I'm working on project 9 of the arduino starter-kit - the motorized pinwheel.

Here is the schematic: enter image description here

The diode in question (and the only one here) is toward the bottom right of the breadboard. It is to prevent the back-voltage from the motor from harming the circuit.

According to the above img, would the backvoltage travel back down the motor's supply of energy (the red wire)?

If so, what is the diode preventing here? That whole energy strip will just have current running through it and the diode's cathode will prevent it from running to the transistor - which would also happen fi we just didn't connect anything to that power strip.

If not, and the back-voltage is running through the black wire (our ground), then I still do not see how the diode is helping. The diode seems to want to pull the extra energy back to our power strip (which is why the anode is facing toward the left, where the motor's output is. But this doesn't prevent the current from still hitting the green wire which leads to the transistor's 'source'.

So how is this diode working?


  • 1
    Think about what the diode would do if it was just the diode and the motor, and you turned the motor in the opposite direction to normal (generated "back EMF"). It's just the same whether the rest of the circuit is there or not. – Majenko Jun 7 '15 at 19:42
  • Thanks for the response. I'm still not understanding this. What happens if we just take the diode away here. Where is the back-voltage going? To the battery? – ZAR Jun 7 '15 at 19:44
  • +1 Excellent question. Something puzzling that seemed not to make sene -> major lesson learned. (Hopefully :-) ) – Russell McMahon Jun 8 '15 at 7:57
  • Please note that only fast diodes are effective there, as most H-Bridges that require external diodes will specify on their data sheets. If you are using a transistor like in your circuit, it's always a good idea to go for the fast version of a diode. For example, don't use a 1N4007, but an UF4007 (Ultra Fast) instead. – nbloqs Feb 12 '17 at 20:14

Under normal operation the + connection of any inductive component (motor, relay, etc) is at a higher potential than the - connection.

During a "back EMF" event (the collapsing of the magnetic field) a large voltage of the opposite potential is generated. That means that the - connection suddenly becomes a much higher potential than the + connection.

That is very bad for anything connected to the - pin of the inductive load.

With the diode connected backwards like that across the inductive load, under normal operation, it does absolutely nothing. However, when the - pin becomes a higher potential than the + pin the diode will become forward biased and start to conduct. That allows a very low resistance path for all the energy stored in the coil of the inductive component to flow from the high potential to the low potential, which is what it wants to do.

Without it, the only way for that energy to get from the high potential to the low potential is through the rest of your circuit, which it won't like.

The excess energy at the - side of your motor wants to get to the + side. That's the only place it wants to go. It'll (crudely speaking*) take the shortest, quickest, route to get there. That means it goes through the diode and nowhere else. Straight from the black wire to the red wire of your motor.

* Note: In actuality the energy will take all possible routes at once, but in varying proportions depending on the resistance of the route. The lower the resistance the greater the proportion. Because the diode is such low resistance it gets the majority of the energy and only a very negligible proportion takes the other possible routes through your circuit. Unless your circuit is incredibly sensitive it won't even notice it's there.

  • This makes sense. Thank you. I have the same question to you as I do with the above answer. Can you check my question in the comments to that answer? After that, I should be good to go. Thank you! – ZAR Jun 7 '15 at 19:55
  • 1
    See my edit. It should cover it for you. – Majenko Jun 7 '15 at 20:00

Flywheel Diode.

When the transistor shuts off, the stopping of the motor generates back EMF (ElectroMotive Force) . This manifests itself as negative potential difference (voltage) across the motor. In plain English the collector of the transistor, connected to the -Ve side of the motor, is raised to a damagingly high voltage. If you include the diode, this voltage is siphoned immediately back to the positive rail, so the damage is averted.

  • This makes sense. I now get that the diode is trying to conduct the extra charge back to the positive rail. Still, won't some of that extra (harmful) voltage still reach the transistor through the greenwire? Or can that simply not happen because of the presence of the diode? – ZAR Jun 7 '15 at 19:55

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