# Why does the characteristic curve (V vs I) for a light bulb backbend?

When teaching Ohm's Law, I have students do an exploration of a small, incandescent light bulb with a low frequency (1-2 Hz) sine wave. It's a simple series circuit of source and light bulb, monitoring the current and voltage. The results are definitely non-linear due to temperature-dependent resistance of the filament. Surprisingly, however, the curvature is not uniformly positive or negative. As seen in the picture, there is a "backbend" in the curve near the extrema of the voltage. If one interpreted this "ohmically", one might say the resistance becomes infinite, then negative, before passing back through infinity before becoming positive again. I understand the basic temperature dependence of a warming filament having an increasing resistance (and hence an increasing slope), but I'm puzzled by the backbend potion of the curve. Is there a physics (condensed matter? thermodynamic?) explanation for it?

Edit: For clarification, the backbends occur on the way toward as the voltage approaches the + and - amplitudes. The smooth curvature occurs as the filament cools as the voltage drops away from the amplitudes toward zero.

Edit 2: In the first quadrant, for example, the path is counterclockwise as time increases. Same in the third quadrant, so the data paths cross over at V=0. The light trace in the third quadrant is the beginning trace when the filament was initially at room temperature and began to warm.

Edit 3: Below is a trace using a triangle voltage input of amplitude 2 V at frequency 0.01 Hz collecting at a 20 Hz rate. There seems to be a sudden change in the resistivity of this filament once it reaches a certain temperature. • Can you re-clarify? :) In the first quadrant, does the trace go clockwise or counterclockwise? Third quadrant (thick line)? What is the faint line in the third quadrant? Oct 27 '15 at 16:22
• @garyp See Edit 2 Oct 27 '15 at 16:28
• The heating power is $P=RI^2=U^2/R$, i.e. when you change the voltage by a factor of 5 (assuming that the resistance stays roughly equal), then the power will change by a factor of 25. The power-temperature dependence of the filament is a complicated function, because you have to take both conducted cooling (trough the leads and the glass body) and radiative cooling into account. The latter has a very steep dependence on the temperature due to Stefan-Boltzman: en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law, which has a $P\propto T^4$ form. Oct 28 '15 at 20:55
• I would guess that at low power/temperature cooling is basically trough the leads and it's more or less proportional to the temperature difference to the environment, but at some point the radiative cooling takes over and there is this steep dependence of temperature, which means one has to put a lot more power in to make the filament even slightly hotter, which means that the resistance will also not change very fast from that point on. Oct 28 '15 at 20:58
• @CuriousOne - Your comment which begins "The heating power is P=RI 2 =U 2 /R" is only partly true, since it assumes that R is constant. In fact, R changes by more than an order of magnitude over the range from zero to full brightness. Oct 28 '15 at 22:03