1. Feb 16, 2008

### kuski

1. The problem statement, all variables and given/known data
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,

2. Relevant 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

3. 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

2. Feb 16, 2008

### Mapes

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.

3. Feb 16, 2008

### kuski

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.