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Neon signs use transformers that can output at 10 kV DC at 30 mA = 300 W. I want to toggle this power with a common 5 V relay, but the DC rating for these are about 28 V DC at 10 A = 280 W. Does the output of the transformer exceed the rating of the relay?

If so, why would the same relay (with a 125 V AC 10 A rating) work on the input of the same transformer if its input is rated at 125 V AC 0.9 A? Am I confused about the power limits of the relay?

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    \$\begingroup\$ Yes 10kV DC is exceeding 28V DC. \$\endgroup\$ – MatsK Feb 12 '19 at 21:28
  • \$\begingroup\$ Indeed I don't think it's a good idea to attempt to control the 10kVDC current getting out of the transfomer (or ballast). Better put a normal relay at the main line, before it. \$\endgroup\$ – Fredled Feb 12 '19 at 22:25
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It's not the power, it is the individual components of power - current and voltage.

The characteristics of the contacts determine the current rating of the relay. Contact material, shape, and pressure affect how much current they can pass, and can break open repeatedly without degrading prematurely to the point where then cannot conduct any current at all. This is why small signal relays have gold contacts (no corrosion even under light pressure) and high current AC power relays have silver-based contacts (lower resistance/lower heating, and the arc that forms when the high current is broken helps "clean off" oxidation).

In your case the problem is the voltage rating. When open, the contacts are too close together for that high a voltage; it will arc continuously across the gap.

Change your way of thinking: A relay does not switch power - it switches a current at a voltage. Neither rating can be exceeded.

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  • \$\begingroup\$ So any power combination of voltage/amperage combination UNDER the rated max voltage would be fine (eg 1V 200A)? Or would the contacts then heat up due to the high amperage (and therefore be the limiting factor)? \$\endgroup\$ – jacob.hanson1010 Feb 12 '19 at 21:08
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    \$\begingroup\$ There are three independent limits, voltage, current and power, you need to work within ALL of them. Also, note that relays (really switches in general) often have different ratings depending on the type of current (AC/DC) and type of load. 200A would exceed the 10A current rating of your relay and would likely weld the contacts shortly before melting the whole thing. \$\endgroup\$ – Dan Mills Feb 12 '19 at 21:13
  • \$\begingroup\$ JH - no. The voltage and current ratings are functionally independent. Neither can be exceeded. Think about it - under what circumstances do you think putting 200 A through something that says 10 A on the cover will work? \$\endgroup\$ – AnalogKid Feb 12 '19 at 21:53
  • \$\begingroup\$ Ah I understand. So a lower voltage would prevent arcing, but introduce load issues. I was under the impression that the total power was the limiting factor and it didn't matter which combination of voltage/amperage you used. I had assumed the 125VAC 10A rating was indicating an example of total power load. Thank you for your explanations! \$\endgroup\$ – jacob.hanson1010 Feb 12 '19 at 21:53
  • \$\begingroup\$ @jacob.hanson1010 - " I had assumed the 125VAC 10A rating was indicating an example of total power load." Well, in one sense it is. If you observe both the voltage and current limits, you can multiply the two to get power. It's just that you can't work backwards. \$\endgroup\$ – WhatRoughBeast Feb 12 '19 at 23:10
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No, the contacts will arc across and it will likely arc to the coil as well. Very, very dangerous.

A commercial relay properly rated to switch 10kVAC will probably cost in excess of $1000. There are some DIY approaches, but better suited for non-novices.

You should consider switching the primary, but being an inductive load that relay may not be acceptable.

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    \$\begingroup\$ +1, this is the only answer right now that emphasizes how dangerous the asker's proposal is. \$\endgroup\$ – Hearth Feb 12 '19 at 21:20
  • \$\begingroup\$ A mechanical ignition distributor switches 20 kV and costs about $50. Than leaves $950 to devise the solenoid to drive it. \$\endgroup\$ – D Duck Feb 13 '19 at 0:23
  • \$\begingroup\$ @DDuck Right, and you can do things with cheap routed-outline PCBs that don't require compression molded phenolics. It's all about getting enough insulation between the contacts in both states and whatever the actuator is. But a suitably rated vacuum relay is $1k+. \$\endgroup\$ – Spehro Pefhany Feb 13 '19 at 0:25
  • \$\begingroup\$ This one (Pickering 68-1-A-12/2D) can only manage 7.5 kV (almost there) but costs £32. \$\endgroup\$ – D Duck Feb 13 '19 at 0:28
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    \$\begingroup\$ Cynergy3 Components DAT70515T is $65 from Allied in single qty, with flying leads, 15 kV isolation, can switch 10 kV and 2A. And it has a 5 V coil. \$\endgroup\$ – D Duck Feb 13 '19 at 0:41
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One major point to consider on relay (or any switch) contact rating is whether it is AC or DC. Generally speaking, switch contacts can handle higher amperes if you are using AC because of zero crossings. With DC, the switch has to break the connection in a manner that will not arc or continue to function reliably with the resulting arc.

You're calculating power and comparing the resultant power (300 vs 280 watts). However, in this case the voltage is part of the reason for the rating (arc distance) as well as current (contact thickness). Finally, the direct or alternating nature of the current determines what the device can ultimately safely (and reliably) switch.

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