DC circuit voltage drop issue

[electic]

I am trying to remember how to carry out a DC voltage drop calculation. I have got a 24V DC led driver ( constant voltage), 14 A supplying 6 number of led luminaries (36w each) and i am trying to check what the cable length which i can achieve for a 24V DC supply.

if i use the typical AC formula then the cable size would be huge

The LED drive output is 350W supply 24V DC.
the driver is about 50 meters away from the first light and the lights connected together with 6 meters away from each other around the building

Does anyone know if the formula is modified for DC.

 

[GoldDigger]

I am trying to remember how to carry out a DC voltage drop calculation. I have got a 24V DC led driver ( constant voltage), 14 A supplying 6 number of led luminaries (36w each) and i am trying to check what the cable length which i can achieve for a 24V DC supply.

if i use the typical AC formula then the cable size would be huge

The LED drive output is 350W supply 24V DC.
the driver is about 50 meters away from the first light and the lights connected together with 6 meters away from each other around the building

Does anyone know if the formula is modified for DC.

It is not modified in any way. But some VD calculators assume that you want them to calculate the two wire (out and back) voltage drop when you enter the one-way distance IF you select DC. And the one wire voltage drop if you select three-phase.
The assumption varies if you select single phase.

What is very different is that for 24V DC a 2.4V VD is a 10% drop while for 240AC a 2.4V VD is only a 1% VD. That happens because of the lower voltage, not because it is DC.

Many LED drivers can accept a wide range of input voltages, and as a result the light output will not go down even with a fairly large voltage drop. But it does reduce the overall efficiency and produce waste heat in the wires.

For a string of lights at regular intervals, you need to calculated the voltage drop for each individual run of wire. Only the first 50 meters will have to carry the full current and the last 6 meter segment will only carry the current of one LED unit.

 

[Besoeker]

I am trying to remember how to carry out a DC voltage drop calculation. I have got a 24V DC led driver ( constant voltage), 14 A supplying 6 number of led luminaries (36w each) and i am trying to check what the cable length which i can achieve for a 24V DC supply.

if i use the typical AC formula then the cable size would be huge

The LED drive output is 350W supply 24V DC.
the driver is about 50 meters away from the first light and the lights connected together with 6 meters away from each other around the building

Does anyone know if the formula is modified for DC.

GD is right. Up to about 16mm^2, around what you need to be, the difference in voltage drop between DC and AC is neglible.

You gave the distance in metres so you are obviously familiar with metric (SI) units but just for completeness 16mm^2 is the equivalent of 5AWG which may not be available to buy so you would have to go up to the next available size.

 

[gar]

170329-0841 EDT

electic:

I need more information.

What is the V-I (current vs voltage curve) characteristic of one light? Does each light have an internal driver?

If a light looked like a resistance, then I = K*V. If K is exactly the same for each light, then the lights could be connected in series with a line current of about 36/24 = 1.5 A. Six lights would require a source voltage of 144 V at 1.5 A which equals 216 W = 6*36. Suppose your total wire resistance was 1 ohm, then wire voltage drop would be 1.5 V, insignificant.

If K was slightly different for each light, then the series connection would have a slightly different intensity for each light.

If each light was more like a constant voltage load, inherently each little LED device is, then it needs to be driven by a constant current source. If each light fixture internally contains a driver that converts its input voltage to a constant current, or closer to a constant current, to the LED elements in the fixture, then you many not care too much about input voltage to a fixture.

The implication is that the fixtures require a DC source. This does not tell us if there is an internal driver or resistor.

Light strings of LEDs internally contain resistors that tend to make the string look more like a resistor, than would just the string of LEDs alone. This is a power wasting approach, but it makes a simple inexpensive light that can be connected to a car battery in automotive application.

If you take a single LED and measure current vs voltage you will see current vary very rapidly with voltage after a certain voltage threshold is reached. For example see http://www.electronics-tutorials.ws/blog/i-v-characteristic-curves.html .

.

 

[Julius Right]

According to BS7671 Table 4D1A Single-core 70oC copper conductor [two wire a.c. or d.c.]
1.5 mm^2 it is the minimum required cross section area.
As GD already said , for 230 V [400/SQRT(3)]- as usual in IEC world- it will be good up to 12 m.
If the supply voltage is only 24 V D.C.- in your case- and the final conductor temperature will be 70oC you need 95 mm^2 copper in order to get no more than 5% voltage drop.
However, luckily, the conductor temperature will be very low[31.2 oC if the ambient air is 30 oC]
so 50 mm^2 will be fair.:happyyes:

 

[gar]

170329-0944 EDT

To continue a little more.

Find out more about how one of your LED fixtures behaves with varying input voltage. Also what is the maximum rated input voltage. If there is an internal driver, then a fixture might look like a constant power load. If this is the case, then light intensity won't change much with input voltage, and your only concern would be maximum input voltage.

Depending upon light fixture characteristics and specifications you might be able to run them in parallel and feed with #14 wire, because voltage drop would not be a real issue. Just the source power supply would need a high enough voltage to over come the full load voltage drop.

.

 

[Besoeker]

According to BS7671 Table 4D1A Single-core 70oC copper conductor [two wire a.c. or d.c.]
1.5 mm^2 it is the minimum required cross section area.
As GD already said , for 230 V [400/SQRT(3)]- as usual in IEC world- it will be good up to 12 m.
If the supply voltage is only 24 V D.C.- in your case- and the final conductor temperature will be 70oC you need 95 mm^2 copper in order to get no more than 5% voltage drop.
However, luckily, the conductor temperature will be very low[31.2 oC if the ambient air is 30 oC]
so 50 mm^2 will be fair.:happyyes:

I'm not sure if this is of relevance to the OP.
I mentioned 16mm^2. That would give about 5% drop using the values in table 4D1B.

 

[Julius Right]

I'm not sure if this is of relevance to the OP.
I mentioned 16mm^2. That would give about 5% drop using the values in table 4D1B.

The o.p. said:
“14 A supplying 6 number of led luminaries
50 meters away from the first light and the lights connected together with 6 meters away from each other”
I checked ones again my -rough-calculation and I think 35 mm^2 will be fair enough indeed.:ashamed:
In my opinion, if you use the 2.8 mV/m/A [=2.8 ohm/km] 50 m up to first LED and 6*5/2=15 m up to last will get 10.6% voltage drop.
You may reduce the conductor temperature from 70oC to 30+(70-30)/16*1.5=33.75oC then the resistance will be 2.8/(234.5+70)*(234.5+20)*(234.5+33.75)/(234.5+20)=2.47 ohm/km.
However the voltage drop up to last LED will be 9.37%.:weeping:

 

[Besoeker]

The o.p. said:
“14 A supplying 6 number of led luminaries
50 meters away from the first light and the lights connected together with 6 meters away from each other”

I think 14A is the rating of the supply. Six 36W fittings would need 9A at 24V.

 

[Phil Corso]

Electric,

Presuming the 350W power supply unit is being fed from an AC source, why don't you install a variable transformer like Variac!

Regards, Phil Corso

 

[Smart $]

Electric,

Presuming the 350W power supply unit is being fed from an AC source, why don't you install a variable transformer like Variac!

Regards, Phil Corso

Most LED drivers outputs are regulated. They do not output more watts simply by increasing the input voltage.

 

[Phil Corso]

Smart_A,

Thaks for the UD on the LeD!

Phil

 

[Electric-Light]

I am trying to remember how to carry out a DC voltage drop calculation. I have got a 24V DC led driver ( constant voltage), 14 A supplying 6 number of led luminaries (36w each) and i am trying to check what the cable length which i can achieve for a 24V DC supply.

if i use the typical AC formula then the cable size would be huge

The LED drive output is 350W supply 24V DC.
the driver is about 50 meters away from the first light and the lights connected together with 6 meters away from each other around the building

Does anyone know if the formula is modified for DC.

Volt drop per length of any given cable is a function of amps. You need 4 AWG if you want 24v 15A 50m away within about 5% drop. If the load is dynamic (flash, blink outdoor light emitting decorative signs), the non blinking portions will flicker considerably. In 12 and 24v circuits, you will run into voltage drop constrained limit long before temperature rise limit. So your prediction is correct. Given the same cable, percent drop per watt at 24v is 100 times what it is at 240v, because you need 10 times the amps to support the same watt.

That design is poor. You should bring the power supply closer to the load and make as much of the 50m run as you can on the AC side.

Most LED drivers outputs are regulated. They do not output more watts simply by increasing the input voltage.

Smart_A,

Thaks for the UD on the LeD!

Phil

There are two general types. The first type is like any DC power supply. 12v or 24v voltage source that supply individually ballasted LED strings.
The other type is an LED ballast, which is a lot more like a neon ballast and provides the same mA within a range of voltage. The cabling on the latter type is only impacted by the temperature rise limit of the cable. Although a long run will lower the efficiency. The LED string used for the latter type is connected in a long chain and generally do not use a load side ballast.

 

[electic]

Hello All

Hello All

Thank you all for your replies
please find attached files for the light arrangements and driver data sheet for a clearer picture

 

[Smart $]

Thank you all for your replies
please find attached files for the light arrangements and driver data sheet for a clearer picture

The data sheet indicates the output voltage is adjustable up to 26.4V. You should be able to use that to compensate for voltage drop.

 

[Electric-Light]

Still a bad layout to feed in from one end like that.

There are three sets of terminals so you can have three sets of cables land on the terminal block.

 

[steve66]

Voltage drop is the big disadvantage with low voltage lights. Is there any way to run 120V line voltage and put the driver much closer to the first light?

If not, you might be better off with multiple runs. Maybe even a dedicated pair of wires for each light.

 

[Electric-Light]

This is a much preferred layout unless you have constraints preventing you from doing this.

Red line is 240v.