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.If one wanted to make hand measurements with a fixed power supply and no function generator, one could either change the current-sense resistor for each measurement (larger resistor for smaller currents) or keep the current-sense resistor and bulb constant, but add extra series resistors to reduce the voltage and current. It would take a fair amount of patience to gather many measurements (or lots of students with identical bulbs).
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.