What is the importance of using the diodes shown in this Bipolar Transistor H-Bridge Motor Driver?

Bipolar H-bridge motor diode with flyback diodes

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    \$\begingroup\$ Perhaps a bit less dependence on assumed telepathic skills of site participants would have helped deliver a more comprehensible question. Voting to close. \$\endgroup\$ – Anindo Ghosh Jan 11 '13 at 17:30
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    \$\begingroup\$ I think the question must be about flyback diodes. If you try posting your question again with a schematic, or a link to one, it will be received more favorably. \$\endgroup\$ – Phil Frost Jan 11 '13 at 17:35
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    \$\begingroup\$ @PhilFrost Ideally the information will be added in to this question and then we can vote to reopen the question \$\endgroup\$ – W5VO Jan 11 '13 at 17:41
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    \$\begingroup\$ I'm astounded that this would be closed. Of course the diodes are flyback diodes, however the poster probably doesn't know that (hence the question). And no, not everyone has a schematic they can link to, or knows how to post it - nor for such an elementary question should they need to. However, if someone wanted to post an answer with a diagram and explanation of an H bridge and properly applied diodes and an explanation, that would be a great contribution from those who have expertise in the topic and the website... \$\endgroup\$ – Chris Stratton Jan 11 '13 at 19:34
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    \$\begingroup\$ @ChrisStratton The perfect solution here was for someone to edit it and clarify what they think the user is asking and then we reopen and they can let us know if it does not match what they need. I do agree that users came off a bit more rude then acceptable when they thought they were being funny. \$\endgroup\$ – Kortuk Jan 11 '13 at 20:28

The motor coils have a significant inductive component. Very loosely and from the point of view of a circuit only (not how the physics really works), inductors can be thought of as giving inertia to current.

Imagine what happens when a current has built up in a inductor, then a switch in series with it is suddenly opened. The current can't stop instantly (hence the inertia analogy), so will initially cause whatever voltage is necessary to keep it going. In this case, that will be causing a arc between the switch contacts. The greater this reverse voltage, the faster the current dies down. Eventually it gets to zero, the arc stops, and the switch if truly open. This may take µs to ms depending on the value of the inductance, the amount of current thru it, how fast the switch contacts can separate, etc.

The same principle applies whether the switch is a mechanical contact or a transistor. Transistors can turn off very quickly, so the high voltage caused by the inertia of the current the transistor is trying to stop builds up very quickly as well. It can easily build up to levels that will damage the transistor, or something else connected to the same line. The diodes provide safe conduction paths when this happens. They give the current a place to flow so that it doesn't have to cause a high voltage to keep flowing.

Another use for the diodes is to deal with permanent magnet motors that can be driven externally. Such motors work backwards as generators. If driven too fast, they can cause too high a voltage just from the generator action. The diodes again provide a path for this current to go so that the voltage can't get much past the power voltage. Note that this makes the motor essentially drive the power rail instead of the other way around. If the motor can be driven like that, this needs to be taken into account in the design of the larger system.

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