Re: [CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
Posted by
Phil Mattison
on 2006-01-24 09:10:47 UTC
Nice tutorial. So this pretty much gets to the bottom of what this was about
to begin with. Torque decreases with speed due to inductive reactance and
iron losses. Torque could be maintained theoretically by increasing voltage
proportional to speed, but that produces diminishing returns because power
output increases linearly with voltage while iron losses increase
exponentially. So there is a point at which iron losses exceed power gains
and any further voltage increase is dissipated as heat.
Since chopper-type current regulators are designed to maintain a constant
current through the coils they will do this as speed increases up to the
point where the supply voltage becomes the limitation. Since friction of
lead screws and other mechanical stuff increases with speed, you really need
more power, not less, as speed increases. So I guess the bottom line is that
it makes sense to use a power supply as near as possible to the limit of the
motor controller's electronics to give it as much head-room as possible.
Beyond that there's not much you can do besides reducing power when idle to
minimize unnecessary heat accumulation, and tweaking your speed to run as
fast as possible without stalling.
--Phil M.
to begin with. Torque decreases with speed due to inductive reactance and
iron losses. Torque could be maintained theoretically by increasing voltage
proportional to speed, but that produces diminishing returns because power
output increases linearly with voltage while iron losses increase
exponentially. So there is a point at which iron losses exceed power gains
and any further voltage increase is dissipated as heat.
Since chopper-type current regulators are designed to maintain a constant
current through the coils they will do this as speed increases up to the
point where the supply voltage becomes the limitation. Since friction of
lead screws and other mechanical stuff increases with speed, you really need
more power, not less, as speed increases. So I guess the bottom line is that
it makes sense to use a power supply as near as possible to the limit of the
motor controller's electronics to give it as much head-room as possible.
Beyond that there's not much you can do besides reducing power when idle to
minimize unnecessary heat accumulation, and tweaking your speed to run as
fast as possible without stalling.
--Phil M.
----- Original Message -----
From: "Mariss Freimanis" <mariss92705@...>
To: <CAD_CAM_EDM_DRO@yahoogroups.com>
Sent: Monday, January 23, 2006 8:18 PM
Subject: [CAD_CAM_EDM_DRO] Re: Stepper Motor Power Calculations
> 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:
> >
> > Well then, that would be a good reason to ignore the advice I saw
> the other
> > day on one of these boards, suggesting that it's ok to run them hot
> until
> > they almost sizzle. I suspect that at high speeds the inductive
> reactance
> > gets so high that the voltage required to overheat the motor would
> be more
> > than the switching components could handle anyway, which is
> probably why we
> > don't see motor drivers with better high speed performance. I guess
> this is
> > one of those cases where there is a good reason nobody has done it
> already.
> >
> > --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
> could be
> > > > substantially exceeded when stepping at high speed without
> damaging the
> > > > motor, since heat dissipation is the main reason for current
> limitation.
> > >
> > >
> > > It is, indeed the main limitation. Out of curiosity I did some
> searching
> > and
> > > came up with this table: http://www.kjmagnetics.com/specs.asp
> > >
> > > Those are some seriously low temperatures! It shows that it does
> not take
> > > much heat to permanently damage a neodymium magnet. Depending on
> the motor
> > > design, I think, the magnets may have a hard time transferring
> heat to the
> > > body of the motor. You may not even realize that your motor is at
> risk
> > until
> > > 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
> found and
> > > maybe save somebody a bit of extra grief)
> > >
> > > Vlad
> > >
> > > [snip]
> > >
> > > --
> > > Vlad's shop
> > > http://www.krupin.net/serendipity/index.php?/categories/2-
> metalworking
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> >
>
>
>
>
>
>
>
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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