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Looking at this web page, i found a simple stepper motor driver circuit https://www.electronicshub.org/stepper-motor-driver-circuit/

The problem is, i just can't understand how it can work when the diodes are reversed. And even if they were not reversed, i still cannot see how any of the coils be energized when one of the pins on the CD4017 counter turns on.

At first i thought the circuit must be flawed until i saw the video showing it does work, with the reversed diodes. Video : https://www.youtube.com/watch?v=ejvTG90kA0k

enter image description here

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2 Answers 2

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The schematic seems to have an error.

The center taps shouldn't just be connected to each other, they should also connect to the +12 V supply.

On your linked page, this is confirmed in the text:

The Stepper Motor is a Unipolar Type in 5 wire configuration. The center pin is shorted internally and is connected to the supply (12V here).

And in another diagram on that page:

enter image description here

And on their breadboard:

enter image description here

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  • \$\begingroup\$ Yes, you are right, i see it now, so what is the purpose of those diodes then ? \$\endgroup\$
    – soundslikefiziks
    Apr 21, 2019 at 5:24
  • \$\begingroup\$ They are flyback protection diodes. You can fin lots of questions about this configuration if you search the site. \$\endgroup\$
    – The Photon
    Apr 21, 2019 at 14:18
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As already pointed out the schematic does not show the phase centers connected to 12V.

But another weakness in the driver configuration is that this does NOT drive the stepper motor to achieve the best torque.
The schematic as shown activates the phase coils in the following manner:

enter image description here

Notice that there is only ever one coil being activated for each position. This severely reduces the torque from the stepper motor which makes it easier to miss steps when driving a load.

The best torque is achieved when two of the winding are activated together:

enter image description here

You can read about this here.

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  • \$\begingroup\$ Also, not all of the 4017 pins are connected , so the timing would be off. The H-bridge configuration on the website you added seems much better. \$\endgroup\$
    – soundslikefiziks
    Apr 21, 2019 at 5:52
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    \$\begingroup\$ @soundslikefiziks NO, the timing is NOT off. The 4017 uses feedback from Q4 to reset the counter so it counts from 0 to 3 continuously. What this very restricted implementation does is only count in one direction, so is of limited value. The 4017 counter cannot be made to count down. A much better implementation is to use a small MCU, which could be a true single chip solution, but you'd need a regulator to drive the MCU since it would not tolerate 12V VCC. \$\endgroup\$
    – Jack Creasey
    Apr 21, 2019 at 14:31
  • \$\begingroup\$ @JackCreasey Some of them can. \$\endgroup\$
    – Hearth
    May 9, 2019 at 1:56
  • \$\begingroup\$ @Hearth I sort of agree ….the MCUs you point to have on board regulators ….they do NOT operate at the high voltage, so I think my comment still stands. \$\endgroup\$
    – Jack Creasey
    May 9, 2019 at 5:18

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