Greg Jacobs in So what does an ohmmeter read when it’s directly connected to a non-ohmic bulb? asked for help in determining how to measure the resistance of a bulb at different low voltages, as his voltage source only goes down to 2V.

I used my FG085 function generator, a resistor, and PteroDAQ to measure the I-vs-V characteristics of a small screw-base bulb rated at 6.3V and 150mA. One would expect a bulb with that rating to have 6.3V/150mA=42Ω resistance, but incandescent bulbs are not ohmic devices—the resistance of the filament depends on its temperature, and at 2700°K, the resistance is going to be much higher than at room temperature.

The thermal coefficient for tungsten is about 0.0045/K, so one would expect a difference of about 2400°K to make a ratio of about (1+0.0045/K 2400K)=11.8. So, if the linear approximation of resistance vs temperature were good over that wide a temperature range, we’d expect about 3.6Ω at room temperature. I measured about 3.8Ω with my ohmmeter, so this seems like a reasonable back-of-the-envelope approximation.

I hooked up the function generator to drive a 10Ω resistor and the bulb in series, and used PteroDAQ (with a Teensy LC board) to measure the voltage across the resistor and across the bulb. I had to play with the amplitude and offset of the function generator a bit, because the 50Ω output impedance means that the voltage across the bulb+resistor was substantially less than the values set (which are the nominal voltage to a high-impedance load).

The plot above is a little hard to interpret, but one can clearly see the range of voltage and current, and that the plot has hysteresis—it follows a different curve when the voltage is increasing than when the voltage is decreasing. (One can also see the really bad digital-to-analog conversion in the FG085 function generator, with big steps and non-monotonic steps.)

Changing the plot to resistance (V/I) versus voltage makes for a clearer plot:

The large noise at the low end of the plot is due to the inherent problem of taking the ratio of two small numbers, each of which has noise added. We can clean up the appearance of the plot by using power (IV) on the x-axis, since if either voltage or power is close to zero, then power is also very close to 0:

Looking at the resistance as a function of time is helpful in interpreting the plots:

The R-vs-V and R-vs-P plots should be interpreted as counter-clockwise hysteresis—the lower curve is the one followed as the filament heats up, and the higher curve is the one followed as the filament cools down.

The curve you get depends very much on how quickly you change the voltage. Using the same voltage range, but making the fluctuations be at 4Hz instead of 1/22 Hz results in a much more constant temperature, and hence a much more constant resistance:

Of course, the question arises whether the standard 60Hz line frequency is high enough to get a constant filament temperature.

The filament temperature is *not *constant, but the variation in resistance is much less. Note that with the higher frequencies, the maximum resistance (and hence maximum temperature) does not occur at the same time as peak power, but substantially later in the cycle. If one wanted to get a really clean reading of resistance vs power, it would be best to change the voltage very slowly, so that the temperature has a chance to reach steady state. Even with a period of 99s, I still saw fairly substantial hysteresis, so I tried a couple cycles with a period of 333s—I expect that the hysteresis will be small except for the low-power end of the curve, as cooling is slower when the temperature difference is small.

Note that I was only looking at the low power range for the bulb—it should take about 945mW when at its full rating, and I stayed below 1/7th of the full rating.

Very nice — thanks for running with this experiment! I’d love to use a problem for a future USIYPT based on something similar to this question.

— Greg Jacobs

Comment by Greg Jacobs — 2016 July 9 @ 10:48 |

That was really interesting, particularly the hysteresis at 60 Hz. I always point out the flicker of fluorescent lights (easily seen by moving your hand really fast) but always assumed incandescent bulbs had enough thermal mass to make that effect negligible. Your experiment shows that I needed to think about how quickly a bulib “turns off” as an indication that the cooling by IR emission is faster than I was thinking it would be. This problem ultimately has thermodynamics involved as well as the resistance as a function of temperature. Since we do this experiment in my lab class, I see I now have one more thing to amuse them with once they have done it.

You do experiments with optical sensors, right? Are they fast enough to see the 60 Hz variation in light output from incandescent and fluorescent bulbs? Can they measure the light curve when you turn on a bulb or turn it off?

Comment by CCPhysicist — 2016 July 9 @ 16:38 |

Phototransistors are way faster than needed to see 120Hz flicker! I can get a pretty good signal at 200kHz from an LED, but noise is a problem at higher frequencies, because my op amps only have a 1MHz gain-bandwidth product, so I don’t get a good transimpedance amplifier at high frequencies.

There is some low-pass filtering of the turn-on or turn off in the phototransistor, but for low frequencies this isn’t a problem. You can even skip the transimpedance amplifier, and just use a resistor (say 51kΩ) to pull up the collector of the NPN phototransistor, using an oscilloscope (or PteroDAQ) to look at the result.

Comment by gasstationwithoutpumps — 2016 July 9 @ 18:43 |

Did you measure the length and/or the cross-sectional radius of the filament? Or is this even possible?

It looks to me like if L is the length and a is the cross sectional radius, then l/a^2 is about 3 per cm.

Comment by Jonathan Keohane — 2016 July 13 @ 13:41 |

Since the filament is inside a small glass bulb and coiled, I have no way to measure the filament without destroying the bulb (and even then it would be difficult).

I don’t see how you are computed the current density, given that there is no information about the dimensions of the filament. One might be able to estimate it from the properties of tungsten, but what would be the use of that? (I’m not trying to be critical here—I just think I missed the point of your comment.)

Comment by gasstationwithoutpumps — 2016 July 13 @ 15:06 |

Sorry, I used a small L rather than a big L — so it looked like an I. My bad … typo.

I just looked up the resistivity of tungsten at room temperatures. I am trying to figure out what can be measured.

It looks like, in terms of proportionalities, you have something like:

R ~ sqrt(P) (Resistance vs. Power)

This is consistent with a high temperature proportionality something like:

R ~ sqrt(T^4) ~ T^2

Hmm now why would that be?

That said, it is interesting that you can actually measure the rate that the fillament seems to heat and cool.

You might want to check out this paper — old papers are really readable.

Click to access temperature_scale_for_tungsten.pdf

Comment by Jonathan Keohane — 2016 July 13 @ 15:30 |

The resistance (or resistivity) vs temperature table in the paper you pointed to seems useful for working with. I’m not seeing the relationship you are seeing between resistance and power, since the resistance at 0W is around 5Ω, so no power law is going to be a good fit.

To measure the rate of heating or cooling, I’d want to use a setup that kept the filament at constant power as its resistance changed. That is a somewhat harder circuit to set up than constant voltage or constant current. One needs to take the product of the current and the voltage (as done, for example, in https://www.maximintegrated.com/en/app-notes/index.mvp/id/4470 ). Alternatively, one can use a voltage source and switched inductor with a fixed on time to deliver constant energy per pulse, adjusting the rate of the pulses would adjust the power.

Comment by gasstationwithoutpumps — 2016 July 13 @ 15:53 |

Yes, I subtracted it off. Here is a question, if you plotted the current as a function of the voltage, does a power-law fit that?

If it does not, is there a different relationship at low and high currents?

The way I see it, you should be able to measure the power-law index between the surface temperature of the filament (from Stefan’s law) and the current flowing through the wire. I wonder what that is?

Comment by Jonathan Keohane — 2016 July 14 @ 15:29 |

I’m not sure that subtracting off the resistance at 0W makes any sense. It should be a scaling factor for whatever rule you come up with. If you want to fit a power law, you’d expect to see a straight line on the R vs. P plot on a log-log scale. I don’t see that at all. I can get an ok fit above 0.02W for R=47.3 P^0.3333,

but resistance doesn’t change much with power for very low power, and is steeper than P^(1/3) for power between 4mW and 20mW.

At low voltages (below 0.1V) I see current as growing almost linearly with voltage (constant 4.78Ω). I can get a bit better fit with I=0.158335 V^0.91379 in that region. But above 0.2V,I see 0.0553757 V^0.461722 or I=0.0558745 sqrt(V). Of course, I was barely lighting the bulb to a dull red glow—it may behave quite differently in the range of power it is intended to be used at. But for a first approximation, low voltage=>constant resistance, higher voltage=>resistance as sqrt(V) seems to be ok. (Note that this would give resistance and cube-root of power, as I saw on the R vs P log-log plot.) The switchover is almost consistent with a constant 4.78Ω resistor in series with a voltage-dependent resistor. Of course, we have a temperature source (about 300°K) other than the power delivered to the bulb, so this sort of crossover between room temperature and temperature related to power is to be expected.

I think that the Stefan-Boltzmann Law says that the temperature should go up as the 4th root of power for a blackbody. Because tungsten resistance is pretty linear with absolute temperature in this range, I’d expect the resistance to go up as the 4th root of power, not the 3rd root. But the bulb is not a vacuum, but a low-pressure gas, so there is undoubtedly convective cooling as well as blackbody radiation. But, unless my thinking is messed up, that should mean that resistance goes up slower than P^0.25, not faster.

Comment by gasstationwithoutpumps — 2016 July 14 @ 17:09 |

That is a good way too. I like it.

Comment by Jonathan Keohane — 2016 July 15 @ 15:48 |

Yes, I see your point. The S-B law states that: P ~ T^4. You observe that P ~ R^3, or another way to see it would be that R ~ T^(4/3) — right, at least for high temperatures. So, like you said, not much different than linear. So all is good with the world.

Thus, if the issue is simply the temperature — probably is for the DC case — then you just measured the temperature dependence of the resistor.

That is pretty cool, and not too hard to do in the lab.

Another way to see it is a little different. Ohm’s law is far from fundamental, rather it is a measured property of constant temperature conductors.

So, the only reason it works is that we apply it to places where it works. These just happen to be very common.

Comment by Jonathan Keohane — 2016 July 15 @ 11:01 |

I teach Ohm’s Law differently—as a definition of resistance (with the alternative definition of dynamic resistance as dV/dI). What makes it a “law” is that for some materials (like metals), resistance is roughly constant. Since the first lab we do with resistors is with thermistors, resistance as a constant property is not a “given” in our class.

We also don’t work much from first principles, but do empirical fits to collected data (calibrating thermistors, looking at I-vs-V plot for electret microphone, looking at impedance vs. freq for a loudspeaker and for electrodes). Physics models are great for the simple cases where they apply, but when you get into measuring chemical and biological systems, the simple physics models are not very useful except for back-of-the-envelope calculations.

What I mainly want students to have gotten from physics classes is the notion of modeling—not a memorized list of equations, which is far too often what they come away from physics classes with.

Comment by gasstationwithoutpumps — 2016 July 15 @ 12:16 |

That is a good way too. I like it.

Comment by Jonathan Keohane — 2016 July 15 @ 15:48 |