Semiconductors — Question 6

Section 26-7 · Physics pages 4282–4283

Question 6 Carrier creation and mobility loss compete in the temperature coefficient An intrinsic model has ni = CT 3/2e−Eg/(2kBT) and µn + µp = MT −s, with constants C,M,Eg and s positive. valence conduction Eg Schematic band energies; vertical separation represents an energy gap. (a) Derive the logarithmic slope of conductivity with respect to temperature. (b) For s=2, find where the model predicts a stationary conductivity and determine whether it is a maximum or minimum.
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Question 6 - Solution Carrier creation and mobility loss compete in the temperature coefficient valence conduction Eg (a) Method and calculation σ = eCMT 3/2−se−Eg/(2kBT). d ln σ d ln T = 3 2 −s + Eg 2kBT . (b) Interpretation and further calculation s = 2 : −1 2 + Eg 2kBT∗ = 0 ⇒T∗= Eg/kB. T < T∗: dσ/dT > 0; T > T∗: dσ/dT < 0. Thus the stationary point is a maximum if the model remains valid that far. Physical check A formal extrapolated extremum need not fall within the temperature range where the assumed laws apply.

Question 6

Carrier creation and mobility loss compete in the temperature coefficient

An intrinsic model has ni=CT3/2e−Eg/(2kBT)n_i=C T^{3/2}e^{-E_g/(2k_BT)} and μn+μp=MT−s\mu_n+\mu_p=M T^{-s}, with constants C,M,Eg and s positive.

Diagram shown in the original SVG worksheet.

  1. Derive the logarithmic slope of conductivity with respect to temperature.

  2. For s=2, find where the model predicts a stationary conductivity and determine whether it is a maximum or minimum.

Question 6 - Solution

Carrier creation and mobility loss compete in the temperature coefficient

Diagram shown in the original SVG worksheet.

(a) Method and calculation

σ=eCMT3/2−se−Eg/(2kBT).\sigma=eCM T^{3/2-s}e^{-E_g/(2k_BT)}.dln⁡σdln⁡T=32−s+Eg2kBT.\frac{d\ln\sigma}{d\ln T}=\frac32-s+\frac{E_g}{2k_BT}.

(b) Interpretation and further calculation

s=2:−12+Eg2kBT*=0⇒T*=Eg/kB.s=2:\quad-\tfrac12+\frac{E_g}{2k_BT_*}=0\Rightarrow T_*=E_g/k_B.T<T*:dσ/dT>0;T>T*:dσ/dT<0.T<T_*:\ d\sigma/dT>0;\quad T>T_*:\ d\sigma/dT<0.Thus the stationary point is a maximum if the model remains valid that far.

Physical check

A formal extrapolated extremum need not fall within the temperature range where the assumed laws apply.