7.7
7.1
(a)
(b)
(c)
(d)
7.2
7.3
7.4
(a)
(b)
(c)
(d)
7.5
(a)
(b)
(c)
(d)
(e)
Exercises
Which of the charge carrier recombination mechanisms below, occurs due to the electronhole recombination
via a defect state in the bandgap?
Radiative recombination.
Auger recombination.
Shockley–Read–Hall (SRH) recombination.
All of the above.
The diffusion length is the average length that a carrier moves between generation and recombination. Calculate
the minority diffusion length of a minority carrier having a lifetime of τ = 10 µs and minority carrier diffusivity
of D = 25.6 cm 2 /s.
The minority carrier lifetime of a material is the average time which a carrier can spend in an excited state after
electron-hole generation before it recombines. Calculate the minority carrier lifetime for a single crystalline
solar cell having diffusion length of L = 200 µm and minority carrier diffusivity of D = 27 cm 2 /s.
We have discussed that indirect band gap materials have a lower absorption coefficient than direct band gap
materials, due to the fact that the charge carriers need a change in energy and momentum in order to be excited.
If both Si and Ge are indirect band gap materials, why does Ge have a much higher absorption coefficient than
Si in the visible wavelength range?
Because Ge has more valence electrons than Si.
Because Ge has a higher band gap than Si.
Because Si has direct transitions in this part of the spectrum.
Because Ge has direct transitions in this part of the spectrum.
Consider a 100 µm thick p-doped c-Si wafer illuminated with a monochromatic light at a wavelength of 650
nm as illustrated in Figure 7.8. The optical complex refractive index (ñ = n − ik) of the c-Si at 650 nm is ñ =
3.84 − 0.015i. The incident irradiance is 1,000 W/m 2 . The absorption coefficient α is given by α = 4πk/λ.
Calculate:
The absorption coefficient at 650 nm.
The reflectance at interface air/Si (assume ñ air = 1).
The photon flux after reflection at x = 0 and x = 50 µm.
The generation rate G L at x = 50 µm.
The excess of minority carriers, Δn at x = 50 µm in the p-doped wafer.
Assume the following steady-state conditions: sample is uniformly illuminated along y direction as shown in
the figure; dominant thermal recombination and generation process and condition of low injection level and
finally there is no current flowing through the wafer, which means:
7.1
(a)
(b)
(c)
(d)
7.2
7.3
7.4
(a)
(b)
(c)
(d)
7.5
(a)
(b)
(c)
(d)
(e)
Exercises
Which of the charge carrier recombination mechanisms below, occurs due to the electronhole recombination
via a defect state in the bandgap?
Radiative recombination.
Auger recombination.
Shockley–Read–Hall (SRH) recombination.
All of the above.
The diffusion length is the average length that a carrier moves between generation and recombination. Calculate
the minority diffusion length of a minority carrier having a lifetime of τ = 10 µs and minority carrier diffusivity
of D = 25.6 cm 2 /s.
The minority carrier lifetime of a material is the average time which a carrier can spend in an excited state after
electron-hole generation before it recombines. Calculate the minority carrier lifetime for a single crystalline
solar cell having diffusion length of L = 200 µm and minority carrier diffusivity of D = 27 cm 2 /s.
We have discussed that indirect band gap materials have a lower absorption coefficient than direct band gap
materials, due to the fact that the charge carriers need a change in energy and momentum in order to be excited.
If both Si and Ge are indirect band gap materials, why does Ge have a much higher absorption coefficient than
Si in the visible wavelength range?
Because Ge has more valence electrons than Si.
Because Ge has a higher band gap than Si.
Because Si has direct transitions in this part of the spectrum.
Because Ge has direct transitions in this part of the spectrum.
Consider a 100 µm thick p-doped c-Si wafer illuminated with a monochromatic light at a wavelength of 650
nm as illustrated in Figure 7.8. The optical complex refractive index (ñ = n − ik) of the c-Si at 650 nm is ñ =
3.84 − 0.015i. The incident irradiance is 1,000 W/m 2 . The absorption coefficient α is given by α = 4πk/λ.
Calculate:
The absorption coefficient at 650 nm.
The reflectance at interface air/Si (assume ñ air = 1).
The photon flux after reflection at x = 0 and x = 50 µm.
The generation rate G L at x = 50 µm.
The excess of minority carriers, Δn at x = 50 µm in the p-doped wafer.
Assume the following steady-state conditions: sample is uniformly illuminated along y direction as shown in
the figure; dominant thermal recombination and generation process and condition of low injection level and
finally there is no current flowing through the wafer, which means: