EL wire brightness, per m not per m\$^2\$

Question

The datasheets and catalog pages for commercial electroluminescent (EL) wire in hobby quantities, e.g., adafruit's, list brightness in candelas per square meter. That's ok for two-dimensional EL sheet, but not for one-dimensional wire.

(I understand that frequency and voltage affect brightness, as well as half-life, power consumption, color, etc.)

How do figures like 46 cd/m\$^2\$ convert to cd/m, which can be meaningfully compared to LED strips or neon tubes? Just consider the circumference of the cladding? Or of the phosphor? If there's multiple layers of phosphor, which one?

Also, the candela measures luminosity only in a particular direction, while EL wire has nontrivial off-axis brightness. Are there specs for lumens per meter of such wire?

Answer 1

Wire does not produce light one-dimensionally - it produces light in all directions and the intensity of the light is maximum right at the surface of the wire.

Imagine the light is emitted in all directions along the length of the wire - there is real power entering into the space between the wire and your eyes and, at any particular distance from the wire, the sum total of that power remains the same (just like it is on a radio antenna).

What happens at a greater distance is that the power per sq metre reduces because the surface area that the power is passing through gets bigger proportional to distance squared.

This is why the lumen per metre does not really mean anything. It's lumens per sq metre.

Answer 2

The \$\frac{\text{cd}}{\text{m}^2}\$ units you are looking at are probably intended to give a measure of the Luminance.

It doesn't matter that the strips are, in some abstract sense, one-dimensional. They are in fact 3D objects which create a surface that emits light, and luminance is the brightness of that surface independent of its size and shape.

Let's make an analogy between luminance and, say, density. We wouldn't use a different measure of density for a steel wire compared to a steel plate or steel ball. That is, not if we had in mind the property of the material itself. There exists the concept of "linear mass": mass per unit length of a wire. But that isn't a property of the material, since it depends on the cross-sectional area of the wire.

Fact is that if these strips are made thicker (or doubled or tripled up, etc), they do give off more light, right? If luminance were given for the strip as \$\frac{\text{cd}}{\text{m}}\$ then it would mean that the thickness of the strip is being ignored: for such a measure to make sense as a constant, a given length of EL wire would have to give off the same amount of light no matter how thick or thin it is.

Answer 3

I regretfully propose to answer that an EL wire's datasheet is useless for comparing its brightness to that of other illuminants; one must buy samples and measure them oneself. (I still hope to be proven wrong.)

Answer 4

the numbers per square meter indicate perceived brightness of the surface, which does not say much about how brightly your EL wire is gonna light an object or surface. the luminous flux per meter varies depending on voltage and frequency of the driver and can reach around 16 lumens for 3 mm high-brightness EL wire.

that should be a good starting figure for your project. Take losses into account when you don't use optics in situations where the light emits only on one side (built-in wires in furniture etc)

Good luck and enjoy the light!

Answer 5

What you care about the is the surface, you don't need to care about which phosphor layer/ It is about the exitance of that surface.

With the OD (outer diameter) = D of the wire:

\$ 46 [\dfrac{cd}{m^2}] * \pi *D [m]\$ i.e. using the circumferential distance around the wire.

The next step is a little more complex because the emission from the surface in the direction along the wire is lambertian whilst around the wire it is isotropic. So your steradian calculation will be a volume that is a toroid, with eth wire running up the Z -axis (wire not shown).

I calculate this as \$ \dfrac{3}{4} \pi^2\$ [steradians}

combining all this...

\$ 46 * D * \dfrac{3}{4} * pi^3\$ [\$\dfrac{lm}{m}\$]