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Density of States Muhammad Ashraful Alam Alam@Purdue.Edu NCN www.nanohub.org EE‐606: Solid State Devices Lecture 8: Density of States Muhammad Ashraful Alam [email protected] Alam ECE‐606 S09 1 Outline 1) Calculation of density of states 2) Density of states for specific materials 3) Characterization of Effective Mass 4) Conclusions Reference: Vol. 6, Ch. 3 (pages 88‐96) Alam ECE‐606 S09 2 Density of States E A single band has total of N‐states 4 OlOnly a ftifraction of stttates are occupidied How many states are occupied upto E? 3 Or equivalently… 2 How many states per unit energy ? (DOS) 1 k Alam ECE‐606 S09 3 Density of States in 1‐D Semiconductors NatomsN atoms a Δk States between E +Δ=×EE & 2 E 11δ k Δk k1+Δk =×2 k1 2π Na E1+ΔE E1 NaΔ k States/unit energy @ E = 1 π ΔE k π 2π δ k = Na a Alam ECE‐606 S09 4 1D‐DOS L N atoms Na Δk States/unit energy @ E = a π ΔE =22k 2mE* ()− E EE−= ⇒= k 0 E 0 2m* =2 dk m* k1+Δk = 2 dE 2= ()EE− 0 k1 E1+ΔE * L m E1 States/unit energy @ E = 2 π 2= ()E − E0 States/unit energy/unit length @ E k π 1 m* 2π ≡=DOS δ k = 2 Na a π 2= (EE− 0 ) Alam ECE‐606 S09 5 1D‐DOS E E 1 m* DOS = 2 π 2= (EE− 0 ) k π DOS 2π Δ=k a Na Conservation of DOS Alam ECE‐606 S09 6 Density of States in 2D Semiconductors b a Show that 2D DOS is a constant independent of energy! Alam ECE‐606 S09 7 Density of States in 3D Semiconductors Macroscopic Sample States between EEE11+Δ & 443 ππ()kdk+− k3 V H ==Δ33kk2 222πππ 2π 2 L W H W L Vk2 Δ States/unit energy @ Ek= 2 2π dE 2π/H 2π/W 22 * * = kdk2mEE( − 0 ) m 2π/L EE−=0 * ⇒= k 22 ⇒ = 22mdEEE==()− 0 k K+dk States/unit energy/unit volume @ E1 m* DOS= 2 m* ( E− E ) 2π 23= 0 Alam ECE‐606 S09 8 3D‐DOS ** mm332 (EE− 0) DOS = 23 E E π = k π DOS − π 2π a Δ=k a Na Conservation of DOS Alam ECE‐606 S09 9 Outline 1) Calculation of density of states 2) Density of states for specific materials 3) Characterization of Effective Mass 4) Conclusions Alam ECE‐606 S09 10 Density of States of GaAs: Conduction/Valence Bands ** mmEEnn2 ( − c) gE()= c 2π 23= ⎧mmEE**2 ()− ⎪ hh hh υ ⎪ 2π 23= ⎪ gEυ ()= ⎨ ⎪ ** ⎪ mmEElh2 lh ( − υ ) ⎩⎪ 2π 23= 11 Four valleys inside BZ for Germanium Alam ECE‐606 S09 12 Ellipsoidal Bands and DOS Effective Mass const. E 22 2 2 22 k2 ==kk12= k3 E − EC =++* * * 222ml mt mt E=const. ellipsoid k 22k k 2 k1 1 =++123 * ** ⎡⎤⎡⎤⎡⎤222ml ()EE−−CCmmt () EEEEC t ()− k3 ⎢⎥⎢⎥⎢⎥222 ⎣⎦⎣⎦⎣⎦=== 2 Transform into … α 2 β 4 4 ⎛⎞2 ≡ π k 3 Vkel= N ⎜⎟παβ eff ⎝⎠3 3 3 *** ⎡⎤* 44222mEE()−−− mEE() mEE() 2meff ()EE− c N ππlctctc ≡ ⎢⎥ el 33===222 = 2 ⎣⎦⎢⎥ ***23 2 13 mNmmeff= el( l t ) Alam ECE‐606 S09 13 DOS Effective Mass for Conduction Band ***23 2 13 ***23 2 13 mmmeff= 4 ( l t ) mmmeff= 6 ( l t ) ** ** mmEEeff2 eff()− c mmEEeff2 eff()− c gE( ) = gEc ( ) = c 2π 23= 2π 23= 14 Outline 1) Calculation of density of states 2) Density of states for specific materials 3) Characterization of Effective Mass 4) Conclusions Alam ECE‐606 S09 15 Measurement of Effective Mass ν0=24 GHz B field variable … (fixed) Iout Iin kz qB m* = con 2πv0 ky kx in -I out I B B con 16 Motion in Real Space and Phase Space Energy=constant. x Liquid He temperature … ky (kx,0) ((,0,ky) y kx (0,-ky) (-kx,0) kz ky kx qB Derive the Cyclotron Formula m* = 0 2πv0 For an particle in (x‐y) plane with B‐field in z‐direction, the Lorentz force is … B m*υ 2 =×=qBqBυυzz r0 qB r υ = 00 m* 2π r 2π m* τ ==0 υ qB0 1 qB ν ≡= 0 0 τ 2π m* Alam ECE‐606 S09 18 Effective mass in Ge B [111] [ 111] [ 111] [ 111] [][][][]111 111 111 111 4 angles between B field and the ellipsoids … Recall the HW1 Alam ECE‐606 S09 19 Derivation for the Cyclotron Formula 22 1 cosθ sin θ θ Show that =+ Given three mc and three , 22 we will Find m , and m mmmmct lt t l The Lorentz force on electrons in a B‐field dυ Fq=×=υ B[] M dt In other words, B * dυx FqBxyzzyt=−=()υυ B m dt dυ y FqB=−=()υυ B m* yzxxztdt * dυz FqBzxyyxl=−=(υυ B) m dt Alam ECE‐606 S09 20 Continued … kz Let (B) make an angle (θ) with longitudinal axis of the ellipsoid (ellipsoids oriented along kz) B0 B ===Bcosθ ,B0 ,B Bsinθ , xyz00() ( ) ky kx Differentiate (vy) and use other equations to find … d 2υ y +=υ ω 222220 with ωθθ ≡⎡ωωw sin + cos ⎤ dt 2 y ⎣ ttl ⎦ qqqqB000qB qB ω0 ≡ ***ωt ≡≡ωl mmmctl 1 sin22θ cos θ = + so that … * 2 mm m2 ()mc lt t Alam ECE‐606 S09 21 Measurement of Effective Mass B=[0.61, 0.61, 0.5] [110] qB1 22 m = 1 cosθ sin θ Three peaks B1, B2, B3 c 2πv =+ Three masses m ,m ,m 0 m 22mmm c1 c2 c3 c t lt Three unique angles: 7, 65, 73 Known θ and mc allows calculation of mt and ml. Alam ECE‐606 S09 22 Valence Band Effective Mass HW. Which peaks relate to valence band? Why are there two valence band peaks? Alam ECE‐606 S09 23 Conclusions 1) Measurement of Effective mass and band gaps define the energy‐band of a material. 2) Only a fraction of the available states are occupied. The number of available states change with energy. DOS captures this variation. 3) DOS is an important and useful characteristic of a material that should be understood carefully. Alam ECE‐606 S09 24.
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