Why does a zener diode have very thin depletion region and high junction voltage? I know that it is heavily doped but can't relate the two. Pls help
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A Zener diode is heavily doped which implies the concentration of impurities is very high, so the carriers. At junction(initially), carriers diffuse from their higher concentration to lower concentration and left behind ions, since the ions formed are more per unit length( since their concentration is high) the depletion layer is very thin. This thin depletion layer forms enough electric field (to oppose carriers), Such that at equilibrium diffusion current is equal to field current(causing net current equal to zero). The junction voltage is high for the same reason and it must be so becuase the carriers on both sides are more so to effectively hinder them the opposing potential( junction potential) must be high.
Why zener diodes need to have such high potentials? This high potential along with thin depletion layer causes very high electric fields and when connected in reverse bias this field is increased such that the electrons bounded to atoms move along the direction of field causing a large amount of field current. This current can increase much more without any significant increase in reverse bias, this property of zener diode is used as voltage regulator, so that it can allow large currents at nearly same potential.
Care must be taken that current doesn't increase beyond the maximum value written on diode because of intense heat the diode will be damaged and become not usable.
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$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– David ZCommented Sep 24, 2016 at 15:49