7.8: Section 3-D Band Diagrams and Tensor Effective Mass Band diagrams characterize the effect of the crystal geometry on the behavior of the electrons within the semiconductor. We discuss 3-D bands and the tensor form of the effective mass. The band edge diagrams plotting energy vs. position provide convenient pictures for device operation such as for diodes. Later sections show the band-edge diagrams owe their existence to the fact that electrons and holes occupy very narrow ranges of energy near the lower and upper edges of the band, respectively. The effective density of states approximation allows us to essentially reduce dispersion curves to two discrete levels. Topic 7.8.1: E-k diagrams for 3-D Crystals Our previous work with bands (such as with the Kronig-Penney model) shows the energy E plotted against the wave vector k. Both positive and negative k-values appear on the same plot. Actual band diagrams do not show the negative kvalues because bands exhibit symmetry in k and theres no need to show redundant information. Instead, actual diagrams show the bands along two different directions. Figure 7.8.1 shows the first Brillouin zone for materials with the Zinc Blende crystal structure. The point sits at the center of the zone (corresponds to k=0 for the maximum of the Figure 7.8.1: Zinc Blende FBZ after Pierret valence band). The line X represents the <100> direction (easy to remember: x stands for the x direction). The line L (L for diagonaL) represents the <111> direction. Figure 7.8.2 shows the band diagram for GaAs which crystallizes in the zinc blende configuration. As mentioned the horizontal axis represents two different directions in this type of diagram. The band diagrams in Figure 7.8.2 do not show the k direction because the bands would look the same as the +k direction superfluous information. The k looks the same as +k because the lattice has inversion symmetry (if R is a lattice vector then so is R not every lattice has rotational symmetry though). Figure 7.8.2 shows a direct band gap at =0. Figure 7.8.2: GaAs Band diagram after Pierret. The minimum in the conduction band and maximum in the valence band look fairly symmetrical about the point. Notice the formation of L and X valleys in the
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conduction band. Under certain conditions, electrons can scatter from the valley into these other valleys. The group velocity and effective mass of the electron in these other valleys must be different from that in the valley. We expect any electrons scattered into these side valleys to decay back to the minimum after a period of time. Actually, because the L valley has low energy for GaAs, a significant number of GaAs electrons have sufficient thermal energy to populate the L valley. The valence band structure consists of the heavy hole, light hole and split-off bands. The heavy hole band gives rise to larger effective masses for the holes than does the light hole band. The split-off band gives roughly the same effective mass as does the light hole band owing to the somewhat similar curvatures. The diagram shows nearly degenerate light and heavy hole bands near the point; that is, they have roughly the same energy at =0. Both the heavy and light hole bands can contribute to current flow and absorption/emission processes. As a point of interest, people sometimes add strain (i.e., strain the lattice a force applied to the atoms in the lattice) to the GaAs lattice by adding Indium. The bandedge shifts to longer wavelengths. More importantly, devices can be made more efficient because the curvature of the heavy hole valence band can be made the same as the curvature of the conduction band. Also, the light hole band moves further away (in energy) from the heavy hole band (and no longer necessarily participates in optical and electronic processes). Topic 7.8.2: Effective Mass for 3-D Band Structure Before discussing the 3-D case, lets first discuss the 1-D case. A 1-D crystal with a direct bandgap (for example, a 1-D version of GaAs) has a dispersion relation (E vs. k) of the form 2 2 kx E Ec = (7.8.1) 2m* near the bottom of the band. In general, we might find a conduction band for a directbandgap semiconductor having the form E E c = Ak 2 x Indirect bandgaps have dispersion relations for the conduction band of the form 2 (7.8.2) E E c = A ( k x k ox )
where k ox gives the center of the parabola in k-space. Obviously, the direct band gap dispersion relation comes from setting k ox = 0 . We see that two wave vectors kx give the same energy E according to
( E Ec ) 2m* We can calculate the effective mass using the relation from Section 7.7 1 1 2E = 2 2 k x m* Taking the second derivative of Equation 7.8.3, we find 1 2A = 2 * m
k x = k ox
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2
(7.8.3)
(7.8.4)
(7.8.5)
The effective mass in Equation 7.8.5 does not depend on the exact location of the center k ox because of the parabolic form of the dispersion curve and the fact that we take two derivatives. The effective mass does not depend on whether k < k ox or k > k ox . Therefore the rate of change of the group velocity due to an applied force F dv 1 = F (7.8.6) dt m* For a 2-D crystal, we expect the dispersion curve to have the form 2 2 2 k k k )2 ( k k )2 (7.8.7) = E Ec = ox y oy * * ( x 2m 2m where k ox , k oy denote the center of the paraboloid in k-space (see Figure 7.8.3). Circles describe a level contour over which E-Ec has a fixed value. However, Equation 7.8.7 describes a crystal with symmetric dispersion curves because the equation has a single effective mass as a coefficient. Rather than discuss the general 2-D case, lets go on to the 3-D one. The 3-D crystal has a dispersion relation of the form (near the minimum) 2 2 2 E = E c + A ( k x k ox ) + B ( k y k oy ) + C ( k z k oz ) (7.8.8)
where the symbol Ec represents the minimum of the band, and kox, koy, koz give the location (in kspace) of the minimum. The surface has the form Figure 7.8.3: Constant energy surfaces for of an ellipsoid for surfaces of constant energy. If (a) Ge, (b) Si and (c) GaAs. Part (d) shows A=B=C then the surface must be a sphere with Ge upto the first Brillouin Zone. Picture from Pierret and Krieger Publishing Co. effective mass independent of direction. Consider the case of non-symmetric bands. We would find the effective mass for motion along the three directions 1 1 2 E 2A 1 2 E 2B 1 2 E 2C 1 1 my ) = 2 2 = 2 mz ) = 2 2 = 2 (7.8.9) mx ) = 2 2 = 2 ( ( ( k x k z k y Apparently the mass and acceleration depend on the direction of an applied force. Depending on the shape of the energy bands, the effective mass can be larger along one direction than another. We need to discuss the effective mass for more than three directions since the particle can move in any direction through the crystal. It seems strange to talk about the effective mass in a given direction since mass should be a scalar quantity measuring inertia. First of all, we really mean that the effective mass depends on wave vector k because the band curvature depends on k. To say that the mass depends on the direction of travel just means that it depends on the values of k x , k y , k z that determine the direction of travel. Next, we should point out that the effective mass really provides a proportionality between the applied force and the Therefore, acceleration. saying the effective mass depends on the direction of motion really means that the relation between the applied force and the acceleration depends on the direction of the applied force.
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Apparently the relation between force and acceleration in a crystal should be written as a tensor product. Some people write Newtons second law in the form of a dyadic equation F = m* a (7.8.10) The mass in this case is a dyad which represents a tensor. Lets leave off the asterisk * for convenience. Technically a dyad like m can be written as m = m xx xx + m xy xy + where x, y, z represent basis vectors. Please refer to Section 3.17 for a review of dyads. The dyad provides a representation for a second rank tensor, both of which can be represented by a 3x3 matrix. m xx m xy m xz m yx m yy m yz m zx m zy m zz
We can now demonstrate the 3-D effective mass and its relation to the band curvature 1 1 +y +z +y +z m 1 = 2 k k E = 2 x x E (7.8.11) k k y k z k x k y k z x where the gradient in k-space k has the form k = x +y +z k x k y k z and neither the dot nor cross product appears between the two gradients. Equation 7.8.11 can be written as matrix elements 2 (7.8.12) ( m1 )ij = 12 k Ek i j
To demonstrate Equation 7.8.11, let F be an arbitrary applied force. The energy supplied by the applied force can be written as dr dE = F dr = F dt = F vg dt (7.8.13a) dt where the vector form of the group velocity can be written as 1 (7.8.13b) vg = k = k E Therefore the rate of change of particle energy must be dE 1 (7.8.13c) = F k E dt Now working with Newtons second law using the dyadic notation for the effective mass dv d 1 (7.8.14a) F = ma = m g = m k E dt dt However, the change in a function G with k can be written as dG = ( k G ) dk (7.8.14b)
In our case, taking the function
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G to be 1 G = k E Therefore Equation 7.8.14a becomes
F = m dG 1 1 1 dk 1 dp 1 = m dG = m k G dk = m k k E = m 2 k k E = m 2 kk E F dt dt dt dt dt
(7.8.14c)
For arbitrary F we conclude that the operator gives 1 1 = m 2 kk E Therefore, as discussed in Section 3.17, we surmise 1 m 1 = 2 k k E
(7.8.15a)
(7.8.15b)
as required for the demonstration. The examples below show how to calculate the effective mass. An average effective mass often appears in formulas such as for the density of states. 1/ 3 The average usually appears as a geometric average such as m = ( m1m 2 m3 ) .
Example 7.11.1: As an example if a = ax then F = xm xx a + ym yx a + zm zx a
Example
7.11.2:
2
Find
2 x
the
effective
mass 1
2
E = A 2k 2 = A
Solution:
(k
+ k2 + k2 ) y y
mij
for
the
isotropic
band
Using 7.8.12, namely
(m )
1
ij
=
2E , we find k i k j
(m )
1
ij
= 2A ij .
Therefore the effective mass m =
1 must be isotropic. 2A
Example 7.11.3: Find the effective mass mij for the band
E = A 2k 2 =
Solution: We use 7.8.12
2
( Ak
1
2
2 x
2 + Bk 2 + Ck z ) y
(m )
1
ij
=
2E k i k j
1 1 to find m11 = 2A m 1 = 2B m33 = 2C and the others are zero. The inverse effective 22 mass matrix and the effective mass matrix must be
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0 2A 0 m = 0 2B 0 0 0 2C
1
1 2A m= 0 0
0 1 2B 0
0 mx 0 = 0 0 1 2C
0 my 0
0 0 mz
Example 7.11.4: acceleration.
Using the last example, show the relation between force and
Solution: Consider just a 2-D case for simplicity. We have F = m a or equivalently Fx m x 0 a x = Fy 0 m y a y
The linearly relations between the force and acceleration become Fx = m x a x
and
Fy = m y a y
Figure 7.8.4: Although acceleration and force are linearly related for each direction, the vector force and acceleration are not parallel when the effective mass depends on direction of motion.
Because the effective mass can be different for motion along different directions, identical forces can produce two different accelerations as illustrated in the upper two panes of Figure 7.8.4. We can combine these two panes to produce the vector diagram in the third pane. Notice the force and acceleration vectors are no longer parallel.
Topic 7.8.3: Introduction to Band Edge Diagrams
The band-edge diagrams (spatial diagrams) can be found from the normal E-k band diagrams (dispersion curves). Recall that a dispersion curve has axes of E vs. k and doesnt give any indication or information on how the energy depends on...