Re: Stepper Power
Posted by
Jon Elson
on 2000-02-25 13:06:56 UTC
stratton@... wrote:
the ratings are
usually written for the voltage and current for EACH winding, so double
these numbers
for the total power.
times the
instantaneous torque AT THAT SPEED! Multiplying the holding torque at
ZERO
speed times some RPM will VERY favorably overstate the power out.
OK, I looked at the graphs. Using the curve for the 34M470, DR8010 and
PWR7205,
They are using half the coils at 8.3 A, 8.2 A ^2 * .42 Ohms = 28.9 W per
winding,
X 2 = 57.9 W input static. That apparently is from the same graph as
your top
number, with the 111 W output, if the conversion factor you used is
correct.
I'd like to see if you can get from MicroKinetics the current draw into
the
driver at this setting. Perhaps their driver draws more current when
the motor
is moving.
Jon
> From: stratton@...First, not that for a 2-phase stepper motor, there are two windings, but
>
> I've attempted some calculations on the stepper motor vs power supply
> size issue.
>
> I took as an example Microkinetics 34M470, nomivally a 470 oz-in motor
>
> in a Nema 34 frame. I haven't bought one yet, so I'm working entirely
>
> off the data on their web site:
>
> http://www.microkinetics.com/34m470.htm
>
> The various wiring voltage/current combinations all indicate that they
>
> rate this motor for just under 15 watts of I^2R type power.
the ratings are
usually written for the voltage and current for EACH winding, so double
these numbers
for the total power.
>These sound like holding torque time RPM. Note that watts requires RPM
> I looked at their graphs for performance with their various drives,
> and used 7.4 x 10^-4 as the conversion from oz-in-RPM to mechanical
> watts (it seems to be right, can anyone verify?). Their data and my
> calculations give results like this:
>
> 500 ozin 300 rpm = 111 watts
> 100 ozin 2300 RPM = 170 watts
> 125 ozin 2100 RPM = 194 watts (1/4 HP)
times the
instantaneous torque AT THAT SPEED! Multiplying the holding torque at
ZERO
speed times some RPM will VERY favorably overstate the power out.
OK, I looked at the graphs. Using the curve for the 34M470, DR8010 and
PWR7205,
They are using half the coils at 8.3 A, 8.2 A ^2 * .42 Ohms = 28.9 W per
winding,
X 2 = 57.9 W input static. That apparently is from the same graph as
your top
number, with the 111 W output, if the conversion factor you used is
correct.
I'd like to see if you can get from MicroKinetics the current draw into
the
driver at this setting. Perhaps their driver draws more current when
the motor
is moving.
Jon