Watt equation applied to incondescence light bulb 2

[mielke]

How could I calculate the minimum voltage required to light up a simple 60W tungsten incandescent light bulb, assuming that I have a wire and a variable voltage source?

I could set up an experiment and connect the bulb to the voltage source and slowly crank up the voltage from zero until I see the light bulb light up, but how can I figure that minimum out from just equations? what are the assumptions?

 

[jraef]

Wattage is the consumption. You cannot control that directly.

Assuming and incandescent lamp here, you can only control the voltage and indirectly by virtue of the applied voltage and the fact that the resistance is relatively fixed, you can control the current, so if you are reducing the voltage, you are reducing the wattage consumed.

LIGHT, however, is relative. The lumens produced by that type of lamp have to do with the fact that the filament is glowing white hot as a result of the resistance and the applied voltage, the term used to describe that is "incandescence", hence the name of the lamp. If you lower the voltage, you lower the temperature of the filament and the electro-magnetic radiation it gives off moves to a lower energy state, becoming radiant heat / infra-red because the yellows and blues in the spectrum are no longer created by incandescence. In the lighting world, this is referred to as the "color temperature"

So at what point do we still call it "light"? Totally subjective. There is ALWAYS going to be some amount of "light", visible or not, coming from an incandescent lamp as long as some amount of voltage is applied to it. What you can see and interpret as "sufficient" is totally up to you, so there is nothing to "calculate" until you determine that first.



"Will work for (the memory of) salami"

 

[crshears]

All good points, jraef.

In the past [think theatre and stage lighting days] I've played around with dimmers of the straight-up ohmic resistive type, the autotransformer [variac?] type, and those of the SCR / triac family, and one thing I've consistently noticed is that regardless of the type, if a dimmer is very slowly and smoothly brought up from zero toward full power output, there comes a point at which the bulb suddenly 'lights,' meaning that even in low-illumination situations I've almost never seen the bulb come from blackness to incipient glow to finally visible light; instead it suddenly is lit, and rather brightly at that.

My guess is that during the run-up the filament is getting hotter and hotter, and as its resistance varies directly with temperature there comes a point when the resistance rapidly spikes up, at which point the voltage developed across the filament rises rapidly, which reduces the current draw across the source, which as a consequence causes its output voltage to rise still further, which which causes the filament to glow even more brightly, which...you get the picture.

Once the bulb would light, it was possible to dim it back down very slowly until it went completely out, but was still giving off infra red.

Fun and games.

Assumptions? I'm ignorant of any...

Calculations? I never had any inclination whatever to try.

But if you want to measure the resistive value of the bulb with a very low-voltage ohmmeter at however many Kelvin degrees room temperature is, then calculate what resistance the bulb must actually have to consume sixty watts of electricity at full voltage and use your material resistivity coefficient tables to calculate how many degrees Kelvin the bulb has to be at to have that resistance...have at 'er; I'm bored already.

 

[Compositepro]

CRshears, the behavior you describe is characteristic of cheap triac dimmers. It has nothing to do with the filament or with variacs, which will indeed go from dark to a barely visible red glow. Variacs are not infinitely variable but change in small steps.

 

[jraef]

Yes, the reason it exhibits that behavior is because tungsten, like most metals, has what is called a Positive Temperature Coefficient of resistivity. That means that as the temperature increases, so does the resistance. So when you first ramp the voltage into a tungsten element it has relatively low resistance, but as it heats up, the resistance increases. The effect, in terms of emissivity of light, is therefore non-linear. That low resistance when cold is also, by the way, why lamps tend to burn out more often when you first turn them on, rather than when they are already on. The lower resistance means a high inrush current, which causes a rapid thermal change which puts physical stress on the filament.

"Will work for (the memory of) salami"

 

[IRstuff]

The eye's response to light is non-linear, so there's that:

http://www.telescope-optics.net/eye_intensity_response.htm

At some point, your eye's responsivity supposedly goes somewhat exponential.

Additionally, your eye's spectral response is highly nonlinear, as alluded to above. At dim settings, a typical bulb is putting out a bunch of red, to which your eyes are less responsive. As the filament glows whiter, your eyes' green and blue responses are stimulated, and your green response is something on the order of 7x that of your eyes' red response.

http://en.wikipedia.org/wiki/Luminous_coefficient#Efficacy_and_efficiency

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[waross]

I remembered the ratio of cold resistance to hot resistance of an incandescent lamp as 10:1 but on checking, Wiki gives it as about 15:1


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[OperaHouse]

Before the days of everything on a chip, lamps were used to make the output level stable of audio oscillators. Lamps are amazing. I built a precision AC source using only resistors and a lamp. Input voltage could vary 20% and output would be rock solid. With dramatic changes in current, you may see a sudden burst of light. These are the results of the circuits lamps are connected to. Electronics always does what it is supposed to do, not so often what you think it will do.

 

[crshears]

Compositepro wrote:

CRshears, the behavior you describe is characteristic of cheap triac dimmers. It has nothing to do with
the filament or with variacs, which will indeed go from dark to a barely visible red glow. Variacs are
not infinitely variable but change in small steps.

To which I respond:

Faulty memory on my part re the behaviour of ohmic and autotransformer dimmers, I guess...

Further to which, I do recall one of my cousins with ties to the music industry telling me that he had
seen in the past how before rock concerts began the stage techs would circulate a low value of current
through all the lights for ten minutes or more, with just enough cuurent flow applied to heat up the
filaments without creating visible light. Apparently this practice not only greatly extended the life of
the bulbs but also reduced the likelihood that one would blow during the performance.

The autotransformer dimmers that I recall working with were wound in a circle and used a sweep finger with
a V tip that travelled the face of the secondary winding. The output voltage may not have been infinitely
variable, but my recollection is that the increments were so small that the end result so closely approached
'infinitely variable' as to be indistinguishable from it.

 

[racookpe1978]

It depends too to what you want from the bulb; Let's pretend you were trying to get some kind emergency light into your basement during a power failure. Hooking a 120 v/60 hz normal bulb to your car 12 volt dc battery directly with a jumper cable to the extension cord prongs would get you "some" light - maybe enough to use after your eyes adjusted to the dim glow. But at 12 volt/120 volts supplied, you'd be at at much less than 10% of what the bulb is designed to generate. A 100 watt bulb "might" create 10 watts, but probably less as shown above for cool bulbs at low amps.

Not efficient, but maybe effective for that emergency.

Putting that same 120 vac normal bulb into a 220/240 dryer voltage circuit would also cause a "glow", but would burn the bulb out quickly

 

[IRstuff]

Anyways, back to the original question; I don't see anything where you stipulated how MUCH light you want. Since output is a continuous function of input, there is no "minimum voltage required to light up" a bulb.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[VEBill]

Animated GIF (from Wikipedia - gasp!) demonstrating Black Body radiation over visible spectrum versus temperature (°K). Filament temperature versus power is certainly a case of YMMV (depends on the bulb design of course). All this is complicated by the non-linear nature of the filament's resistance with temperature, the non-linear curve of triac dimmers with who knows what curve potentiometer in them, etc. in summary, you need ABOUT 1000°K to be visible. But colour temperature would be extremely dull red.

 

[ijl]

The light output from an incandescent lamp depends on the temperature of the filament, as stated above by others. The calculation of the voltage needed for a given filament temperature is straightforward, in principle (but not in practice . In steady state the temperature assumes a value where the power In and power Out are in balance.

The power In is the electrical power, Pe = V^2 / R, where V is voltage and R is the resistance of the filament. The resistance depends on the temperature, as R = R0+k(T-T0), as a first approximation. T is temperature, T0 a reference temperature and k is a coefficient that depends on the material of the filament.

The heat is transferred out of the filament mainly by radiation at these temperatures. The power Out = the radiated power is proportional to the fourth power of the absolute temperature of the filament, Pr = h T^4, where Pr is the radiated power, and h is a coefficient.

In steady state, Pe = Pr, or V^2 /(R0+k(T-T0)) = h T^4, from which the voltage can be calculated, in principle. The hard part is the determination of the coefficients.

 

[VEBill]

Happy New Year.

The physical size (surface area) of the filament has to be included; obviously a filament on the scale of about one meter (instead of about one cm) dissipating for example 100 watts is going to be barely warm. Another example: a typical human body emits about 80 watts and we make poor light bulbs because we're too big. Is size (surface area) hidden inside one of those coefficients?

Or perhaps you've taken proportionality and accidentally carried it forward as equality.

 

[ijl]

The coefficient h contains all factors, such as Stefan-Boltzmann constant, area, emissivity. (I just picked "h" to mean all this.) The equations can certainly be refined. I only wanted to explain the basic idea.