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I came across a stepper motor with 10.2kgcm torque, which is ideal for my application. However it says Voltage rating: 3.3V, Phase current: 2Amp

If the current rating is the maximum allowed current. Then how come a 3.3V is supposed to provide such a high torque.

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Torque is a function only of current. The spec is telling you that with 2A, the motor has 10.2 kg-cm of holding torque, and that the available torque for motion will be somewhat less than that. At increased speeds, the current will not reach the desired point (see below), so the torque falls off.

The 3.3V rating tells you how much resistance the windings have (1.65 ohms), specifically that there will be 3.3V volts across the winding with 2A current.

The other important parameter you need to know is the winding inductance, which is the limiting factor (in conjunction with your power supply voltage) on how rapidly the current can be changed, and therefore how quickly the motor can be stepped: $$ dI/dt = V/L $$

A stepper motor requires a current-controlled stepper driver which applies a (perhaps much) higher voltage momentarily to bring the current up to the design point (say 2A), and then uses PWM to regulate the current to that level. When you step the motor, the current in one coil must be reversed, so the high voltage is applied in reverse until the desired current is reached in the opposite direction, and then the controller resumes PWMing again to maintain the current.

You can think of the controller as operating like a buck-mode synchronous switching converter where the motor winding is the inductor and it has a full H-bridge topology so that the current can be driven in either direction.

While it's not stepping, the current is constant and the voltage across the winding will be 3.3V. Say you're using a 24V supply, the PWM duty cycle at constant current would be 13.75%, the coil current 2A and the supply current 275mA.

During a step, there will be significant voltage across one winding (100% duty cycle), as necessary to overcome the inductance and flip the current to the other direction.

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  • \$\begingroup\$ Thank you, I a got it. Just one thing. At some places only voltage rating is given along with coil resistance. Does that mean I am allowed to reach up to voltage/resistance (V/R) current value. \$\endgroup\$ – Nazish Feb 16 '16 at 7:30
  • \$\begingroup\$ Probably. The main limit is thermal. You can put a lot of current in if you don't do it for long but if drawing current continuously then the allowable limit is lower. There is also a limit due to saturation of the core, which is why it's good to see a detailed datasheet. \$\endgroup\$ – William Brodie-Tyrrell Feb 16 '16 at 23:52
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Sounds like you are quoting holding torque- that's a plausible number for a NEMA 23 motor.

Holding torque is the most torque you'll get from the motor- you need a torque-speed curve to evaluate it for most real applications:

enter image description here

(for the above graph, 1000mN-m = 10.2kg-cm)

Note that normally you'd run the motor at a much higher voltage than 3.3V and chop the current (to maintain an average current within ratings) so that it will have acceptable torque when spinning. It's not unusual to use 24V or higher supply to a '3.3V' motor driver. See, for example, this datasheet.

Generally, the curves are like this:

enter image description here

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  • \$\begingroup\$ Thank you. Yes it is NEMA 23. The answer has even cleared my other doubts too. \$\endgroup\$ – Nazish Feb 16 '16 at 7:27

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