If a DC coil was just a coil, it would saturate. If an AC coil was just a coil, the magnetic field would drop to zero at 120 Hz and the relay would chatter. Presumably the designs in some way prevent these problems. How? And does it prevent me from putting DC on an AC-rated coil?
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asked Jun 26, 2015 at 14:09
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\$\begingroup\$ why would the field collapse at 120Hz \$\endgroup\$– user16222Commented Jun 26, 2015 at 14:16
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\$\begingroup\$ @JonRB Because the AC voltage passes through zero twice per 60 Hz cycle (assuming 60 Hz line frequency). No voltage, no current, no magnetic field. Or so I understand. If that's not accurate, I'd love to know about that! \$\endgroup\$– Stephen CollingsCommented Jun 26, 2015 at 14:18
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\$\begingroup\$ sorry, I misread part of the question (ie it works at 60hz but not at 120). Yes part of the field collapses, there is usually another coil that produces a field out of phase of the main excitation field. This contributes to the retaining force and thus allows the relay to ride through zero crossings of the AC britannica.com/technology/shading-coil \$\endgroup\$– user16222Commented Jun 26, 2015 at 14:21
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5 Answers
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The DC coil relay has a resistance from the copper wire typically used. The current is limited by that resistance.
AC coil relays have inductance as well (the DC relays have inductance as well, of course, but it does not affect the 'on' current). They also typically have a shading ring that acts as a shorted turn in order to cause a magnetic field 90° out of phase with that from the coil, so that the total magnitude of flux does not drop to zero at the zero crossings.
Edit: As Andy says, an AC relay will work on (greatly reduced) DC. Of course you can also make a DC relay work on AC by adding stuff to it (if you use a capacitor filter you might need only 18VAC (RMS) to operate a 24V relay, and 24VAC would cause it to overheat and fail early).
answered Jun 26, 2015 at 14:52
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1\$\begingroup\$ +1 from me for the link. Never thought about driving an AC relay from DC or vice-versa. \$\endgroup\$ Commented Jun 26, 2015 at 15:24
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1\$\begingroup\$ After putting shader ring still it is chattaring. there is special type of material in core in solonoid Ac coil \$\endgroup\$– PowerCommented Sep 21, 2016 at 9:31
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\$\begingroup\$ The shader ring is used to slow down the closing of the relay armature. The current induced in it opposes the increasing magnetic flux as the armature closes and causes a 'soft' closing. The current in the shader is indeed out of phase with the coil, but at some point in the cycle, the 2 currents will cancel \$\endgroup\$– jp314Commented Sep 2 at 0:56
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\$\begingroup\$ @jp314 There's something of a subtle point there- because of the physical construction, the attraction forces add, not the magnetic fields, otherwise they would indeed cancel out as you say (compare (sin(x)+ sin(x_theta))^2 to sin(x)^2 + sin(x+theta)^2. For a graphic illustration of the chattering you get when the shading coil is broken in an industrial contactor so it does not conduct see this video. The shading coil does affect the closing, but that's another matter. \$\endgroup\$ Commented Sep 2 at 1:13
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\$\begingroup\$ Also some good information in Chapter 2 of "Electric Relays, Principles and Applications" by Thurston (2006). \$\endgroup\$ Commented Sep 2 at 1:32
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A Dc relay coil has resistance that limits the dc current. An AC coil relies on its impedance for governing the current. An AC relay will remain contact closed due to mechanical inertia and a little mechanical hysteresis and, the fact that an alternating north and south pole both attract the relay armature.
Putting dc on an ac coil should work fine but be prepared for the resistance to be low. In other words, If relay is rated at 24 volts ac don't use 24 volts dc.
answered Jun 26, 2015 at 14:38
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In theory AC coils can be driven with DC as long as you limit the DC coil current to the level of the AC holding current (to keep the coil from overheating). For larger relays i.e. contactors, a DC current limited to the equivalent AC holding current is often insufficient to actuate the contactor effectively. Since the coil inductance of the contactor increases once the contactor is closed the AC holding current is less than the AC pull in current. DC contactors usually have some mechanism to shift between a coil pull in current and the holding current. This is the difference between an AC coil and a DC coil in large relays. The flux shunt (shader ring) is external to the coil and only shorts a portion of the core.
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Recently, I had a problem with a batch of four 2-pole plug-in relays that had coils marked as 240V AC. When connected to the supply, three of the four merely 'chattered' and failed to 'pull-in' the contacts; the fourth operated correctly.
Believing the problem to be one of poor manufacturing quality control, I received some replacements from the same supplier which subsequently worked OK.
I assumed that the three faulty relays had been fitted in error with DC coils. Consequently, rather than binning them, I overcame the problem by internally fitting 1000 PIV DIL bridge rectifiers allowing the relay coils to operate normally on raw DC (i.e. no filtering).
Perhaps the use of 1000PIV devices was overkill, but I wanted the rectifiers to have a sufficiently high withstand voltage to the back emf generated when the relay released.
Additionally, as the relay coils were now operating on DC (albeit raw DC), I could have fitted a single rectifier wired in inverse parallel to the polarity of the raw DC to snub the back emf, thus permitting use of a bridge rectifier with a lower PIV, but a lack of sufficient space within the relay precluded this solution.
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1\$\begingroup\$ a bridge rectifier would act as a DC snubber (free wheeling), so 350V diodes would have been fine, but 1N4007 are so cheap that there's nothing gained by using weaker diodes. \$\endgroup\$ Commented Mar 24, 2018 at 0:42
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If you use a full wave bridge, the diodes in the bridge should act as free-wheeling diodes, and should snub the back EMF.
