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

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

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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

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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.
 
494
176
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

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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.
 
494
176
OP: Same bulb, half the voltage, I believe.
 

Baluncore

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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

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...
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]

voltsKelvin
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.
 
494
176
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

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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

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At T = 2800K, rated power = 100W, rated voltage = 220V
Notice how the current is limited at higher powers.
Code:
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
Code:
  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
 
494
176
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

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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.
 
494
176
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?
 
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 dont 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.
 
Last edited:

OmCheeto

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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

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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
 
494
176
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

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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 modelled 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.
 
Last edited:

OmCheeto

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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.

2019.08.15.tungsten.vs.flat.resistor.png


Fascinating.

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

sophiecentaur

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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

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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?
 
494
176
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


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

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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]
 
Last edited by a moderator:

OmCheeto

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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.

2019.08.16.Baluncore.vs.OmCheeto.png


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?
 

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