I'm trying to figure out how dense of a strip to buy for my project. Everybody knows that the white LED cobs in bulbs are far more efficient than incandescent bulbs but I'm having a hard time finding information specific to the RGB LEDs in strips.

The closest I came was looking at this data sheet for WS2812B: http://www.seeedstudio.com/document/pdf/WS2812B%20Datasheet.pdf which states an approximate 2.1cd of luminous intensity for its combined RGB output.

Assuming I get 5m of 30LED/m strip, that would be 315cd total. Trying to compare it to a bulb, I assumed its sri to be 180 degrees which would make it ~2000lm total for 45W.

The approximate average output of a 10W white LED bulb is 900lm so 45W worth would be 4050lm. That would mean that the RGB strip is less than half as efficient as its cousin white LED cob.

Is this conclusion anywhere near correct?

  • 1
    \$\begingroup\$ How can you compare the luminous intensity of white bulb and an RGB LED? This intensity is wavelength (read - color)-weighted. I.e. if you put the same amount of power into red LED and a green one, you will get far more bright light from the green one (as it is seen by human eye). \$\endgroup\$
    – Eugene Sh.
    Jun 1, 2018 at 21:30
  • 1
    \$\begingroup\$ @EugeneSh. I'd do it by assuming the RGB is driven to provide an exact D60 white point for human color perception via the CIE 1976 color space. Given specific LED details, that forces the current distributions. So it's quite doable. \$\endgroup\$
    – jonk
    Jun 1, 2018 at 21:54
  • \$\begingroup\$ @jonk - yes, but the resulting light isn't equivalent to white light of the same apparent colour, due to having a lower colour rendering index. Therefore any comparison in amount of power used is not particularly meaningful. \$\endgroup\$
    – Jules
    Jun 3, 2018 at 16:23
  • \$\begingroup\$ @Jules CRI is about "conscious or subconscious comparisons" of "color appearance." I don't think it refutes the approach I proposed. It just says different lighting spectra interact in complicated ways for the standard group of colors used in that test. Worse, from Edwin Land's results from the late 1970's, human perception changes profoundly if you place two or more of those colors next to each other under the same test lighting conditions -- something the index doesn't capture. In any case, I provided one defensible approach. Not the only one. \$\endgroup\$
    – jonk
    Jun 3, 2018 at 17:05

4 Answers 4


Costs of COB, RGB and phosphor

Those particular LEDs are individually addressible, and that brings a host of efficiency problems along.

  • The on-LED controller needs its own power.
  • Supply power is not a voltage that is ideal for LEDs - the LEDs want ~3V but the supply is ~5V. That's most likely a linear driver (I really doubt that chip includes the chokes for 3 buck converters), you are chopping 5V down to 3V linearly, discarding 40% of the energy before it even gets to the emitter proper.
  • Power distribution is at only 5V, which is extremely vulnerable to transmission losses. I've seen designs that called for 350kcmil + and - buses. You gotta be kidding.

The upshot is that individually addressable LEDs are optimized for building billboards and stadium TVs, not illumination.

Now if you want to go for the non-addressible RGB LED strips, they mostly cure these problems. No controller to waste power. Distribution is at 12V (somewhat better) or 24V (much better). 12V is better for powering a stack of 3 LEDs, you maybe lose 20% to the limiting resistor.

That said, the notchy R/G/B light is not going to look quite right to everyone, and won't have an excellent CRI. It's not white light, it's just designed to trick the eye into looking white. I would expect it to give the same kind of eye fatigue you get from looking at a screen for a long time.

On the upside, since the LED is internally doped to make that specific color, it's not using a phosphor to make the color, so it's not paying phosphor losses.

Now if you go with single-color white LEDs, those actually make blue internally and use phosphors to convert the blue into a full spectrum. Phosphors have an efficiency loss (and they cost money). That's why bluer light LEDs (6000k) are cheaper and more efficient than less blue LEDs (2700k).

Don't forget aiming

There's one more angle (heh) to efficiency. An old bulb emits spherical light. It even illuminates the bottom of its own socket. But nobody wants to light a sphere. They typically want a cone or wedge ranging from 8-120 degrees. Historically you solved that with inefficient reflectors.

LEDs are the first to give a cone of light, of roughly 140-160 degrees. And that lends itself to optical (lens) aiming, which is near 100% efficient. Now you don't have to light up the wall, the sky, the neighbor's bedroom window - you can just light the subject. That is a gigantic efficiency advantage that you should make full use of.

  • \$\begingroup\$ Siemens recommended the use of separate switching supply rails for each of R, G, and B, to match closely to the required forward voltage at maximum current ("100%" being determined by the D60 white point analysis and thus implying the forward voltage.) So 3 separate supply rails. Modest current control overhead voltage was also required, of course. (These were in 16x16 RGB modules. I remember the wattage at 80 W, or so, at 100% D60 white. But I don't recall the human perceived lumens. Probably not so good, I bet.) Nice answer, though. +1. \$\endgroup\$
    – jonk
    Jun 2, 2018 at 4:50

I assumed its sri to be 180 degrees which would make it ~2000lm total for 45W.

Given your assumptions ≈2,000 lm would be correct. What you do not have is the efficacy (lm/w) and you did not include the losses due to current regulation or light bulb diffuser transmission loss.

which states an approximate 2.1cd of luminous intensity for its combined RGB output.

Not sure where you got 2.1 cd.

enter image description here

Assuming I get 5m of 30LED/m strip, that would be 315cd total.

Using the average cd values I get 3.05 cd (0.900 + 1.250 + 0.300) per LED. 150 LEDs would then be 457.5 cd.

The problem here is they do not state at what current or view angle theses values are valid. The safe assumption is between 20 and 50 mA.

Converting candela to lumens requires the apex (view) angle of the spectral radiation distribution.

enter image description here

I assumed its sri to be 180 degrees

High and Mid Power Lighting LEDs are typically rated at a view angle of 120°. The RGB LEDs used in this strip are likely indicator LEDs where the angle is much less like maybe 10° (like those shown below).

enter image description here

180° may be the correct apex angle, maybe not.

Using 120°

457.5 × (2π(1 - cos(120°/2))) = 1,437 lm

Using 12°

457.5 × (2π(1 - cos(12°/2))) = 15.7 lm

LINK: cd to lm calculator

You datasheet is only for the LEDs and not how the manufacturer of the tape used them (e.g. LED resistor values).

Using the tape wattage use a loss of 46% for the resistors. The blue and green lose 40% to their resistors and the red 60%.

At 20 mA the LEDs would use about 14 watts, and the resistors 6.44 watts or 20 watts. So if your strip is 45 Watts the LEDs are drawing about 45 mA. If this is true the resistors would be about 45Ω.

The roundabout average output of a 10W white LED bulb is 900lm

This presents another problem. Incandescent light bulbs are isotropic (360°) and LEDs are directional. Usually 30° is subtracted for the base obstruction. Even though LED bulbs are not isotropic their luminosity is measured the same.

Bottom Line

There is NO WAY you can compare the efficacy of these RGB LEDs to white LEDs used in a quality light bulb.

You have insufficient data to guesstimate the lumens.

Even if you had all the data, you could still only guesstimate.
You would also need the transmission loss in the light bulb's diffuser cover.

The best way to determine the luminosity of the tape is to light up a dark room with the tape then light up the room with a light bulbs and compare perceived brightness.


Everybody knows that the white LED cobs in bulbs are far more efficient than incandescent bulbs

Most light bulbs use mid-power LEDs because they have been more efficient than CoBs. The highest efficacy LED currently available is the Samsung LM301B @ 229 lm/W. OSRAM just announced, in the past month, the highest efficacy CoB @ 201 lm/W. Currently available highest is a Samsung COB-D Gen3 @ 182 lm/W.

I removed the disffuser covers from some light bulbs I use around my house.

Notice the heatsink the CoB bulb uses.

enter image description here

  • \$\begingroup\$ This answer alone worth joining Electrical Engineering SE. \$\endgroup\$
    – Art
    Feb 20, 2019 at 22:47

They're different types of LEDs and not necessarily as efficient at producing visually-weighted light as the typical blue LED + phosphor combination.

There is another major factor- the forward voltage of the LEDs (depending on color) is around 2-3V, so about half the power is wasted in the (linear) driver circuit.

LED light bulbs use more efficient drivers, in part because they just need one big driver rather than hundreds of little ones.


48V ledstrips are more efficient than 24V which are more efficient than 12V which operate from dim at 9V to 14.2 on cars so that roughly 1/4 of the power is wasted on the series R .

Other than this supplier quality specs define the lumens per watt where 120 lpw is nominal +-50%


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