VIII.2. A Semiconductor Device Primer

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Using the above results n i = N −( E −E ) / k T which one can solve to obtain EF c e E c F B = N e + E 2 k T 2 −E ) / k T If the band structure is symmetrical (Nc=Nv), the intrinsic Fermi level lies in the middle of the band gap. Even rather substantial deviations from a symmetrical band structure will not affect this result significantly, as Nc /Nv enters logarithmically and kBT is much smaller than the band gap. A remarkable result is that the product of the electron and hole concentrations np = n depends only on the band gap Eg and not on the Fermi level. This result, the law of mass action, is very useful in semiconductor device analysis. It requires only that the Boltzmann approximation holds. Qualitatively, it says that if one carrier type exceeds this equilibrium concentration, recombination will decrease the concentrations of both electrons and holes to maintain np= ni 2 . Introduction to Radiation Detectors and Electronics Copyright © 1998 by Helmuth Spieler VIII.2.a. A Semiconductor Device Primer, Doping and Diodes v −( E F v B ( N N ) c v B E F = − log c / 2 i = N c N v e −( E −E ) / k T c v B = N N e c v v −E / k T g B
2. Carrier Concentrations in Doped Crystals The equality ne = nh only holds for pure crystals, where all of the electrons in the conduction band have been thermally excited from the valence band. In practical semiconductors the presence of impurities tips the balance towards either the electrons or holes. Impurities are an unavoidable byproduct of the crystal growth process, although special techniques can achieve astounding results. For example, in the purest semiconductor crystals – “ultrapure” Ge – the net impurity concentration is about 3 . 10 10 cm -3 . In semiconductor device technology impurities are introduced intentionally to control the conductivity of the semiconductor. Let Nd + be the concentration of ionized donors and Na - the concentration of ionized acceptors. Overall charge neutrality is preserved, as each ionized dopant introduces a charged carrier and an oppositely charged atom, but the net carrier concentration is now or Assume that the activation energy of the donors and acceptors is sufficiently small so that they are fully ionized Then + D + − Δn = n − p = N d − Na p + − + N D = n + N A D Introduction to Radiation Detectors and Electronics Copyright © 1998 by Helmuth Spieler VIII.2.a. A Semiconductor Device Primer, Doping and Diodes − A N ≈ N and N ≈ p + N = n + N D A , N A
Page 1 and 2: VIII.2. A Semiconductor Device Prim
Page 3 and 4: The number of occupied electron sta
Page 5: With these simplifications the numb
Page 9 and 10: It is often convenient to refer all
Page 11 and 12: Equilibrium is attained when the tw
Page 13 and 14: As either NA or ND increases relati
Page 15 and 16: 4. The Forward-Biased p-n Junction
Page 17 and 18: This yields the solution n p ( x) =
Page 19 and 20: Since the two electron currents mus
Page 21 and 22: Energy band diagrams showing the in
Page 23 and 24: Since the equilibrium concentration
Page 25 and 26: Note that in the diode equation a)
Page 27 and 28: However, on a logarithmic scale it
Page 29: The discrepancies between the measu

VIII.2. A Semiconductor Device Primer

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