The contacts of the relay (not the coil of the relay) are connected in series with water boiler (2kW) and AC power socket (220V, 50Hz). Suddenly, relay stops its operation due to the electric arc which continuously appeared at commutation of high power electrical device.

The question is following. What is approximate voltage (EMF), which appears on the commutation contacts of the relay?

Additionally, the datasheet for the relay may be found here.

P.S. My colleague tells me that wire has no inductance, but I am not agree. The wire is of 5m length, so the approximate inductance is equal to L=8444nH, then U=L*dI/dt. dI and dt are unknown and I cannot even imagine approximate values.

P.S. If someone took the measurement for such kind of situation using oscilloscope, please share your results!

  • \$\begingroup\$ approximately 220V \$\endgroup\$
    – PlasmaHH
    Jun 30, 2016 at 13:11
  • \$\begingroup\$ If you're concerned about the wire inductance, then you have to know the maximum current in the system. Assuming a 2kW resistive load, you can figure the peak current from 220VAC (RMS). Relay opening will be on the millisecond timescale. However, your inductance calculation (too many sig. fig.!) seems not to take into account the fact that AC power lines are usually very close to each other and carrying total zero current. \$\endgroup\$ Jun 30, 2016 at 13:20
  • \$\begingroup\$ I'm not sure about "Relay opening will be on the millisecond timescale". I think it is shorter. If to calculate EMF considering your assumption, it is equal to 8444e-9H*(2000W/220V)/1e-6s = 76.8 V. Don't you think it is very small? \$\endgroup\$ Jun 30, 2016 at 13:28
  • \$\begingroup\$ An approximate inductance for 5m of wire would be 8uH, not something with 10 significant figures! \$\endgroup\$
    – Neil_UK
    Jun 30, 2016 at 13:30
  • 1
    \$\begingroup\$ The question is not in the representation of the number ;) \$\endgroup\$ Jun 30, 2016 at 13:36

2 Answers 2


If you're worried about inductive EMF we can calculate that. Let's assume that we go from 10 A to zero in 1 ms. This seems a reasonable minimum time value as the contactor is mechanical and the mains is only at peak value for a fraction of each half-cycle anyway. Let's also assume that your value of 8 uH is correct. (I'm not going to check.)

$$ V = L \frac {di}{dt} = 8\times 10^{-6} \frac{10}{10^{-3}} = 8 \times 10 ^{-2} = 80~mV $$

I don't think you need worry about that.

enter image description here

Figure 1. (left) Pristine contacts from a relay (right) The nearly destroyed contacts from a relay operated under power for nearly 100,000 cycles. Source: Wikipedia Arc Suppression.

The contact wear will come from the arcing and decreased contact area and pressure as the contacts open and close. Additional heat will be generated at this time.

Update (after OP supplied datasheet):

enter image description here

Figure 2. Extract from datasheet.

That relay is too small despite the manufacturer's specification. How do I know? It failed!

Note that with 100 mΩ contact resistance (and not 100 MΩ as stated) that at 10 A you will have \$ P = I^2R = 10^2 \times 0.1 = 10~W \$ dissipated by the contacts. That contact resistance is quoted at 6 V so that's probably worst case but it gives some idea of the hardship the device endures

Secondly, since it's a DC coil I presume you are switching it electronically and have a snubber diode across the coil. This will slow down the release and increase the problem dramatically.

Tell your colleague that inductance is not the problem here.

OP's comment:

Anyway, I cannot agree with thansistor's [sic] prove, because calculations hasn't took into account the real value of inductance, which is unknown for the complete electrical circuit...

OK. What is the supply impedance? A quick web search brought me to an audio enthusiast site, Acoustica. (These guys worry about everything.)

One of the factors frequently overlooked by the hifi comics is that the electrical mains is not infinitely powerful; more specifically it can exhibit significant reluctance to supply enough electrons. Sometimes the impression is left that you could have the entire output of SIzewell B [a British nuclear power station] on tap if you buy a big enough power cord. In this year's colour, of course....

It ain't necessarily so; it ain't even close. IEC725:1981 models the European domestic mains supply as having an impedance of (0.4+j0.25)ohms. Surprisingly, measurement shows that the UK agrees with or somewhat betters the model on the whole, at something like (0.25+j0.23) ohms. What does that really mean? It's a definition which include the real resistance of the distribution network and wiring up to the socket, estimated as 0.25ohms, plus the reactive (j) component, allowing for stray capacitance and inductance of the wiring. The impedance is dominated by inductance, and equates to roughly 0.23ohms in the UK, for a net supply impedance of nearly 0.5ohms at 50Hz.

I haven't verified these numbers but let's take the inductance of the mains at your water heater supply as having a 0.25 Ω inductive impedance. We can calculate the inductance as follows:

$$ L = \frac {X_L}{2 \pi f} = \frac {0.25}{2 \pi 50} = 8 \times 10^{-4}~H $$

Plugging these back into our first equation and ignoring the \$ 8 \times 10^{-6} ~H \$ of the heater wiring we get:

$$ V = L \frac {di}{dt} = 8 \times 10^{-4} \frac{10}{10^{-3}} = 8~V $$

I think you can tell your colleague that we're still OK. Inductance isn't the problem.

I've never looked at this problem before so please let me know if I have missed something.


There is another reference to IEC 725:1981 in The Influence of Source Impedance in Electrical Characterization of Solid State LOighting Sources by D. Zhao and G. Rietveld, VSL, Dutch Metrology Institute below equation (6) on page 3. This uses the same 0.4+0.25j Ω as the Acoustica source.

  • \$\begingroup\$ I found, that dielectric strength for air is 20...75 (kV/inch). Then, in turn, if to suppose that arc length is of 1(mm), then voltage for this arc is 0.78...2.95(kV). What do you think about such assumption? \$\endgroup\$ Jun 30, 2016 at 13:58
  • \$\begingroup\$ @VictorSignaevskyi You have it the wrong way around. You've selected an arbitary (unknown) distance (1mm) and then worked out what the voltage to breakdown air would be. You need to start with the (known) voltage you have and then work out the distance it could strike the arc. Its clear from transistor's calculation that the back emf (80mV) is not the issue so the only voltage involved in any arcing is the supply. Assuming the break comes at the very peak of the cycle that gives 308V max (220*1.4). Taking your worst case value (20kV/inch) gives an arc length of 0.015 inchs / 0.4mm \$\endgroup\$ Jun 30, 2016 at 15:01
  • \$\begingroup\$ ... at which point the arc will extinguish. \$\endgroup\$
    – Transistor
    Jun 30, 2016 at 15:20
  • \$\begingroup\$ @JImDearden Your explanation looks correct, but it doesn't take into account inductance (taking into account EMF we should get voltage larger than 220*sqrt(2)). Moreover, I think, inductance is of much higher level than I thought before. My colleague told me, that we should take into account not only the length of wire walking through the socket, relay and water boiler... Also we must include all wire through which current flows. I mean we should take into account whole wire (for example, from transformer to the customer). I think I should measure inductance in the socket... \$\endgroup\$ Jun 30, 2016 at 20:18
  • \$\begingroup\$ What are you trying to solve here? Are you trying to figure out why a real relay stopped working? What stopped - the armature stopped moving or the contacts stopped conducting? Link to datasheet for relay? \$\endgroup\$
    – Transistor
    Jun 30, 2016 at 20:32

Relay sounds sticky, unless your load is inductive as well as resistive (wire wound). Also, check the driver circuit for proper operation w/respect to voltage & current to the coil. Other possibility is the coil on relay could have a shorted winding.


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