Max circuit length for LEDs powered by constant current driver.

[Pretty Mouth]

Hi all. Constant current LED lighting is a new concept to me. Not that difficult to get the head around, but I cannot find an answer to the following question:

How do I determine the maximum length of a ELV LED lighting circuit powered by a constant current driver?

For a constant voltage ELV circuit, a 5% voltage drop is the limiting factor - straightforward enough. For constant current, the voltage will rise (within its limits) to maintain the steady current, so a voltage drop calculation cannot be made.

I thought maybe this could be calculated by taking any surplus power from the driver, and allowing this be lost on the cable. From that, a resistance could be calculated from which a cable length could be determined.

For example:

Driver: 350mA, 9W
Load: 6W
Therefore surplus available power = 3W.

R=V/I

V cannot be determined, so let's use V=P/I, giving:

R = P/I²

R = 3/.35² = 24.5 ohms, which can be converted to a length of conductor of a chosen cross section.

Please let me know what you think, am I barking up the very wrong tree?

 

[ForkEnigma]

Hi all. Constant current LED lighting is a new concept to me. Not that difficult to get the head around, but I cannot find an answer to the following question:

How do I determine the maximum length of a ELV LED lighting circuit powered by a constant current driver?

For a constant voltage ELV circuit, a 5% voltage drop is the limiting factor - straightforward enough. For constant current, the voltage will rise (within its limits) to maintain the steady current, so a voltage drop calculation cannot be made.

I thought maybe this could be calculated by taking any surplus power from the driver, and allowing this be lost on the cable. From that, a resistance could be calculated from which a cable length could be determined.

For example:

Driver: 350mA, 9W
Load: 6W
Therefore surplus available power = 3W.

R=V/I

V cannot be determined, so let's use V=P/I, giving:

R = P/I²

R = 3/.35² = 24.5 ohms, which can be converted to a length of conductor of a chosen cross section.

Please let me know what you think, am I barking up the very wrong tree?

Yep barking up wrong tree.
A practical experiment would give you better answers. Keep adding cable until the lights fail. Also over certain lengths you can put a resistor in parrralel to make them work. You can also use bigger cable. Constant current, means connected in series right? Something else to think about. Also the voltage will only go down, can't see it going up.

 

[davesparks]

Constant current, means connected in series right? Something else to think about. Also the voltage will only go down, can't see it going up.

Constant current means that the source of power delivers a constant current and varying voltage, which is the opposite to our normal supplies which are constant voltage. Loads are usually connected in series yes, but that is not the same as saying 'constant current means connected in series'

The voltage will increase to maintain the current as the length of cable increases.

 

[Pretty Mouth]

For constant current, the voltage will rise (within its limits) to maintain the steady current, so a voltage drop calculation cannot be made.

I didn't word this very well in my OP, @davesparks explains it much better:

Constant current means that the source of power delivers a constant current and varying voltage, which is the opposite to our normal supplies which are constant voltage. Loads are usually connected in series yes, but that is not the same as saying 'constant current means connected in series'

The voltage will increase to maintain the current as the length of cable increases.

 

[Baddegg]

I’ve just read your original post at 0700 and now my head hurts ?

 

[James]

A 350mA constant current driver will deliver that current all the time.
So as you say, the voltage will rise and fall to provide that current exactly.
The power supply should have a voltage output range like 12v to 36v
If you know the load applied, you could work out it’s resistance.
You could then consider the cable to be a resistor in series.
If you need 12v across the load then you have a spare 24v to drop along the cable length.
V=I x r
R= v / i
= 24 / 0.35
=68.6 ohm max cable resistance

My head hurts too, need to sketch this out to be sure but I think I am on the right track.

 

[Pretty Mouth]

I’ve just read your original post at 0700 and now my head hurts ?

Is that due to the enormous complexity of the subject at hand, or the fact that it's clearly written by a simpleton trying to understand something beyond his skill set?

A 350mA constant current driver will deliver that current all the time.
So as you say, the voltage will rise and fall to provide that current exactly.
The power supply should have a voltage output range like 12v to 36v
If you know the load applied, you could work out it’s resistance.
You could then consider the cable to be a resistor in series.
If you need 12v across the load then you have a spare 24v to drop along the cable length.
V=I x r
R= v / i
= 24 / 0.35
=68.6 ohm max cable resistance

My head hurts too, need to sketch this out to be sure but I think I am on the right track.

Thanks for the reply @James . I'm glad it's not just me getting a headache from this, it's rather tricky to get your head around isn't it? What you say seems to make sense, I'm going to attempt an example:

These are the specs for a meanwell APC-8-350:

Rated current 350mA
Operating voltage range 11~23V
Rated power 8.05W.

Using P = VI,
P = 23X.35 = 8.05W, which confirms the above rated power.

Let's use 6W of lighting, and find the voltage across it:

V = P/I
V = 6/0.35 = 17.14V across the lighting

Now max. driver voltage, minus lighting voltage = 'spare' voltage
23V - 17.14V = 5.86V

Now the max. resistance of the cable:
R = V/I
R = 5.86/.35 = 16.74 ohms.

Can get quite a nice long cable out of that!

 

[Baddegg]

No mate it’s due to the simpleton reading the complex formula with a hangover ?

 

[Pretty Mouth]

Ok, I'm going to try the method I proposed in my OP, with the same driver and 6W lights, see if I get the same result.

So first find the surplus available power:

8.05 - 6 = 2.05W

R = V/I, and V = P/I, so:
R = P/I²
R = 2.05/.35² = 16.73 ohms!

Same result, just a different way of getting there. Thanks again @James , I'm fairly happy with this now

 

[ForkEnigma]

Ok, I'm going to try the method I proposed in my OP, with the same driver and 6W lights, see if I get the same result.

So first find the surplus available power:

8.05 - 6 = 2.05W

R = V/I, and V = P/I, so:
R = P/I²
R = 2.05/.35² = 16.73 ohms!

Same result, just a different way of getting there. Thanks again @James , I'm fairly happy with this now

Well done
I'm struggling to see how a spark would know this. Certainly way above the thinking of any installer.