Is it possible to calculate an incandescent bulb's temperature from V?

[HomeExperiement]

Hi!

I have a question:
Is it possible to calculate incandescent light color temperature based on it''s input voltage? For example, if for example 100w light bulb rated for 220V gives light with color temperature of 2800k then what would color temperature be at 110v? As voltage is now 2 times lower, the power (in watts) should be 4 times less (25W). But can it be directly applied to color temperature and say that it now gives 700k light or does it change some other way?

PS sorry if it's in wrong forum, didnt really know what subforum exactly to choose.

 

[Drakkith]

I don't believe you can calculate color temperature based on the voltage and power of the bulb. Two bulbs can be the same color temperature despite being very different in terms of wattage and voltage, and two bulbs of the same wattage and voltage can be very different in color temperature.

That being said, if you're varying the voltage on the same bulb, you might be able to calculate the new color temp since you already know the properties of the bulb, but I don't really know how to do that.

 

[phinds]

I don't think so. Color temperature is a function of how the bulb is constructed and the coating used, not the voltage at which it operates. Lowering the voltage lowers the intensity, not the color. I could have 60Watt, 100Watt, 150Watt, all running at 110volts and all at the same color temp but I could just as well buy a different type bulb and have a different color temp at the same voltage and wattage.

EDIT: I see Drakkith beat me to it.

 

[hutchphd]

Hi!

I have a question:
Is it possible to calculate incandescent light color temperature based on it''s input voltage? For example, if for example 100w light bulb rated for 220V gives light with color temperature of 2800k then what would color temperature be at 110v? As voltage is now 2 times lower, the power (in watts) should be 4 times less (25W). But can it be directly applied to color temperature and say that it now gives 700k light or does it change some other way?

PS sorry if it's in wrong forum, didnt really know what subforum exactly to choose.

I think your method could be made to work but two problems with your approach:

1. the energy output of a blackbody goes like T⁴
2. the resistance of the filament is not constant but is (very roughly) linear with T

with those assumptions what do you get?

 

[sophiecentaur]

The problem here is that resistance does not 'scale'. The resisance R of a filament will depend on its length, cross sectional area and the resistivity ρ.
R=ρl/A
The operating temperature of a filament will depend on its surface area (not A, btw) and the Power it dissipates.
The operating Power will be V²/R.
Two filaments may have the same Resistance (hence the same Power) but their surface areas can be different so the surface temperature can be different.
The actual structure of a filament (coil vs coiled coil etc) will affect the surface temperature for a given Power. It will also affect the life of the filament.
On an experimental note: you could always do some measurements on the bulbs of interest and get some calibration points for a set of V vs T and a Fourth Power curve of best fit. It would depend on the context of the OP's actual requirement.

 

[hutchphd]

OP: Same bulb, half the voltage, I believe.

 

[Baluncore]

It will be difficult to accurately predict temperature from voltage because the thermal transmission through the envelope is dependent on wavelength, and the filament tends to be current regulating due to the thermal coefficient of resistance.

Start by realising that the electrical resistance of an elemental metal filament is almost linear with absolute temperature. Knowing the voltage and measuring the current gives the resistance, which is the most reliable way to know colour temperature.

First use a low power digital ohmmeter to measure the cold resistance, Rr, of the filament at room temperature. Also measure room temperature, Tr. From that compute the linear term of the thermal coefficient of resistance a = Rr / Tr. You then know the critical detail about the individual bulb you are using, which allows you to compute actual colour temperature from measured I and applied V.

For example, if for example 100w light bulb rated for 220V gives light with color temperature of 2800k then what would color temperature be at 110v?

At T = 2800K, rated power = 100W, rated voltage = 220V; ∴ current I = 100W / 220V = 0.4545 amp; and resistance R = 220 / 0.4545 = 484.0 ohms. Tempco, a = 484 / 2800 = 0.1728 ohms per kelvin.

Assume room temperature is 20°C = 293 K. Cold filament resistance will be;
Rcold = 484 ohms * 293K / 2800K = 50.65 ohms. You can check that with a digital ohmmeter.

We know that at 220V, most of the 100W filament power is lost as heat with a temperature difference of 2800K - 293K = 2507K.
The thermal resistance of the globe envelope is about 2507K / 100W = 25.07K per watt.

Now you must solve two equations, or guess at the numbers.
Assume 25 watt. Filament temp = ( 25W * 25.07K ) + 293K = 919.75K
Filament resistance = 919.75 * 0.1728 = 158.93 ohms.

Guess an applied voltage = 100V. compute I = 100V / 158.93R = 0.6292 amp.
Power = 100V * 0.6292A = 62.92W.
Repeat the guessing game;
2'nd guess; V = 75, I = 75 / 158.93 = 0.4719 amp; Power = 75V * 0.4719A = 35.39 W.
3'rd guess; V = 63, I = 63 / 158.93 = 0.3964 A; Power = 63 * 0.3964 = 24.97W. Which is close to the 25W assumed earlier, that will produce a colour temp of about 920K at 63 volts.

Now using the same relationships you can solve for the approximate colour temp at 110V.
Remember that the thermal resistance of the envelope is confoundingly dependent on wavelength and resistive thermal feedback.

 

[OmCheeto]

...
On an experimental note: you could always do some measurements on the bulbs of interest and get some calibration points for a set of V vs T and a Fourth Power curve of best fit.
...

That's kind of what I did this morning. Kind of, as in, I just googled and looked at other peoples data.

My 4th power curve fit: Temp = 343.03 * voltage^0.38
[for a 250 volt bulb]

volts	Kelvin
250.0​	2800.0​
125.0​	2150.0​

Of course, as everyone has alluded to, this equation is good for only "my" hypothetical light bulb.

 

[hutchphd]

Like I said: if R is proportional to T and the loss is radiative σT ⁴ it is not difficult to show

(T₂/T₁)=(V₂ / V₁)^(2/5)​

This is a black body radiation problem...

 

[Baluncore]

This is a black body radiation problem...

Heat is lost to room temperature, while the filament resistance and colour temperature are proportional to absolute temperature. How does that change your equation?

 

[Baluncore]

At T = 2800K, rated power = 100W, rated voltage = 220V
Notice how the current is limited at higher powers.

thermal variables
Troom = 273 + 20    ' abs room temperature assumed 20C
Tfil = 2800         ' abs filament temperature
w = 100             ' rated wattage
ta = ( Tfil - Troom ) / w   ' thermal resistance, kelvin per watt

electrical variables
v = 220             ' rated voltage
i = w / v           ' current
r = v / i           ' resistance
ra = r / Tfil       ' ohms per kelvin

given power w, compute voltage.
Tfil = ( ta * w ) + Troom   ' filament temp
r = Tfil * ra       ' filament resistance
v = Sqr( w * r )    ' voltage
i = v / r           ' current

  power W   temp K    R ohms   voltage    I amp
     0.0      293      50.6       0.0     0.000
     2.5      356      61.5      12.4     0.202
     5.0      418      72.3      19.0     0.263
     7.5      481      83.1      25.0     0.300
    10.0      544      94.0      30.7     0.326
    12.5      606     104.8      36.2     0.345
    15.0      669     115.7      41.7     0.360
    17.5      732     126.5      47.0     0.372
    20.0      794     137.3      52.4     0.382
    22.5      857     148.2      57.7     0.390
    25.0      920     159.0      63.0     0.397
    27.5      982     169.8      68.3     0.402
    30.0     1045     180.7      73.6     0.408
    32.5     1108     191.5      78.9     0.412
    35.0     1170     202.3      84.2     0.416
    37.5     1233     213.2      89.4     0.419
    40.0     1296     224.0      94.7     0.423
    42.5     1358     234.8      99.9     0.425
    45.0     1421     245.7     105.1     0.428
    47.5     1484     256.5     110.4     0.430
    50.0     1547     267.3     115.6     0.432
    52.5     1609     278.2     120.8     0.434
    55.0     1672     289.0     126.1     0.436
    57.5     1735     299.8     131.3     0.438
    60.0     1797     310.7     136.5     0.439
    62.5     1860     321.5     141.8     0.441
    65.0     1923     332.3     147.0     0.442
    67.5     1985     343.2     152.2     0.444
    70.0     2048     354.0     157.4     0.445
    72.5     2111     364.8     162.6     0.446
    75.0     2173     375.7     167.9     0.447
    77.5     2236     386.5     173.1     0.448
    80.0     2299     397.3     178.3     0.449
    82.5     2361     408.2     183.5     0.450
    85.0     2424     419.0     188.7     0.450
    87.5     2487     429.8     193.9     0.451
    90.0     2549     440.7     199.1     0.452
    92.5     2612     451.5     204.4     0.453
    95.0     2675     462.3     209.6     0.453
    97.5     2737     473.2     214.8     0.454
   100.0     2800     484.0     220.0     0.455

 

[hutchphd]

Heat is lost to room temperature, while the filament resistance and colour temperature are proportional to absolute temperature. How does that change your equation?

The filament is very much isolated thermally...it is enclosed in an evacuated sphere suspended on thin wires.
The room will radiate back onto the filament but it is at 300K and so this effect is very small (≈10^-3) .
This is a deliciously simple example of a black body...

 

[Baluncore]

The filament is very much isolated thermally...it is enclosed in an evacuated sphere suspended on thin wires.

I disagree. The envelope is filled with an inert gas. Gas pressure when cold is less than 1 atm, when operating hot, pressure is greater than 1 atm. That increases lifetime by reducing filament evaporation and the deposition of metal on the inside of the glass envelope.

 

[hutchphd]

The filament is quite small and argon is not easy to heat radiatively. I think the effect will be very small. How did you choose the value for thermal resistance for the calculation?

 

[HomeExperiement]

Thanks. I knew that filament would change the resistance. But the difficulty for me calculate the resistance at given voltage. for example, my 42W halogen bulb (I don't have access to 100W incandescent right now) that I tested has resistance of 84 Ohm, which at 238V would draw 674W power but it clearly isnt. But I just didnt know before what formula to exactly use for resistance.

I am also not 100% sure, if temperature of filament matches the color temperature of emitted light (anyone can confirm this?). That table posted here by Baluncore looks great. Only problem is that that to verify it I would need expensive spectrometer as DSLR and phone apps can easily lie +/- 300k, and with ultra warm lights, DSLR can even lie with +/-500k. I asked this because I was wondering if I could somehow easily get reference for different light colors - since incandescent seemed easy and predictable, I hoped I could use it to easily get color within +/- 100k from desired point.

 

[OmCheeto]

I am also not 100% sure, if temperature of filament matches the color temperature of emitted light (anyone can confirm this?).

According to one graph I looked at and digitized this morning, they are pretty close.
Color Temp = 1.0495 * (actual temp) - 41.914
Their color temp axis ended at 3500 K, where the discrepancy was 3.8%

[ref: USHIO pdf page 22 figure 48]

 

[Baluncore]

How did you choose the value for thermal resistance for the calculation?

thermal variables
Troom = 273 + 20 ' abs room temperature assumed 20C
Tfil = 2800 ' abs filament temperature
w = 100 ' rated wattage
ta = ( Tfil - Troom ) / w ' thermal resistance, kelvin per watt

 

[hutchphd]

So where does the energy for the light come from.??...if I understand you are just conducting all the energy away. That is clearly not the physics. Light bulbs radiate.

 

[Baluncore]

So where does the energy for the light come from.??...if I understand you are just conducting all the energy away. That is clearly not the physics. Light bulbs radiate.

The optical and some IR is radiated, but that is only a small percentage of the total energy that heats the filament. The majority of energy lost from a gas filled globe is not conducted, it is convected. There is very little net outward radiation at the bottom end of the model.

My model is based on only three published parameters; T, W and V. In effect the model interpolates between zero at the bottom end, and the rated parameters at the top end. Based on only three parameters, it cannot be simpler or more accurate.

The electrical resistance of a tungsten filament due to temperature has a slight curvature that can be modeled as a T squared term added to a T term. That would give only a slight improvement to the model, while it would make it too complex for beginners.
Also, the lack of the T squared term is swamped by the major confounding effect which is the assumed constant thermal resistance. Interpolation between specified end-points has a significant accuracy improvement over extrapolation using guessed parameters.

The poor efficiency of light production from a hot filament suggests the temperature of the filament will be determined by bulk IR rather than optical transmission. Last time I measured and plotted the characteristics of a tungsten filament, (circa 1972), I found slight ripples, (only a few percent), in the current against voltage graph. They only appeared near the knee in the graph. I assumed they were due to the IR transmission of the gas filling and/or the glass envelope. The black body radiation from a hot filament is too broad-band to resolve the sharp spectral transmission characteristics expected for an unspecified glass and gas fill. That explains why the plotted ripples are small.

Again, I am interpolating between end-points. With unknown materials, and only three model parameters, there is really nothing further that can be done about estimating thermal resistance.

As I see it, at the top end we have about 5% black body radiation, maybe 5% thermal conduction and 90% thermal convection. I agree the model does not have a perfect interpolaton transition, but it is the best 3 parameter approximation available, and I am prepared to live with it.

 

[OmCheeto]

Notice how the current is limited at higher powers.

Not really. Gobs of numbers are somewhat confusing to me.
Your comment didn't make sense until I compared it graphically to a fixed resistance.

Fascinating.

ps. The title of a previous PF post now makes more sense: Stupid Light Bulbs

 

[sophiecentaur]

Gobs of numbers are somewhat confusing to me.

Me too. Especially when the units are not familiar. A good old graph delivers the message - and your two are great.

 

[Baluncore]

Your comment didn't make sense until I compared it graphically to a fixed resistance.

I notice for the fixed resistance, your green Temp K line curves differently to the blue Power line. The tungsten filament model has the opposite characteristic.
How did you compute the temperature of the fixed resistor?

 

[hutchphd]

The optical and some IR is radiated, but that is only a small percentage of the total energy that heats the filament. The majority of energy lost from a gas filled globe is not conducted, it is convected. There is very little net outward radiation at the bottom end of the model.

This is incorrect.

For a 100W incandescent light at 2800K the luminous efficacy is ~15 lm/W

https://en.wikipedia.org/wiki/Incandescent_light_bulb
The luminous output from a typical 100W bulb ~1200 lm (its on the package). The radiation luminous efficiency of the bulb is therefore 80% . A lot does radiate in the IR.

Therefore only 20% of the input power goes out via conduction and convection.

Can you quote a source for your claim ??

 

[Baluncore]

For a 100W incandescent light at 2800K the luminous efficacy is ~15 lm/W

A comparison requires the units have the same dimensions. [edited by moderator] The question should be "How many watts of light do you get from the 100 watts dissipated in the filament?". That figure is usually quoted as about 5%.

[edited by moderator] If you think you have a better numerical model then please present it here. [edited by moderator]

[edited by moderator]

 

[OmCheeto]

I notice for the fixed resistance, your green Temp K line curves differently to the blue Power line. The tungsten filament model has the opposite characteristic.
How did you compute the temperature of the fixed resistor?

It's computed using the Stefan-Boltzmann equation: j* = σT⁴
Since the area of the filament doesn't change(much), I simply redefined a new sigma type constant at 100 watts @ 2800 Kelvin.
I labeled it "s".
s = 100/(2800^4) = 1.63E-12
Then I reversed the Stefan-Boltzmann equation to derive the temperature from the given power.
T=(p/s)^(1/4)

Interesting. If I re-plot your data using my method, I get a different graph.

It appears we have a difference of opinion on how to approach this problem.
Perhaps we've gone off topic. Are we still discussing "filament color temperature vs voltage"?

Is it possible to calculate incandescent light color temperature based on it''s input voltage?

 

[hutchphd]

If you think you have a better numerical model then please present it here so we can all benefit from your wisdom.

Like I said: if R is proportional to T and the loss is radiative σT ⁴ it is not difficult to show

(T₂/T₁)=(V₂ / V₁)^(2/5)​

This is a black body radiation problem...

 

[anorlunda]

It seems that this thread got overheated (pun intended) without good reason.

The complete Wikipedia quote is this

Incandescent bulbs are much less efficient than other types of electric lighting, conveting less than 5% of the energy they use into visible light.[1] The remaining energy is lost as heat. The luminous efficacy of a typical incandescent bulb for 120 V operation is 16 lumens per watt, compared with 60 lm/W for a compact fluorescent bulb or 150 lm/W for some white LED lamps.[2]

We have both energy efficiency and luminous efficacy to consider.

Luminous efficacy is a measure of how well a light source produces visible light. It is the ratio of luminous flux to power, measured in lumens per watt

So it seems to me that the luminous efficacy applies to the 5% of the energy that goes to light, and ignores the 95% that goes to heat.

It also seems that there is opportunity for misestimation if we blend gas filled and vacuum bulbs, especially the convective component of heat removal. That Wikipedia quote fails to specify gas or vacuum incandescent bulb.

 

[Baluncore]

It appears we have a difference of opinion on how to approach this problem.Perhaps we've gone off topic. Are we still discussing "filament color temperature vs voltage"?

We are on topic, but the question has now become “how can you best estimate colour temp based on filament voltage.

I contend that you need to measure both filament current and voltage. From that you can compute resistance to maybe three decimal places. Then using the polynomial relationship between temperature and the resistivity of tungsten, you can accurately calculate the filament temperature.

An old method used to measure the temperature of a furnace was to view the furnace through a filament globe. That showed the outline of the filament against the hot background. The filament voltage was then adjusted until the filament became invisible against the background, neither darker nor lighter. At that point the filament resistance was computed from V and I, so the temperature could be known. That method cancels most of the spectral band variation in the temperature comparison.

But the method does not actually measure the colour temperature of a black body, which is a broad weighted spectrum, much of which is unseen by the eye.
We need to define why we need to measure or know, and what we mean by colour temperature.
Maybe with narrow band LEDs and three colour cameras and displays colour temperature has become an irrelevant concept.

 

[hutchphd]

So it seems to me that the luminous efficacy applies to the 5% of the energy that goes to light, and ignores the 95% that goes to heat.

If you put the word "visible" in front of light we are in agreement. And <10% of the radiant energy is visible light for a 2800K black body so >50% of the total input energy is radiation., mostly IR.
But IR and conducted heat are not the same, and that was my point. It was not my purpose to offend, but the physics is quite different..

 

[Eric Bretschneider]

This is incorrect.

For a 100W incandescent light at 2800K the luminous efficacy is ~15 lm/W

https://en.wikipedia.org/wiki/Incandescent_light_bulb
The luminous output from a typical 100W bulb ~1200 lm (its on the package). The radiation luminous efficiency of the bulb is therefore 80% . A lot does radiate in the IR.

Therefore only 20% of the input power goes out via conduction and convection.

Can you quote a source for your claim ??

You can calculate the theoretical efficacy of an incandescent bulb as a function of temperature. This isn't overly difficult. It does give you an upper limit on what can be achieved.

1. Choose a temperature and calculate the spectral power distribution using Planck's Law
2. Use the photopic response curve to calculate the LER (Luminous Efficacy of Radiation) of the SPD

This gives the following:
2200 K = 3.676 lm/W
2400 K = 6.691 lm/W
2600 K = 10.863 lm/W
2800 K = 16.141 lm/W
3000 K = 22.365 lm/W

Of course you should take into account the emissivity of tungsten as a function of wavelength and temperature.

And all of the energy goes into heating the filament. That's the input. The question is the output divided up.

 

[Eric Bretschneider]

Of course rather than arguing who has the best theoretical model/knowledge, you could take advantage of the fact that Illuminating Engineering as a field is over 100 years old and take advantage of the information base of that field.

The following is from "Lighting Handbook: Reference and Application" (8th ed.) from the IES (Illuminating Engineering Society) p.186

Life1/Life0 = (Volt0/Volt1)^13
Lumen1/Lumen0 = (Volt0/Volt1)^1.9
Efficacy0/Efficacy1 = (Volt0/Volt1)^3.4
Watt1/Watt0 = (Volt1/Volt0)^1.6
CCT1/CCT0 = (Volt1/Volt0)^0.42

Subscript 0 is rated and subscript 1 = adjusted value

There is of course some variation between manufacturers. The exponents are not intended to be applicable to huge differences in input voltage. In the absence of experimental data this should suffice.

 

[Eric Bretschneider]

We are on topic, but the question has now become “how can you best estimate colour temp based on filament voltage.

I contend that you need to measure both filament current and voltage. From that you can compute resistance to maybe three decimal places. Then using the polynomial relationship between temperature and the resistivity of tungsten, you can accurately calculate the filament temperature.

An old method used to measure the temperature of a furnace was to view the furnace through a filament globe. That showed the outline of the filament against the hot background. The filament voltage was then adjusted until the filament became invisible against the background, neither darker nor lighter. At that point the filament resistance was computed from V and I, so the temperature could be known. That method cancels most of the spectral band variation in the temperature comparison.

But the method does not actually measure the colour temperature of a black body, which is a broad weighted spectrum, much of which is unseen by the eye.
We need to define why we need to measure or know, and what we mean by colour temperature.
Maybe with narrow band LEDs and three colour cameras and displays colour temperature has become an irrelevant concept.

Color Temperature is only valid for light sources with chromaticity coordinates that lie on the Planckian Locus (the locus of points for all emitters whose spectra follows Planck's Law).

For light sources with chromaticity coordinates that are near the Planckian locus, a different metric: Correlated Color Temperature (CCT) is used. Correlated Color Temperature is defined as the minimum distance between the chromaticity coordinates of a given light source and the Planckian locus as measured using CIE 1960 uv chromaticity coordinates.

CCT is still used and is not an irrelevant concept. However many people incorrectly use color temperature when they actually mean CCT. (a nit picking detail)

Color Temperature and CCT are relevant because they are used to define "shades of white". White light is most often characterized by appearance. Different spectral power distributions can have the same appearance due to the way the LMS cones our visual system process light. CCT is important for displays as it relates to the white point of the display.

While outside of the scope of this thread, color rendering (how objects appear under a light) is a critical issue in the lighting industry.

 

[hutchphd]

CCT1/CCT0 = (Volt1/Volt0)^0.42

Again (now for the third time) this exponent follows because the filament is a very good black body emitter...please see #9 and #25 above.
I have designed (and redesigned) several commercial colorimeters, the first of which used a tungsten light bulb as a source (and yes I am that old!)

 

[Eric Bretschneider]

Again (now for the third time) this exponent follows because the filament is a very good black body emitter...please see #9 and #25 above.
I have designed (and redesigned) several commercial colorimeters, the first of which used a tungsten light bulb as a source (and yes I am that old!)

Your model is a good approximation. However my point was to utilize something based on decades of DATA.

Theoretical derivations are nice, but the OP didn't really require that.

Although for extra credit you should factor in the change in spectral power distribution to shift your exponent from 0.40 to 0.42.

 

[hutchphd]

I seem to be having some trouble making my primary point.
My "model"(your word) can be written down in three lines if you know the physics. It requires no spreadsheet or even knowing the numerical constants.
So with one pencil and a small scrap of paper the original question

Hi!

I have a question:
Is it possible to calculate incandescent light color temperature based on it''s input voltage?

can be answered simply with an emphatic "yes, if you know a little physics"....

 

[Eric Bretschneider]

I seem to be having some trouble making my primary point.
My "model"(your word) can be written down in three lines if you know the physics. It requires no spreadsheet or even knowing the numerical constants.
So with one pencil and a small scrap of paper the original questioncan be answered simply with an emphatic "yes, if you know a little physics"....

The answer is also an emphatic "yes, if you do a little research"

You can always try first principles calculations, but Mother Nature makes no assumptions.

And the tristimulus response curves can't be derived from first principles . . .

 

[hutchphd]

The tristimulus response curves have nothing to do with the Planck temperature.
They do tell us what the surface temperature of the sun is (and has likely been for some time!)
But the sun would shine the same whether we can name its colors or not (falling tree in the forest notwithstanding!)

 

[Eric Bretschneider]

The tristimulus response curves have nothing to do with the Planck temperature.
They do tell us what the surface temperature of the sun is (and has likely been for some time!)
But the sun would shine the same whether we can name its colors or not (falling tree in the forest notwithstanding!)

"Planck temperature" is a very different concept - it is the highest possible temperature (about 1.4e+32 K). I think you were referring to the temperature of a Planckian source which is different.

Tristimulus response curves are used to calculate chromaticity. Without those you can't calculate color temperature. The concept of color temperature includes the human visual system which in turn brings in the tristimulus response curves.

They have nothing to do with the surface temperature of the sun. They allow us to calculate the chromaticity coordinates of sunlight (which does not lie on the Planckian locus).

 

[hutchphd]

Yes the term "Planck Temperature" was absolutely incorrect. I was unaware of that definition and was using it as a shorthand for the temperature of the blackbody radiator. My bad.

Tristimulus response curves are used to calculate chromaticity. Without those you can't calculate color temperature. The concept of color temperature includes the human visual system which in turn brings in the tristimulus response curves.

They have nothing to do with the surface temperature of the sun. They allow us to calculate the chromaticity coordinates of sunlight (which does not lie on the Planckian locus).

The entire response of the human eye has evolved to maximize the use of solar radiation. Compare the response curve of the eye to a 6000K blackbody. Serendipity? Do you really think that if we lived near a red dwarf we would have the same color response?

No Mas...

 

[weirdoguy]

"Planck temperature" is a very different concept - it is the highest possible temperature (about 1.4e+32 K).

No it is not. The same goes with Plancks lenght, time and other units. They don't have any physical meaning, except that they can define some sort of usable scale. But not always, e.g. Plancks resistivity is about 30 ohms and it's not the highest nor the lowest resistivity possible.

 

[pinball1970]

I did some similar work on this with LEDs What changed increasing wattage on measurement was the intensity not the SPD or CCT. More photons but at the same wavelengths.

 

[Baluncore]

I did some similar work on this with LEDs
What changed increasing wattage on measurement was the intensity not the SPD or CCT.
More photons but at the same wavelengths.

Unlike a filament lamp, LEDs are fixed voltage devices, with light output dependent on forward current, and forward voltage dependent on material = fixed colour.
The blank lines in your post make it look like you used an MS windows based editor, then pasted your text ?

 

[pinball1970]

Unlike a filament lamp, LEDs are fixed voltage devices, with light output dependent on forward current, and forward voltage dependent on material = fixed colour.
The blank lines in your post make it look like you used an MS windows based editor, then pasted your text ?

Yes I am lazy, I am always missing words out or something then noticing after,so I type in a blank email then paste. I am usually on pf at lunch at work while I am answering mails.
You can get dimmable LEDs, does the SPD change on those?

 

[Baluncore]

You can get dimmable LEDs, does the SPD change on those?

We see Spectral Power Distribution as colour, which is set by quantum chemistry of LED materials.
A dimmable LED will probably be a violet LED, with three or more fluorescent chemicals to change the colour balance. It should not change colour with power.

 

[martin314159]

Regarding the 220 volt incandescent 2800 cct running at 110 volts. Omcheeto had It correct. The cct goes to the 0.38 power of voltage. The new cct is 2125 K. The power drops from 100 watts to 33.3 ; the lumen output from 1550 to 95; the operating resistance from 482 ohms to 363.
best regards,
Martin314159

 

[martin314159]

I would also like to say Eric Bretschneider reply below in his August 19, 2019 #31 is correct too. The power of the voltage can vary from .38 to .42 depending on the temperature of the filament whether and whether the filament is in a vacuum or inert gas.

The following is from "Lighting Handbook: Reference and Application" (8th ed.) from the IES (Illuminating Engineering Society) p.186

Life1/Life0 = (Volt0/Volt1)^13
Lumen1/Lumen0 = (Volt0/Volt1)^1.9
Efficacy0/Efficacy1 = (Volt0/Volt1)^3.4
Watt1/Watt0 = (Volt1/Volt0)^1.6
CCT1/CCT0 = (Volt1/Volt0)^0.42