So in an application like this:

enter image description here

The coil of the motor will, by Lenz's Law, induce a very large negative voltage across its terminals (without the diode in place). Would this voltage be very negative at the left terminal of the inductor and very positive at the right? In this way the diode will conduct, but the negative voltage in this case would be directly connected to the power supply.

Also, how are the flyback diodes in this way utilized[L293 - H Bridge IC]:

enter image description here

It seems that the two left diodes will always be reversed biased, but what about the right-side two? Why are the left-side diodes even present if they are always reversed bias? It seems that the large negative voltage would forward bias one of the diodes to the right (whichever terminal is induced to a large negative potential), but what benefit does that have? Also, current flow back into the H-Bridge circuit does not appear very practical.

Perhaps I am looking at this all wrong.


In your first example, before the switch opens, there is a current flowing through L. When the switch opens the current wants to continue flowing in the same direction so this has to create a positive voltage on the right-hand side of L relative to the left hand side (which is fixed at +Vs).

This positive voltage causes current to continue to flow and this current finds the path of least resistance through diode D. It continues to flow until all the magnetic energy in the inductor is gone then current = 0 and the voltage on the right hand side of L is the same as the left hand side i.e. +Vs.

In the 2nd example with the H bridge, all the diodes are needed because it has to cater for scenarios when the motor is deactivated i.e. pin 3 and pin 6 are open circuit. This is the simplest way to explain it I think. The motor current may have previously been upwards or downwards so all four diodes are needed to catch the back emf when the motor is switched off.

Also, current flow back into the H-Bridge circuit does not appear very practical.

That's how they work.

  • \$\begingroup\$ When you say a large positive voltage has to be generated at the right terminal of L, then current would actually be flowing from right to left - unless of course you are talking about electron flow. But even then, the conventional current was flowing left to right. When the switch is opened and the current tends to continue along this path, why is it not that a large negative voltage be generated at the right terminal to facilitate this conventional current flow as it was before the switch? \$\endgroup\$ – sherrellbc Aug 11 '13 at 21:14
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    \$\begingroup\$ @sherrellbc I didn't say large... Anyway, when the switch opens, the current continues to flow through L (the same way) and it finds a path thru D - the current thru D flows from right to left and the current thru L from left to right - it forms a little loop. I don't mean electron flow - I mean conventional current flow. The right side of L has to produce a small positive voltage to overcome the diode drop. Just think that current doesn't want to change direction in the inductor but it has stored energy that it must get rid of and it does so by pushing a current through D. Make sense? \$\endgroup\$ – Andy aka Aug 11 '13 at 21:29
  • \$\begingroup\$ Yes, I was not thinking about the diode loop for a moment. Thank you for your answer! \$\endgroup\$ – sherrellbc Aug 11 '13 at 21:30
  • \$\begingroup\$ With respect to what you had said regarding the first depiction, what will happen, in terms of induced voltage and current, in the second schematic given pins 3 and 6 were open? \$\endgroup\$ – sherrellbc Aug 11 '13 at 21:37
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    \$\begingroup\$ @sherrellbc when pins 3 and 6 open, if current was previously flowing upwards, it will continue to flow upwards through the motor and forward bias diodes top-left and bottom right. The current will flow through the power source and possibly raise the power source voltage a small amount unless it was a battery. \$\endgroup\$ – Andy aka Aug 11 '13 at 21:56

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