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When a (semi)conductor is connected to a voltage bias, a charge is injected by the electrodes. When steady state is reached (constant current flow), does the total charge of the (semi)conductor differ when compared to when a voltage bias is zero? If yes, where does this extra charge come from?

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  • $\begingroup$ Basically, a semiconductor is a resistor that does not follow ohm's law, so physics.stackexchange.com/questions/144444/… , physics.stackexchange.com/questions/143300/… $\endgroup$ – philip_0008 Jun 14 '16 at 0:12
  • $\begingroup$ @philip_0008 what about filling up interface trap states of a heterojunction or bulk defects? $\endgroup$ – Sparkler Jun 14 '16 at 0:42
  • $\begingroup$ You have to give more details of your semiconductor, in particular, what kind of contacts and doping profile you are considering. Is it an ohmic contacts with a layer of high doping in the semiconductor or is it a Schottky contact on a homogeneously doped material? Which region of the semiconductor do you then include in your charge balance consideration? Are the highly doped contact regions, should you have such, included or exclude? $\endgroup$ – freecharly Sep 7 '16 at 17:48
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Depending on the boundary conditions doping profile, current density voltage characteristic, etc., of a semiconductor sample you can get a total non-zero charge in a piece of semiconductor upon applying a voltage which is different from the zero voltage and current case. Due to charge conservation this charge must of course come from outside the semiconductor region considered. On the other hand, when you include the metal contacts in a Gauss surface enclosing the semiconductor, the total charge in this region will be zero because the electric field in the metal contacts are negligible even with a current is flowing.

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