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Electrical Conductivity of p‐GaSe Single Crystals in Strong Electric Fields
Author(s) -
Abdullaev G. B.,
Guseinova E. S.,
Tagiev B. G.
Publication year - 1966
Publication title -
physica status solidi (b)
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.51
H-Index - 109
eISSN - 1521-3951
pISSN - 0370-1972
DOI - 10.1002/pssb.19660160119
Subject(s) - physics , electrical resistivity and conductivity , condensed matter physics , dielectric , energy (signal processing) , electric field , field (mathematics) , conductivity , thermal conductivity , atomic physics , thermodynamics , quantum mechanics , mathematics , pure mathematics
The electrical conductivity of p‐GaSe single crystals is measured in strong electric fields up to 3 × 10 4 V/cm. Pulse duration is used with 4 μs and a repetition rate of 50 Hz. The critical field for p‐GaSe is about 1600 V/cm, above which Ohm's law fails and the conductivity increases with field according to Frenkel's law: σ = σ 0 e β \documentclass{article}\pagestyle{empty}\begin{document}$\sqrt E $\end{document} . From the deduced temperature dependence of β \documentclass{article}\pagestyle{empty}\begin{document}$\\({\beta = \sqrt {e^3 /} k{\rm}T\sqrt \varepsilon}$\end{document} the dielectric constant ϵ of p‐GaSe single crystals is found to be approximately five. The hole activation energy determined from the temperature dependence of the conductivity, decreases with increasing field, which agrees with Frenkel's theory. By extrapolating the lg σ = f (1/ T ) straight lines a thermal breakdown temperature of about 25000 °K is found, which can be referred to as the metallization temperature ( T m ). It is found that k T m ≈ 2 eV is the thermal energy gap at absolute zero. By extrapolating the lg σ = f ( \documentclass{article}\pagestyle{empty}\begin{document}$\sqrt E $\end{document} ) straight lines a breakdown intensity, E b , of about 10 7 V/cm is found, which characterizes the electrical strength. The width of the energy gap Δ E 0 , obtained from the expression: Δ E 0 = 2 e \documentclass{article}\pagestyle{empty}\begin{document}$\sqrt {eE_b /\varepsilon} $\end{document} appears to be about 1.9 eV.