Question

A5)

i) The journey of free electrons from the emitter to the collector in a NPN BJT consists of two parts. What are they ?

ii)  A NPN BJT is made of two pn junctions, base/emitter junction and base/collector junction. Which pn junction is thicker and why ?

iii) How can you eliminate a pn junction?

iv) How can you increase or decrease the width of a pn junction?

Answer 1

1) In npn bjt, drift and diffusion are two main components for electron flow from emitter to collector, where diffusion current means diffusion of electron- holes and drift current is due to applied source.

ii) in BJT, two junction are formed one is base emitter junction which is forward biased and another is collector base junction which is reversed biased, doping concentration of these three elements are as Emitter>Base>Collector i.e emitter is highly doped as compared to base and collector, therefore base collector junction when formed it is thicker or wider due to low doping concentration.

iii) PN junction is made up of sandwiching p and n semiconductor material, we can eliminate barrier junction by applying forward biased, when applied voltage is greater it increases current flow due to recombination of electron hole pair. Hence it reduces on junction barrier.

iv) Pn junction width depends on biasing of the diode, when pn junction is forward biased then due to repelling of electrons and holes towards junction current flow increase and due to high recombination on junction width reduces. When diode is reversed biased, at that time electron and holes get attracted towards positive and negative terminal of applied voltage, which increase width due to low recombination if electron and holes, hence in reverse biase on junction width Increases.

Note- please find me in comment box if you find any difficulty. I am little unsure about part ii) and part iii), because it is not mentioned In which respect it is asked?. I request you please let me know if any discrepancy, I will update it as per your recommendation. Good luck