Hi, I have a couple of questions on photoelectrons. When a photoelectron of about 3-eV (varies) interacts within 0.2-um depletion region of silicon, what happens? I know, it will generate an electron-hole pair with an efficiency of 1 for 3.6-eV photoelectron. But what happens if the photoelectron energy is lower than 3.6-eV? Now, what happens to the current following this generation? Since, this pair is in a reverse biased p-n diode it will be pulled to form a signal. The question is what would be the magnitude and how much time would they be available before being lost or filled up (sorry, I am little rusty with the terminology - have moved away from electronics, but have come across this problem). I believe, current (I) = # electrons*charge/time Where, # electrons are the number of electrons; charge- 1.6e-19 C; time - the amount of time the electron is present to generate/affect the signal before they are lost. So, if 25 photoelectrons (3.6-eV) with 100% efficiency then we see a generation of 25 pairs. Let's say, they each are available for a time of 1-us. So, current will be 25*1.6e-19/1e-6 (A). Am I wrong? I don't know what happens when the energy is lower than 3.6-eV and the time this electron-hole pair are available. Please share your valuable thoughts. Regards, Manit.