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The specs of this Stepper Motor say at a voltage of 3.36 V and phase current 2.1 A, the maximum holding torque is 65Ncm. I'm confused though as on the datasheet there is no voltage mentioned, only a max phase current of 2.1A and same holding torque. I know that voltage applied to a stepper, directly affects torque though. Say I have a 12 V supply to power this stepper, will the stepper draw the same 2.1 A but give a greater torque output or is it saying that 3.36 V is the maximum voltage for this motor? (3.36 V seems way too low to me)

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    \$\begingroup\$ Driving this stepper with raw 12V would yield a great deal of torque for a short time, until the coils burned up from over-current. Some stepper driver circuits might include a current regulator circuit, which could be set to 2.1A - allowing a 2A, 12V DC supply. \$\endgroup\$ – glen_geek Apr 11 at 16:58
  • \$\begingroup\$ @glen_geek is right. The stepper driver circuits usually have current limiters. Some are settable via dip switches. You would program it to never drive more than the 2.1A, but you could still use 12V or even higher so that you can step faster. The inductance of the coils will take up most of the voltage at the beginning of a step so that the current won't exceed 2.1A. \$\endgroup\$ – IanJ Apr 11 at 18:51
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I'm confused though as on the datasheet there is no voltage mentioned, only a max phase current of 2.1A and same holding torque.

But on the line right under the amps/phase of 2.1A, there's a line that says the coil resistance is \$1.6\Omega\$. \$2.1\mathrm A \cdot 1.6\Omega = 3.36 \mathrm V\$, so there you go.

I know that voltage applied to a stepper, directly affects torque though.

This is only correct if the motor is at rest. Stepper motors, like any other permanent-magnet motor, generate a torque that has a component that just comes from the motor construction, and a component that's proportional to current, not voltage. They also generate back EMF. So if you apply a constant-voltage step and move the motor rapidly, the available torque would go down.


How much back EMF you ask? Well, fortunately, you can figure this out. If a motor is perfect, then its torque constant is exactly equal to its speed constant -- this actually falls out of conservation of energy, because voltage * amps = speed * torque. If the holding torque really is only from coil current, then the torque constant is 0.65 N-m / Amp, which means the speed constant is 0.65 Volt / (radians/second) -- so you can work out what the back EMF is at any speed you may be driving the motor.

Note that this only gives you a rough estimate -- stepper motor datasheets usually leave a lot to you to figure out. I'd figure that number is correct to +0%, -20% or so, and if I really needed it to be something specific I'd get a sample motor and test.


Say I have a 12 V supply to power this stepper, will the stepper draw the same 2.1 A but give a greater torque output

That depends. If you use a simple motor driver that just switches the supply voltage, and the motor is moving slowly, you'll drive way more current (around \$12\mathrm V / 1.6 \Omega\$, in fact) through the motor. It'll be way more than it's designed for and the motor will burn up.

Torque will be greater, though -- briefly. Then it'll be much less -- permanently.

or is it saying that 3.36 V is the maximum voltage for this motor? (3.36 V seems way too low to me)

No. The equation isn't "either 2.1A or 3.36V", because the motor generates back-EMF. If you drive the motor with a constant-current stepper driver (which you can design yourself, build from chips designed to do the job, or buy as standalone boxes) then the driver will pulse the motor with 2.1A pulses to the best of its ability. "The best of its ability" means that it'll drive 2.1A up to some voltage that's limited by your supply voltage and the capability of the driver.

That, in turn, means that you'll get maximum torque at any speed up to some maximum limited by the driver and its supply voltage, at which point the torque will drop off.

Using my back-EMF estimate from above, if you're supplying 12V to a driver that has zero overhead voltage (i.e., it can deliver 12V pulses at 2.1A), then you could drive the motor at 120 RPM at maximum torque.

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  1. All torque is proportional to current
  2. BEMF reduces applied V from velocity so holding torque reduces slightly with rising step rates.
  3. Voltage affects current from I= V/DCR(=1.6ohm) V=L(=3mH)dI/dt or I as the integral of voltage.
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