Electrical model for incandescent lamp - Page 1
Forum section: electronics / projects

author: michael2
author_rank: Contributor
timestamp: August 07, 2015, 07:27:33 am
content: Hello all,for the sake of interests I measured the I/U characteristic of incandescent lamps. To be exactly it is an assembly of eight lamps connected parallel I built a few years ago. However, after some quick&dirty measurements I decided to invest some more time. I made 7 measurements in 1mV decrements and calculated the mean. It took about 15 hours...I tried to fit a curve to the measured data. After some investigation on the net I learned about the exponential relationships of incandescent lamps. It worked quite well for the electrical power of the lamp (fitted exponent is rounded 1.577, literature mentions about 1.6)However the exponential relationship is poorly working on the lower side of the supply voltage. In fact it is especially totally wrong for the resistance in the current range below about 0.25 Amps. (To made it more visible I plotted the resistance against the current.)So my question is, what would be a good function to fit for the resistance? Fitting an exponent in a limited range, say 4 to 12 volts for the resistance works quite acceptable. But I would like to fit a function for the whole range. But my mathematical experience are limited. I tried a bunch of guessed functions but nothing worked well. How can I find a "perfect" function to fit to the data?

author: michael2
author_rank: Contributor
timestamp: August 07, 2015, 07:31:38 am
content: The measured data for anybody interested in.

author: TimFox
author_rank: Super Contributor
timestamp: August 09, 2015, 08:40:04 pm
content: Note that the resistance of the incandescent filament is a function of its temperature, which, in turn, depends on the past history in time of the current, not only on the present value of the current.

author: IanB
author_rank: Super Contributor
timestamp: August 09, 2015, 09:13:27 pm
content: Expanding on Tim's answer, and also disagreeing with Tim a little:Firstly, to get a good fit to the data you need an approximate physical model of the filament with parameters you can fit.Secondly, the resistance of the filament is a function of its temperature, and there is indeed a dynamic component to the temperature change. If you suddenly change the voltage, it will take a certain time for transient effect to settle out and for the new temperature to stabilize.In order to simplify your modelling you will probably want to ignore the dynamics and consider only the steady state response. To do that you need to wait for the system to settle and for the numbers to stop changing before recording each measurement.That said, temperature, and specifically heat losses, are the reason you have trouble at lower voltages. There is a power balance at play. Electrical power is fed into the filament, heating it up, but the filament is also dissipating heat, cooling it down. At low power levels the filament can lose heat by conduction and convection as well as radiation (most bulbs are gas filled). This keeps the filament cooler than it would be if it were losing heat by radiation alone.At higher power levels the conductive and convective elements become less significant and radiation dominates. Once radiation dominates the exponential or power law function will take over.So you will need a physical model that accounts for the different heat loss mechanisms. Convective and conductive heat losses will be proportional to filament temperature, and radiation losses are proportional to the 4th power of temperature.

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: August 09, 2015, 09:17:57 pm
content: As TimFox mentions, then the problem is the extreme temperature dependence of the resistance of a Tungsten filament.The problem you have in fitting the curves at low DC power is due to a change-over in the most important heat transport mechanism from the filament to the outside world. At low power the heat transfer is likely dominated by convective heat flow through the Argon gas filling the lamp envelope.As the power - and thus temperature - increases, direct radiant heat transfer as infrared energy begins to dominate (the filament begins to glow), and the fit start to match closer and closer to the theory.If the envelope had been evacuated, then you'd only have two competing mechanisms for heat loss: Direct radiant energy plus a bif of conduction through the filament connections to the lamp base.Edit: Beaten to the punch by IanB.

author: michael2
author_rank: Contributor
timestamp: August 09, 2015, 10:11:02 pm
content: Thank you for your comments so far.I'm only interested for the static model. I stepped in millivolt steps and started at 12 volts. Before I took the first sample, I let the lamps light for 10 seconds I think. Then, each step took 0.75 seconds: Reduce the voltage for 1 millivolt, wait 0.75 seconds, measure the current. A full round from 12 volts down to zero took 2.5 hours. I made seven rounds and averaged the data.So I think, I canceled most parts of the dynamics a little bit... However, if you zoom into the 12 volt end, you can see small current settling. Maybe the 10 seconds startup was to short.My main problem is to find a mathematical "basic function" to fit. I experimented with 1 / (e^x + 1) functions, but i can't get the "crossover" range correct.

author: IanB
author_rank: Super Contributor
timestamp: August 09, 2015, 10:50:21 pm
content: I suggest you might try a function of the form:R = a(U) * R0 + [1-a(U)] * R(U)Where R0 is the cold filament resistance and R(U) is the hot filament resistance function.a(U) is a cross-over function varying between 0 and 1. For this you might try a function like:a(U) = exp(-U/Ux)When U = 0, a(U) = 1. As U increases, a(U) tends to 0. Ux controls how fast this happens.You would choose the value of Ux to get a best fit to the data.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: August 09, 2015, 11:10:27 pm
content: The behavior of the lamp, and therefore your model will also depend on ambient temperature. Also will vary with lamp types etc. Not sure you want to collect the data required to do an empirical model that incorporates all of that. In looking at your data I believe your first two or three data points are influenced by transient behavior, but other than that don't see anything obvious that indicates the data isn't good enough for a static model.I have used your data generate an empirical model with significantly smaller errors. The model may or may not be useful for your purpose. It may be more useful in demonstrating how to use Excel to generate empirical fits for a wide variety of data.The empirical model is based on the power relationship for currents high enough to be dominated by radiative heat transfer. It then recognizes that resistance is relatively constant at low current readings. An initial try using that model gave errors roughly the same as your initial model - the exponential term did not dominate quickly enough. So a second version modeling an exponential decrease of the constant term with current was tried, which gives the results shown.The key is using the Excel Solver function to minimize the square error between the model and the modelled function. Solver is not included in a standard Excel install, but is included in the distribution as a group called Engineering Add-Ins. In the attached spreadsheet based on your data has four added columns. Column E has an estimate of the current flow as a function of voltage that applies for high voltages. It is a two parameter model, the parameters are in cells J1 and J2. Column F estimates resistance as the sum of Voltage divided by the estimated current and another exponentially decaying value, parameterized in cells J3 and J4. Column G is the error between this estimate and the measured E/I. Column H is the squared error. Cell H1 contains the sum of the squared errors (the first few cells are not summed due to the possible dynamic effects in these cells).The spreadsheet attached had most of the data deleted to fit the BLOG size limits. Those wanting to experiment with this will have to add the original data back in, and extend the equations in the four columns as well as changing the summing function in cell H2. The graph shows the result of the model with the full data set.Models other than the exponential decay might give better results. A reciprocal of voltage for example. Better fits can require a lot of experimentation. Any time you do something like this remember the limits of an empirical model. Since it does not represent underlying physics it is unlikely to extrapolate to new situations.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: August 09, 2015, 11:12:11 pm
content: Ianb posted while I was posting. His suggestion is essentially the same as mine. The Excel tools help you select appropriate values.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 09, 2015, 11:22:38 pm
content: How is the zero-intercept current nonzero???Tim

author: IanB
author_rank: Super Contributor
timestamp: August 09, 2015, 11:44:51 pm
content: Working independently of CatalinaWOW, I came up with the following model fitting filament resistance to applied voltage:R = a * R0 + (1-a) * R(U)whereR0 = 1.2212a(U) = exp(-U/0.45673)R(U) = 3.1473 * U^0.43010I also used the Excel solver to minimize the sum of squares of relative errors.It's interesting that the cold filament resistance seems to decrease initially before increasing. I suspect this is an artifact of measurement errors at low currents. Most likely the cold filament resistance is actually about 1.18 ohms.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: August 10, 2015, 03:48:59 am
content: I suspect all of the readings below 4 to 6 milliamps.The use of Excel (or the Open Office equivalent) to minimize model errors in the least squares sense is a potent tool for any ones tool box, but one that doesn't seem to be very widely used. While there are closed form approaches to doing this, the math gets messy for all but the simplest systems.The biggest problem with these numerical solutions is stability of the solution. If your model is "stiff" the answers won't converge to anything reasonable. The program will not necessarily report this to you. There are some knobs you can twist in Excel to help, setting constraints can help and thinking is always a useful tool.

author: IanB
author_rank: Super Contributor
timestamp: August 10, 2015, 04:30:23 am
content: Quote from: CatalinaWOW on August 10, 2015, 03:48:59 amI generally plot the raw data and the correlating function on a chart so I can visually eyeball how good the fit is. Sometimes the solver can find different solutions depending on the initial guess, so I typically adjust the parameters by hand until I have a reasonably close fit and then I let the Excel solver do the final polishing.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: August 10, 2015, 03:33:10 pm
content: Solver is a gradient descent optimizer so will find the local minima. Hand selecting can help in making sure that this is the global minimum.

author: unitedatoms
author_rank: Frequent Contributor
timestamp: August 11, 2015, 12:05:42 am
content: Quote from: IanB on August 09, 2015, 11:44:51 pmNice find with low extremity on resistance curve! Is curie point of tungsten somewhere at warm glow temperature level ? This is the point when material looses the paramagnetism. Like ordinary magnets loose ferromagnetism when heated. After this point the Tempco is reversed. I can not find the tungsten curie point no matter how hard I google.Example for curie point for PTC thermistors:http://precisionsensors.meas-spec.com/ptc-engineering.aspIn my primitive understanding the metals are mixture of electrons like a liquid and atoms bound into magnetic domains with boundaries turning fuzzy and disappearing after some heating. When domain boundaries disappear - the resistance is lowest, and the higher the temperature, the more mechanical obstacles for electrons, so resistance grows.When boundaries are intact the higher the temperature, the more opportunities for electrons to jump across the boundaries, so the temco is negative.On unrelated note: An unattended person has an incadescent bulb in Europe in personal posession !... Did not all tungsten bulb disappear with ban to save electricity ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 11, 2015, 02:22:14 am
content: Curie temp is specific to magnetic (and electric) materials, which tungsten and other ordinary metals aren't. That reference appears to be abusing the term, with whatever material their PTCs are made of. Resistivity does not change suddenly in materials that do exhibit a Curie temp.PTCs used for fusing applications, as far as I know, can't be more different, physically: they use conductive particles in a polymer matrix, which expands and becomes highly resistive when heated.As far as I know, normal metals (and alloys) exhibit various sorts of humps and plateaus at lower temperatures, but always have a general R ~ T^2, or other generally rising power or polynomial shape, especially at higher temperatures (including up to, and beyond, the melting point). I don't know the mechanics of it, but some alloys have dips and humps, which can be controlled by composition, resulting in a low tempco around room temperature, or over a specified range. But always PTC in the grander scheme of things.Tim

author: unitedatoms
author_rank: Frequent Contributor
timestamp: August 11, 2015, 02:43:01 am
content: Of some paramagnetic elements I only could find that Terbium has Curie point. May be this thread is a first observation (discovery) of Tungsten's Curie point, since it is paramagnetic.Quote from: T3sl4co1l on August 11, 2015, 02:22:14 amInteresting. If PTC is made of discrete particles, then every joins/disjoins during heat induced movement should cause stepped changes. Like some sort of very high frequency sound of multiple tiny switches. We should plan PTC "ladder noise" study after incandescent bulb study is done.

author: michael2
author_rank: Contributor
timestamp: August 11, 2015, 10:59:04 pm
content: Thank you again for all your ideas and suggestions.I made some progress and reduced the resistance fitting error in the order of about 100. My goal is not to find a physical model for the resistance but just simply fit the measured data:I measured the current using my power supply. But it looks like I need more digits, accuracy and precision for the current. The relative error on the low voltage range is rather noisy:Even predicting the power of the lamps using the fitted resistance works nicely:The fitting function consists of a quadratic function for the first part and a power function for the second part crossed over using two independent weight functions of hyperbolic secant:Code:f0 = a0 + b0*u + c0*u**2f1 = a1 * u**b1 + c1w0 = 1/cosh(x0 * u)w1 = 1/cosh(x1 * u)return w0*f0 + w1*f1 + r0Parameters are:Code:a0 = -8.1948229b0 = 0.32115525c0 = -0.12208141a1 = -2.00108962b1 = 2.38133535c1 = -2.48010758x0 = 0.23150578x1 = 2.04460732r0 = 11.83335988I fixed the first 9 resistance measurements to the value of sample #10.All work done in SciPy and fitted usingcurve_fitwithsigma = 5 * resistance.So next task is to make the measurement again with better instruments.

author: Kalvin
author_rank: Super Contributor
timestamp: August 12, 2015, 01:39:12 pm
content: Model looks nice. But tweaking the model's accuracy based on one incandescent lamp might be quite useless, unless you can verify the model using a few lamps [from different manufacturing lots]. I have my doubts that the lamps are produced using tight tolerances. It has been interesting to read about the model fitting process.