5.1 ThepinPhotodetector.
A common semiconductor photodetector is thepin photodiode, shown schemati-
cally in Fig.5.1[ 1 – 8 ]. The device structure consists of apn junction, which is the
interface between p-doped and n-doped materials within a continuous crystal. As a
result of free electron diffusion across the pn junction, an insulating zone orintrinsic
region(i region) that contains no freecharge carriers(electrons or holes) is built up
between the p and n regions. This i region results because free electrons from the n
region will diffuse across the pn junction and be captured by the holes in the p
region, and similarly free holes from the p region will cross the junction interface
and be captured by electrons in the n region. In normal operation a sufficiently large
reverse-bias voltage is applied across the device through a load resistor RLso that
the intrinsic region is fully depleted of carriers. That is, the intrinsic n and p carrier
concentrations are negligibly small in comparison with the impurity concentration
in this region. Thus the i region is referred to as thedepletion region.
As a photonflux penetrates into a photodiode, it will be absorbed as it progresses
through the material. Suppose Pinis the optical power level falling on the pho-
todetector at the device surface where x = 0 and let P(x) be the power level at a
distance x into the material. Then the incremental change dP(x) in the optical power
level as this photonflux passes through an incremental distance dx in the semi-
conductor is given by dP(x) =−αs(λ)P(x)dx, whereαs(λ) is thephoton absorption
coefficientat a wavelengthλ. Integrating this relationship gives the power level at a
distance x into the material as
PðxÞ¼PinexpðÞðasx 5 : 1 ÞThe lower part of Fig.5.1gives an example of the power level as a function of
the penetration depth into the intrinsic region, which has a width w. The width of
the p region typically is very thin so that little radiation is absorbed there.
Example 5.1Consider an InGaAs photodiode in which the absorption
coefficient is 0.8μm−^1 at 1550 nm. What is the penetration depth x at which
P(x)/Pin= 1/e = 0.368?Solution: From Eq. (5.1) the power ratio isPðxÞ
Pin¼expðÞ¼asx exp½ð 0 : 8 Þx¼ 0 : 368Therefore 0 :8x¼ln 0: 368 ¼ 0 : 9997which yields x = 1.25μm.120 5 Fundamentals of Optical Detectors