Re: [CAD_CAM_EDM_DRO] I need some help with servo's PLZ!!!! RETROFIT

Chris Marek wrote:

>I need more information than that, getting information from Dan is like pulling teeth
>
>
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Ok, as I said in an earlier post, these Ametek motors have about 4.5 Ohms
of armature resistance. That's the killer. As the resistance goes up, the
amount of time the motor can deliver peak torque goes down. On these
motors, the peak torque is available for one second every minute, at
most. If you are really jogging that big table around, you will smoke
such a motor. Small, high performance servo motors in the same
size should have an armature resistance of 1.5 Ohms or less, or forced
air cooling.

Here's a big difference between steppers and servo motors. Steppers deliver
peak torque near zero speed, and lose torque rapidly at higher speeds.
Servo
motors have a peak torque rating (the 600 Oz-In of the Ametek motors)
and a continuous torque rating (much lower, maybe 200 Oz-In on the
same motors, or even less). They can deliver up to the peak torque rating
up to some speed at which the back EMF plus the IR drop equals the
power supply voltage. The back EMF is the voltage produced by the
windings spinning in the magnetic field, and is the velocity constant
Kv times the speed in thousand RPM. So, if the Kv is 50 V/KRPM,
and the motor is turning 500 RPM, the back EMF is 25 V. The IR drop
is calculated from Ohm's law, E=IR. E is voltage, I is current, R is
resistance. If the motor is delivering some torque that requires 5 Amps
of current, and the resistance is 4.5 Ohms, 5 * 4.5 = 22.5 V.
So, at the conditions stated above, the motor needs at least 25 + 22.5 =
47.5 V available from the power supply to develop this torque.
If full torque were needed, that would raise I to, say, 10 Amps, then
the IR drop would be 45 V, and the power supply would need to
deliver at least 70 V, even though the speed has not changed.

Also note that power dissipated is I^2 *R, (only the input power that
doesn't leave the motor shaft mechanically is dissipated as heat).
So, in the above peak load case, 10 A and 4.5 Ohms, that is 450 W!
Obviously a motor not much bigger than a soda can can't dissipate
450 W of heat without a huge temperature rise.

Jon