I am planning to use a DC drived SPST reed relay. Below I include the spec for voltage, current and power ratings.

Contact Rating: 200W
Carry Current: 5A
Switching Current: 3A
Switching Voltage: 3500V

I want to use this relay for 2000V - 3 Amp power source switching. I know it makes 6000W which is out of spec for contact rating.

However, my question is if I turn off the power before switching off the relay and turn on the power after I switch on the the relay can I use this relay with this voltage and current?

This is the spec link of the relay I want to use (model 5503 more precisely). Here, another one similar to it.

  • \$\begingroup\$ If you switch off the power before switching the relay, and turn on the relay before turning power back on, what purpose is the relay fulfilling? \$\endgroup\$ Nov 1, 2012 at 12:46
  • \$\begingroup\$ A datasheet of the specific relay would help in providing meaningful answers. \$\endgroup\$ Nov 1, 2012 at 12:47
  • 1
    \$\begingroup\$ Hi @AnindoGhosh there is one source, I want to connect it to one of the 20 different nodes at a time. All nodes has one SPST relay between the source. Switch power off, turn off all relays, turn on the relay I need, switch power on. Reverse logic is similar. I need something like this. \$\endgroup\$
    – rsa
    Nov 1, 2012 at 13:12
  • \$\begingroup\$ Isn't contact rating specified as the maximum power that the physical copper parts are able to conduct? If so, it doesn't matter what kind of electric component you are dealing with - the switching characteristics of a relay would have nothing to do with it. \$\endgroup\$
    – Lundin
    Nov 1, 2012 at 13:47
  • \$\begingroup\$ Short answer: No data sheet = you're merely going to get speculation, not real answers. Long answer: No, the power rating is defined for the relay's contact resistance, however small. Exceeding that will cause overheating, relay damage, loss of UL certification, and thereby no insurance cover on your site. \$\endgroup\$ Nov 1, 2012 at 14:19

2 Answers 2


Yes, you can do this. The important thing is that you aren't switching 6kW. When the relay switches the voltage is zero, so the wattage is zero. And when power is applied, the relay sees either 3 amps and zero volts, or 2000V and zero amps, all within spec.

However, you need to be very sure the relay doesn't switch under power! The first switch event under power may be the relay's last, and the 6kW will make a spectacular failure.

Edit: In response to comments, the relay's safe operating area looks like this:

enter image description here

The area is bounded by I=3A, V=3500V, and VI=200W.

  • \$\begingroup\$ Relay with 6Kw is hard to find. Even if I can find probably it will be too expensive to use in 20 units. So I think main power switch (may be an IGBT design) with 6Kw or even higher rating together with these relays configuration would work well. \$\endgroup\$
    – rsa
    Nov 1, 2012 at 18:18
  • \$\begingroup\$ By my interpretation, "switching power" is the worst-case amount of instantaneous power the relay could see if its resistance were to increase non-instantaneously from 1 microohm to 1 picosiemens. Depending upon the nature of the load, that could be anything up to (open-circuit voltage) times (closed-circuit current), but in the situation at hand it would be essentially zero. Would you agree with that interpretation? \$\endgroup\$
    – supercat
    Nov 2, 2012 at 15:09
  • \$\begingroup\$ For purposes of that diagram, how should one figure voltage and current? For example, suppose one has a 300 volt source driving a 150 ohm resistor. Should one count the power in that case as being 600 watts (current of two amps, times 300 volts), or should one figure that the worst-case maximum power dissipation that the relay itself could possibly see would be if it for a brief moment had a resistance of exactly 150 ohms, in which case it would drop 150 volts across it while carrying one amp, for a maximum power of 150 watts? \$\endgroup\$
    – supercat
    Nov 2, 2012 at 18:51
  • \$\begingroup\$ My inclination in that scenario would be to draw a line from (0 volts 2 amps) to (300 volts 0 amps) and figure that the relay would always be operating somewhere on that line while it opens and closes (assuming a purely resistive load), and figure that if the line doesn't intersect the edge of the safe operating region, all is good. Would that be the right approach? \$\endgroup\$
    – supercat
    Nov 2, 2012 at 18:58
  • \$\begingroup\$ That's a valid approach, but you'll need a much better plot. Mine is clearly not-to-scale, and the 1/x curve is too shallow. \$\endgroup\$
    – markrages
    Nov 2, 2012 at 19:00

The ratings of any relay are mutually independent maximum values, and are not a combination or a best effort recommendation.

Thus, each specific rating must be met. In fact, for certain types of loads, the recommended actual loads are significantly lower than the rated value.

For instance, when driving an incandescent lamp as a load, the filament's initial resistance when cold can be as little as 15% of the resistance when hot. This means, without a resistor in series, said lamp would draw nearly 8 times it's rated current at start-up. Experts will thus often suggest using a relay rated at 8 times the current that the incandescent lamp is nominally supposed to carry.

Similarly, for inductive loads, the voltage spike at circuit break can be several times as much as the operating voltage. The solution in such cases is to use a reverse biased diode across the relay contacts, to shunt the voltage spike harmlessly past the relay contacts.

The contact wattage too is an independent maximum rating, regardless of the values of the other parameters. This is because the rating is computed against the actual resistance of the contacts and their prongs, however small this resistance might be. Exceeding this rating can cause overheating of the relay, insulation meltdown of the driving coil, contacts latching closed, and potentially fire.

Another important point to keep in mind for any electrical equipment is that exceeding any single rating inherently violates Underwriters Laboratories (UL) certification, which means the site of such an installation, be it domestic, commercial or industrial, is no longer covered by insurance. Even an unrelated mishap would thus be denied insurance recourse. Similar rules apply in many countries.

Specific to the two devices mentioned in the question: One of them has a 200 watt contact rating, and the second is even lower at 50 watts. Thus, the requirement far exceeds this rating in both cases.

In sum: No, you can not use those relays for the purpose described.

  • \$\begingroup\$ If the relay is designed to carry 5A, its resistance is almost certainly less than 0.1 ohm. If its designed to switch 3500 volts, its resistance is almost certainly greater than 25 megohms. Even assuming the "worst case" values above, if it's never switched under load, how would the contacts ever see more than half a watt? \$\endgroup\$
    – supercat
    Nov 2, 2012 at 15:05

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