Optoelectronics problem. Please assist

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SUMMARY

The discussion focuses on calculating the intensity of far infrared (IR) radiation and the carrier lifetime in a phosphorus-doped silicon (Si) photoconductor with a steady-state resistance of 550 Ω. The dimensions of the photoconductor are 6 mm × 25 mm × 12 mm, and the electron and hole mobilities are 2500 cm² V⁻¹ s⁻¹ and 1800 cm² V⁻¹ s⁻¹, respectively. The resistance rises to 1 kΩ in 20 μs after the IR radiation is switched off. Participants suggest estimating photon energy to calculate radiant power per unit area incident on the sample.

PREREQUISITES
  • Understanding of photoconductors and doping, specifically phosphorus-doped silicon.
  • Familiarity with resistance calculations in semiconductor physics.
  • Knowledge of electron and hole mobility in semiconductors.
  • Basic principles of photon energy and its relation to carrier generation.
NEXT STEPS
  • Calculate the intensity of far IR radiation using the relationship between resistance and carrier concentration.
  • Determine the carrier lifetime based on the rise time of resistance after IR exposure.
  • Explore the relationship between photon energy and the conductivity of photoconductors.
  • Investigate the I-V characteristics of photoconductors under varying illumination and voltage bias.
USEFUL FOR

Students and professionals in optoelectronics, semiconductor physics, and electrical engineering, particularly those working with photoconductors and IR radiation applications.

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Homework Statement


A phosphorus (P) doped Si photoconductor is exposed to far IR radiation of unknown intensity. Its
steady state resistance is found to be 550 Ω. The far IR radiation is then
switched off, and the resistance of the photoconductor takes 20 μs to rise to 1
kΩ. Calculate the
i) intensity of the far IR radiation,
ii) carrier lifetime


Homework Equations


Dimensions of photoconductor = (Lx × Ly × Lz) = 6 mm × 25 mm × 12 mm
(N.B. current is flowing in the x-direction, while the far IR radiation is incident
on the x-y plane of the photoconductor).
Wavelength of IR, λ = 12 μm
Electron mobility μ
e = 2500 cm2 V−1 s−1
Hole mobility μh = 1800 cm2 V−1 s−1


The Attempt at a Solution


My attempt
Since the photoconductor was doped with P, the electron would be the majority carrier. And i equate
I = Area * (eVn + eVp)
= Area * (eVn)
And from this i would get

R= (length/area) 1/(eμn)
Since i have the resistance for steady state and during no illumination
I can apply the above formula with the resistance to find the n created by photons and the intrinsic n.

After which i have no idea what so ever on how to look for the intensity. Please help
 
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Perhaps you could assume that every excited carrier corresponds to a photon passing through the sample. You should be able to estimate the photon energy quite precisely. Then calculate the radiant power per unit area incident on the sample.
 
thanks.. will definitely try that out. Another question that came to my mind is a photoconductor's conductivity changes according to the absorbed incident photons. So the i-v curve would be linear? in the dark? And what happens under voltage bias? I keep thinking of a diode's i-v curve.
 

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