When an electron distribution is subjected to an electric field, the electrons tend to move in the field direction (opposite to the field E) and gain velocity from the field. However, because of imperfections in the crystal potential, they suffer scattering. A steady state is established in which the electrons have some net drift velocity in the field direction. The response of the electrons to the field can be represented by a velocity-field relation. We will briefly discuss the velocity-field relationships at low electric fields and moderately high electric fields.
4.4. Macroscopic transport properties 163
Low field response: mobility
At low electric fields, the macroscopic transport properties of the material (mobility, conductivity) can be related to the microscopic properties (scattering rate or relaxation time) by simple arguments. We will not solve the Boltzmann transport equation, but we will use simple conceptual arguments to understand this relationship. In this approach we make the following assumptions:
(i) The electrons in the semiconductor do not interact with each other. This approximation is called the independent electron approximation.
(ii) Electrons suffer collisions from various scattering sources and the time rsc describes the mean time between successive collisions,
(iii) The electrons move according to the free electron equation
= E ( 4 , 4 ,
in between collisions. After a collision, the electrons lose all their excess energy (on the average) so that the electron gas is essentially at thermal equilibrium. This assumption is really valid only at very low electric fields.
According to these assumptions, immediately after a collision the electron ve- locity is the same as that given by the thermal equilibrium conditions. This average velocity is thus zero after collisions. The electron gains a velocity in between collisions;
i.e., only for the time rsc.
This average velocity gain is then that of an electron with mass m*, traveling in a field E, for a time rsc
e¥T
vd (4.25)
a v g ^ d
6 ra*
where vd is the drift velocity. The current density is now
J = -nevd = ^ J i E (4.26)
ra* v J
Comparing this with the Ohm's law result for conductivity a
J = <JE (4.27) we have
ne2rsc
a ( 4 2 8
>
The resistivity of the semiconductor is simply the inverse of the conductivity. From the definition of mobility //, for electrons
vd = /iE (4.29)
we have
Semiconductor C
GaN Ge Si a-SiC GaSb GaAs GaP InSb InAs InP CdTe PbTe
Ino.53Gao.47As
Bandgap (eV) 300 K
5.47 3.4 0.66 1.12 3.00 0.72 1.42 2.26 0.17 0.36 1.35 1.48 0.31 0.8
0 K 5.48 3.5 0.74 1.17 3.30 0.81 1.52 2.34 0.23 0.42 1.42 1.61 0.19 0.88
Mobility (cm2 Elec.
1800 1400 3900 1500 400 5000 8500 110 80000 33000 4600 1050 6000 11000
at 300 K /V-s)
Holes 1200 350 1900 450 50 850 400 75 1250 460 150 100 4000 400
Table 4.1: Bandgaps along with electron and hole mobilities in several semiconductors. Prop- erties of large bandgap materials (C, GaN, SiC) are continuously changing (mobility is improv- ing), due to progress in crystal growth. Zero temperature bandgap is extrapolated.
If both electrons and holes are present, the conductivity of the material becomes a - nefjLn +pefAp (4.31) where jin and fip are the electron and hole mobilities and n and p are their densities.
Notice that the mobility has an explicit ^~ dependence in it. Additionally rsc
also decreases with m*. Thus the mobility has a strong dependence on the carrier mass.
In Table 4.1 we show the mobilities of several important semiconductors at room tem- perature. The results are shown for pure materials. If the semiconductors are doped, the mobility decreases. Note that Ge has the best hole mobility among all semiconductors.
In Appendix E we have derived mobility limited by ionized impurity and by alloy scattering.
The total relaxation time due to ionized impurity scattering is 1
((r))
In 1 + (8m*kBT
(kBT) 3/2
1 + 8m*k
BT
(4.32)
4.4. Macroscopic transport properties 165 The mobility limited from ionized impurity scattering is
fi = m*
The mobility limited by ionized dopant has the special feature that it decreases with temperature (/i ~ T3/2). This temperature dependence is quite unique to ionized impurity scattering. One can understand this behavior physically by saying that at higher temperatures, the electrons are traveling faster and are less affected by the ionized impurities.
After doing the proper ensemble averaging the relaxation time for the alloy scattering is
m(kBT) 1
- x) ^ 2 h 3 — (4.33) according to which the mobility due to alloy scattering is
The temperature dependence of mobility is in contrast to the situation for the ionized impurity scattering. The value of Uau is usually in the range of 1.0 eV.
E X A M P L E 4.2 Consider a semiconductor with effective mass m* = 0.26 mo. The optical phonon energy is 50 meV. The carrier scattering relaxation time is 10~13 sec at 300 K. Calculate the electric field at which the electron can emit optical phonons on the average.
In this problem we have to remember that an electron can emit an optical phonon only if its energy is equal to (or greater than) the phonon energy. According to the transport theory, the average energy of the electrons is (vd is the drift velocity)
E = ^kBT-\- -m vd
In our case, this has to be 50 meV at 300 K. Since ksT ~ 26 meV at 300 K, we have -m*v2d = 50 - 39 = 11 meV
2 _ 2 x (11 x 10~3 x 1.6 x 10~19 J)
Vd ~ (0.91 x 10~30 x 0.26 kg) vd = 1.22 x 105 m/s
Also we should note that the symbol E is being used for the electric field and energy vd = — -erE
m*
Substituting for Vd, we get (for the average electrons) for the electric field (0.26 x 0.91 x 10"30 kg)(1.22 x 105 m/s) E = (4.8 x 10"10 esu)(1013 s)
= 18.04 kV/cm
The results discussed correspond approximately to silicon. Of course, since the distribution function has a spread, electrons start emitting optical phonons at a field lower than the one calculated above for the average electron.
E X A M P L E 4.3 The mobility of electrons in pure GaAs at 300 K is 8500 cm2/V-s. Calculate the relaxation time. If the GaAs sample is doped at N^ = 1017 cm"3, the mobility decreases to 5000 cm2/V-s. Calculate the relaxation time due to ionized impurity scattering.
The relaxation time is related to the mobility by
( 1 ) _ m*fi _ (0.067 x 0.91 x 10~30 kg)(8500 x 10"4 m2/V • s)
Tsc ~ e ~~ 1.6 x 10-1 9 C
= 3.24 x 10~13 s
If the ionized impurities are present, the time is
= 1.9 x 10"
e
The total scattering rate is the sum of individual scattering rates. Since the scattering rate is inverse of scattering time we find that (this is called Mathieson's rule) the impurity-related
, • (imp) •
time TIC J IS given
which gives
by
1 r( 2 )
'sc
r(i mp ) 1 1
= 4.6 x 1
(imp)
Tsc
10 s
E X A M P L E 4.4 The mobility of electrons in pure silicon at 300 K is 1500 cm2/Vs. Calculate the time between scattering events using the conductivity effective mass.
The conductivity mass for indirect semiconductors, such as Si, is given by
—+ - ^
m* m
IN"1
26rao
( 2 I X "1
= 3 1 = 0.
V0.19mo 0.98mo; The scattering time is then
_ fiml _ (0.26 x 0.91 x 10~30)(1500 x 10~4)
Tac ~ e ~ 1.6 x lO"19
= 2.2 x 10"13 s
E X A M P L E 4.5 Consider two semiconductor samples, one Si and one GaAs. Both materials are doped n-type at Nd = 1017 cm"3. Assume 50 % of the donors are ionized at 300 K.
Calculate the conductivity of the samples. Compare this conductivity to the conductivity of undoped samples.
You may assume the following values:
pn(Si) = 1000 c m2/ V - s fip(Si) = 350 c m2/ V - s /zn(GaAs) = 8000 c m2/ V - s
= 400 c m2/ V - s
4.4. Macroscopic transport properties 167 In the doped semiconductors, the electron density is (50 % of 101T cm"3)
ridoped = 5 x 1016 c m "3
and hole density can be found from
n?
Pdoped ^doped
For silicon we have
3 - 3
- 4.5 x 10 cm
PdoPed = 5 x 1 Q 1 6
which is negligible for the conductivity calculation.
The conductivity is
crdoped = riefin + pefip = 8 (Q c m ) "1
In the case of undoped silicon we get (n = rii — p = 1.5 x 1010 cm"3)
For GaAs we get
= 3.24 x 10 6 (Ocm)
<TdoPed = 5 x 1016 x 1.6 x 10"19 x 8000 = 64 (Q cm)"1 For undoped GaAs we get (m = 1.84 x 106 cm"3)
^undoped = nxe\in + PiCfip = 2.47 x 1 0 " (£1 c m ) "
You can see the very large difTerence in the conductivities of the doped and undoped samples.
Also there is a large difference between GaAs and Si.
E X A M P L E 4.6 Consider a semiconductor in equilibrium in which the position of the Fermi level can be placed anywhere within the bandgap.
What is the maximum and minimum conductivity for Si and GaAs at 300 K? You can use the data given in the problem above.
The maximum carrier density occurs when the Fermi level coincides with the conduc- tion bandedge if Nc > Nv or with the valence bandedge if Nv > A^c. If Nc > Nv; the Boltzmann approximation gives
Wmax = Nc
while if Nv > Nc we get
P m a x —- I* v
This gives us for the maximum density: i) for Si, 2.78 xlO19 cm"3 ii) for GaAs, 7.72 xlO18 cm"3. Based on these numbers we can calculate the maximum conducitvity:
For Si
<Tmax = 2.78 x 1019 x 1.6 x 10"19 x 1000 = 4.45 x 103 (ft cm)"1 For GaAs
<rmax = 7.72 x 1018 x 1.6 x 10"19 x 400 = 4.9 x 102 (Q cm)"1
To find the minimum conductivity we need to find the minima of the expression
<T =
To find the minimum we take the derivative with respect to p and equate the result to zero.
This gives
This then gives for the minimum conductivity
r llh , //^M
Cmin = Tlie[fln* h fXp , / J y fin \j fip
For Si this gives upon plugging in numbers
<Tmin = 2.8 x 10"6 (Q cm)"1 and for GaAs
crmin = 1.05 x 10"9 (Q cm)"1
Note that these values are lower than the values we get in the the previous problem for the undoped cases. This example shows the tremendous variation in conductivity that can be obtained in a semiconductor.
High field transport: velocity—field relations
In most electronic devices a significant portion of the electronic transport occurs under strong electric fields. This is especially true of field effect transistors. At such high fields (~ 1— 100 kV/cm) the electrons get "hot" and aquire a high average energy. The extra energy comes due to the strong electric fields. The drift velocities are also quite high. The description of electrons at such high electric fields is quite complex and requires either numerical techniques or computer simulations. We will only summarize the results.
At high electric field as the carriers gain energy from the field they suffer greater rates of scattering, i.e., rsc decreases. The mobility thus starts to decrease. It is usual to represent the response of the carriers to the electric field by velocity-field relations. In Fig. 4.10 we show such relations for several semiconductors. At very high fields the drift velocity becomes saturated; i.e., becomes independent of the electric field. The drift velocity for carriers in most materials saturates to a value of ~ 107 cm/s. The fact that the velocity saturates is very important in understanding current flow in semiconductor devices.
EXAMPLE 4.7 The mobility of electrons in a semiconductor decreases as the electric field is increased. This is because the scattering rate increases as electrons become hotter due to the applied field. Calculate the relaxation time of electrons in silicon at 1 kV/cm and 100 kV/cm at 300 K.
The velocity of the silicon electrons at 1 kV/cm and 100 kV/cm is approximately 1.4 x 106 cm s and 1.0 X 107 cm/s, respectively, from the v-F curves given in Fig. 4.10. The mobilities are then
fi(l kV/cm) = -^ = 1400 cm2/V • s
^(100 kV/cm) = 100 cm2/V • s
4.4. Macroscopic transport properties 169
108 F
107
s
>
u<
Electrons
102 103 104
ELECTRIC FIELD (V/cm)
105 106
Figure 4.10: Velocity-field relations for several semiconductors at 300 K.
The corresponding relaxation times are
< 0 ' 26 X °' 91
MlkV/cm) = j1
J.. D X J. U
= 2.1 x
Thus the scattering rate has dramatically increased at the higher field.
E X A M P L E 4.8 The average electric field in a particular 2.0 //m GaAs device is 5 kV/cm.
Calculate the transit time of an electron through the device (a) if the low field mobility value of 8000 cm2/V-s is used; (b) if the saturation velocity value of 10r c m / s is used.
If the low field mobility is used, the average velocity of the electron is v = fiE = (8000 cm2/Vs) x (5 x 103 V / cm) = 4 x 107 c m / s The transit time through the device becomes
_ L _ 2.0 x 10~4 cm _
TtI~ v ~ 4 x 107 c m / s ~ 5 P S
The transit time, if the saturation velocity (which is the correct velocity value) is used, is _ L_ _ 2 x 10~4
In our discussion on MESFET and MOSFET devices later in the text, we use a simple analytical model, and use the constant mobility model for electron velocity. As this example shows, this can cause an error in transit time.
Conduction band
Final state has two electrons + one hole
Initial state has one electron
Valence band
Figure 4.11: The impact ionization process where a high energy conduction-band electron scatters from a valence-band electron, producing two conduction-band electrons and a hole.
Very high field transport: breakdown phenomena
When the electric field becomes extremely high (~100 kV cm"1), the semiconductor suffers a "breakdown" in which the current has a "runaway" behavior. The breakdown occurs due to carrier multiplication, which arises from the two sources discussed below.
By carrier multiplication we mean that the number of electrons and holes that can participate in current flow increases. Of course, the total number of electrons is always conserved.
Avalanche breakdown
In the transport considered in the previous subsections, the electron (hole) remains in the same band during the transport. At very high electric fields, this does not hold true.
In the impact ionization process shown schematically in Fig. 4.11, an electron, which is
"very hot" (i.e., has a very high energy due to the applied field) scatters with an electron in the valence band via coulombic interaction, and knocks it into the conduction band.
The initial electron must provide enough energy to bring the valence-band electron up into the conduction band. Thus the initial electron should have energy slightly larger than the bandgap (measured from the conduction-band minimum). In the final state we now have two electrons in the conduction band and one hole in the valence band.
Thus the number of current carrying charges have multiplied, and the process is often called avalanching. Note that the same could happen to "hot holes" and thus could then trigger the avalanche.
4.4. Macroscopic transport properties 171
Material
GaAs Ge InP Si
Ino.53Gao.47As C
SiC SiO2
S13N4 GaN
Bandgap (eV)
1.43 0.664 1.34 1.1 0.8
5.5
2.9 9 5 3.4
Breakdown electric field (V/cm)
4 x l O5 105
3 x l 05 2 x l O5 107 2-3 x 106 -107 -107
~5x106 Table 4.2: Breakdown electric fields in some materials.
Once avalanching starts, the carrier density in a device changes as dn{z)
dz
ximp i (4.34)
where n is the carrier density and a im p represents the average rate of ionization per unit distance.
The coefficients aimp for electrons and f3-imp for holes depend upon the bandgap of the material in a very strong manner. This is because, as discussed above, the process can start only if the initial electron has a kinetic energy equal to a certain threshold (roughly equal to the bandgap). This is achieved for lower electric fields in narrow gap materials.
If the electric field is constant so that a? m p is constant, the number of times an initial electron will suffer impact ionization after traveling a distance x is
n(x) - exp (ai (4.35)
A critical breakdown field Ecrit is defined where ai m p or $m p approaches 104 cm"1. When a im p (Amp) approaches 104 cm"1, there is about one impact ion- ization when a carrier travels a distance of one micron. Values of the critical field are
Electrons in conduction band Available empty
states (holes) in valence band
(a)
T
1
1 1
-Xj
K
\
(b)
Figure 4.12: (a) A schematic showing the E-x and E-k diagram for a p-n junction. An electron in the conduction band can tunnel into an unoccupied state in the valence band or vice versa, (b) The potential profile seen by the electron during the tunneling process.
given for several semiconductors in Table 4.2. The avalanche process places an important limitation on the power output of devices. Once the process starts, the current rapidly increases due to carrier multiplication and the control over the device is lost. The push for high-power devices is one of the reasons for research in large gap semiconductor devices. It must be noted that in certain devices, such as avalanche photodetectors, the process is exploited for high gain detection. The process is also exploited in special microwave devices.
Band-to-band tunneling breakdown
In quantum mechanics electrons behave as waves and one of the outcomes of this is that electrons can tunnel through regions where classically they are forbidden. Thus they can penetrate regions where the potential energy is larger than their total energy.
This process is described by the tunneling theory. This theory is invoked to understand another phenomenon responsible for high field breakdown. Consider a semiconductor under a strong field, as shown in Fig. 4.12a. At strong electric fields, the electrons in the valence band can tunnel into an unoccupied state in the conduction band. As the electron tunnels, it sees the potential profile shown in Fig. 4.12b.