A p type semiconductor is brought into contact with an n type semiconductor to form p - n junction. Which statement is false?
A. During forward bias condition, if applied p.d. overcomes the junction potential, electron will cross steadily from n - type side to p - type side
B. During reverse bias condition, the width of the depletion region becomes larger as the externally applied p.d. adds to the junction potential
C. During reverse bias condition, the p type semiconductor becomes less negative
D. During reverse bias condition, the p type semiconductor becomes less positive
E. Under increasingly high reverse bias p.d., current can increase sharply through the p-n junction
The Attempt at a Solution
I think option A is correct but I am not sure about "if the applied p.d. overcomes thr junction potential". Let say I have a diode connected to dc source having a small emf (maybe 1 mV) in forward bias condition, is it possible that current not flowing because 1 mV can not overcome junction potential?
Option B is correct because holes in p type will be attracted to negative terminal of the power source and electrons in n type will be attracted to positive terminal of power source so potential barrier increases.
Option C and D, not sure. Can n type or p type of p-n junction become less positive / negative in forward / reverse bias condition? If it is possible, it means that in reverse bias condition p type becomes less positive because electrons from the negative terminal of power source fill the holes in p type?
I think option E correct, breakdown of p-n junction happens but how to explain it with regards to depletion layer? In reverse bias condition, the width of depletion layer increses and if the p.d. of power source is increased again and again there will be a point where the depletion layer decreases until zero instead of keep increasing?