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Diode on relay coil - downside.

Diver300

Well-Known Member
Most Helpful Member
I've got an automotive relay that contains a resistor in parallel with the coil, and I measured what effect having a freewheel diode would have.

The relay is standard cube one. The coil is about 120 Ohms, and has a 680 Ohm resistor in parallel, so the total resistance is around 98 Ohms and it takes about 115 mA when supplied with 12 V. I supplied the relay through a pushbutton switch and I had a 1 kOhm pull-up on the contacts to monitor how long the contacts took to open.

Here is the waveform:-
relay_resistor.PNG

The voltage peaked at around 65 V and the relay took about 2.6 ms to open. It's interesting to see that the current increased slightly as the armature moved.

With a diode in parallel with the coil, it's very different:-
relay_diode.PNG

The voltage stayed at 0.7 V and the current took 25 - 30 ms to die down. It took around 8 ms before the contacts started to open. I found it interesting that the contacts bounced for around 2 ms with at least two false openings before the finally opening. The bounce was quite consistent and happened much like that every time the relay released.

My take from this is not to use a diode to suppress the inductive spike on a relay if the contacts are taking anywhere near their maximum rating. Any sort of heavy load or inductive load on the relay is going to damage the contacts more than turning the relay off quickly. The coil suppression almost inevitably slows the relay release, so the amount of suppression is a compromise between protecting whatever is controlling the coil and protecting the contacts.

Of course, some suppression is vital in most applications. This is what it looked like when I disconnected the resistor:-
relay_nothing.PNG

It's even faster than with the resistor but the inductive spikes are huge.
 
Say you have a 12V relay that takes 12V and 10mS to magnetize the relay. Then it will take 120VmS to undo the magnet. With a diode (0.7V) it will take a long time. Letting the coil fly up to 120V will take only 1mS.

It could be the car was designed for a 100 ohm relay, but someone make a 120 ohm relay, so they added a 680 resistor.
If you were to add the diode with a 100-to-120-ohm resistor in series, the coil will flyback to 12V over the supply. (approx. 12V across the resistor + 0.7 of diode) This gives you fast response and limited high voltage.
1738005333747.png
 
Indeed; for the fastest opening of contacts, the diode-relay combination is the best, as long as the transistor can handle the additional voltage stress.
Like everything in engineering, there are trade offs.
 
Indeed; for the fastest opening of contacts, the diode-relay combination is the best, as long as the transistor can handle the additional voltage stress.
Like everything in engineering, there are trade offs.

Actually, a diode-Zener in series is the fastest.

Below is an LTspice sim for a 1H, 120Ω inductor for three voltage suppression configurations: a diode, a 130Ω resistor-diode, and a 12V Zener-diode:
The diode give about a -0.7V transient, and both the resistor-diode and the Zener-diode give about a 15V peak transient.

So for the current to drop to 10% of the initial value requires 25.7ms for the diode (red trace), 18.3ms for the resistor-diode (purple trace), and 14.7ms for the Zener-diode (amber trace).

So there's the trade-offs of cost/complexity vs. decay/turn-off time.

1738014794176.png
 
A compromise is to put a zener diode across the coil. This lets the coil free-wheel for a bit before being clamped, so the contacts open faster but the voltage spike is not high enough to damage anything. An example would be a 24 V zener in a 12 V circuit.

P&B has an app notes on this.

ak

ps. Carl beat me to this, but only because a honey-do came up.
 

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This lets the coil free-wheel for a bit before being clamped
Don't understand what you mean by "free-wheel". :confused:
To me that's what just the diode across it does by clamping it to a near zero voltage (hence the term "free-wheel diode").

In contrast the Zener immediately clamps the inductor to a near constant Zener-plus-the-diode voltage to slow the inductive current as rapidly as possible for that value of peak clamp voltage.
Notice the near linear reduction in current versus time for that clamp in the sim.

Those honey-do's can be pesky.
 
Last edited:
We use VDRs on inductive loads than need a fast shut-off.

You can select a clamp voltage to stay within the driver transistor limits, while dissipating the stored energy rapidly.

(I'm well aware of claims that VDRs have limited life - but as long as the peak energy is well below their ratings so they never heat up much, they seem to last forever - we've got units in service for 30+ years without failures other than from external overloads).


A lot of older machine tools use wirewound resistors across electromagnetic clutch coils as snubbers to limit the peak voltage and reduce contact arcing, which unfortunately many maintenance people do not understand and ignore or even disconnect. That can cause clutch failure from insulation breakdown, as well as contact arcing.
 
I've got an automotive relay that contains a resistor in parallel with the coil, and I measured what effect having a freewheel diode would have.

The relay is standard cube one. The coil is about 120 Ohms, and has a 680 Ohm resistor in parallel, so the total resistance is around 98 Ohms and it takes about 115 mA when supplied with 12 V. I supplied the relay through a pushbutton switch and I had a 1 kOhm pull-up on the contacts to monitor how long the contacts took to open.

Here is the waveform:-
View attachment 148662
The voltage peaked at around 65 V and the relay took about 2.6 ms to open. It's interesting to see that the current increased slightly as the armature moved.

With a diode in parallel with the coil, it's very different:-
View attachment 148664
The voltage stayed at 0.7 V and the current took 25 - 30 ms to die down. It took around 8 ms before the contacts started to open. I found it interesting that the contacts bounced for around 2 ms with at least two false openings before the finally opening. The bounce was quite consistent and happened much like that every time the relay released.

My take from this is not to use a diode to suppress the inductive spike on a relay if the contacts are taking anywhere near their maximum rating. Any sort of heavy load or inductive load on the relay is going to damage the contacts more than turning the relay off quickly. The coil suppression almost inevitably slows the relay release, so the amount of suppression is a compromise between protecting whatever is controlling the coil and protecting the contacts.

Of course, some suppression is vital in most applications. This is what it looked like when I disconnected the resistor:-
View attachment 148665
It's even faster than with the resistor but the inductive spikes are huge.
Hello there,

You've stumbled across one of the behaviors of relays that is not always properly understood.
It starts by looking at what we call the "volt seconds". That is simply the volts times the seconds.

If you charge an inductance with 10 volt seconds and the current levels off, then you must discharge it with 10 volt seconds to get all the energy out of it. To get a relay to open, that means the contacts will start to open somewhere between 0 and 10 volts seconds, although more likely between maybe 5 and 10 volt seconds. The coil will be completely discharged after 10 volt seconds.

10 volt seconds means 10 volts over a period of 1 second, or 1 volt over a period of 10 seconds. You can see right off that the lower the voltage, the longer it takes to discharge the inductance of the coil. If you allow the voltage to go to 100 volts then the discharge will only take 0.1 seconds, and that is because 10 volts seconds is also equal to 100 volts times 0.1 seconds. If you wanted that to discharge in 0.01 seconds, you'd have to let the voltage go up to 1000 volts.

So you can see why they may have used a resistor rather than a diode. The resistor, in this case, will allow the voltage to go up to around 60 volts while the diode only around 1 volt, and that means a lot longer with just 1 volt. The math may be slightly different because the coil may start to open before the full volt seconds is reached, so it could be less than what we calculate.

Just to make this even more clear, if we had 68 volts for 2ms to open but we instead tried to short out the coil to get the relay to open, the time to open would go up toward infinity. That's because 68 volts for 3ms is around 0.200 volt seconds, and 0.200 volt seconds divided by 0 seconds is infinity. It would never open. That's pure theory though because real life coils have resistance of their own.

So the bottom line is, the higher you let the voltage go, the faster the contacts open.
The limiting factor is how high is the voltage that the drive transistor can take. If it can take 100 volts, then a 680 Ohm resistor might be ok because the voltage stays well under 100 volts. If that works out ok then there is little reason to change it.

You do have to be a little careful when measuring spikes with a scope though. The voltage could go up higher than what the scope could take, but you may not see it on the display because the scope did not catch that faster signal.
 
I tried with a pair of back-to-back 33 V zener diodes in parallel with the coil. This is the result:-

1738577669817.png

That is nearly as fast as with no suppression, and I think that a zener is better than all the alternatives.

With the zener the current has completely ceased about 1 ms before the contacts open. With a 680 Ohm resistor, the current is only a small fraction of the normal current when the contacts open.
 
I also measured the current in the coil when there is a diode in parallel with the coil:-
1738580130593.png

When the armature starts to move, there is a rapid increase in coil current, which slows the armature and allows the bounce of the contacts.

The only other case where there is any coil current when the contacts open is with a resistor in parallel with the coil. At that point the voltage is around 12 V so the coil current is around 18 mA, and the current does not change nearly as fast as it does when there is just a diode, so the armature is not slowed as much.
 
I tried with a pair of back-to-back 33 V zener diodes in parallel with the coil.
A diode and a Zener will give the same results of course, as the flyback voltage and current are only in one direction.
 
The only other case where there is any coil current when the contacts open is with a resistor in parallel with the coil. At that point the voltage is around 12 V so the coil current is around 18 mA, and the current does not change nearly as fast as it does when there is just a diode, so the armature is not slowed as much.
That's the reverse of what happens.
The current at which the contacts start to open is basically the same whether it is a diode, resistor, or Zener suppressor.
The current changes faster with a resistor, and the fastest with a Zener.
Look at my post #5 simulation.
 
Actually, a diode-Zener in series is the fastest.

Below is an LTspice sim for a 1H, 120Ω inductor for three voltage suppression configurations: a diode, a 130Ω resistor-diode, and a 12V Zener-diode:
The diode give about a -0.7V transient, and both the resistor-diode and the Zener-diode give about a 15V peak transient.

So for the current to drop to 10% of the initial value requires 25.7ms for the diode (red trace), 18.3ms for the resistor-diode (purple trace), and 14.7ms for the Zener-diode (amber trace).

So there's the trade-offs of cost/complexity vs. decay/turn-off time.

View attachment 148672

Real question is using a spice model w/o mechanical elements, eg. mass, inertia and the like, what dominates
the response ? Not sure although I know the bigger contactors this is a real issue, small signal relays normally
not but seeing as how the discussion is centered around single digit mS delays might still be a consideration.
 
That's the reverse of what happens.
The current at which the contacts start to open is basically the same whether it is a diode, resistor, or Zener suppressor.
The current changes faster with a resistor, and the fastest with a Zener.
Look at my post #5 simulation.
The limitation of the simulation is that it models the coil as a pure inductor. It is not a pure inductor because of the change of reluctance of the magnetic circuit as the armature moves.

If you look at the waveforms (except the voltage waveform with a diode) there is an increase in voltage or current as the armature starts to move. On the one with the current waveform with the diode, the current increases from about 25 mA to 35 mA at around 7 ms after the supply is removed.

I have found that a small voltage is generated by just moving the armature by hand with no power supply present.

As the armature moves, that generates a voltage. Where there is no suppressor, or where there is a zener and the coil voltage is less than the zener voltage, there is a significant voltage across the coil around the time that the contacts open. That voltage is generated by the magnetic field being changed by the armature moving. When there is a diode in parallel, the current increases as the armature moves, and that current slows the armature. It is a bit like where a motor is shorted out and it becomes harder to move quickly.
 

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