How NPN transistor works? Explain by suitable diagram.

Explanation

n-p-n Transistor: In this figure, negative terminal of the battery `V_(EB)` is connected to the emitter and positive terminal of this battery is connected to the base. Similarly, positive terminal of the battery `V_(CB)` is connected to the collector and negative terminal of the battery is connected to the base. These connections make E-B junction forward biased and C-B junction reverse biased. Since emitter-base junction is forward biased, electrons from emitter region move towards base and holes from base region move towards emitter. As thickness of the base is very small and also it is lightly doped, only few `(lt 5%)` of the electrons, entering the base region, recombine with the holes there. As soon as an electron combines with a hole in the base region, an electron leaves the negative terminal of the supply `V_(EB)` and enters teh emitter region, at the same time an electron from the base is pulled up by the positive terminal of the battery `V_(EB)`. This gives rise to base current, `I_(B)`. Most of the electrons which pass through the base are pulled by the collector region which is reverse biased and is at high positive potential. Finally, these electrons reach the positive termina of the battery `V_(CB)` giving rise to collector current `I_(C)`. Hence, `I_(E) = I_(B) + I_(C)`

In a n-p-n transistor, current is due to flow of electrons (inside as well as outside). Remember that due to the forward bias, a large current enters the emitter-base junction. Most of this current is diverted to adjacent reverse biased base collector junction and hence current coming out of the base is very small as compared to the current entering the base. If hole current and electron current crossing the forward biased junction are represented by `I_(h) and I_(e)` respectively, then total current in the forward biased diode is `I_(E)= I_(h) + I_(e)` but base current `I_(B) lt lt I_(e) + I_(h)` because major part in n-p-n transistor, goes to collector instead of coming out of the base terminal. Hence base current is a small part i.e., only 5% of emitter current.