Semiconductors — Question 4

Section 41-5 · Physics pages 6878–6879

Question 4 The density-of-states prefactor modifies thermal activation Over a specified range take ni(T) = CT 3/2e−Eg/(2kBT), with constant gap Eg = 1.00 eV and kB = 8.617 × 10−5 eV/K. 1/T ln(ni/T 3/2) Schematic; labels, not drawing lengths, specify the data. (a) Derive the ratio ni(400 K)/ni(300 K). (b) If both mobilities scale as T −3/2, find the corresponding intrinsic conductivity ratio.
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Question 4 - Solution The density-of-states prefactor modifies thermal activation 1/T ln(ni/T 3/2) (a) Method and calculation ni(T2) ni(T1) = (T2 T1 )3/2 exp [ Eg 2kB ( 1 T1 −1 T2 )] . ni(400)/ni(300) = (4/3)3/2e4.8354 = 193.8 . (b) Interpretation and further calculation σi = eni(µn + µp) ∝e−Eg/(2kBT). σi(400)/σi(300) = e4.8354 = 125.9 . Physical check The increasing carrier density outweighs the falling mobility in this model. Gap variation with temperature would require an additional correction.

Question 4

The density-of-states prefactor modifies thermal activation

Over a specified range take ni(T)=CT3/2e−Eg/(2kBT)n_i(T)=C T^{3/2}e^{-E_g/(2k_BT)}, with constant gap Eg=1.00eVE_g=1.00\,\mathrm{eV} and kB=8.617×10−5eV/Kk_B=8.617\times10^{-5}\,\mathrm{eV/K}.

Diagram shown in the original SVG worksheet.

  1. Derive the ratio ni(400K)/ni(300K)n_i(400\,\mathrm K)/n_i(300\,\mathrm K).

  2. If both mobilities scale as T−3/2T^{-3/2}, find the corresponding intrinsic conductivity ratio.

Question 4 - Solution

The density-of-states prefactor modifies thermal activation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ni(T2)ni(T1)=(T2T1)3/2exp⁡[Eg2kB(1T1−1T2)].\frac{n_i(T_2)}{n_i(T_1)}=\left(\frac{T_2}{T_1}\right)^{3/2}\exp\left[\frac{E_g}{2k_B}\left(\frac1{T_1}-\frac1{T_2}\right)\right].ni(400)/ni(300)=(4/3)3/2e4.8354=193.8.\boxed{n_i(400)/n_i(300)=(4/3)^{3/2}e^{4.8354}=193.8}.

(b) Interpretation and further calculation

σi=eni(μn+μp)∝e−Eg/(2kBT).\sigma_i=en_i(\mu_n+\mu_p)\propto e^{-E_g/(2k_BT)}.σi(400)/σi(300)=e4.8354=125.9.\boxed{\sigma_i(400)/\sigma_i(300)=e^{4.8354}=125.9}.

Physical check

The increasing carrier density outweighs the falling mobility in this model. Gap variation with temperature would require an additional correction.