1 OR 1.5mm twin & earth cable for bathroom LED downlighters... discuss
- Thread starter rob4beer
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What did you expect to find?
The answer is in Tabe 4D5 of the Wiring Regulations.
You will notice that even under the most adverse installation conditions 1mm² is still adequate for a 6A circuit.
You may use cable larger than necessary if you wish.
This seems to have become a fashion - like using 10mm² for cookers where 4mm² would suffice.
As you mention downlights, you may find it better to use flexible cable smaller than 1mm².
I'd forgotten that we need to talk about downlights....
Rob - are you aware of (and planning to deal with) the issues surrounding installing downlights in bathrooms?
And are you aware of the fact that you always see them (and are probably planning to use them) in multiples arrayed across the ceiling because they aren't actually any good at the job of lighting up rooms?
Lastly, anybody lying in a bath to relax will not thank you for installing that type of lighting.
Rob - are you aware of (and planning to deal with) the issues surrounding installing downlights in bathrooms?
And are you aware of the fact that you always see them (and are probably planning to use them) in multiples arrayed across the ceiling because they aren't actually any good at the job of lighting up rooms?
Lastly, anybody lying in a bath to relax will not thank you for installing that type of lighting.
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Does volt drop not depend on the current? How can you quote lengths without knowing how many lamps are used and the wattage?
J
jg321
I fully agree with you here, however... What type of light fitting would be more appropriate, and not look rubbish in a modern bathroom? This is a genuine question, not least because I'm about to get a new bathroom fitted in my house, after spending many years living in "trendy" city centre flats with annoying downlighters in the bathroom. Annoying downlighters everywhere in fact.
It does - and the current reducing after each light.
I assumed (never assume) Eric was using 6A but it doesn't compute.
Also, if lights evenly spaced, an average of 3A throughout the circuit ???
Anyway, the difference should be at the ratio of 29:44, or 2:3, or (2x1)to(2x1.5)
Where am I going wrong?
6.9V (3% of 230). 1mm² is 44mV/A/m
6900mV ÷ 44 ÷ 3 = 52m. (52 ÷ 29 x 44 = 78)
6A would only be 26m.
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Yes, I also would have imagined/suspected that he was working with the latter. But, as you go on to say ....
Quite. No matter what current he was assuming, eric's figures of 39m (with 1.0mm²) and 44m (with 1.5mm²) are far too close to be correct.
Other than that I presume you meant 44 mV/A/m, I see nothing wrong with that. The implication would be that eric was assuming a current of about 4.0A (to get 39m) - could that be (for some reason) 2A at the end of the circuit and the other 4A evenly spaced throughout the rest of the circuit?
However, even if that's what he was doing, it doesn't explain his figure for 1.5mm² cable.
Kind Regards, John
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You need to read, I quoted 6 amp.
As per exchange between EFLI and myself above, could you please explain how you arrived at the circuit lengths you reported?
Kind Regards, John
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Using the formula given in BS7671 to work out the correction and then going by the volt drop of 6.9 volts, so for a typical Ze of 0.35Ω I would expect a Zs of 1.5Ω installation method 100 Max permitted operating temp tp 70°C Rating factor Cg and Ca both 1 and as said Design current for circuit Ib 6 amp then Corrected mV/A/m = 39.3830374753452 and Volt drop = 6.9 and the Cable length meters = 29.2003886373653 I use a java script program so have been able to check the readings over the years so know it is not far out if any thing in fact it was you John many years ago who helped me test the program to ensure no errors.
In real terms I suppose with a 6 amp overload we should assume the lights are even spaced so we could be considering the Design current for circuit Ib to be 3 amp rather than 6 for working out volt drop, so at 3 amp we get 59.8245434024215 meters of 1 mm² cable before we hit the 6.9 volt drop. It is better than double because of the correction factor changes so Corrected mV/A/m = 38.4457593688363 and the Zs would be 2.65Ω.
In days gone by with fluorescent fittings being so dependent on voltage the 3% volt drop for lighting was important, but with the move from wire wound ballast to electronic and the change to LED lighting don't really need the 6.9 volt drop figure any more. And although we are told traditionally to allow for 60W per lamp that also is now not the case, my living room which is quite large is 50 watt total and the kitchen is 98W (A 6 foot tube 70W plus LED tube 28W) dinning room 30W total bedrooms 10W each and landing 18W bathroom 25W plus a couple of wall lights so whole house everything switched on is around 300 watt so 1.3 amp approx with every lamp turned on. So in real terms the design current of even 3 amp is over egging it, work it out at 1.5 amp and looking at 120 meters max length and still using 6.9 volt maximum volt drop.
Which means unless the owner does something really silly like convert a room into a planetarium with little lights set into the ceiling then 1 mm² is good enough for most houses even without splitting upper and lower floor.
In real terms I suppose with a 6 amp overload we should assume the lights are even spaced so we could be considering the Design current for circuit Ib to be 3 amp rather than 6 for working out volt drop, so at 3 amp we get 59.8245434024215 meters of 1 mm² cable before we hit the 6.9 volt drop. It is better than double because of the correction factor changes so Corrected mV/A/m = 38.4457593688363 and the Zs would be 2.65Ω.
In days gone by with fluorescent fittings being so dependent on voltage the 3% volt drop for lighting was important, but with the move from wire wound ballast to electronic and the change to LED lighting don't really need the 6.9 volt drop figure any more. And although we are told traditionally to allow for 60W per lamp that also is now not the case, my living room which is quite large is 50 watt total and the kitchen is 98W (A 6 foot tube 70W plus LED tube 28W) dinning room 30W total bedrooms 10W each and landing 18W bathroom 25W plus a couple of wall lights so whole house everything switched on is around 300 watt so 1.3 amp approx with every lamp turned on. So in real terms the design current of even 3 amp is over egging it, work it out at 1.5 amp and looking at 120 meters max length and still using 6.9 volt maximum volt drop.
Which means unless the owner does something really silly like convert a room into a planetarium with little lights set into the ceiling then 1 mm² is good enough for most houses even without splitting upper and lower floor.
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What if your Ze is 0.8?
You calculate the size of the cable required and use that size. FYI, a lot of us just fit 1.5 as a default.
You appeared to be talking about the maximum cable lengths in terms of voltage drop (hence assuming that Zs was OK) and, since voltage drop is measured from the origin of the installation, Ze is surely irrelevant?
Kind Regards, John
That would be the sensible/'proper' course, but it seems that an awful lot of people don't do it - since I imagine that it would be pretty rare (in any 'ordinary' house) for that calculation to indicate that 1.0mm² cable was not adequate, yet it quite often gets installed. As you go on to say ...
... and, as above, it seems that you are far from alone. When you say "as a default", I presume you mean that they just install 1.5mm² cable, without any consderation of whether or not 1.0mm² would have been adequate?
Kind Regards, John
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