Experiment 1:Don't touch the adjustments. Simply press the button "Start".
Comment:There are two P/N junctions inside of a bipolar transistor. Without being attached to a voltage source, the movable charges inside are diffusing through the crystal lattice. If an extra electron hits a hole, both charges vanish during a recombination process, leaving immobile charges at the impurity inside of the crystal lattice (see simulation of a diode for details). The two zones of the middle layer get more and more negatively charged, the N-doped zones to the left and to the right become positively charged. The electric field caused by the charges inhibits holes from moving away from the middle layer and electrons from moving into the P-doped region. Some minutes pass by until the state of equilibrium is reached
Stop the simulation and have a look at the distribution of charges at the 10 different zones.
Experiment 2:Adjust the collector voltage to 50% and start the simulation once again.
Comment:In this simulation, the attached voltage leads to the injection of electrons at the negative terminal, until the left zone is 5 times negatively charged (50%). At the positive terminal, electrons are removed until the right zone is 5 times positively charged (the electrons have to move to the positive terminal before they are extracted). Those disequilibrium of charges generates an electric field which causes the widening of the right depletion layer. The left depletion layer is weakened, by what recombination processes occur once again and holes are removed from the middle layer. Wait until no more recombination processes occur
The middle layer is more negatively charged than before. Once again, holes can't leave the P-doped layer and electrons can't enter those region.
Increase the collector voltage to 100% and start the simulation once again.
Experiment 3:Adjust the base voltage to 40%, leave the collector voltage at 100%.
Comment:By attaching a voltage to the base, additional holes are created inside of the middle layer, to keep the total charge of this region on a constant level. For each hole, an electron is injected at the negative terminal, to keep the total charge of the crystal at a level of zero. The holes start drifting to the negative terminal, the electrons are moving away from the negative terminal. An electric current is running from the emitter (left layer) to the base (middle layer). Stop the simulation and have a look at the distribution of charges:
The P-doped region is less negatively charges. Some electrons are entering the middle layer, because the repelling electric field is weakened.
Click the "start" button and observe the movement of some electrons from the left, N-doped region of the crystal to the right, N-doped region. An electric current is running between emitter and collector. The ratio of emitter-collector-current to emitter-base-current is clearly below 1.
Inside of a real transistor, those ratio is clearly above 1:
The number of electrons running from the emitter through the transistor to the collector is significantly higher than those running from the emitter to the base. Why is the ratio resulting from or simulation different from those in practice? Well, the probability that an electron can cross the P-doped layer increases if the gaps between the holes increase, too. To do so, the concentration of impurity atoms has to be decreased. At this simulation, the concentration of impurity atoms is significantly higher than inside of a real semiconductor devices. Without the "trick" of lowering the interaction between extra electrons and holes inside of a certain radius around those particles, not a single electron could pass the middle layer at this simulation. Besides the concentration of impurity atoms, the size of the middle layer is critical for the electric properties of a bipolar junction transistor. If those layer is too thin, electrons can move from the emitter to the collector as soon as a certain voltage level is reached, even without a base voltage attached to the device. If the middle layer becomes too thick, the electrons can't pass the device as soon as the emitter base voltage becomes too low, even if a base voltage is attached to the transistor. The size of the middle layer hat to meet the specifications of the operational voltage.
Experiment 4:Click at the button "PNP" and redo experiment 1-3
Comment:Besides using a NPN sandwich, transistors can also be manufactured in PNP order. PNP transistors work in the same way, but the base has to be charged negatively by injecting electrons to make the emitter-collector line conductive. Electrons are removed at the emitter and they are injected at the collector, hence the current is running in the opposite direction. One difference between the movement of extra electrons and holes is the mobility of those charges. Holes are moving slower through the crystal than electrons do. Hence the time, the holes reside in the middle, N-doped layer is larger than those of electrons inside of the P-doped layer in an NPN transistor. The larger transit time causes a higher probability for the holes to recombine with an electron and vanish on it's way through the barrier. Without altering the parameters of the software, the number of holes moving from the emitter to the collector is zero while simulating a PNP transistor. To facilitate the movement of at least some holes from the emitter to the collector, the size of the N-doped layer has to be reduced.