As you probably know, in the applications that the turn off speed of the solenoid valves are crucial, the simple flyback diode is not effective. Some people put a resistor in series with the flyback diode to alleviate the problem, but for real fast applications the Zener diode is suggested.

You can see it in the picture (third one from the left). Zener Diode protecting a solenoid valve

I think (but I am not sure and please correct me if I am wrong) that the current flows through the loop only when the voltage is higher than the Zener voltage V_z.

What I don't understand is:

  1. What happens to the voltage in the coil that is lower than the V_z? Is it going to remain there? I mean at some point, the voltage drops under V_z and the leg that contains the diode is out! But how can the remaining voltage is going to affect everything in the circuit? and the next turn on command?

  2. The most important question: Is it going to affect the next turn on command in a negative way? For my application I need to turn it on and off 10 times per second (about 5 cycle of on/off)

  3. And what is the trade off between choosing a higher value of V_z against lower value?! Assume it never reaches the switch (MOSFET) safe voltage? Does lower V_z means slower turn off? How can V_z affect everything in positive/negative ways?

FYI, I want to turn Airtec 2P025-08 on/off with an Arduino. 12Vdc, 0.5 Ampere, Don't know the inductance/resistance of the coil!

  • \$\begingroup\$ I'm not an expert of such electromechanical devices, but are you sure that valve can be switched at such an high (for an electromechanical device) rate? I see from the datasheet that it has a minimum activating time of 0.05s. I'm not sure of the exact technical meaning, but it could only mean that when you energize the coil the valve takes 50ms to respond, but this doesn't mean that you can switch it on/off with a period of 50ms. In other words, are you sure the valve can tolerate the wear and tear of so many commutations per seconds? \$\endgroup\$ May 23, 2015 at 13:27

3 Answers 3


Just a bit of preliminary theory.

As you probably know, without any flyback diode, be it a rectifier or a Zener, you'll have a (theoretically infinite) kickback voltage from the inductor (valve coil, relay winding or whatever) whenever you try to interrupt its current abruptly. In reality the kickback won't be infinite because the spike will trigger any sort of nasty effects in the circuit it is connected: it will generate electric arcs, it will drive semiconductors in destructive breakdown, it will fry resistors or punch through capacitors dielectric, etc.

All this in the attempt of get rid of the energy stored in the inductor, which is

\$ E_L = \frac 1 2\, L\, I_L^2 \$

where \$I_L\$ is the instantaneous current at the time immediately before the (attempted) switch-off.

Putting a rectifier in parallel with the coil is the standard low-speed countermeasure, as you know. Assuming the diode can stand the inrush current pulse generated by the kickback, it will clamp the voltage across the coil to a safe ~0.7V. Why is it slow? Because at that voltage level (a diode forward drop) and with usual forward resistance values the power dissipated is low, so it takes more time to convert \$E_L\$ into heat.

Using a Zener is faster essentially because it allows the kickback voltage to rise more before clamping it. Of course the Zener voltage must be chosen not to be dangerous for the rest of the circuit. Since the clamp happens at higher voltage, and the breakdown dynamic resistance of a Zener may also be lower, the dissipated power is bigger, hence it takes less time to convert \$E_L\$ into heat.

If you wonder what happens when the clamp action ceases because the current is not enough to keep the Zener (or the clamp diode) in breakdown (conduction), well the answer is that it will probably oscillate, because the energy MUST be converted, since the power source of the coil has been cut-off, and the stored energy depends on the current in the coil. The coil won't "hold the energy" as a capacitor would do, because for that to be possible a current should flow into the coil itself. Therefore the remaining energy will find other ways to get converted: stray capacitance and leakage current of the diodes and parasitic capacitance of the coil itself (for example). It is sort of a non-ideal non-linear tank circuit, which will exhibit damped oscillations until the energy is completely converted into heat.


(In response to a comment from @supercat)

Here's some results from a hastily conceived circuit simulation using LTspice showing the damped oscillation that may arise in a situation similar to the one described above.

enter image description here

The transient analysis produces the following plots:

enter image description here

If we zoom in the interesting parts we have:

enter image description here enter image description here

In the following extremely zoomed-in plot you may notice the estimated frequency of the oscillations (I've enhanced the image to show where LTspice cursors are placed).

enter image description here

  • \$\begingroup\$ Thank you very much for the excellent explanation. Do you think adding a resistor to the Zener diode and normal diode going to improve its performance? And last question: How do I choose a proper Zener Diode? any rule of thumb!? \$\endgroup\$ May 23, 2015 at 15:19
  • \$\begingroup\$ @arudino.tyro The transistor will be biased "upside down" until Zener opens, so the voltage should not exceed maximal Vce.reverse for the transistor, base-collector current should not exceed maximal allowed etc. Power supply circuit should be strong enough to not be disturbed by the pulse etc. \$\endgroup\$
    – ilkhd
    May 23, 2015 at 16:01
  • \$\begingroup\$ Why would the system oscillate? If the Zener has enough pre-breakdown leakage that there's not enough current to get the voltage up to the breakdown voltage, that means that the Zener is letting current through at a lower voltage. That won't cause the current to fall as fast as it would at a higher voltage, but if the current is that low I don't think it will really matter how long it takes to dissipate the last of it. \$\endgroup\$
    – supercat
    May 23, 2015 at 22:22
  • 1
    \$\begingroup\$ @supercat It may oscillate, or it may not (a difficult problem to analyze), it all depends on how much are the "resistive" leakage effects prevailing on capacitive effects and the effective Q of the tank circuit. That's why I said "...it will probably oscillate...". I concede I should probably have said "it might oscillate". \$\endgroup\$ May 23, 2015 at 22:58
  • \$\begingroup\$ Thanks for all the great questions and answers... How can the Zener voltage vale affect the oscillation?! And is there any way to make this oscillation last sooner? \$\endgroup\$ May 24, 2015 at 14:46

Aaaah, electronics, it is a confusing and cruel mistress.

Makes it fun though.

The thing here is the reaction speed of different components of the problem and/or solution.

First: A diode's forward voltage and forward current are linked. The higher the voltage you can supply across it, the easier the current will flow.

Second: A coil that has a current flowing and then is switched off reacts incredibly fast. If the current can go nowhere within fractions of fractions of a microsecond it can spike up to unbearable voltages (100's, if not 1000's).

So adding a resistor in series is a nice little trick, to slightly tweak the response, it allows the coil voltage to increase a bit further before the diode starts dribbling away the power. But then, the resistor is also in the current path, impeding its own help, so it really is an inferior solution.

The zener diode, however, oh they are magical. Once you reach the breakdown voltage, it really... well.. breaks down! The voltage-current curve of a zener diode at breakdown is much more impressive, this is to do with the compression of the blocking field once current is capable of flowing, if I'm allowed to very badly paraphrase a 380 page book.

So once you reach zener conductance, the current can truly be gone in an instant and as I mentioned, for the coil reaching zener conductance is a piece of cake.

With regards to the zener voltage, the difference in this application between 3V and 6V is more pronounced than the difference between 6V and 12V and so on. Usually the rule of Vz > 2*VCC is good enough to guarantee a fast switch off. More important is that your zener can handle the current spike.

The reason zeners aren't as popular as normal diodes for protection is their current handling capability and destroying your protection device is kind of defeating the purpose a little.

I'll round off now, since I still have to do shopping before venturing into Germany.

EDIT: PS: 10 times a second is not a high-speed requirement. High speed switch off for a relay is in the order of single mili seconds or less. Forgot to make this point at the top before posting. And high-speed switch off will not interfere with a new switch on.

  • \$\begingroup\$ Thank you very much for your response. But what will happen for the remained voltage that is lower than V_z? \$\endgroup\$ May 23, 2015 at 12:00
  • \$\begingroup\$ Also I'd like to know how can higher or lower V_z can affect different things in the performance (both in a good / bad way) ? \$\endgroup\$ May 23, 2015 at 12:01
  • \$\begingroup\$ @arudino.tyro there is no "remaining voltage", but some "remaining energy" (see my answer). \$\endgroup\$ May 23, 2015 at 12:24

In order, your questions:

  1. It will very quickly decay, milliseconds at most. In fact the voltage doesn't go to zero instantly because it is an LC tank circuit mostly with coil distributed capacitance but also stray and transistor capacitance so it will 'ring' at high frequency. The coil has significant resistance so the Q is low and the ringing quickly damps out.

  2. If you wait more than say 10ms it will not affect the next operation in any practical way.

  3. A higher Vz is harder on the transistor but faster turn-off. Turn-on is not affected appeciably (there are other tricks for improving turn-on speed). If you go lower Vz than the maximum possible power supply voltage (worst case) plus a diode drop, the zener diode will conduct when the coil is 'on', probably destroying the zener and transistor. The right-hand circuit does not have that problem (but a sustained overvoltage could cause the Zener diode to overheat).

  • \$\begingroup\$ Thank you very much, Could you please elaborate on number 2 !? \$\endgroup\$ May 23, 2015 at 13:11
  • 1
    \$\begingroup\$ Number two is really a repetition of one. The current dies off very quickly so you're dealing with a starting state of zero current and voltage. The 10ms is a number based on experience of many valves and coils. If course if you had a coil the size of a refrigerator or some other unusual conditions it might have a different behavior. \$\endgroup\$ May 23, 2015 at 13:22
  • \$\begingroup\$ What about the remaining energy inside the coil? Because of the voltage lower than V_z? \$\endgroup\$ May 23, 2015 at 14:06
  • \$\begingroup\$ As I said the energy is dissipated very quickly in the resistance of the coil as it rings at many kHz. It will not hang around like voltage on a charged capacitor. The energy would be gone completely when current = 0 if not for that capacitance. Energy stored in 100pF at 16V (say) is only 0.01J which will power the coil for about 0.002 seconds. In a tank circuit that energy sloshes back and forth about Q times until it's gone. \$\endgroup\$ May 23, 2015 at 14:16

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