# Why does collector current increase with base current in BJT?

The base current in the BJT is due to the recombination of holes and electrons. To neutralize the p region an electron leaves the p region. If we take it the other way round, the larger the base current the greater is the recombination, so the collector current should reduce, but $$\beta$$ $$I_c/I_b$$. $$I_c=\beta I_b$$, which shows that $$I_b$$ is directly proportional to $$I_c$$. I am searching for the physical essence of this thing. As we know that in MOSFET voltage controls the drain current by changing the channel width, similarly the BJT is said to be a current-controlled device, so I am eager to know how exactly the base current controls the collector current.

• Wouldn't this be more suitable to electronics stack exchange? Jul 29 '19 at 15:36
• Crossposted to electronics.stackexchange.com/questions/450712/… Jul 31 '19 at 21:54
• There is something wrong in you equation Ib=betaxIc. Aug 1 '19 at 16:41

Additionally electrons are injected from the p-type base to the emitter which are called minority carriers. As indicated in the figure below in steady state the excess minority carrier concentration will have a straight line profile. The concentration of electrons will be lowest or zero at the collector side and highest at the emitter side. This is denoted by $$n_p(0)$$ in the below figure.
The positive collector voltage $$V_{CB}$$ is the cause for zero concentration at the collector side of the base. As a result the electrons are swept away across from the Collector Base depletion region.