When an NPN transistor is used as a switch as shown below, is it operating in the Forward Active Region or the Saturation region when it is turned ON? I believe its working in the Forward Active Region since V_BC = 0V < 0.4V and V_BE = 0.7V. I read somewhere that an NPN is in saturation when operated as a switch in the shown configuration. I'd appreciate if someone can clarify.
A fast calculation: ON-state Ib is about 4.3mA. (= (5V-0.7V)/1kOhm). If the transistor has current gain =50, then the Ic should be 50 * 4.3mA = 215mA.
R2 and +5VDC supply limits the Ic under 5V/100Ohm = 50mA. So there's at least 4x exessive Ib, the transistor is heavily saturated.
Due led's voltage drop (often about 1.5V) the theoretical max current is probably only 35mA. This gives more reason to believe the saturation to be true.
The current gain =50 is only a quess, but be it even as low as 25, the transistor is still heavily saturated.
This was not asked, but maybe it's interesting: The saturation is resulted from exessive Ib. This is done to be sure that different individual transistors of same type surely have enough Ib for proper(= low voltage drop) switching. The saturation makes the return to off-state slow and delayed, but often this is not harmful. High power high frequency pulse circuits suffer much if the action is slow or delayed. It causes exessive heating. Even short circuits are possible if there is a temporary current path through transistors right from +supply to GND or -supply.