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I have on hand a simple 12V/350mA stepper motor

enter image description here

the datasheet is lacking, so I can only assume 12V is the maximum sustained voltage and 350mA is again, the maximum associated current. The datasheet also lists the winding resistances as 35 Ohms, which I have verified.

What I found is that while driving the stepper at a rate of about 0.5 revolutions per second that it would draw 350mA at 12V. However, if I were to decrease or increase the speed of rotation, the motor would draw more or less current, respectively. Why is this? I am planning on testing this later today, but is it then necessary to drive the motor at a lower voltage to abide by the current constraints?

In summary for 12V supply:

0.25 rev/sec => 450mA

0.5 rev/sec => 350mA

3 rev/sec => 150mA

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  • \$\begingroup\$ How are you measuring the current? How many phases it has? Maybe the current is specified per phase? If the resistance of single phase is 350 Ohm, it will actually draw around 350mA per phase when stall. But if you have two of these, obviously you will get more. BTW, the link you are giving is giving error 403. \$\endgroup\$ – Eugene Sh. Jan 21 '15 at 17:06
  • \$\begingroup\$ If the voltage is staying constant at 12V, the current has to increase/decrease to change the speed. Voltage and current have an inverse linear relationship, so if you decrease the voltage you'll need more current not less (for same speed + torque)... @Eugene Sh. - reload the page and the error will clear and page will load. Also it does say 350mA per phase current with 2 phases \$\endgroup\$ – I. Wolfe Jan 21 '15 at 17:08
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    \$\begingroup\$ At higher step rates, inductance will reduce the current. Your numbers indicate that your motor has a high winding resistance and thus certainly high inductance as well. It will perform quite poorly at speed. \$\endgroup\$ – Chris Stratton Jan 21 '15 at 17:44
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When driving a stepper motor, the maximum torque is proportional to current, and it is your job to regulate the current. What you're doing is providing a fixed-voltage drive to the motor, with unregulated current. With the motor commanded to a stop, only the resistance of the windings will determine the current.

When the motor is not stationary, as the rotational speed increases, given a fixed voltage the currents will decrease. This happens since the motor windings are inductances: the currents will decrease the faster you switch them.

The only workaround is to use a higher drive voltage with a current-mode controller. In most cases you don't want to exceed 48V drive voltage so as not to break down the insulation on the motor's windings. The current will be limited in all cases by your current-mode controller. A current mode controller can be a simple linear current source, or it can be a more efficient switched-mode current source.

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  • \$\begingroup\$ I plan on a DRV8811 stepper driver to control the motor in the end. These tests were to determine the appropriate current limiting and ramp times in order to select passive components to configure the operating point. \$\endgroup\$ – sherrellbc Jan 21 '15 at 17:59
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    \$\begingroup\$ It's not worth spending money on drive electronics until you get a lower inductance motor. That one was intended for fairly static positioning. \$\endgroup\$ – Chris Stratton Jan 21 '15 at 18:39

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