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I have just been cogitating on the tutorial at http://www.electronics-tutorials.ws/io/io_5.html, and in the discussion of flywheel diodes it includes this sentence without further elaboration:

As well as using flywheel Diodes for protection of semiconductor components, other devices used for protection include RC Snubber Networks, Metal Oxide Varistors or MOV and Zener Diodes.

I can kind of see how an RC network might be needed if it is a large device and therefore the coil could be kicking back more current than you want to dissipate through a single diode. (Please correct me if that's not the reason.)

I don't have a clue what an MOV is so for the moment I'll ignore that one. :-)

I have read a bit about Zener diodes, but I don't understand why their lower reverse breakdown voltage might be desirable here?

Edit: I'm also puzzled by the following diagram from the tutorial above:

enter image description here

Wouldn't this take any flyback voltage and dump it into the Vcc net? Would it not be a better idea to have the relay coil be between TR1 and ground, and the diode dissipating the flyback voltage to ground?

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The current from the relay opening doesn't go into the Vcc rail at all. It follows the path shown here:

enter image description here

The stored energy is dissipated in the diode drop and the coil resistance of the relay.

In the Zener diode configuration, the stored energy is dissipated in the full Zener voltage of the diode. V*I is a lot higher power, so the current will fall faster and the relay might open a little faster:

enter image description here

MOVs are different than Zeners, but fulfill a similar circuit function: They absorb energy when the voltage exceeds a certain level. They are used for overvoltage protection, not for precision things like voltage regulators.

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    \$\begingroup\$ Nice explanation of the Zener config… Makes perfect sense when it's spelled out! I still don't understand why (in a larger circuit with more components) current from the de-energizing coil wouldn't go all over the board on the Vcc net though. Isn't it seeking a better path to ground? \$\endgroup\$
    – Kaelin Colclasure
    Feb 21, 2012 at 0:59
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    \$\begingroup\$ No, remember ground is just a convenient name for "zero volts". If you like to anthropomorphize current, consider the current in the inductor when the transistor opens: it wants to keep going even after the switch opens and will generate as much voltage as necessary to make that happen. The various snubbers are just providing convenient paths for that current. \$\endgroup\$
    – markrages
    Feb 21, 2012 at 4:14
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    \$\begingroup\$ A good way to study snubber circuits is to study switching regulator circuits. \$\endgroup\$
    – markrages
    Feb 21, 2012 at 4:15
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    \$\begingroup\$ +1 for drawing currents into the schematics, in red. Getting used to thinking in currents rather than voltages seems to be less common and may need some effort, but it is often so much easier... \$\endgroup\$
    – zebonaut
    Feb 21, 2012 at 6:53
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    \$\begingroup\$ @zebonaut: When considering current, it's often helpful to think in terms of electrons. If a buck mode switcher is stepping voltage down 3:1, ignoring inefficiencies, the reason it can output three times as much current as it's taking in is that each electron coming from the supply will, on average, get to go through the load three times (going through the recirculating diode twice to skip trips through the supply). \$\endgroup\$
    – supercat
    Nov 18, 2012 at 15:50
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The rate at which the magnetic field will collapse in a solenoid, electromagnet, or similar device when power is removed will be proportional to the voltage which is allowed to appear across the device. If one operates a 12-volt solenoid or relay with a push button and no flyback protection, releasing the button may cause hundreds or thousands of volts to appear across the coil until the field collapses; because of the large voltage on the coil, however, the field would collapse almost instantly.

Adding a simple catch diode will prevent any significant voltage from appearing on the solenoid or relay when it is released. It will also, however, cause the coil to remain magnetized for much longer than it otherwise would. If it would take 5ms for the magnetic field in a relay coil to reach full strength at 12 volts, it will take about 17 times that long, (i.e. 85ms) for it to dissipate through a catch diode. In some situations, that could be a problem. Adding some other circuitry to drop voltage can allow the coil to de-energize much faster.

BTW, if one is switching many 12V relays frequently, I would expect that one could save a fair amount of energy by having the clamp diodes charge a cap and then taking energy from that cap for some other purpose. I'm not sure whether or where that's done, but in something like a pinball machine it would seem like it might be a useful concept.

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    \$\begingroup\$ The idea of storing the energy through the clamp diode is brilliant. \$\endgroup\$
    – abdullah kahraman
    May 1, 2012 at 20:11
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    \$\begingroup\$ a handy trick is to use the relay coil as the inductor of a flyback convertor. e.g. use the 5V relay to make a 12V supply. Another trick is to use a small signal relay attached to a battery operated microprocessor pins, to pull the processors VDD up, so that there is enough voltage to reliably switch the relay itself. \$\endgroup\$
    – Henry Crun
    Jul 27, 2018 at 23:50
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    \$\begingroup\$ It actually takes the same time, not 17x. Most of the dissipation being in the coil R, not in the diode. see simulation in my answer. The real issue is that its an RL exponential decay, and the relay may not release until 20% current is reached \$\endgroup\$
    – Henry Crun
    Jul 31, 2018 at 11:04
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The zener diode would normally go in series with the freewheel diode, cathode to cathode (pointing at each other). This causes the voltage to collapse faster and therefore the coil field will collapse faster and therefore the relay/solenoid will open faster. In switch mode power supplies (SMPS) this is also known as a zener snubber.

schematic

simulate this circuit – Schematic created using CircuitLab

See also this question/answer: zener diode question

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  • \$\begingroup\$ @Henry Crun, How did you shrink the drawing? \$\endgroup\$
    – Aaron
    Oct 9, 2019 at 20:54
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Some of these answers are confused about what happens with a simple diode. The energy is dissipated primarily in Rcoil, not in the diode.

Key is that when using a diode, the dissipation is an RL exponential decay (like RC). It is being exponential that makes it take so long (especially since release current might be only 20%). With a zener it is a linear drop to zero.

This simulates a real relay from its datasheet values of R and L.

schematic

simulate this circuit – Schematic created using CircuitLab

You will notice that the ON (current rise) time is longer that the off time using a diode (L1,D1).

This is not correct as the inductance is greater (0.74H) when the relay armature is closed (better magnetic circuit) that when open (0.49H). The real On time (with 0.49H) and the off-time with a diode are almost the same.

enter image description here

L2,L4 currents are the same, as there is the same drop in both cases (and the same Vdrain on the fet.

ignore this

schematic

simulate this circuit

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Here is an app note about using normal + Zener diodes to protect components and still de-energize quickly. It shows decay time and voltage values for several methods.

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    \$\begingroup\$ Link-only answers aren't great. It would be a better answer if you could add the key points to the actual answer. \$\endgroup\$
    – pipe
    Mar 31, 2017 at 20:15

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