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Snubber circuit

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Frozenguy

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I have read that snubber circuits are good when running inductive loads on a rely to prevent problems on the other side when switching off.

I have read that diodes can be used instead of an RC if you don't care about about preserving energy in the core?

I'll be running a pair of 40watt fluorescent tubes off the relay.

Thanks for checking this out.
 

crutschow

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Are you talking about protecting the circuit driving the relay coil or protecting the relay contacts?
 

MikeMl

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Pommie

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It's not clear what you are trying to do. Relays normally have a diode across the coil to stop back emf from damaging the circuit. A snubber is normally across the contacts to stop arching. Which do you need?

Mike.
 

MikeMl

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I added some verbage to describe how the snubber works in the link above...
 

MaxHeadRoom78

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I have read that snubber circuits are good when running inductive loads on a rely to prevent problems on the other side when switching off.

I have read that diodes can be used instead of an RC if you don't care about about preserving energy in the core?
RC snubbers are used on AC coils but diodes can only be used with DC coils.
Max.
 

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Diver300

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A diode across the coil can make the relay open more slowly than intended. It can be better to put a resistor in series with the diode, of about 5 - 10 times the relay resistance. That will limit the voltage to 5 - 10 times the supply voltage, but will kill the coil current quickly, ensuring the contact open quickly.
 

spec

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I have never come to terms with this whole catching/snubbing business and there seems to be different theories depending on where you look, even though there are some good tutorials, like MaxHR linked in post #7.

These are just some thoughts.

The item being controlled is either:
(1) An inductor, including motors.
(2) A relay (with mechanical contacts).

The objectives of catching/snubbing are to:
(1) Protect the device (coil breakdown etc) (in the case of a relay protect the contacts).
(2) Protect the controlling element (BJT, MOSFET, etc).
(3) Ensure the circuit performance.

Am I right is saying that some of the above objectives conflict and that there is no universal catching/snubbing approach that can be used?

I am trying to understand (not oppose) the theory of what Driver is saying in post #8 above.

By the way, I think it has been mentioned that you cannot use catching diodes on AC driven devices, but I suspect that you can get a similar effect by using back to back catching zenner diodes which have a Zenner voltage 10% to 15% higher than any expected AC peak voltage.

spec
 
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Diver300

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I am trying to understand (not oppose) the theory of what Driver is saying in post #8 above.
spec
The issue with a relay is that if the coil current decays slowly, the contacts can open slowly. If the contact current is near the relay's rating, and the relay is rated for a fast opening, then opening slowly can increase the contact arcing and wear. Adding a diode makes the current decay slowly. Resistance in series with the diode will increase the back voltage, so the current in the coil will decay faster. The voltage is very predicable with a resistor, so the voltage rating of the coil driver is easy to calculate.

At least one car manufacture has design rules that prevent such diodes being used on their own. Normally a relay has just a resistor in parallel, often inside the relay housing, so that it doesn't matter which way round the coil is connected. Diodes are fine on solenoids or motors where there is no downside of turning the current off slowly.

I have seen a refrigerator board where a relay had welded shut. The relays were all 24 V relays, run from transformerless power supplies, made with a capacitor and a bridge rectifier. With those the current would decay slowly as it could go through the bridge. It was the NO contact of the relay that had welded. This last paragraph is just anecdotal. I have no idea if that failure is common.
 

spec

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Thanks for the clear explanation, but there is one thing I cannot get my head around.
Adding a diode makes the current decay slowly. Resistance in series with the diode will increase the back voltage, so the current in the coil will decay faster.
Isn't the quickest way of dissipating energy in a coil by a short circuit across the the coil (like a diode). I would have thought that a higher resistance across the relay coil would take longer to dissipate the energy in the coil. This is the bit that I do not understand.

It is a conundrum, because I accept that putting a resistor in series with the diode is recommended practice in some areas, as you say.

And most off-the-shelf AC snubbers have relatively high value resistors in series with the capacitor.

spec
 

Diver300

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Thanks for the clear explanation, but there is one thing I cannot get my head around.

Isn't the quickest way of dissipating energy in a coil by a short circuit across the the coil (like a diode). I would have thought that a higher resistance across the relay coil would take longer to dissipate the energy in the coil. This is the bit that I do not understand.

It is a conundrum, because I accept that putting a resistor in series with the diode is recommended practice in some areas, as you say.

spec
The coil has a high inductance. The voltage across the coil is the inductance times the rate of change of current. If the decaying coil current goes through a diode only, the only resistance in the way of the current is the coil resistance, so the current decays slowly. If there is a big resistance in series, the voltage is larger and the current decays faster.

If there is no resistor, the rate of change of current is huge, so the voltage is huge, and that does damage to whatever was trying to turn the circuit off.

Some figures. A 12 V automotive relay takes around 50 mA, so has a resistance of 240 Ω. It will take about 5 ms to turn on, so the time constant is about that, so the inductance is around 1 H. If there is no series resistance, the current will decay with the same time constant. The initial rate of change of current is 12 A/s

If there is a series resistance of 2400 Ω, the relay current of 50 mA will initially be the same and will generate 120 V across the resistor, and therefore across the coil. The rate of change of current is 120 A/s. The coil energy mainly ends up in the resistor.

Using just a diode is a very good way of dissipating the energy, if you don't care how slowly the relay turns off. The coil power is no bigger while turning off than when running, and the peak voltage on the switching device is only 0.7 V bigger than the supply, so it is very good to the rest of the circuit. It is the contacts that can suffer.

And most off-the-shelf AC snubbers have relatively high value resistors in series with the capacitor.
spec
That can stop unintended resonance with the coil, and it can stop a large current inrush to charge the capacitor, which can be as harmful as a voltage surge when trying to turn off the coil.
 

spec

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Driver, the penny has dropped and I think I understand how the high resistance in series with the snubbing diode works.

The instant that the driving element (transistor etc) turns off there is still a full charge in the coil. With a short circuit across that coil (diode) that energy generates sufficient current to hold the relay in its existing state and the current does decay fast but not fast enough to prevent relay chatter, especially as the coil forms a tuned circuit with a relatively high Q.

On the other hand, a high value resistor of the correct value, as you say, limits the coil current to below the holding current of the relay instantly and because the Q of the coil is low, by virtue of the high resistance, any oscillations will not generate enough current to operate the coil.

I think the current will, in fact last longer but because it is below the holding current of the relay it will have no effect.

The end result is that the relay will instantly change state and will stay there- very clever.

spec

(our posts crossed)
 

Diver300

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I wouldn't put it like that.

I don't think that you can easily get relay chatter, but you can easily get slow opening of the contacts.

The high resistance does not initially reduce the current. The coil has inductance, so the voltage across it is proportional to the rate of change of current, so you can't get a step change of current, as that would need an infinite voltage. The high resistance increases voltage, and therefore the rate of change of current, so that the relay current falls to faster.

It is a bit like putting a brake on flywheel. At the time the brake goes on, the flywheel was doing whatever speed it had been before. A bit later, a brake with more torque will have slowed the flywheel more, but that takes time as well as the brake torque.
 

crutschow

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There's obviously a tradeoff between the value of the resistor you use in series with the suppression diode across a relay coil, the relay release time, and the amount of inductive turn-off voltage the drive circuit can tolerate.

The spike voltage is is approximately equal to the series resistor divided by the relay coil resistance times the relay operating voltage.

This is shown in the simulation below of the coil voltage and current for a typical small Omron 12V relay with a diode series resistor values of 1mΩ (essentially no resistor), 288Ω (equal to coil resistance), 500Ω and 1kΩ.
(Note I used a current source instead of a voltage source, since it's easy to simulate an ideal switch that way.)

It difficult to definitively say how much resistance is needed to not significantly affect the rated 3ms relay turn-off time but it would appear that a value of 1kΩ or greater would likely be sufficient.
1kΩ gives a transient spike of about 41V.

upload_2017-1-18_16-41-42.png
 

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Pommie

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An interesting discussion (post #15). However, it begs the question, why use a diode at all? Wouldn't a resistor that is 4 times the relay coil resistance do the job better? How does the 0.7V add anything to the 41V across the coil?

Mike.
 

crutschow

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why use a diode at all? Wouldn't a resistor that is 4 times the relay coil resistance do the job better?
To avoid waste current through the resistor when the relay is on.
If you don't mind this extra current, then you don't need the diode. :)
 

MaxHeadRoom78

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The issue with a relay is that if the coil current decays slowly, the contacts can open slowly. .
In all the decades I have used reverse diodes across coils I have never seen any adverse effects or signs of slow opening, It generally is noticeable in large inductive devices such as brakes and clutches.
Many relays suppliers such as Omron, Idec etc supply DC relays with diodes built in.
Max.
 

spec

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The coil has inductance, so the voltage across it is proportional to the rate of change of current, so you can't get a step change of current, as that would need an infinite voltage.
Yes, of course, you cant get a step change in current with an inductor, as you say, just like you can't get a step change in voltage across a capacitor. The former implies infinite voltage, as you say and the later implies infinite current.

Thanks for the explanation.

spec
 
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