Another 12v Volt Drop Query

[Adam Withers]

Hi Guys,

I realise you have probably seen several of these previously, but just want to confirm as volt drop calcs are not my strong point

I would like to get lighting to the end of my garden where I cannot run 240v without ripping up a newly landscaped garden - confirmed by electrician (it is a long story not for this topic as to why this came about!).

So my thoughts have come to using 12v ELV to run LED lighting. I have calculated the length to be a generous 20 meters from the DC adapter.

From what I have seen on a calculator (eg from www.calculator.net/voltage-drop-calculator.htm) if I was to use a 2.5mm cable at max 10amps (from a 150w adapter) there would be a drop of ~5.3v, leaving ~6.7v? Is this correct?

I was also looking at using the Techmar lighting system AC-DC adapter, cabling and maybe some of their lighting (but may run some cheaper non-techmar items from it also), but there is no mention in their guides on any length. They just advise total up the watts for the lights, pick the adapter higher than total, add the cabling and your good. Their cabling does look quite thick, but not sure what they are. Is this right?

What are my options?

Cheers,

Adam.

 

[EFLImpudence]

I get 2.4V drop with the calculator ???
What have we done differently?

I get nearer 3V working it out myself using -
Resistance of 2 x 2.5mm² is 14.64mΩ/m (you wont need to add anything for conductor temperature rise; might be a bit less for cooler)
x 20m. x 10A /1000 = 2.928V drop, so 25%.

However, if your lights are evenly spaced from beginning to end of the cable, then you can halve that - or allow for whatever spacing the lights have.

 

[Adam Withers]

Ah- thanks for the reply. I selected 18 awg thinking the kcmil was the mm2.

So is a 3v drop acceptible, or does it depend on light I am connecting?

I will have a seperate circuit to run spaced lights, but i want them seperately switched.

Thanks again

 

[Adam Withers]

Having a search around I have seen a 5% (0.6v) drop is acceptable?
If this is the case, for a 20m cable run, I can have a maximum of 2.2amps or 26watts - is this correct
?

 

[ban-all-sheds]

I realise you have probably seen several of these previously, but just want to confirm as volt drop calcs are not my strong point

In any given length of cable, voltage drop is proportional to current.

So the total drop of any cable depends on its length, and how much current is flowing - the longer the cable, the greater the drop. The higher the current, the greater the drop.

It follows that one characteristic of a cable is its voltage drop for a particular current over a particular length and at the sorts of currents and lengths found in low and extra-low voltage final circuits, millivolts (of drop) per amp per metre are the most sensible unit.

So that table shows that, to use your scenario, 2.5mm² cable will have a drop of 18mV for every amp of current for every metre of length.

If you've got a 10A load, over your 20m the voltage drop will be

(18 x 10 x 20)mV = 3.6V

 

[oldbutnotdead]

You might be better off looking at higher voltage for the long run (48v AC) then local transformers, rectifiers etc to drop down to a voltage that won't pop your LED stuff.

 

[ban-all-sheds]

Or using a constant-current setup.

 

[FrodoOne]

Hi Guys,

I realise you have probably seen several of these previously, but just want to confirm as volt drop calcs are not my strong point

I would like to get lighting to the end of my garden where I cannot run 240v without ripping up a newly landscaped garden - confirmed by electrician (it is a long story not for this topic as to why this came about!).

So my thoughts have come to using 12v ELV to run LED lighting. I have calculated the length to be a generous 20 meters from the DC adapter.

From what I have seen on a calculator (eg from www.calculator.net/voltage-drop-calculator.htm) if I was to use a 2.5mm cable at max 10amps (from a 150w adapter) there would be a drop of ~5.3v, leaving ~6.7v? Is this correct?

I could be wrong but I believe that the answers given so far may assume that the 10 A load is along the entire length of the 20 Metre cable, which is unlikely to be so.
It is probable that you may wish to have a number of 12 V lamps distributed along this length of cable, with a total of 10 A from the power supply.

If you assume 10 lamps each drawing 1 A placed at 2 metre intervals and the use of 2.5 mm CSA paired cable (with a resistance per wire per single metre slightly higher than the 6.571 mΩ/m quoted for 13 AWG with 2.62 mm CSA) - say, 6.9 mΩ/m per single wire or 13.8 mΩ/m per pair, one can derive the following table (which I have not been able to display too well.)

………Distance (m)………Current (I)………Voltage Drop (V)
Sector……Cumulative…………………………Sector………Cumulative
2…………………2…………………10………………0.276……………0.276
2…………………4……………………9………………0.2484…………0.5244
2…………………6……………………8………………0.2208…………0.7452
2…………………8……………………7………………0.1932…………0.9384
2………………10……………………6………………0.1656…………1.104
2………………12……………………5………………0.138……………1.242
2………………14……………………4………………0.1104…………1.3524
2………………16……………………3………………0.0828…………1.4352
2………………18……………………2………………0.0552…………1.4904
2………………20……………………1………………0.0276…………1.518

Suffice to say that using this methodology, the voltage drop at the end of the line will be shown to be down by 1.518 V

Should you care to run a second cable from the power supply to the furthest point (and set up a "ring" supply), the voltage drops at each point - from the house to the 20 metre point - will become

0.272193
0.510465
0.715913
0.88878
1.028513
1.133476
1.200061
1.220127
1.172291
0.979355

 

[bernardgreen]

Another option would be a power supply with remote sensing,

This link saves me typing out an explanation

https://community.keysight.com/comm...te-sensing-is-important-for-your-power-supply

 

[bernardgreen]

Or using a constant-current setup.

Which one of the best solutions for powering up a set of lights spread out along a drive way ( or airport runway )

 

[JohnW2]

Which [a constant current setup] one of the best solutions for powering up a set of lights spread out along a drive way ....

Indeed (with only one operational downside - see below) - but would require the use of DC (or some approximation thereto). However, much as we often see that suggested (sometimes even by myself), I wonder how easy it is to find (I think not in 'consumer/retail' outlets) the 'bare LED element' lamps/bulbs/fittings that such a system requires - I certainly cannot recall often seeing them on offer, even in catalogues/websites of electrical wholesalers (although perhaps that's because I've never looked for them!)

... ( or airport runway )

The one 'operational' downside of putting a number of 'bare' LEDs (or incandescent bulbs/lamps) is series is that if one dies open-circuit, then all lights will go out. As has been discussed before, in the case of runway lights (when loss of all lights could be catastrophic), it seems that rather than put the lamps/bulbs (of whatever type) in series, they use AC and put a whole pile of (wirewound) transformers in series (one for each lamp). That's straightforward (even if expensive) with incandescent, but with LEDs would require some rectification (internal or external to the lamp) for each lamp.

Kind Regards, John

 

[ban-all-sheds]

the following table (which I have not been able to display too well.)

1. Change the extension of your spreadsheet to .pdf
2. Drag'n'drop it into your post
3. Instruct people to change the extension back to .xls, or .xlsx. or whatever after they download it

 

[ban-all-sheds]

Or f*rt about using [ CODE] tags and a monospaced font

 

[Adam Withers]

wow - thanks for all the replies on this.

So, in answer to FrodoOne, this specific run will be to power one a few lights at the top end of the garden - security lights on a PIR a switchable light in a shed a few others close by. I will have a separate circuit for some spaced-out lights, but this will be installed later once we know where we want them. Interesting point on making a ring to reduce the voltage drop further as this may work with where we want them.

The 48v solution looks like it may be a goer. Using the calcs to stay within 5% drop, I can get max 10a @ 48V over the 20m. I cannot locate a 240vAC-48vAC power supply, but I have seen a few to 48v DC and then use another to step down to 12V at the remote end. For example:
https://www.amazon.co.uk/Switching-...srs=6247905031&ie=UTF8&qid=1548600567&sr=8-27
or
https://www.amazon.co.uk/COLM-100-2...548604723&sr=8-9&keywords=48v+dc+power+supply

with
https://www.amazon.co.uk/DIGITEN-Wa...86&sr=8-4&keywords=48v+to+12v+dc-dc+converter

Am I barking up the wrong tree with these items or would I need something else.

I read the bernardgreen remote sensing, but couldn't locate and PSUs with it. Are you able to recommend somewhere sells them and also the shielded twisted cable required?

Thanks All,

Adam.

 

[JohnW2]

I read the bernardgreen remote sensing, but couldn't locate and PSUs with it. Are you able to recommend somewhere sells them and also the shielded twisted cable required?

I would personally have thought that that approach would be 'technological overkill' for garden lights!

Also, do you really need ~120W worth of LED lighting at the end of your garden?

I also suspect that you could probably get away with a lot more than a 5% voltage drop for such an application - but you might also consider using a source producing a bit more than 12V (if you could find one).

Kind Regards, John