What do zener diodes facing opposite each other do? 1

[Thinksalot]

I have a solenoid circuit with two zener diodes faced opposite in parallel with the coil. I don't understand the function in a DC circuit of this arrangement. Any insight would be helpful. I know in an AC circuit, two zeners facing each other, rather than opposite each other, will clip the AC voltage. I do not know if the order of zeners makes a difference in this application, either. Insights would be appreciated.

 

[nbucska]

I think what you mean the two diodes -- connected in series
back to back -- are parallel to the coil.

Well, one diode clamps the high voltage generated when the current of the coil is turned off while the other one is
forward biased with about .7 volt drop. The coil voltage
can reach only the Vz + Vpowersupply .

When the coil is energized, the first zener is forward,
and the second one reverse biased. The second one is selected so its breakdown voltage is HIGHER then the
coil voltage. <nbucska@pcperipherals.com>

 

[melone]

However, if it is two zeners facing opposite directions, it could be an attempt to clamp the flyback voltage at a specific point. It just sounds like either the person who wired the circuit either had extra zeners sitting around, or ran out of standard diodes.

Regardless, the purpose for these diodes is for inductive flyback. Whenever you energize an inductor (a solenoid coil is an inductor), and quickly remove either the high or low side power connection, a voltage spike occurs. The spike characteristics are based primarily on a couple of factors: which side was dropped out, and how big is the inductor. If you drop out the low side of the inductor/coil, you will get a positive spike and if you drop the high side, you will get a negative spike.

Spike size is based primarliy on inductor/coil size. The larger the inductor, the more energy must be disapated once &quot;power&quot; is removed. The neat thing about these inductors is that current can not change instantaneously though them. Therefore, when the current path has been broken, the inductor has stored charge and must get rid of it. Since no current path exists, the inductor changes the extra charge into voltage & heat to disapte the energy. Unfortunately, any other devices that are connected to this coil will see the same voltage spike. Therefore, some protection is needed. That is where the diodes come ine.

If you place a diode in the opposite polarization of the coil and drop out the power, a current path through the diode exists. Therefore, the energy stored in the inductor has a path to discharge. Also, the voltage spike is clamped to the turn-on voltage of the diode (~0.7V). This protects the additional circuitry from excess voltage.

Why the circuit designer decided to clamp the voltage spike at a specific voltage is a mystery. Perhaps another circuit is connected to monitor whether the coil ever turns off (using the spike as a trigger). Or possibly both sides of the solenoid are being controlled (don't know which side is going to drop out) and must protect for both positive or negative spikes. Take a look at the additional circuitry and see if you find anything.

I hope I haven't insulted anyones intellegence and told you what you already know. But some of this stuff isn't always taught in school, and if you have never played with it before, it can be a little confusing!

 

[nbucska]

About the &quot;mistery &quot;:
The clamping diode slows down the turning off the solenoid -- with the Zener, the higher voltage Zener gives higher speed.

<nbucska@pcperipherals.com>