CAD CAM EDM DRO - Yahoo Group Archive

Re: Stepper Motor Power Calculations

on 2006-01-24 11:53:55 UTC
I don't completely agree with your conclusion.

You should pick your supply voltage to be just adequate to the task
and no more. Anything above that results in a hotter than necessary
motor with nothing to show for it.

What supply voltage is adequate? Empirically, run your mechanism with
a 24VDC supply initially. Note at what percentage of the desired
speed the motor stalls. Divide 24VDC by that percentage to arrive at
your optimal supply voltage.

Example: Say your mechanism reaches 70% of your design speed at
24VDC. Divide 24V by 0.70 and you get 34VDC for the optimal supply
voltage.

Mariss


--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Phil Mattison"
<mattison20@c...> wrote:
>
> 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.
>
> ----- Original Message -----
> From: "Mariss Freimanis" <mariss92705@y...>
> 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