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Magnetic blowout on relay contacts 1

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jraef

Electrical
May 29, 2002
11,346
I have a request for supplying a 30A relay with a "magnetic blowout" option on the contacts. Does anyone know why someone would ask for that feature? I suspect that it is for faster arc quenching, but what kind of application would make that be so important on something so small? I have asked the user and he has no idea, the original design decision process is unknown, all we know is that it is being used on a small DC motor.
 
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That is strange. I would hazard the guess that the other machinery involved or the required wiring scheme may have forced the requirement. Could've come out of experience too.

Keith Cress
Flamin Systems, Inc.-
 
In DC Circuits, metallic arcs can be initiated at pretty low voltages and currents. The metallic arcs destroy the relay contacts (burned beyond recognition). The magnetic blowouts blow out the metallic arcs.
 
OK, so I found some additional info and it is starting to make sense. The motor draws 15A at 24VDC. The relay they selected years ago when this was engineered comes with or without a magnetic blowout, but the only difference seems to be that without it, the DC switching voltage is limited to 28V. With it, the current is derated but the voltage can go as high as 350VDC. So that follows what you said sried; the mag-blow allows the use of a higher DC voltage.

But it appears as though they don't really need it since they are only switching 24VDC. They may have selected it anyway just for reliability, but that is the information that is lost in time unfortunately. That may come back to what you said Keith,; experience... the harshest of teachers.
 
Mag blowouts are useful on highly inductive circuits where the arc voltage can be far higher than the supply voltage. They use the load current flowing through a coil to create a magnetic field which drives the arc up the widening arc chute where it eventually breaks. Clever idea.



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Question for Sreid: how does a ‘metallic arc’ differ from an ‘arc’?

It is my understanding that:

[a] ‘Blowouts’ are generally used to break the arc upon contact opening.

Basically, the nature of the arc will be determined by the circuit, contacts & atmospheric conditions.

[c] When there is sufficient energy in the system (usually inductive), the energy is dissipated, in the form of an arc, across the opening contacts.

At low voltages (6-48V) and a given load (W), the relay may be designed with a contact/air gap sufficient enough to break the arc when the contacts open. When the same relay is required to switch greater loads, or when the voltages exceed 48V, ‘magnetic blowouts’ can be added to increase the effective air gap. If the magnetic field is aligned correctly, the arc will be bent outwards, and when the length of the gap is sufficient enough to break to arc, the material in the arc will be blown outwards (thus, called blowouts). Of course, if the magnetic field is the wrong way around then the arc will be blown inwards and this can destroy the relay.

The small added cost of the ‘blowout’s may double, or more, the life of the contacts/relay – load dependant.

My comments above are related to DC applications.

Regards,
 
An arc is usually thought of as a high voltage jumping an air gap. The arc resulting from a relay contact opening is different in that a metallic vapor is the conducting path.
 
an alternative to the use of the magnetic blowout for DC loads would be to connect a diode across the contacts. this diode should be connected so that it does not conduct current when the contact is open. when the contact opens with an inductive load, the voltage is reversed across the diode and it conducts until the energy stored in the load inductance is dissipated. This does a great job in protecting the contacts. if the load is a relay, the open time will be slightly increased, and i would expect that the motor turn off time would be a little longer, but probably not enough to be noticed. this is normal industrial practice for DC coil relays and many come standard with the diode.
 
Diode connected in reverse across a coil is common practice; a diode connected in reverse across a contact wouldn't buy you anything. The voltage doesn't reverse, it goes from zero across the contact to the open circuit voltage and the current through the inductor continues to flow until dissipated; since the current couldn't flow through the reverse diode the diode is of no use. On a coil, the diode allows current in the coil to flow through the diode and back into the coil rather than trying to develop a high voltage across an opening contact somewhere.
 
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