<p>I just got my first 3D printer today, QIDI X-ONE[2], and so far so good with the setup and getting my 1<sup>st</sup> print. I wanted to power off the printer, but I don't see any instructions on how to properly power off the machine.</p>

<p>Does anyone know how long I should wait, or what the minimum temperature would be safe to power down the machine? </p>

[{'answer_id': 5556, 'author': 'Tom van der Zanden', 'author_id': 26, 'author_profile': 'https://3dprinting.stackexchange.com/users/26', 'pm_score': 2, 'selected': False, 'text': '<p>Kinda, sort of, but not really. I\'ll look at the A4988 (<a href="https://www.pololu.com/file/0J450/a4988_DMOS_microstepping_driver_with_translator.pdf" rel="nofollow noreferrer">datasheet</a>).</p>\n\n<p>The motor pins are connected by diodes to ground and Vbb (the motor suppply voltage). Essentially, they act as a bridge rectifier making any back EMF or inductive spikes appear (rectified) on Vbb. If you were to suddenly power down the driver this could cause a rather large spike on Vbb.</p>\n\n<p>According to the datasheet, there is a 40&nbsp;V Zener on Vbb which will clamp the voltage to that level. (Another popular stepper driver, the DRV8825, does not appear to have this Zener - always check your datasheet!)</p>\n\n<p>So, yes, there is inductive kickback protection. However, it only clamps the voltage to 40&nbsp;V. Depending on the rest of your circuit, this could be quite damaging.</p>\n\n<p>The datasheet recommends that a 100&nbsp;&mu;F capacitor be placed on Vbb. If you are driving a typical stepper motor with 2&nbsp;A and 4&nbsp;mH coil inductance, the energy stored in the coil is 8&nbsp;mJ. This energy is only enough to take the capacitor up from 12&nbsp;V to ~17.5&nbsp;V, so if you have a large enough capacitor on your stepper driver (as you should!) then you\'re protected against inductive kickback.</p>\n\n<p>Note that if you move the motors by hand then you can still build up a higher voltage on Vbb. I\'ve heard anecdotes of people who damaged their printers like that.</p>\n'}, {'answer_id': 10297, 'author': 'cmm', 'author_id': 2082, 'author_profile': 'https://3dprinting.stackexchange.com/users/2082', 'pm_score': 0, 'selected': False, 'text': '<p>"Inductive kickback" from motor coils is caused by the inductance of the coils and the wiring.  It is not some strange effect.  Inductance is charged with current, just as capacitance is changed with voltage.  Most bipolar motor drivers use an H-bridge on the output.  By driving the H-bridge correctly, the driver can continue to provide a path for the motor current when it is no longer applying voltage, such as by connecting both wires of a motor to ground.</p>\n\n<p>Although some may find it counter-intuitive, connecting the Vmot side of the motor to Gmot, while the other side remains at Gmot, keeps current flowing in the coil more smoothly and with lower losses.  The voltage across the motor coil is near zero, so there is little voltage trying to change the motor current.  If the coil is undriven and subject to the clamping diodes, the voltage across the motor will me much higher, and the current will stop faster.  </p>\n\n<p>The higher ripple current in the motor increases the coil heating, may increase audible noise, and decreases the efficiency of the drive.</p>\n'}]
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