where an is the capture cross-section of the EL2+ level for electrons and Vth ( = y'8kBT/1rm*) is the mean thermal velocity of electrons. By allowing an to be "thermally activated", i.e., O"n (T) = O"n ( oo )e-E0.fk5T, where Eu=0.07 e V for undoped SI GaAs and 0.25 eV for Cr-doped SI GaAs [3.86], we have
(A4) where we use Eu=0.07 eV and "le= 2 x 10-a cm3 /sec at T=300K (see Table 3.3).
c) Number density of EL2+ in the dark: NEL2+ (Io=0)
Since EL2 is considered as an impurity donor, NEL2+ is obtained by means of the usual theoretical treatment seen in basic solid-state physics textbooks, i.e., by considering the Fermi-Dirac statistics or "the so-called maximum probability method" [3.86]. The result is given by
NEL2+
=
NEL2 - NEL20_ N _ NEL2
- EL2 1
+
(go/gi)e(Ep-Eo)/k5T_ NEL2
- 1
+
(gif go)e(Eo-Ep)/kBT' (AS)where go and g1 are the degeneracy factors for the EL2+ and EL2° levels, respectively, and EF and En are the Fermi energy and the energy of the EL2°
level, respectively, measured from the conduction band. The value of gif go in Eq.
(AS) is 2 because of the two spin states at the EL2° level [3.86]. The temperature dependence of the energy En is approximated by En = 0.76 - 2.5 x 10-4T (eV) for SI GaAs [3.96], but this temperature dependence is negligible in the range of our interest ( 200K<T<400K). So we assume that En and EF are constant in 200k<T<400K. The energy difference EF - En at room temperature is found by using Eq.(AS) together with the data of NEL20 and NEL2+ at T=300K (see Table 3 .4). The result is
(A6)
where NEL2 = 1.704 x 1016 cm-3.
d) Thermal ionization rate: f3e
At thermal equilibrium in the dark, Eq. (2.2a) is written as 8NEL2+
at
= 0 = f3e(N EL2 - NEL2+) - ,enNEL2+.Thus
NEL2 NEL2+ = 1
+
n,e/ f3e.(A1)
(AB) We neglect the recombination of electrons· in the conduction band with holes in the valence band so that n is thermally excited from the EL2° level to the conduction band. We also assume that the Fermi energy is larger than kB T, which is appropriate in SI GaAs. Then it is easy to show that
(A9) where Ne= 2(21rm*kBT)312/h3. Substitution of Eq. (A9) into Eq. (A8) yields to
N NEL2
EL2+ = 1
+
(Nc,e/ f3e)e-Ep/ksT. (AlO) Comparison of Eq.(Al0) with Eq.(AS) immediately gives(All) where C is constant. Using the fact that the dark conductivity O'd = 1 x 10-8(Ocm)-1 at room temperature and f3e/se = ud[se(NEL2 - NEL2+ )rReeµe]-1 at thermal eq\lilibrium in the dark, we obtain
nwf3e = 5 X 107T 2e-0·83/ksT mW/cm2 t \ a
"'= .
1 09 µm,Se
(A12) where Se is assumed to be constant in the range of our interest (200K
<
T<
400K) and TRe = l/(,eNEL2+) = 3 X 10-s sec. at T=300K is used. When
f3e
=
selo/nw, i.e., the number of photoexcited electrons per unit time is the same as that of thermally excited electrons per unit time, it is found to be I0=3 x 1014 photons/ cm2 .sec = 0.06 mW/ cm2 at A = 1.09 µm at room temperature.For BaTi03, using Nn = 1017 - 1018 cm- 3, NA = 2 x 1016 cm- 3,
,e
=5 x 10-8 cm3 /sec, Se = 10-19 - 10-18 cm2 and ud = 1.3 x 10-12 sec [3.97], we find Io = 2 x 1015 - 2 x 1017 photons/ cm2 .sec = 0.8-80 mW/ cm2 at A = 0.5 µm at room temperature.
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4.1 Introduction