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Asked in GSEB Board July 2016 · Depletion region and barrier potential
Given: barrier potential V = 0.50 V and depletion width w = 5.0×10⁻⁷ m.
Idea: the barrier is built up by the field of the uncovered donor and acceptor ions in the depletion layer. Taking that field as uniform across the layer gives V = E w, so E = V/w.
E = (0.50)/(5.0×10⁻⁷)
E = 1.0×10⁶ V m⁻¹
Half a volt dropped across half a micrometre really does mean a field of a million volts per metre; that is why the junction is so sensitive to the applied bias.
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