Photoelectric effect experiment: is charging time proportional to current? I'm not really sure what tags to add to this question, but definitely this is just an introductory physics experiment. Let me refresh you first with the basics of the photoelectric effect.
We all know that in the photoelectric effect the stopping voltage is just the kinetic energy obtained by the electrons displaced by the photons that struck the photodiode. Usually we experimentally measure the maximum stopping voltage because we only know the work function of the metal, which is the minimum energy needed to displace an electron from the surface. 
The main result of the experiment is that the stopping voltage is proportional to the frequency of the incident radiation, and not correlated whatsoever to its intensity. Instead, it is the current generated that should be directly proportional to intensity, since intensity is simply the amount of photons carried by a light beam (and more photons incident on the metal means more electrons displaced --> higher current).
Also, the experiment includes measuring the charging time needed to achieve the stopping potential. This is the apparatus we used: ( http://www.pha.jhu.edu/~c173_608/photoelectric/he9370.pdf ). I'm pretty sure most photoelectric effect experiments use something similar, such that the stopping voltage is achieved after a nonzero charging time in the circuits of the apparatus.
My question is this: Is the charging time proportional to the current? One thing for sure is that the current is correlated to the charging time. More electrons received by the anode by the displacing effect of the photons should mean a faster time to achieve the stopping potential. I'm afraid of my hunches, but I suspect the relationship is probably not linear, even disregarding noise and random error, however I can't come up with a solid argument for it. (By the way I'm hoping the relationship is linear, or at least has a simple mathematical form.)
Can anyone give me a definite yes or no? Thank you in advance!
 A: In order to answer your question you must think through the entire measurement. In this case you are measuring the photoelectric effect by placing a photodiode into a circuit. What kind of circuit element is the photodiode? It is a source. And as you have pointed out above, it is a constant voltage source (with voltage determined by the energy of the photons and the work function of the metal). It is also current limited (with no load it still cannot draw more current than there are incoming photons). I think you may be slightly confusing yourself because current is charge/time so it depends on how fast the electrons are moving (i.e. the voltage). Always recall that it is a constant voltage source. 
To proceed with understanding your lab you need to understand your circuit. There is a schematic diagram of the photodiode, op-amp, and associated circuitry in your pdf. I am sure you are familiar with the basic idea: the circuit acts as a capacitor which builds up voltage until the current is 0 (or approaches 0). The 'current-limited' feature of the diode power supply effects the capacitor charging only at the start of a measurement. The trick to this equipment is understanding the op-amp. There is plenty of literature out there on op-amps. Wiki is always a good place to start.
If you have done your lab by now you will have noticed that there IS some intensity dependence. How can you explain this?  (as an aside... ideally you have recorded voltage vs time for each measurement and fit it to an exponential--this assumes it is capacitor charging at constant voltage) I usually tell my students to chalk it up to the finite input impedance of the amp causing the current at the input to the amp to be non-zero and changing the feedback mechanism. I got this explanation from a professor a long time ago, and I am not sure I entirely buy it (Pasco claims there is an enormous input impedance). I never came up with a better answer while I was teaching the lab though.
