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Let's say we buildt a common PN-Junction Diode into a circuit. Inside the Diode there is a PN-Junction. But outside, there will also be junctions between the metal of the wires and the semiconductor material. So technically, we have 1 PN-Junction and 2 Schottky-Contacts.

Is this true, and if so, why don't the additional Schottky-Contacts have an influence on the current-voltage-characteristics of the diode?

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  • $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$
    – Qmechanic
    Oct 9, 2016 at 7:37
  • $\begingroup$ I see why it would fit in there. But I don't see why this question doesn't fit in here. $\endgroup$ Oct 9, 2016 at 7:47
  • $\begingroup$ You want the metal-semiconductor contacts to be ohmic. This is achieved by careful doping and alloying of both materials. $\endgroup$
    – polwel
    Oct 9, 2016 at 8:38

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In an pn-junction diode you have indeed the pn-junction itself and two metal contacts, one to the p-type semiconductor region and one to the n-type region. These metal-semiconductor contacts, however, are so-called ohmic contacts that sustain no voltage drop from the current flowing. They are not rectifying Schottky contacts. Usually you get these ohmic contacts by depositing the metal on a highly doped zone on the surface of the semiconductor. This makes the depletion zone and Schottky barrier so thin that electrons can tunnel through this barrier easily resulting in a very low resistance transition from the contact metal to the semiconductor of the pn-junction.

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