What prevents lead attached to a piece of semiconductor (for example in BJTs and FETs) from forming a Shottky junction?
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1\$\begingroup\$ I can't answer your question, but I can tell you that the technical name for these is called "Ohmic contacts" or "Ohmic junctions". You can look up those phrases for some rather technical material about the technology. Apparently it's an entire field unto itself just like PN junctions. \$\endgroup\$– DKNguyenAug 14, 2020 at 15:58
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\$\begingroup\$ FYI, in the integrated circuit industry, this "lead wire" (not Pb but like leading-in) is called a "bond wire", and the process of welding the bond wire is called "wire bonding"; see electronics.stackexchange.com/questions/302551/… \$\endgroup\$– MarkUAug 15, 2020 at 4:45
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2 Answers
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There is usually an intermediary material between the semiconductor and the lead wires that does not form a Schottky barrier.
Either the metal-side material is chosen so that the resultant potential energy barrier at the junction is low enough that it doesn't significantly impede the flow of charge carriers or the semiconductor near the junction is so highly doped that the barrier is thin enough that charge carriers can tunnel through it.
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1\$\begingroup\$ Thanks, I'll edit the answer, for your information there is an edit button and you can edit your answer \$\endgroup\$– Voltage Spike ♦Aug 14, 2020 at 21:26
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In VLSI design we have billions of contacts from metal to silicon, and to my knowledge this does not typically require any additional layers or materials to prevent the formation of a Schottky barrier. However, these contacts are always made directly to a heavily doped n or p region such as the source/drain regions of a transistor and the resulting "junction" is effectively ohmic.
answered Aug 14, 2020 at 23:52