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Joined: Mar 2004
Posts: 9
samster Offline OP
Junior Member
Hi Guys,
I've been asked by a landscape contractor to provide power for a cord-connected transformer and some low voltage lights. When I arrived on site it was obvious that their 170'(1st light at 92’) run and #12 wire would not cut it. I've run some VD calcs and have come to the conclusion that the transformer needs to be within 10' to 20', however, I don't know what the acceptable VD on a 12v lighting circuit could be?
7 lights @ 35w each
12v transformer w/two poles 350w each
70' one way length from 1st light to last
Could some give me some advice on this please?

Thanks Sam

Joined: Jan 2003
Posts: 4,391
Hi Sam, I question the 350 watts per pole on the transformer.

The NEC limits the current on LV lighting systems to 25 amps, 350 watts at 12 volts is 29 amps.

The voltage drop issue is not an NEC concern but it will be the customers concern.

Even 1 volt loss on 12 volts is almost 10% and will result in a noticeable drop in brightness.

Your 7 lamps @ 35 watts = 245 watts, or 20 amps.

They are going to need something a good deal larger than 12 AWG for this.

Bob Badger
Construction & Maintenance Electrician
Joined: Jan 2003
Posts: 1,429
LK Offline
I just repaired a job close to your run length, and i used #8 all the way out, the owner was told by the landscape contractor that an electrician could fix it for about $150 well, the wire cost was up there using #8 and i had to dig up all the #14 that the landscape contractor buried, he told the owner they had electrical problems, and that,s why the lights were dim. OH YA!!
Don't forget to check that transformer rating
you may have to size up.

[This message has been edited by LK (edited 04-07-2004).]

Joined: Apr 2002
Posts: 7,387
Likes: 7
Basic Vd=KIL/A or A=KIL/Vd

Firstoff, is the xfr a true 12 volt, as some are 14/16 at the terminals, the factory "calculated" this in for Vd purposes

10.4 x 20 x 340 / 6530 = 10.83 Vd

Split the 'load'

10.4 x 11 x340 /6530 = 5.96 Vd

Increase to # 10 LV cabling....

10.4 x 20 x 340 / 10380 = 6.81 Vd


10.4 x 11 x 340 /10380 = 3.75 Vd

"IF" the xfr has 14 volts, you will squeek by with 10+ volts at the last fixture.

Setting the xfr in the center of the run...
like around 85' will help....

10.4 x 20 x 170 /10380 (#10) = 3.4 VD

Splitting the load....

10.4 x 11 * 170 /10380 = 1.87 VD
14 at xfr, 12+ at end.....


Joined: Mar 2004
Posts: 9
samster Offline OP
Junior Member
Thanks guys for helping me on this.
John, the only voltage calculation that I¡¦ve ever used are out of my Ugly¡¦s Book.
Vd=2KLI/Cm or Cm=2kLI/Vd with K=12.9 for copper and 21.2 for aluminum
However, I found a Vd calculator on the web for 12v car stereo systems and found that its calculations and yours are very close.
So if you could please help educate me on this I would be thankful. Yes it is a true 12v transformer.
Why did you choose 10.4 for your K and not 12.9
Why do you not multiply K by 2 in this example.
What would be a reasonable Vd percentage on this 12v circuit?


By the way I just got $.85/pound for un-striped wire at the scrap yard today. ļ

Joined: Jan 2003
Posts: 1,429
LK Offline
John's recommendation on center feed location for the transformer, would give you good results, and help keep the wire size down.

Joined: Mar 2004
Posts: 9
samster Offline OP
Junior Member
OK John I’m getting it,

10.4 for K is believed by most to be more accurate with 12.9 being to conservative. Right?

Doubling the length is the same as using 2k in my calculation. Sorry for being so dense sometime I rely on formulas to much and forget basic math.

My problem is that I have 170’ run with the first 100’ to a heavy stone deck. The 7 lights and 70’ of #12 wire are all underneath. They even have a wire coming out of the last fixture so more lights can be added later! I didn’t create this problem and would normally pass on a job like this (you know messed up by others), however, this landscape contractor is good for twenty to thirty thousand gross a year.
I’ve told them we may want to install transformers at each end (the tail may be useful after all). They (home owner and job foreman), however, think it is just a bad transformer and will replace it first. On things like this I usually just wait then someone 2 or 3 tears up the rung will call and I will finally be listened to.

I prefer putting a transformer at each end. If I install just one at the 1st light with the #12 wire the drop will be over 5v.
Would anyone consider putting a larger transformer on a circuit like this? I know they will ask. If yes how large would you go? Even if I get 12v to the first light then the last light would still be noticeably dimmer!


Joined: Jan 2004
Posts: 615
I would consider using a 24V transformer (with 24V bulbs). The bulbs will be pricier and probably special order. (I'm thinking of MR16)

As far as larger transformer:

In dealing with a similar situation, my brother (also electrician) contacted the transformer manufaturer. They told him to size the transformer as close to the load as possible. They told him too small of a transformer will cause premature failure (expectedly so) as will an oversized transformer (which was a suprise).

As with your situation, we were called in to make the lanscapers setup work after everything was laid. For some reason they only wanted to supply 300 & 1000 watt transformers. So we to re-spec. everything. Of course the landscape company rolled their eyes about our concerns (they do this type of install all the time).

Joined: Oct 2000
Posts: 2,723
Likes: 1
Broom Pusher and
A few things to consider for this installation:

[*] Larger Size Transformer,

[*] Split-Coil Transformer - with a slightly higher output Voltage,

[*] Placing the Transformer in the center of the runs - as mentioned before,

[*] Running several low voltage branch circuits to different areas, to reduce total load current on each circuit,

[*] Running low voltage multi-wire branch circuits.

For the first option, a larger Transformer, or multiple Transformers will reduce the loading effects on the Secondary Winding.

The second option would be to use a typical Isolated Transformer (like one used for Buck/Boost applications), which has Secondary Voltages of 16/32 VAC.

Third option already mentioned by members.

Fourth option would be to run several two-wire low voltage circuits out from the Transformer to the Lighting Fixtures, and break up the load currents into multiple runs.

Fifth option combines options #2 and #4, but the low voltage branch circuitry would be several 3-wire runs, which would be 16/32 VAC 1Ø 3 Wire.

Possibly, a combination of all listed items will work best.

My suggestions would be to go with two Split-Coil Transformers - sized larger than calculated loads, with 16/32 VAC Secondaries, and connect the low voltage Lighting to several 3-wire branch circuits - using #12 cu minimum.

Feel free to ask for more information if needed.


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Joined: Sep 2003
Posts: 650
The low end solution to consider: replace the bulbs with lower wattage bulbs.

As I figure it, including the efficiency change in tungsten filament lamps when the voltage drops, you will get _more_ light and more _even_ distribution if you simply change to 10W lamps. If you go to 10W bulbs without moving the transformer, and go to a transformer with a 14V output, then you might even be up to acceptable light output, about 2x your current output at the beginning of the string, and about 4x at the end.

If you can move the transformer to the end of the deck (so that you only have 70 feet of wire to deal with) and go to 15W lamps, then you will get about 2.5x your current light output.

If you can move the transformer _and_ go to a 24V system, then you will get essentially full light output.

I don't think that feeding both ends of the string is needed, if you get the transformer to 1 end of the string and go to lower wattage lamps or higher voltage.

Note: before you go to the various rewiring proposals, you mention that the transformer has two there any chance that this is a 24V center tapped transformer? Is there any chance that the wire run was with 12-3 rather than 12-2?

How much space is available at the light fixtures?

Here are some 'rule of thumb' equations for light output for Tungsten filament lamps:
Vn is the 'normalized rated voltage' where Vn=1 means 'full voltage' for the lamp in question.
Va is the 'normalized applied voltage', if the normal voltage for the lamp is 10V, and I apply 9V, then Va=0.9
In is the normalized rated current
Ia is the normalized current draw
Ln is the normalized rated light output
La is the normalized actual light output

Ia = Va^0.55
this means that as you reduce the voltage, the current goes down more slowly than Ohm's law would predict. Apply 5V to a 10V 1A lamp, and you will get a current flow of approximately 0.7A
Pa = Va^1.55
so rather than power being proportional to voltage squared as in a resistor, power remains the product of voltage and current, and current is not following Ohm's law.
The reason for this is that as the power dissipated in the filament changes, the temperature of the filament changes, and with the temperature change the resistance of the filament changes. Over short time periods where the temperature of the filament is essentially constant, the current does follow Ohm's law.

La =Va^3.55
This says that small changes in voltage lead to _big_ changes in light output.
If the voltage drops by 10%, then the light output will drop by 30%. The light color will also change noticeably. But the current will have only dropped by 6%, so even though you are getting lots less light output, you still get lots of voltage drop.


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