Welcome! Log In Create A New Profile


nema17 motor power

Posted by ntar827 
nema17 motor power
July 03, 2012 05:00PM
What is a reasonalble total weight for an assembly that a nema17 motor can move at the required speed of a printer?
Re: nema17 motor power
July 03, 2012 05:51PM
The weight determines the acceleration, not the speed.

Re: nema17 motor power
July 03, 2012 07:35PM
speed is determined by pulses per second.look at the motor torque curve for pulses per second. higher voltage always increases pulses per second, because there is enough potential to overcome the motors resistance at higher speeds. think of the motor as an inductor or a synchronous AC motor.[en.wikipedia.org]

my preference is to get a lower voltage high current motor, rather than a high voltage lower current motor. the reason is 12v-19v is enough to compensate for the change in inductance. the danger with this is reprap electronics are not high enough current, and you really need to know what you are doing or the 200-400$ electronics will not work anymore. curiosity can kill your budget here.

what i will state is one fact: increased voltage beyond voltage rating of motor, with proper chopper circuitry improves performance of motors, as long as heat is properly dissipated in the motor and in the controller circuitry.

Other than that your heading into a theoretical train wreck, so i would just purchase the standard motor that everyone agrees works OK. here is a reference. [reprap.org], then experiment later on in life when you know everything works as it should.

with that said, one of the main drawbacks to speed is overcoming the inertia properly. mass motors need to move, tension of belts, straightness and smoothness of rods used, type of grease and types of bearings used make as much a difference for performance as increasing torque would.
Re: nema17 motor power
July 04, 2012 03:46AM
I can attest to the low V vs high V motors. I bought some cheap 12v low amp motors, the only way to get enough torque out of them is to turn up the current pots and run them so hot the plastic mounts melt. I assumed that the voltage wasnt as much of an issue as the total watts, aparently i was wrong. High voltage and low amps can make the same amount of watts as low voltage and high amp. Im no electrics guru so i still dont get why low volts and high amps makes less heat than high volts and low amps. Watts is watts i assume but what do i know.
Re: nema17 motor power
July 04, 2012 04:43AM
The temperature rise is directly proportional to the power regardless. A low torque motor has to be run at a higher temperature to get the same torque as a high torque motor. High torque motors are bigger and so can use thicker wire for the same number of ampere turns so they have lower resistive losses and hence are more efficient.

Re: nema17 motor power
July 04, 2012 05:00AM
@GITRDUN: Microstepping delivers switched current to motor coils. As many reprapers use 12V power supply for their motors, a lower than 12V rating is needed for the motors. This way, current can actually be regulated by the switching action. If you use a 12V motor then this limits the maximum current that can be achieved. As far as the power supply voltage is 12V you cannot exceed the current stated in the motor specs.

However if you use a lower voltage motor (lower coil resistance) the switching action on the motor driver can be adapted to the desired current (more time on than off) all the time.

Anyway, a lower power motor will heat up more than a higher power rated one for a given delivered power. It is said that you want at least your power supply voltage to be four times your stepper nominal voltage (when using constant current drivers).
Re: nema17 motor power
July 04, 2012 12:24PM
nophead Wrote:
> The temperature rise is directly proportional to
> the power regardless.

The temperature rise is proportional to the dissipated power (I^2*R), but not the input electrical power (I*V), because a lot of the voltage drop comes from the back-emf of the motor as it is spinning. A higher voltage source can drive a motor at the same current as a lower voltage source (so the same torque), but at a higher speed, with the same resistive losses.
Sorry, only registered users may post in this forum.

Click here to login