Integrated Systems Laboratory
October 27, 2009
The total current J in a semiconductor sample due to an applied electric field E can be written as the sum of the electron and hole current components
J = J n
+ J p
= ( qnµ n
+ qpµ p
) E.
(1) q denotes the elementary charge, n and p are the carrier densities and µ n
, µ p denote the electron and hole mobilities, respectively. The quantity in parentheses is known as conductivity (in 1 / (Ω cm )):
σ = qnµ n
+ qpµ p
(2)
Note that the electron and hole contributions to conductivity are simply additive. The corresponding resistivity of the semiconductor is the reciprocal of σ , i.e., ρ = 1 /σ (in
Ω cm ).
Figure 1 shows the electron and hole mobilities for Si and GaAs at room temperature as a function of doping density. Calculate the conductivity and the resistivity under the assumption of complete ionization (doping corresponds to majority carrier density) for
• n-doped Silicon and GaAs for doping levels N
D
= 1e15, 1e16, 1e17, 1e18 and 1e19 cm
−
3 .
• p-doped Silicon and GaAs for doping levels N
A
= 1e15, 1e16, 1e17, 1e18 and 1e19 cm
−
3 .
What happens with conductivity and resistivity as function of doping density?

Figure 1: Doping dependent mobility of Si and GaAs at room temperature.
The resistance of a material is given by
R =
L
A
·
ρ (3) where L , A denote the length and the area of the material, respectively.
ρ labels the materials resistivity (in Ωcm). Figure 2 shows the schematic of a n-doped device. The area of the device is A = 100 µ m 2 and the length is L = 100 µ m.
2

Figure 2: Current conduction in a unifomly doped bar with length L and cross-sectional area A .
Tasks
•
Calculate the resistance of a silicon device with L and A for n-doping 1e17, 1e18 cm
−
3 .
•
Calculate the resistance of a silicon device with L and A for p-doping 1e17, 1e18 cm
−
3 .
•
Organic materials have mobilities much lower than the typical semiconductors Si or GaAs. In fact, Alq
3
(Aluminum 8-hydroxyquinoline) has mobilities µ n
= 1e-4 cm 2 / Vs and µ p
= 1e-6 cm 2 / Vs. Assume that the Alq
3 is n-doped with 1e19 cm
−
What is the resistance of a device with A = 100 µ m 2 and length L = 100 µ m?
3 .
•
Metals have very low resistivities. For example, gold has a resistivity of ρ = 2e-6
Ωcm. Calculate the resistance of a device with A = 100 µ m 2 and length L = 100
µ m.
A device made of organic material with length L = 0 .
5 mm and cross-section area
A = 100 µ m 2 is n-doped with a unknown doping concentration N d
[ 1 / cm
3
]. The mobilities are µ n
= 1 e
−
4 the material is n i cm
2 /
Vs and µ p
= 1 e
−
6 cm
2 /
Vs
. The intrinsic carrier density of
= 5 e 5 1 / cm 3 and the resistance of this device is R = 4 .
2 e 7 Ω. Calculate the conductivity σ of the material (in 1 /
Ωcm
).
Use the expression which relates the conductivity σ with the carrier densities and the carrier mobilities to derive a formula for N d
. You can assume full ionization and N d
>> n i
. But this time do not neglect the influence of the minority carriers. Solve your
3

equation analytically for N d and show if the minority carrier mobility is influencing the resulting doping concentration. Finally, calculate the doping concentration N d
(in 1 / cm 3
) to achieve the conductivity σ in this material.
The carrier concentration of a doped material may be different that the impurity concentration, because the ionized impurity density depends on the temperature and the impurity energy level. To measure the carrier concentration directly, the most commonly used method is the Hall effect. Figure 3 shows an electric field applied along the x-axis and a magnetic field applied along the z-axis. Consider a p-type semiconductor sample. The Lorentz force
Figure 3: Basic setup to measure carrier concentration using the Hall effect.
q ( v
×
B ) (= qv x
B z
) (4) due to the magnetic field will exert an average upward force on the holes flowing in the x-direction. The upward directed current causes an accumulation of holes at the top of the sample that gives rise to a downward-directed electric field E y
. Since there is no net
4

current flow along the y-direction in the steady state, the electric field along the y-axis exactly balances the Lorentz force: qE y
= qv x
B z
.
(5)
The establishment of the electric field is known as the Hall effect. This electric field is called Hall field and the terminal voltage V
H
Hall field E y can be written as
= E y
W is called the Hall voltage. The
E y
=
J p qp
B z
= R
H
J p
B z
(6) where R
H
=1 / ( qp ) denotes the Hall coefficient. Thus p =
1 qR
H
=
J p
B z qE y
=
IB z
W
.
qV
H
A
(7)
Tasks
A sample of Si is doped with 1e16 phosphorus atoms/cm 3 . Find the Hall voltage in a sample with W = 500 µ m, A = 2e-3 cm 2 , I = 1 mA, and B z
= 1e-4 Wb/cm 2 .
5