A NPN bipolar junction transistor is made by two heavily N doted pieces (Collector and Emitter) of the semiconductor which are spectated by a smaller layer of a lighter doted P substrate (Base). Thus two PN junctions are created.
(Voltages are in respect to the order of the given nodes in brackets. -> U(BC)=-U(CB))
~Cutoff~
Neither of the two junctions are saturated.
Requirements: U(BE)<0.5V and U(BC)<0.4V
—> I(CE)=I(B)=I(BE)=I(BC)=0A
~Active~
The B-E junction is saturated but the B-C junction is reverse biased.
Requirements: U(BE)=0.7V and U(BC)<0.4V
Electrons can move freely between emitter and base. Some holes at the Base have recombined but most electrons will go to the collector due to the reverse biasing of the B-C junction.
Conventional current opposes the flow of electrons thus the conventional current flows from collector to emitter.
—> I(B)=I(BE)>0A; I(BC)=0A and I(CE)=beta*I(B)
—> U(CE)>0.3V
~Saturation~
Both of the junctions are now saturated.
Requirements: U(BE)=0.7V and
U(BC)=approx. 0.5V
All of the holes in the P-substrate have recombined with electrons now. The entire BJT is now a piece of electron saturated semiconductor (hence the name).
—>I(C)=beta(forced)*I(B)
—> U(CE)=approx. 0.2V
~Reverse Operation~
(Coming soon...maybe)
This is not meant to be any kind of lesson, it’s just a revision for me. Still I want to share it, so maybe someone else can benefit from it as well. The given information are not necessarily right so please correct me if I understood something wrong.
