Page 84 - Photonics Essentials an introduction with experiments

Photoconductivity

78 Photonic Devices

electron is different from the mobility of holes, and for the vast major-
ity of semiconductors, it turns out that e > h . The mobilities for elec-
trons and holes are different for each semiconductor material. For ex-
ample, in silicon,
–1
2
2
–1
e = 1500 cm -V -sec –1 and h = 600 cm -V -sec –1
whereas for GaAs,
–1
2
2
–1
e = 8000 cm -V -sec –1 and h = 600 cm -V -sec –1
The mobilities in any given material will depend also on the tempera-
ture and on the level of impurities, and this feature can be exploited
to optimize the performance of a photoconductive device. For simplici-
ty, however, we will concentrate first on the behavior of undoped
semiconductor materials at room temperature.
The mobility is a key parameter for charge transport. It relates the
velocity of charge propagation to the electric field:
v = · cm-sec –1 (5.2)

Equation 5.2 implies some important assumptions. A free electron in
a vacuum is accelerated by an electric field, which provides a con-
stant force. In Eq. 5.2, the application of a constant force produces a
constant velocity. This kind of relationship is typically used to de-
scribe resistive or viscous fluid flow. Skydivers speak of terminal ve-
locity in free-fall conditions. This is the velocity produced by gravita-
tional acceleration opposed by air resistance. In analogy, a constant
“terminal” drift velocity of an electric charge is the result of the op-
posing forces of acceleration by an electric field and the resistive
force of the semiconductor material. The mobility is the constant of
proportionality, reducing to a single number the complex movement
of electronic charge through the semiconductor material. The unusu-
al units attributed to the mobility are needed to relate electric field
to velocity.

Example 5.1
Determine the transit time of an electron and a hole across a photo-
conductive detector made of GaAs with an electrode separation of 10
microns, and a bias voltage of 1 V.
The photoconductive device structure is often an interdigitated ar-
ray, as shown in Fig. 5.1.
First, determine the drift velocity:
1
6
v = · = v e = 8000 · = 8 × 10 cm-sec –1 for electrons
10 × 10 –4

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2004 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.