Greetings berkeman:
Being that this conversation is about photo diodes, the junction of a semiconductor device determines the characteristics of such a device just as the amounts of 'dopant' and elements used to make the conducting P or N substrates. In a liner device if forward voltage is increased current drop will remain constant, positive junction resistance. If the junction is thick, maybe over a couple of thousand angstroms or more, maybe a couple of hundred atoms thick and capable of quantum tunneling, the junction when it reaches avalanche voltage will exhibit negative resistance in that the current drop across junction will decrease weather this is caused by quantum tunneling or the photo voltaic effect. It is the junction of the device that makes it a semiconductor. Even if that junction is formed by a crude quartz crystal being jabbed by an iron whisker.
As for the V-I diagram the voltage applied is measured by a volt meter at the source, neglecting resistance of the wires, and the current, junction current drop, is measured across the device by an ampere meter, there would be a resistive load in series with the device to prevent destruction of the device, the resistance of which and the voltage drop would be determined in advance mathematically.
I hope that this clears this up in that I had to make this post quickly.
Eimacman.