
stays constant.
1) Let's take care of inductance first. Inductance has a property
called inductive reactance. Inductive reactance is measured in Ohms
and is proportional to frequency.
To double the speed of a step motor you must double the driving
frequency of the winding currents. Inductive reactance follows Ohm's
Law, so inductive current must be halved.
Torque is proportional to ampere-turns, or simplified, proportional
to current. This means if speed is doubled, torque is halved. This
means the motor speed-torque curve is the 1/x function with speed.
Power is torque times speed. Speed doubles, torque is halved, power
stays constant. This means the power output of a step motor is
independent of speed.
1a) Still following Ohm's Law from (1), inductive current is
proportional to supply voltage. This makes torque proportional to
supply voltage as well. By extension, output power is proportional as
well, V/sqrt L specifically.
2) I^2*R losses have been covered previously except for the motor's
rated current is not a thermal limit but rather a magnetic saturation
limit.
A 1A, 5V motor has 10W of I^2 dissipation (5W / winding), yet the
motor can output 80 Watts of mechanical power. Step motor conversion
efficiency tops-out in the 70% range; 80 Watts output generates 34W
of heat at that efficiency.
I^2*R losses drop very rapidly with increasing speed. Current being
the inverse of speed makes copper losses drop as a 1/x^2 function of
speed. They essentially become insignificant at higher speeds.
3) Iron losses (eddy-current and hysterisis) generates the majority
of heat dissipated at high speeds. These losses are proportional to
speed (frequency of flux-reversals per second)and are also
proportional to the supply voltage (inductive current amplitude).
Conclusion: Motor output power increases proportional to supply
voltage while iron losses go up with the square of the supply voltage.
Please see "Losses and Power vs. Voltage.pdf" in the files section of
this group, Circuits / Miscellaneous Circuits folder. This is some
dyno data I took a few years ago that deals with these subjects.
Mariss
--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Phil Mattison"
<mattison20@c...> wrote:
stays constant.
1) Let's take care of inductance first. Inductance has a property
called inductive reactance. Inductive reactance is measured in Ohms
and is proportional to frequency.
To double the speed of a step motor you must double the driving
frequency of the winding currents. Inductive reactance follows Ohm's
Law, so inductive current must be halved.
Torque is proportional to ampere-turns, or simplified, proportional
to current. This means if speed is doubled, torque is halved. This
means the motor speed-torque curve is the 1/x function with speed.
Power is torque times speed. Speed doubles, torque is halved, power
stays constant. This means the power output of a step motor is
independent of speed.
1a) Still following Ohm's Law from (1), inductive current is
proportional to supply voltage. This makes torque proportional to
supply voltage as well. By extension, output power is proportional as
well, V/sqrt L specifically.
2) I^2*R losses have been covered previously except for the motor's
rated current is not a thermal limit but rather a magnetic saturation
limit.
A 1A, 5V motor has 10W of I^2 dissipation (5W / winding), yet the
motor can output 80 Watts of mechanical power. Step motor conversion
efficiency tops-out in the 70% range; 80 Watts output generates 34W
of heat at that efficiency.
I^2*R losses drop very rapidly with increasing speed. Current being
the inverse of speed makes copper losses drop as a 1/x^2 function of
speed. They essentially become insignificant at higher speeds.
3) Iron losses (eddy-current and hysterisis) generates the majority
of heat dissipated at high speeds. These losses are proportional to
speed (frequency of flux-reversals per second)and are also
proportional to the supply voltage (inductive current amplitude).
Conclusion: Motor output power increases proportional to supply
voltage while iron losses go up with the square of the supply voltage.
Please see "Losses and Power vs. Voltage.pdf" in the files section of
this group, Circuits / Miscellaneous Circuits folder. This is some
dyno data I took a few years ago that deals with these subjects.
Mariss
--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Phil Mattison"
<mattison20@c...> wrote:
>the other
> Well then, that would be a good reason to ignore the advice I saw
> day on one of these boards, suggesting that it's ok to run them hotuntil
> they almost sizzle. I suspect that at high speeds the inductivereactance
> gets so high that the voltage required to overheat the motor wouldbe more
> than the switching components could handle anyway, which isprobably why we
> don't see motor drivers with better high speed performance. I guessthis is
> one of those cases where there is a good reason nobody has done italready.
>could be
> --Phil M.
>
> ----- Original Message -----
> From: "Vlad Krupin" <vlad.cnc@g...>
> To: <CAD_CAM_EDM_DRO@yahoogroups.com>
> Sent: Monday, January 23, 2006 5:32 PM
> Subject: Re: [CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
>
>
> > On 1/23/06, Phil Mattison <mattison20@c...> wrote:
> > [snip]
> >
> > > I would think
> > > that the motor current ratings are for holding current, and
> > > substantially exceeded when stepping at high speed withoutdamaging the
> > > motor, since heat dissipation is the main reason for currentlimitation.
> >searching
> >
> > It is, indeed the main limitation. Out of curiosity I did some
> andnot take
> > came up with this table: http://www.kjmagnetics.com/specs.asp
> >
> > Those are some seriously low temperatures! It shows that it does
> > much heat to permanently damage a neodymium magnet. Depending onthe motor
> > design, I think, the magnets may have a hard time transferringheat to the
> > body of the motor. You may not even realize that your motor is atrisk
> untilfound and
> > it is too late.
> >
> > (probably 99% of you knew that already, but just how low the safe
> operating
> > temperature can be was news to me. So I figured I'd share what I
> > maybe save somebody a bit of extra grief)metalworking
> >
> > Vlad
> >
> > [snip]
> >
> > --
> > Vlad's shop
> > http://www.krupin.net/serendipity/index.php?/categories/2-
> >
> >
> > [Non-text portions of this message have been removed]
> >
> >
>
Discussion Thread
Phil Mattison
2006-01-22 12:01:51 UTC
Stepper Motor Power Calculations
caudlet
2006-01-22 13:54:49 UTC
Re: Stepper Motor Power Calculations
JanRwl@A...
2006-01-22 23:41:23 UTC
Re: [CAD_CAM_EDM_DRO] Stepper Motor Power Calculations
Phil Mattison
2006-01-23 07:53:09 UTC
[CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
caudlet
2006-01-23 09:08:24 UTC
Re: Stepper Motor Power Calculations
Phil Mattison
2006-01-23 16:13:08 UTC
[CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
Vlad Krupin
2006-01-23 16:33:13 UTC
Re: [CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
Phil Mattison
2006-01-23 17:43:51 UTC
[CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
Mariss Freimanis
2006-01-23 19:20:05 UTC
Re: Stepper Motor Power Calculations
Phil Mattison
2006-01-24 09:10:47 UTC
Re: [CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
Mariss Freimanis
2006-01-24 11:53:55 UTC
Re: Stepper Motor Power Calculations