Photovoltaic cell as an instrument for measuring intensity of light I teach a lab in which my students measure the transmission of light through polarizing films oriented at an angle with respect to one another. To measure the intensity of the light, we use a silicon photovoltaic cell with a home-made collimator attached on the front.
When I first made up the lab, I needed to find out how the voltage from the cell related to the power of the light. I expected the relationship to be of the form $P\propto V^2$, based on experience with ohmic circuits. I tested this by exposing a cell simultaneously to two sources of light and turning the sources on and off independently. I was surprised to find that it actually seemed to be $V$, not $V^2$, that behaved additively. This is with relatively weak light sources, which produce voltages of ~10-100 mV.
For someone like me who's not an expert on semiconductor physics, is there any simple way to understand why this should be so?
Conservation of energy would seem to tell me that when I hook the thing up to a voltmeter, it's acting like a fixed source of current, which seems odd. (This is assuming that the efficiency is independent of intensity.)
 A: The simplest way to think about this is that 1 photon absorbed in the semiconductor generates 1 excited electron-hole pair. Ideally, each electron will contribute to the current in the circuit (if it can be extracted before recombination). Since current $I\propto$ (# excited electrons/time), absorbed optical power $P\propto$ (# photons/time), and (# excited electrons) $=$ (# absorbed photons), then $I\propto P$.
Remember that the optical power absorbed into the semiconductor will not equal the electrical power dissipated in your electrical circuit. Indeed, when measuring a solar cell in the open-circuit condition (like with an ideal voltmeter), the power in the outside circuit is zero because there is no current flowing (i.e. the efficiency of the solar cell is trivially zero). In this case, the current generated by the light is entirely shunted back through the diode itself. This, of course, is not to mention the various inefficiencies which lead to loss of power into heat, etc.
A: I'd suggest to measure current, not emf. See 
here
It's the short-circuit current that's proportional to light intensity. So you must care that for your light levels the load line (meter's resistance) remains above the characteristic's knee. 
A: Your photovoltaic cell is a silicon p-n juncion. To describe current - voltage charakteristic use the Shockley equation https://en.m.wikipedia.org/wiki/Shockley_diode_equation in first approximation. Next using the related P = UI derive the I vs P curve.
