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We are using this stepper motor driver A3967(https://www.sparkfun.com/products/12779). If we increase the input current to the driver to 2Amp, will it cutoff the current to the motor at 750mA everytime since it is a chopper driver?

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You can't increase the input current, the driver will sink the needed amount which in your case is 750mA (per phase IMO, so 1.5A total). You have to provide a stable voltage source that can source at least that amount of current, higher is better.

Q: Will the chopper driver limit the current efficiently anytime?

A: Well, it depends on the setpoint speed and supply voltage. The motor is always producing back EMF voltage - generator mode. As long the supplied voltage is greater than motor BEMF voltage, then it will be able to maintain the current at setpoint level. At higher speeds the motor BEMF voltage becomes high, almost the same as supply voltage. In this situation the potential difference is zero and it is impossible to feed the motor with such current. The motor looses torque at high speed. Higher voltage -> higher max. motor speed.

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  • \$\begingroup\$ This is incorrect. The BEMF may not have any impact at all. Consider that the stepper motor is stationary. The current limit set for the driver (through the detection resistor) is used to ensure the peak current does not exceed the setting. The current supply is via a PWM and the average current through the phase is going to be much less than the peak setting. Marko is correct that there is an effect when the motor is rotating at high speed, but that doe snot impact the peak current detection ...only the ability of the current to reach the peak at the switching frequency. \$\endgroup\$ Jul 20, 2017 at 19:09
  • \$\begingroup\$ @JackCreasey A stepper motor driver has a setting on current, not a peak current. When the motor is standstill (0 rpm) and switched on the current is nominal (as set), meanwhile the voltage is \$I_{set}\cdot R_{winding}\$. When the motor rotates the current is still the same (as long it can be sustained), but the voltage on the motor has to be\$I_{set}\cdot R_{winding}+ V_{BEMF}\$. \$\endgroup\$ Jul 20, 2017 at 19:22
  • \$\begingroup\$ Incorrect, for the A3967, the current detection is the PEAK current for the switcher. Read page 5 of the datasheet ...Internal PWM current control. If you were to set the trip current at 700 mA, the average or RMS current through the winding would be less than the current value you set. When the motor is stationary there is no BEMF. \$\endgroup\$ Jul 21, 2017 at 0:02
  • \$\begingroup\$ You can also look at Table 2 which shows the winding (phase) current relative to the Itrip (peak current or 100% current depending on language you want to use). \$\endgroup\$ Jul 21, 2017 at 0:11
  • \$\begingroup\$ @JackCreasey It's the way the chopper works, at certain point (Itrip) the transistors are switched off and than back on when the current falls under threshold value. Keep in mind that current through the winding is always continuous, not discrete, so the RMS value is almost the same as Itrip. The datasheet doesn't mention it , so we can say that current is Itrip - this is the goal of the driver IC. Table 2 doesn't represent RMS current vs Itrip, rather phase current vs step. The stator current is split in phase A and phase B, such that the.... \$\endgroup\$ Jul 21, 2017 at 7:35

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