Lecture 21 : Graphene Bandstructure Ref. Chapter 6.1 Net work f or - PowerPoint PPT Presentation

Fundam entals of Nanoelectronics Nanoelectronics Fundam entals of Prof. Supriyo Datta ECE 453 Purdue University 10.20.2004 Lecture 21 : Graphene Bandstructure Ref. Chapter 6.1 Net work f or Comput at ional Nanot echnology

00: 05 Review of Reciprocal Lattice • In the last class we learned how to • Similarly any point in the reciprocal lattice construct the reciprocal lattice. can be written as: � � � � • For 1D w have: = + + K M A N A P A 1 2 3 Real-Space: x � • How are the vectors “A” related to = a ˆ R x a vectors “a”? The defining condition is: � ⋅ � δ = ≠ 0 for i j = πδ 2 A j a � ij i ij ˆ π = δ = = 2 / K x a 1 for i j BZ k-Space: ij x • The significance of reciprocal lattice - p /a 0 p /a vectors “A” is that points in k space which are apart from each other by an integer • In general for periodic structures we can multiple of “Ai’s”, give is the same write 3 basis vectors such that any point in wavefunction solution. the lattice can be written as a linear combination of them with the condition that � the coefficients must be integers. � � � = + + R m a n a p a 10.20.2004 1 2 3

06: 50 Graphene Graphene is made up of carbon atoms bonded in a hexagonal 2D plane. Graphite is 3D structure that is made up of weakly coupled Graphene sheets. This is of particular importance because carbon nanotubes are made up of a Graphene sheet that is rolled up like cylinder. Carbon nanotubes themselves are of interest because people believe they can make all kinds of Nano devices with them. 10.20.2004

08: 03 Reciprocal Lattice in 3D • Semiconductors of interest to us have what is called a diamond structure. The diamond structure is composed of to interpenetrating FCC lattices the following way: Imagine two FCC lattices such that each atom of each lattice is on top of the corresponding atom of the other lattice. You should only be seeing 1 FCC lattice as of now. Then fix one lattice and move the other one in the direction of the body diagonal of the fixed one by ¼ of the body diagonal. Now you’ve yourself a diamond lattice. If the two FCC lattices are made up of two different types of atoms, the structure is then called a Zinchblend lattice. • To visualize the reciprocal lattice focus only on one FCC lattice in the diamond structure. BCC in Brillouin Zone in Reciprocal FCC in Real Space Reciprocal Space Lattice 111 100 10.20.2004 110

11: 16 E-k Diagrams for 3D Reciprocal Lattices • Some useful information: • Since the reciprocal space is now 3 • The top of the valence band usually dimensional, to draw the E-k diagram we occurs at the Gamma point (k=0). The have choose particular directions and draw bottom of conduction band however does E-k diagram along those directions: not always lie at k=0. For example consider Silicon: E E k k Γ X L • If both conduction band minimum and the valence band maximum lie at the same value of k, the material is called a direction bandgap semiconductor. Other wise the 10.20.2004 material is indirect like Si.

17: 18 Parabolic Approximation • Usually, it is necessary to derive an expression for E( k x , k y, k z ) about the Silicon Parabolic Conduction Band conduction points of a bulk solid Approximation • For silicon, use the parabolic approximation E Approximation + + 2 2 2 � 2 � 2 2 ( ) k k k k = + = x y z E E c 2 * 2 * m m where m* is the effective mass. k y • For nanotubes we can derive a similar parabolic expression via a Taylor series expansion that approximates the subbands near the conduction valleys 10.20.2004

21: 15 E-k Relation for Graphene • Let’s get back to Graphene. First • Remember the general result of principle identify the basic unit cell of bandstructure: � [ ] { } ( ) { } φ = φ E h k Basic 0 0 � [ ] ( ) Unit Cell � � � ( ) [ ] ∑ • − = i k d d h k H e m n nm m The lattice • To write h(k) consider one unit cell an its structure nearest neighbors. Figure shows that there only repeats will be 5 terms in the summation for h(k). in pairs of 2! � a b 1 a � a 2 10.20.2004

23: 45 Graphene E-k Diagram • Remember the general result of principle • Writing the summation terms and adding � ε   0 * of bandstructure: h them up we get: � [ ] { } = ( ) ( ) h k   { } ε φ = φ   h E h k 0 ( ) 0 0 • Where � � ⋅ + � � ⋅ = + [ � ] ( ) i k a i k a 1 � � � h t e e [ ] ( ) 1 2 ∑ • − = i k d d 0 h k H e m n nm • The eigenvalues of this matrix are given m � � ( ) by: = ε ± ( ) E k h 0 k •To write h(k) consider one unit cell an its nearest neighbors. Figure shows that there will be 5 terms in the summation for h(k). E Conduction Conduction � Point Point a b 1 e a k { � a 2 filled states 10.20.2004

29: 15 Magnitude of h(k) • Next we like to locate the conduction points in the 2 dimensional k space: � y = + = + ˆ ˆ ˆ ˆ 3 a a x b y 3 a x a y x � 2 2 1 0 0 a � 1 = − = − Unit ˆ ˆ ˆ ˆ 3 a a x b y a x a y 3 � 2 2 2 0 0 Cell � a 2 a ˆ + = 0 ˆ k k x k y x y ( ) ( ) ( ) ( ) ( ) � ⋅ + � � � ik x cos + − = 1 + = 1 + + = + ⋅ i k a k b i k a k b a 2 i k a i k a t e k b 1 t e x y e x y h t e e 1 2 y 0 • To find the conduction points we need to set |h(k)|=0. So we need to find |h(k)|: ( ) ∴ 2 = = + + * 2 2 1 4 cos cos 4 cos h h h t k a k b k b 0 0 0 x y y so, � = + + 2 ( ) 1 4 cos cos 4 cos h k t k a k b k b 0 x y y 10.20.2004

38: 35 Conduction Valleys � = + + = • Now let 2 ( ) 1 4 cos cos 4 cos 0 h k t k a k b k b 0 x y y • Let kxa=0 and investigate h(k) as a function of ky. π ( ) 2 = + = ⇒ = = 1 2 cos for 0 to get h (k) 0 h t k b k k y b 0 0 y x 3 • Let kxa=pi and investigate h(k) as a function of ky. = π ( ) = − = π ⇒ = 1 2 cos for to get h (k) 0 h t k b k k y b 0 0 y x 3 (0, 2 p /3b) Conduction Valley ( p /a, p /3b) k x ( p /a,- p /3b) Conduction k y Valley 10.20.2004

43: 45 Two Full Valleys k y k y Translating two • The six Brillouin valleys really of the corners in only give 2 independent valleys, each group of 3 1 1 3 3 e.g. in each group of 3 that are in the picture two of the valleys k x k x are away form the other by a reciprocal lattice unit vector; hence represent the same state. One can think that each 1 3 corner in the 1 st Brillouin zone contributes 1/3 rd .1/3 x 6 = 2(left figure). Alternatively we can � translate two of the corners in = ε ± ( ) E h 0 k each group to get the full E valleys on the right. Conduction Conduction e +3|t| Valley Valley e • Dispersion relation along ky. k y ( ) = + = 1 2 cos for 0 h t k b k 0 y x e -3|t| 10.20.2004