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On a circuit taken from this page http://www.opend.co.za/tutorials/steptut1.htm

The last circuit shows two diodes positioned in reverse polarity to the current.

It is supposed to protect against the inductive kickback of each coil when its driving transistor is turned off.

But when one transistor is closed and the diode is not connected in parallel to the coil (as most flyback diodes I've seen) how can a closed circuit be formed between the two sides of the coil to discharge it?

enter image description here

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  • \$\begingroup\$ The answer is on the website: D2 is protecting Q1. \$\endgroup\$ – Huisman Apr 21 at 13:49
  • \$\begingroup\$ @Huisman Yes, i read is before posting, but i still don't understand how can a closed circuit be formed between the two sides of the coil to discharge it \$\endgroup\$ – soundslikefiziks Apr 21 at 15:08
  • \$\begingroup\$ You CANNOT put the diode in series with the coil. The schematic is correct and the closed circuit INCLUDES the power supply. \$\endgroup\$ – Jack Creasey Apr 21 at 15:40
  • \$\begingroup\$ @JackCreasey , if coil A is energized when Q1 is open, then when it's closed, coil A would be in series with the diode D1, still energized by the magnetic field, in that case where would that current go if the diode is connected in such a way ? \$\endgroup\$ – soundslikefiziks Apr 21 at 15:49
  • \$\begingroup\$ @soundslikefiziks If coil A is energized then Q1 is ON ...it cannot be energized otherwise. When Q1 is turned OFF, the back EMF is reflected on BOTH D1 and D2 (the two coils are effectively a transformer and usually bifilar wound). So as the voltage across D1 rises to VCC (and beyond) D1 will not conduct …..but the voltage across D2 will drop (transformer action) and drive the cathode of D2 below Gnd where it will conduct. Coil B and the diode are in series only for the back EMF with the circuit completed by the power supply (usually with large output capacitance. \$\endgroup\$ – Jack Creasey Apr 21 at 16:05
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The answer to your question really depends on what type of stepping motor you are using.
In this case from the web link you provided they are discussing a unipolar stepping motor where the phase coils are bifilar wound and act very much as a transformer.

Here is a schematic with the simulation so you can see the waveforms.

schematic

simulate this circuit – Schematic created using CircuitLab

Waveform

Notice that the back EMF is clipped by the diode on the OPPOSITE transistor to the one just turned off. So D2 clips the back EMF caused by Q1 turning off and D1 clips the back EMF when Q2 turns off.

The website is a very poor tutorial since no attempt is made to clarify the differences between stepper motor configurations. The technique for example could not be applied to a 2 phase single winding stepper motor. It uniquely applies to a dual wound (or center tapped) winding configuration driven as a unipolar stepper.

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  • \$\begingroup\$ Yup...the circuit only applies to 4-phase step motors with center-tapped windings. \$\endgroup\$ – whitegreg56 Apr 21 at 20:32
  • \$\begingroup\$ Phases A and C are 100% coupled. Note the dot convention! \$\endgroup\$ – whitegreg56 Apr 21 at 20:40
  • \$\begingroup\$ When phase A shuts off, the current in phase C jumps to the same value....except negative! Eventually, the current in phase C goes positive. On the next step, this is repeated except with phases B and D. \$\endgroup\$ – whitegreg56 Apr 21 at 20:50
  • \$\begingroup\$ @JackCreasey i tried to simulate this circuit on an online simulator falstad.com/circuit/circuitjs.html , because it shows a live simulation of the current flow. i used two coils instead of a transformer, and when Q1 is turned off, current flows from ground through D1 and Coil 1 , and not D2. i then removed Q2 L2(the 2nd coil) and D2 from the circuit and the same thing happened. when Q1 is turned off, current flows through D1 , from ground to 12v. this is the opposite from what i know about inductors and diodes .. current flowing from low to high ? \$\endgroup\$ – soundslikefiziks Apr 23 at 21:29
  • \$\begingroup\$ @soundslikefiziks You cannot use two independent coils as falstad know nothing about the coupling coefficients. \$\endgroup\$ – Jack Creasey Apr 24 at 3:02

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