Coil Flyback Current / Voltage

Thread Starter

ben sorenson

Joined Feb 28, 2022
180
Let's say I have a 2, 12 Volt solinoids. They each have a resistance of 45 Ohms and placed in parellel and are powered by 2, 18.V Batteries and driven through a 10 ohm resistor via pulsed DC.

Without a freewheeling diode, there would be a high arc that would destroy the contacts and cause many more issues because of the "emf kickback" when the pulse is in its "off" state and creates an "open" in the circuit.

Adding the freewheeling diode, or diode resistor in series across the solinoid prevents this condition. My question is that if only a single diode is used across the coils and not needing to be Accurate how much current would generally be dissipated across the diode? 10ma 50ma 100ma ? Just a general idea would be ok.
 

Ian0

Joined Aug 7, 2020
9,847
Immediately after the switch opens the current through the diode will be identical to the current through the solenoid.
(Current passes through a component, current is not dissipated by the component. Power and energy are dissipated)
The current will decay exponentially to zero at a rate determined by the resistance and inductance.
The total amount of energy stored will be (I^2. L)/2
Some of that energy will be dissipated in the diode, but most will be dissipated in the resistance of the solenoid.
 

crutschow

Joined Mar 14, 2008
34,470
I have a 2, 12 Volt solinoids. They each have a resistance of 45 Ohms and placed in parellel and are powered by 2, 18.V Batteries and driven through a 10 ohm resistor via pulsed DC.
.................................
how much current would generally be dissipated across the diode?
The peak diode current will be 18V / (10+22.5)Ω = 554mA

Note that such a diode will slow down the release time of the solenoids.
If that's a problem, then you can add a resistor in series with the diode.
The resistor value determines the peak flyback voltage, V = IR, where I is ON solenoid current.
 

MisterBill2

Joined Jan 23, 2018
18,604
Quite an interesting simulation. How was the inductance determined? Or is that just a guess? Certainly the pulse is very narrow, and that affects the heating power of the spark when the contacts open, a reality seldom mentioned in this forum. So while the voltage spike can damage semiconductors by over-voltage breakdown, it is much less likely to deliver enough heat energy to damage adequately sized contacts. That is seldom considered.
 
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crutschow

Joined Mar 14, 2008
34,470
much less likely to deliver enough heat energy to damage adequately sized contacts. That is seldom considered.
So let's consider it.
How do you determine what are adequately sized contacts to tolerate the arc generated when the contacts open, without damaging the contacts?
The inductive energy that contacts can safely absorb is not usually stated in the relay specs.
 

MisterBill2

Joined Jan 23, 2018
18,604
So let's consider it.
How do you determine what are adequately sized contacts to tolerate the arc generated when the contacts open, without damaging the contacts?
The inductive energy that contacts can safely absorb is not usually stated in the relay specs.
Usually I look at the rating, when I am looking at an actual supplier's catalog. Information that amazon would not provide even if they understood it perfectly. At other times it is just a matter of inspecting the candidate device considering the application. On the other side, it takes considering the amount of arc energy the inductive device could possibly deliver.Coil resistance and operating current are part of that consideration.
Besides that, considering the intended application of the switching device helps determine how much, if any, protection will be required.. That is how determining if contacts are adequately sized.
 

crutschow

Joined Mar 14, 2008
34,470
That is how determining if contacts are adequately sized.
Sounds mostly like educated guess-work and not necessarily sufficient to insure a reliable design.
Adding a suppression circuit seems to be well-worth doing to insure that any damage to the contacts is minimal.
 

MisterBill2

Joined Jan 23, 2018
18,604
(Response to post #8)
Probably more than would be spent in a spark, although that is unknown.. BUT I have cured the buzz of a lighting contactor, operating on AC, by shunting it with a diode and having a second diode apply the opposite polarity. That is, two diodes back to back, with the contactor coil across one of them . So that shows that a diode across a coil slows the release time. Slowing the release time fora relay in some applications may not be a benefit.

And certainly educated choices, based on a circuit evaluation, can be worth a bit. How much does the debugging time spent finding one reversed diode connection cost??? Quite a bit, it turns out.
 
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Thread Starter

ben sorenson

Joined Feb 28, 2022
180
Quite an interesting simulation. How was the inductance determined? Or is that just a guess? Certainly the pulse is very narrow, and that affects the heating power of the spark when the contacts open, a reality seldom mentioned in this forum. So while the voltage spike can damage semiconductors by over-voltage breakdown, it is much less likely to deliver enough heat energy to damage adequately sized contacts. That is seldom considered.
 

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crutschow

Joined Mar 14, 2008
34,470
Slowing the release time fora relay in some applications may not be a benefit.
If a slow release time is a problem, than adding a suitable resistor in series with the diode will reduce that time.
How much does the debugging time spent finding one reversed diode connection cost???
So you want to not use a part because it might be incorrectly installed?
Don't see the relation of that to doing a reliable design.
 

MisterBill2

Joined Jan 23, 2018
18,604
My major point is that especially here, where many of the posts are from beginners, that the religion of a diode across every coil every time is adding to the complexity of every suggested circuit. It is a great way to make a beginner see more complexity than is needed for somebody just starting out, building a circuit that may not be used for very many operations, even. And especially with low power, how much benefit do all of those diodes actually provide??? Not very much, based on my experience. We are not creating production systems that operate every ten seconds every day.
 

Thread Starter

ben sorenson

Joined Feb 28, 2022
180
Immediately after the switch opens the current through the diode will be identical to the current through the solenoid.
(Current passes through a component, current is not dissipated by the component. Power and energy are dissipated)
The current will decay exponentially to zero at a rate determined by the resistance and inductance.
The total amount of energy stored will be (I^2. L)/2
Some of that energy will be dissipated in the diode, but most will be dissipated in the resistance of the solenoid.
You said some of the energy will dissipate through the diode, but most would be dissipated through the coil. Is that because the "resistance" of the diode is low when forward biased correct? But if the "flyback voltage" is above the pos supply, wouldn't that voltage want to then dissipate back through the 10 ohm resistor toward the pos supply and not the solinoid "resistance" witch would be nearly double than the 10 ohms resistance? Or doesthe "flyback voltage" become lower than the supply if faced with the 10ohm resistance, so instead it pushes through the solenoid.?
 

crutschow

Joined Mar 14, 2008
34,470
But if the "flyback voltage" is above the pos supply
No.
The "flyback" voltage will be equal to the diode forward voltage drop, as that will be the maximum across the inductor.
So the inductive energy will be dissipated in the diode forward drop voltage, and the coil resistance.
And especially with low power, how much benefit do all of those diodes actually provide??? Not very much, based on my experience.
So, in your experience, you've built a lot of systems switching inductors and seldom or never used any flyback protection?
 

Thread Starter

ben sorenson

Joined Feb 28, 2022
180
You said some of the energy will dissipate through the diode, but most would be dissipated through the coil. Is that because the "resistance" of the diode is low when forward biased correct? But if the "flyback voltage" is above the pos supply, wouldn't that voltage want to then dissipate back through the 10 ohm resistor toward the pos supply and not the solinoid "resistance" witch would be nearly double than the 10 ohms resistance? Or doesthe "flyback voltage" become lower than the supply if faced with the 10ohm resistance, so instead it pushes through the solenoid.?
No.
The "flyback" voltage will be equal to the diode forward voltage drop, as that will be the maximum across the inductor.
So the inductive energy will be dissipated in the diode forward drop voltage, and the coil resistance.
So, in your experience, you've built a lot of systems switching inductors and seldom or never used any flyback protection?
So in my scenario approx 0.6v (forward voltage drop) @ 600ma? That is what would be "recirculating" so to speak?
 

Thread Starter

ben sorenson

Joined Feb 28, 2022
180
A little more than that.
My sim below shows nearly -0.9Vpk with a 1A diode.

View attachment 313980
Thank you so much that was really actually helpful!

So in my set up/experiment I have the flyback diode, but also ( in series) with the negative side of this LED array to the Positive side of the diode, Positive side of LED Array to ground.

The Array of LED Strips are wired in parellel, I don't have any information on them, as I couldn't find them online, but im assuming the individual diodes in the strips are wired in series. I recovered them from a 120v LED Flood Light.

During the "open" I can get the entire LED array to light up very intensly. The LED's will flash rapidly, however as the frequency increases they will stay on without any noticeable flashing.

If the " flyback voltage " is limited to the 0.9 volts, how would it be possible to have this entire, LED array light up so intense. ?
 

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MrAl

Joined Jun 17, 2014
11,496
Let's say I have a 2, 12 Volt solinoids. They each have a resistance of 45 Ohms and placed in parellel and are powered by 2, 18.V Batteries and driven through a 10 ohm resistor via pulsed DC.

Without a freewheeling diode, there would be a high arc that would destroy the contacts and cause many more issues because of the "emf kickback" when the pulse is in its "off" state and creates an "open" in the circuit.

Adding the freewheeling diode, or diode resistor in series across the solinoid prevents this condition. My question is that if only a single diode is used across the coils and not needing to be Accurate how much current would generally be dissipated across the diode? 10ma 50ma 100ma ? Just a general idea would be ok.
Hello,

I take it you are driving the solenoids with a relay that has contacts that could arc when they open.

The preferred method is shown in the diagram, but note the added resistor Rx.
Rx represents the load on the 18v supply without considering the load introduced by the two coils when they are turned on. What this means is that the 18v supply has to have enough of a load on it already in order to absorb the spike energy that will come from the two coils when they are turned off. The voltage of the 18v supply may increase due to this spike energy, so there has to be enough of a load on it to keep this to an acceptable level, or else a zener or something added so if the voltage goes over 18v (like maybe 20v) the zener catches the current from the two coils.

The reason this works so well (given the above requirements) is because then the coils can turn off quickly.
Coil energy can be viewed as a pseudo energy in volt seconds and that means the higher the voltage the faster the coil dissipates all its energy. By utilizing the 18v supply line, we have a higher voltage than say the (removed) diode and that discharges the coils much faster.
For a simple example, say we energize the coils for 0.1 seconds at 18 volts. The pseudo energy is 18 times 0.1 which is 1.8 volt seconds. That means we need to take 1.8 volt seconds from the coil to fully discharge it. By using a diode that drops just 1 volt we would see it would theoretically take 1.8 seconds to discharge completely because 1 times 1.8 is of course 1.8, but with an 18 volt drop it would only take 0.1 seconds because 0.1 times 18 is 1.8 also. If we had a 36 volt power supply to use as an energy return path, then it would only take 0.05 seconds to discharge completely. Conversely, a 9 volt power supply would lead to a discharge time of 0.2 seconds.
Note this is considering that the coil(s) are only turned on for a short time. If they are turned on for a long time, then the volt seconds would be the maximum volt seconds at whatever voltage they were driven at. In any case, an 18 volt power source dissipates the energy much faster than a 1 volt diode, and if the normal load on the circuit (shown as Rx) is enough to dissipate the energy, then it's just a matter of adding one diode to the circuit (as shown in the diagram).
I got the nice diagram from one of Cruts's posts and just modified it a little for the illustration.

BTW this technique is used in power converters in order to keep the efficiency higher. If the coil (or transformer) energy is dissipated as heat in a resistance or diode, that energy is lost forever. By returning some of it back into the power supply (often rectifier and filter caps) the efficiency can be raised. Since this energy is usually a fraction of the total input power and the filter capacitors are fairly large, the buss voltage does not go up very much.
When higher speed switching is involved though a fast recovery diode has to be used rather than an ordinary rectifier diode like the 1N400x series.
 

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