A laser beam of 0.005 W with photon energy of 1.6 eV is incident on a GaAs PIN phot-detector. The detector i-region width is 6 jum , its cross sectional area is A = 0.8 cm, and its absorption coefficient at the incident photon energy is agaAs =1 x 10 cm'. If 20\% of the incident photon flux is reflected back at the detector surface, calculate the detector photo-current I: Select one: O A. IL = 0.00230 A O B. IL = 9.02e-4 A O C. IL = 8.00e-5 A OD. IL = 8.00e-4 A OE. IL = 0.00130 A Clear my choice
A laser beam of 0.005 W with photon energy of 1.6 eV is incident on a GaAs PIN phot-detector. The detector i-region width is 6 jum , its cross sectional area is A = 0.8 cm, and its absorption coefficient at the incident photon energy is agaAs =1 x 10 cm'. If 20\% of the incident photon flux is reflected back at the detector surface, calculate the detector photo-current I: Select one: O A. IL = 0.00230 A O B. IL = 9.02e-4 A O C. IL = 8.00e-5 A OD. IL = 8.00e-4 A OE. IL = 0.00130 A Clear my choice
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![A laser beam of 0.005 W with photon energy of 1.6 eV is incident on a GaAs PIN phot-detector. The detector i-region width
is 6 pum, its cross sectional area is A= 0.8 cm, and its absorption coefficient at the incident photon energy is agaAs
cm'. If 20\% of the incident photon flux is reflected back at the detector surface, calculate the detector photo-current IL:
1x 10
Select one:
O A. IL = 0.00230 A
O B. IL = 9.02e-4 A
O C. IL = 8.00e-5 A
O D. IL = 8.00e-4 A
OE. IL = 0.00130 A
Clear my choice
that will absorh 60 %% of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03fa6e16-ab76-4e78-9ee8-9e9d789cf68c%2F5b54a84a-0a1d-424e-8a57-45b951c04b63%2Fxzumiyb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A laser beam of 0.005 W with photon energy of 1.6 eV is incident on a GaAs PIN phot-detector. The detector i-region width
is 6 pum, its cross sectional area is A= 0.8 cm, and its absorption coefficient at the incident photon energy is agaAs
cm'. If 20\% of the incident photon flux is reflected back at the detector surface, calculate the detector photo-current IL:
1x 10
Select one:
O A. IL = 0.00230 A
O B. IL = 9.02e-4 A
O C. IL = 8.00e-5 A
O D. IL = 8.00e-4 A
OE. IL = 0.00130 A
Clear my choice
that will absorh 60 %% of
![: Si, Ge, GaAs Parameters and Universal Constants
UNIVERSAL CONSTANTS
Properties
SEMICONDUCTOR
h
6.63 × 10-34 J.s
Si
Ge
m.
9.11 x 10-31 Kg
Eg (eV)
п, (ст-3)
Hn (cm²/V – s)
Hp (cm² /V – s)
Ne (cm-8)
N, (cm-³)
m/m.
mi,/m.
€, (F/m)
1.1
0.67
1.42
3.14
1.5 x 1010
2.3 x 1013
1.8 x 106
1.602 × 10–19 C
1500
3900
8500
8.85 x 10-12 F/m
450
1900
400
1.05 × 10¬34 J – s
2.78 x 1019
1.04 x 1019
4.45 × 1017
8.6 x 10-5 eV/K
26 mV (T = 300 K)
26 теV (T %3 300 К)
3 × 10® m/s
K
9.84 × 1018
6 x 1018
7.72 × 1018
KT/q
0.082
0.98
0.067
ΚΤ
0.28
0.49
0.45
11.7
16
13.1
Nair = 1
NGaAs = 3.66
Some useful relations
E ALGA1-,As (x) = 1.424 + 1.247x
EgIn.Ga1-2As(x) = 0.36 + 1.064x
1 eV = 1.602 × 10¬19 J
1 KG = 1 × 10-5 Wb/cm²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03fa6e16-ab76-4e78-9ee8-9e9d789cf68c%2F5b54a84a-0a1d-424e-8a57-45b951c04b63%2Finoxeb9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:: Si, Ge, GaAs Parameters and Universal Constants
UNIVERSAL CONSTANTS
Properties
SEMICONDUCTOR
h
6.63 × 10-34 J.s
Si
Ge
m.
9.11 x 10-31 Kg
Eg (eV)
п, (ст-3)
Hn (cm²/V – s)
Hp (cm² /V – s)
Ne (cm-8)
N, (cm-³)
m/m.
mi,/m.
€, (F/m)
1.1
0.67
1.42
3.14
1.5 x 1010
2.3 x 1013
1.8 x 106
1.602 × 10–19 C
1500
3900
8500
8.85 x 10-12 F/m
450
1900
400
1.05 × 10¬34 J – s
2.78 x 1019
1.04 x 1019
4.45 × 1017
8.6 x 10-5 eV/K
26 mV (T = 300 K)
26 теV (T %3 300 К)
3 × 10® m/s
K
9.84 × 1018
6 x 1018
7.72 × 1018
KT/q
0.082
0.98
0.067
ΚΤ
0.28
0.49
0.45
11.7
16
13.1
Nair = 1
NGaAs = 3.66
Some useful relations
E ALGA1-,As (x) = 1.424 + 1.247x
EgIn.Ga1-2As(x) = 0.36 + 1.064x
1 eV = 1.602 × 10¬19 J
1 KG = 1 × 10-5 Wb/cm²
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