2 GaAs at T-300K is uniformly doped with donor impurity atoms to a concentration of N, 2 x 10" cm. Assume an excess carrier lifetime of 5 x 10"s. 10 a) Determine the excess electron-hole recombination rate if the excess hole concentration is õp- S x 10" cm'. b) What is the lifetime of electrons? c) Determine the thermal equilibrium recombination rate of holes.

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Solve a , b and c . In the solution provided before as attached only a and b are solved .
2
GaAs at T=300K is uniformly doped with donor impurity atoms to a concentration of Ng = 2
10
x 106 cm. Assume an excess carrier lifetime of 5 x 10" s.
a) Determine the excess electron-hole recombination rate if the excess hole
concentration is õp= 5 x 10" cm,
b) What is the lifetime of electrons?
c) Determine the thermal equilibrium recombination rate of holes.
Transcribed Image Text:2 GaAs at T=300K is uniformly doped with donor impurity atoms to a concentration of Ng = 2 10 x 106 cm. Assume an excess carrier lifetime of 5 x 10" s. a) Determine the excess electron-hole recombination rate if the excess hole concentration is õp= 5 x 10" cm, b) What is the lifetime of electrons? c) Determine the thermal equilibrium recombination rate of holes.
7:56
(а)
Equation for electron hole recombination rate in terms of excess electron concentration and
excess carrier lifetime is
R =
Here Sn(1) is excess electron concentration, R is electron hole recombination rate and r is
excess carrier lifetime.
Substitute 5x10" cm' for Sn and 5x10 s for r
R_ =-
(5x10“ cm)
(5x10" s)
= 10" cm's
Therefore electron hole recombination rate is R = 10" cm's"
Step2
b)
(b)
In semiconductor
"Po = n}
P.
Here n, and P, are the thermal-equilibrium concentrations of electrons and holes.
Again
P. = N.
= 2x10" cm
Now n,=
P.
Substitute 1,8 x 10° cm for n, and 2x10" cm"for Po.
P.
(1.8×10* cm")
(2x10" cm")
=1.62 x10 cm
Recombination rates are equal for both electrons and holes
Therefore
- P.
PaxT
II
Transcribed Image Text:7:56 (а) Equation for electron hole recombination rate in terms of excess electron concentration and excess carrier lifetime is R = Here Sn(1) is excess electron concentration, R is electron hole recombination rate and r is excess carrier lifetime. Substitute 5x10" cm' for Sn and 5x10 s for r R_ =- (5x10“ cm) (5x10" s) = 10" cm's Therefore electron hole recombination rate is R = 10" cm's" Step2 b) (b) In semiconductor "Po = n} P. Here n, and P, are the thermal-equilibrium concentrations of electrons and holes. Again P. = N. = 2x10" cm Now n,= P. Substitute 1,8 x 10° cm for n, and 2x10" cm"for Po. P. (1.8×10* cm") (2x10" cm") =1.62 x10 cm Recombination rates are equal for both electrons and holes Therefore - P. PaxT II
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