This question is probably more of a "please explain whatever concept I'm missing." But I'll start with my specific question.

I'm building an electric brewery and am using a 5500W 240V (23A) heating element. I'm using 10 gauge wire for most of the circuit that services the element. I have a small section that I'm yet to design/build that is basically a mechanical relay that feeds a solid state relay. The mechanical relay (model: SLA-05VDC-SL-C) has very small leads, although it is rated for 30 amps. Connecting 10 gauge wire to this relay is fairly impractical.

It would be a lot easier to connect a short run of 18 gauge (could be only a few inches long). I expect this is a bad idea... is it?

If it isn't a bad idea, this rather confuses me... why then could I not get away with daisy-chaining dozens of short 18 gauge runs for the entire circuit?

  • \$\begingroup\$ A thinner wire has a higher resistance which creates a voltage drop across it and the power losses are dissipated as heat on the wire. For extremes the wire can even melt, catch fire etc. When the majority of the wiring is with a proper thick wire and you only use a thinner wire for a small section then it can work but the 18gauge seems on the thin side. \$\endgroup\$ – alexan_e Jan 5 '14 at 19:35
  • \$\begingroup\$ Indeed - 18 gauge is too small for 23A. \$\endgroup\$ – Michael Karas Jan 5 '14 at 20:05
  • \$\begingroup\$ @alexan_e That seems to answer my question: voltage drop is still a problem even if heat dissipation is not. Perhaps you'd like to make that a "real" answer? \$\endgroup\$ – notlesh Jan 5 '14 at 20:24
  • \$\begingroup\$ @MichaelKaras the voltage drop would be pretty negligible on such a short run (under half a volt, if this javascript calculator is correct), so I should think it would work (at least that I would "get away" with it). \$\endgroup\$ – notlesh Jan 5 '14 at 20:25
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    \$\begingroup\$ Just as an aside: In industrial installations where a similar issue arises, i.e. relay contacts too narrow to cope with the designed wire gauge, one solution I have seen is the use of a very short tail of multi-strand silver wire between the relay contact and the thicker gauge copper cabling. Typical tail lengths I have seen are 1-3 centimeters, and the silver to copper wire connection is usually crimped using a thick lug for heat management. \$\endgroup\$ – Anindo Ghosh Jan 6 '14 at 3:55

There are two related but not totally inter-dependant issues:

  • Maximum acceptable current able to be carried by a given wire gauge.

  • Maximum acceptable voltage drop between source and load.

Wire maximum current capability is set by regulations and is based on temperature rise which is a function of energy dissipation per length which is a function of resistance per length which is a function of wire diameter. Other factors which affect allowed regulatory values include sheathing type, application, environment (open air, metal conduit, ...).

Maximum acceptable voltage drop for the load is based on the sum total drop of the circuit components feeding it. An eg 10A load may be fed by a 10A rated conductor until the length of the conductor is such that maximum allowed voltage drop is reached. If you wish to use a longer conductor run you will need a higher rated conductor, say 15A or 20A, not because of conductor current rating per se, but because the heavier conductor allows less total voltage drop.

In your case, as long as total voltage drop on the 10 gauge circuit when fully loaded is below the maximum allowed by regulations, then use of "a few inches" of lighter wire would be acceptable. The voltage drop per length in the lighter conductor will be higher but the additional absolute voltage drop will be minimal.

How light the short length of conductor can be is a matter of regulations and common sense. A few inches of 18 gauge may not burn out even if it is not rated for the current carried, as heat transfer to the device terminals and adjacent thicker conductors may allow lower temperatures than would occur with a longer run. HOWEVER if you have a fire for any reason and investigators determine that you have used an under-rated short link of wire anywhere this may affect insurance payout even if the non-compliant wiring was not the cause.

If the relay is equipped with fixed leads then they are almost certainly rated for the maximum current that the relay can carry. While it is possible to get out of spec equipment, all 'reputable' manufacturers will be well aware of requirements and will meet them. It may well be that the short lengths involved are acceptable for the reasons I mentioned above.

I don't know if your brewery is a home or commercial venture. I also do not know what your local regulations allow wrt wiring of mains powered equipment. It's outside the scope of the question, but you need to be sure that such issues do not affect your insurance coverage - or your chances of needing it.


Using Wikipedia link supplied by alexan_e
Calculated values below are rounded but "=" is used.

For say 1 metre loop-length tail, R = 21 mOhm.
At at say 25A = loss of I^2R = 625 x 0.021 = 13 Watts. That's non trivial.

Voltage drop = IR = 25 * .021 = 0.5V.

10 gauge = 3.3 mOhm/m
or about 2W dissipation
and 0.1V voltage drop.

So extra voltage drop going from 10 gauge to 18 gauge is not very important but the dissipation is significant. 18 Watts in 1 metre loop = 18 Watts in 500mm 2 conductor linear.
That's 3.6 W in a 100mm = 4 inch tail.
Say 4 Watt is not liable to melt insulation, but it will feel warm.

Worst is that the 18 gauge dissipation is > 25% of its conservative fusing value. That's high. Actual dissipation at fusing is about 16x higher, and that fusing value is a very conservative one from Wikipedia's list of alternatives, but I'd aim for a bigger wire diameter for a tail if possible.

  • \$\begingroup\$ Thank you, this clears things up a lot. I think I was mislead by wire size charts I found on the internet after wondering how I could use these relays. I'm guessing they relate voltage drop to wire size and length without consideration of temperature rise. \$\endgroup\$ – notlesh Jan 5 '14 at 23:54
  • \$\begingroup\$ For the record, I'm using 10 gauge wire for all 30 amp runs and 14 gauge for any other 120V runs. I've been very careful to avoid short cuts and to make sure everything is to standards... hence this question. I'm using appropriate circuit breakers and a GFI. This is a home brewery :) \$\endgroup\$ – notlesh Jan 5 '14 at 23:56
  • \$\begingroup\$ @stephelton In your position, if the relay was suitable but 10 gauge was too thick for it and I needed a "tail" I'd use the thickest wire that worked OK and sensibly minimise length used. You are very unlikely to go far astray with that approach. A relay maker may have had brain fade or it may be badly designed by people who do not care BUT in almost all cases, using the thickest wire that fits properly will be fine. [YMMV / IANAL/ ACNR / ... :-) ] \$\endgroup\$ – Russell McMahon Jan 6 '14 at 0:07
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    \$\begingroup\$ Better yet -- get a different relay that is more suited for the application... which is what I've decided to do. \$\endgroup\$ – notlesh Jan 6 '14 at 2:55
  • \$\begingroup\$ If it interests you, I've been using theelectricbrewery.com as a reference. Kal has documented there how to build a similar brewery with lots of safety precautions and 100% to code. \$\endgroup\$ – notlesh Jan 6 '14 at 2:56

A thinner wire has a higher resistance which creates a voltage drop across it and the power losses are dissipated as heat on the wire. For extremes the wire can even melt, catch fire etc.
When the majority of the wiring is with a proper thick wire and you only use a thinner wire for a small section then it can work but the 18gauge seems on the thin side.

That seems to answer my question: voltage drop is still a problem even if heat dissipation is not

The resistance and the power dissipated as heat are related and calculated by the power equation

\$ P= I^2 \times R \$ where P is the power, I the current and R the resistance

The shorter the piece of wire the smaller the resistance (and voltage drop).
According to wikipedia the resistance of a 18 gauge wire is 21m Ohm per meter

To extend my reply I would suggest a solid state relay which offers several advantages over mechanical relays.

  • \$\begingroup\$ As I mentioned in my question, the mechanical relay in question feeds a solid state relay. The reason for the redundancy is because SSR's have a small amount of leakage current which could harm my heating elements. I've also heard that SSR's fail in a closed state (when they fail.) \$\endgroup\$ – notlesh Jan 6 '14 at 2:58
  • \$\begingroup\$ @stephelton Sorry, I missed that. I think it is better to get a proper screw type contact relay that will allow the secure connection of thick wires. \$\endgroup\$ – alexan_e Jan 6 '14 at 10:30

Sometimes wire size is determined more by volt drop than the safe current carrying capacity of the cable. In this case a short run of cable that is thick enough to safely carry the current but thinner than the long distance cable may be ok.

Sometimes thinner cable can be used by accepting a higher operating temperature. Some types of silicone or even fiberglass insulated wire may be able to operate at much higher temperatures than normal PVC wire. Of course before relying on this you need to be sure that the things the wire is connected to can tolerate the higher temperature.

Sometimes for very short runs thinner wire can be used because the heat will be dissipated into the connections.

But all of this is ignoring the elephant in the room.

Your real problem is that the relay you have picked out is designed for PCB mounting, not for connecting to individual wires. Unless you have experiance desinging PCBs for high current applications I strongly recommend you find a relay that is designed for connecting directly to wires of appropriate size.


If you’re relay connections don’t seem to accommodate the larger wire that likely indicates that the relay wouldn’t handle that current anyway. I will bet you that the relay contacts will weld under those kinds of current conditions

  • \$\begingroup\$ Welcome to EE.SE, Peter. You might want to note the dates on the earlier posts. The OP (original poster) accepted an answer over four years ago! On the other hand someone else may find your answer helpful. \$\endgroup\$ – Transistor Apr 29 '18 at 20:03

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