If the emitter and base of NPN transitor have same doping conecentration, how will the collector and base currents be affected?
Answers
The emitter, as the name suggests, emits carriers. In case of an NPN transistor those will be electrons.
A higher doping level means more of these carriers will be generated (per volume and time) compared to a region with less doping. These "shallow doped" areas are the base and the collector.
A BJT works by the carriers from the emitter "overwhelming" the amount of (opposite) carriers (in case of an NPN: holes) in the base region. Also, the base region is narrow. This makes the chance that a carrier originating from the emitter recombines in the base, small. If that chance was large (due to a highly doped base) then the base current (to fill up all those used-up holes) would be much larger !
So: the Base needs to have a lower doping level (compared to the emitter) so that the beta (current amplification) will be large, which is what we want.
The collector could be doped more and the beta could then still be high. Remember that beta is related to the doping ratios of emitter and base. In an NPN under non-saturated operation the collector is at a high voltage so all the carrier electrons will be "sucked out" and travel to the positive supply. The electrons originating from the collector therefore cannot influence what happens at the emitter-base junction