How is a pn reverse bias junction is formed in a NPN transistor? In a normal pn diode during a reverse bias junction an 'internal battery ' is formed which counteracts the movement of electrons . But to become a battery we need same amount of donor ions and acceptor ions formed and because a silicon hole is neutralized when it meets a free electron , it needs more silicon atoms at the n type side . But in the NPN transistor the p type region (base) is very thin so the silicon atoms at the base are fewer than the silicon atoms of the emmiter.

  • $\begingroup$ The base only needs to be (thick enough) and (have enough dopants) to allow it to form depletion regions on either side of itself. $\endgroup$ – Jon Custer Feb 11 at 13:37
  • $\begingroup$ And whomever came up with the 'internal battery' analogy for depletion regions should be shot, but I digress... $\endgroup$ – Jon Custer Feb 11 at 13:49
  • $\begingroup$ So emmiter and collector have less silicon atoms than a normal n type semiconductor and base is thick but emmiter is more doped than the base $\endgroup$ – Max Destiny Feb 11 at 13:56
  • $\begingroup$ It has nothing at all to do with how many silicon atoms are around. It is all about the doping levels which drive the Fermi levels and electron and hole concentrations. Now, the width of the depletion region will be driven by the dopant concentrations, but once there is enough integrated charge in the space charge region to counteract the difference in Fermi levels across the device, you are good. $\endgroup$ – Jon Custer Feb 11 at 13:58
  • $\begingroup$ The depletion region of emmiter base junction isnt increased as we increase voltage hmm? $\endgroup$ – Max Destiny Feb 11 at 14:42

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