Hi all, I'm having a little bit of trouble understanding photodiodes in reverse bias. Say for instance I have the circuit shown, If I'm asked for a certain voltage required to switch on RA I understand that this corresponds to a certain current being allowed through the diode which corresponds to a certain light intensity from the graph. But it's the voltage that I'm unclear about. Say for instance the RA requires a Voltage drop of 6V to operate. This corresponds to the diode allowing 2mA of current through. Does this mean that the voltage drop across the diode in this case is 9V ? And subsequently if the Emf was reduced to say 10V does that mean that the RA would still operate with a V drop of 6v at the same light intensity but the voltage drop across the diode would now be 4V. Many thanks in advance.