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- Title: handout16
- Type: Notes
- School: Cornell
- Course: ECE 4070
- Term: Spring
Handout 16 Conductivity of Electrons in Energy Bands In this lecture you will learn: Inversion symmetry of energy bands The conductivity of electrons in energy bands The electron-hole transformation The conductivity tensor Examples Bloch oscillations ECE 407 Spring 2009 Farhan Rana Cornell University Inversion Symmetry of Energy Bands - I Recall that a Bloch function can be written as: r r n,k (r ) = e i k . r un,k (r ) r rr r Where the periodic part satisfied a Schrodinger-like equation: r r P2 P r h 2k 2 r r r r r r + . hk + + V r un ,k (r ) = E n k un,k (r ) 2m m 2m r r Now let k go to k in the above equation: r r P2 P r h2k 2 r r r r r r . hk + + V r un , k (r ) = E n k un, k (r ) 2m m 2m () () (1) () () And then take the complex conjugate of the whole equation to get: r r P2 P r h 2k 2 r r rr rr + . hk + + V r u * n , k (r ) = E n k u * n , k (r ) 2m m 2m r rr r u * n , k (r ) = un ,k (r ) Comparing (1) and (2) we get the results: r r En k = En k () () (1) () () ECE 407 Spring 2009 Farhan Rana Cornell University 1 Inversion Symmetry of Energy Bands - II r rr r u * n , k (r ) = un ,k (r ) One can also write: r * n, k (r ) = n,k (r ) r r r En k = En k () () () Energy r r r r En k = En k () Recall that: Now let k go to k in the above equation: r r r rr 1r v n k = k En k h () () () () () a r r r 1 r v n k = k En k h r 1r = k En k h r 1r = k En k h rr = v n k r r rr v n k = v n k () () () () kx a ECE 407 Spring 2009 Farhan Rana Cornell University Current Density for Energy Bands In Drude model, the electron current density was given as: Energy r r J = n ( e ) v For a free electron gas the current density was given as: r r r rr r rr 2 d 3k J = ( e ) rf k v k = 2 e fk vk 3 V all k (2 ) ()() ()() Now we want to find the current density due to electrons in energy bands The current density due to electrons in the n-th band can be written in a manner similar to the free-electron case: a kx a r rr r 2 J n = ( e ) r fn k v n k V k in FBZ r rr r d 3k = 2 e f k vn k 3n FBZ (2 ) () () () () ECE 407 Spring 2009 Farhan Rana Cornell University 2 Current Density for a Completely Filled or Empty Bands Consider a completely filled band for which fn k = 1 for r all k in FBZ: () r Energy r J n = 2 e r d 3k FBZ (2 ) 3 rr r fn k v n k = 2 e () () r d 3k FBZ (2 ) 3 rr vn k = 0 Ef () where I have used the fact: r r rr v n k = v n k () () Completely filled bands do not contribute to electrical current or to electrical conductivity Of course, if fn k = 0 for all k in FBZ: () r r r J n = 2 e r d 3k 3 FBZ (2 ) rr r fn k v n k = 0 () () a kx a Completely empty bands do not contribute to electrical current or to electrical conductivity Only partially filled bands contribute to electrical current and to electrical conductivity ECE 407 Spring 2009 Farhan Rana Cornell University Current Density and Electron-Hole Transformation Consider the expression for the current density: r J n = 2 e = 2e r d 3k FBZ r d 3k (2 ) 3 f 3n r (k ) v n (k ) r r (1) 0 r r rrr d 3k r r d 3k = 2 e 1 fn k v n k vn k + 2 e 3 3 FBZ (2 ) FBZ (2 ) r 3 rrr dk = +2 e 1 fn k v n k (2 )3 FBZ FBZ (2 ) r rr [1 fn (k ) 1] v n (k ) () [ ( )] ( ) (2) [ ( )] ( ) The final result implies that since the current density of a filled band is zero, the current density for any band can always be expressed in two ways: a) As an integral over all the occupied states assuming negatively charged particles (as in (1) above) b) As an integral over all the unoccupied states assuming positively charged particles (as in (2) above) ECE 407 Spring 2009 Farhan Rana Cornell University 3 Current Density and Electron-Hole Transformation Consider the current density for a completely filled band with just r one missing electron with wavevector k ' and spin up, as shown. The Electron Choice: The current density is given by: 1 r rr r 1 J n = 2 e r fn k v n k V k in FBZ r 1 = e vn k' V r 1 = e vn k' V Energy r ko r k () () r k' r k' () () The Hole Choice: The current density is given by: 1 r rrr 1 Jn = 2 e r 1 fn k v n k V k in FBZ r 1 = e vn k' V [ ( )] ( ) () ECE 407 Spring 2009 Farhan Rana Cornell University Metals, Semiconductors, and Insulators Materials can be classified into three main categories w.r.t. their electrical properties: Metals: In metals, the highest filled band is partially filled (usually half-filled) Semiconductors: In semiconductors, the highest filled band is completely filled (at least at zero temperature) Insulators: Insulators are like semiconductors but usually have a much larger bandgap Energy Energy Energy Ef Ef Ef kx FBZ Metal FBZ Semiconductor kx FBZ Insulator kx ECE 407 Spring 2009 Farhan Rana Cornell University 4 Inclusion of Scattering in the Dynamical Equation In the presence of a uniform electric field the crystal momentum satisfies the dynamical equation: Energy r r d hk (t ) = e E dt r r r r h k (t ) k d hk (t ) = e E dt r r The boundary condition is that: k (t = 0 ) = k Now we need to add the effect of electron scattering. As in the free-electron case, we assume that scattering adds damping: Conduction band [ ] hh valence band kx Note: the damping term ensures that when the field is turned off, the crystal momentum of the electron goes back to its original value Steady State Solution: h valence band Ex r r e r k (t = ) = k E h In the presence of an electric field, the crystal momentum of every electron is shifted by an equal amount that is determined by the scattering time and the field strength ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity : Conduction Band Consider a solid in which the energy dispersion for conduction band near a band minimum is given by: Energy Conduction band r r h2 r r Ec k = E c k o + k ko 2 () () ( )T . M 1 . (k ko ) r r The velocity of electrons is: rr rr v c k = M 1 . h k ko The current density is: () ( ) r r r ko r k r Jc = 2 e r near ko r d 3k (2 ) f 3c r (k ) v c (k ) In equilibrium, for every state with crystal momentum k ko that is occupied, the rr state k ko is also occupied and these two states have opposite velocities. ( ) ( r r ) Therefore in equilibrium: r Jc = 2 e r near ko r d 3k (2 ) f 3c r (k ) v c (k ) = 0 r r ECE 407 Spring 2009 Farhan Rana Cornell University 5 electrical conductivity : Conduction Band Now assume that an electric field is present that shifts the crystal momentum of all electrons: Conduction band Energy ky r fc k () r ko r r e r k (t = ) = k E h kx e r E h ky r ko r ko kx r E r k r E = Ex x Electron distribution in k-space when E-field is zero Distribution function: fc k () r Electron distribution is shifted in k-space when E-field is not zero Distribution function: fc k + r e r E h Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution is shifted as shown ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity : Conduction Band Current Density: r Jc = 2 e r near ko r d 3k r e r r r fc k + E vc k h (2 )3 r d 3k () e r E h ky r E = Ex x r ko Do a shift in the integration variable: r Jc = 2 e r Jc = 2 e r near ko (2 )3 r d 3k r r r e r fc k v c k E h () () kx r e r r r f k M 1 . h k ko E r 3c h near ko (2 ) r r r v d 3k Jc = e 2 2 f k M 1 . E r 3c near ko (2 ) r 2 1 v Jc = n e M . E r = .E r e r fc k + E h () Where the conductivity is now a tensor given by: = n e 2 M 1 ECE 407 Spring 2009 Farhan Rana Cornell University 6 electrical conductivity Example: Conduction Band of GaAs Consider the conduction band of GaAs near the -point: 1 me M 1 = 0 0 This implies: 0 1 me 0 0 0 1 me Isotropic! r v Jc = n e 2 M 1 . E J x ,c 2 J y ,c = n e J z ,c = 1 me 0 0 0 1 me 0 0 0 1 me E x E y Ez E x r n e 2 Ey = E me Ez = n e 2 me ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity Example: Conduction Band of Silicon In Silicon there are six conduction band minima (valleys) that occur along the six -X directions. For the one that occurs along the -X(2 /a,0,0) direction: r 2 ko = 0.85 ,0,0 a M 1 0 1 ml 1 mt = 0 0 0 0 0 1 mt Not isotropic! m = 0.92 m mt = 0.19 m This implies that for this valley: r v n Jc = e 2 M 1 . E 6 J x ,c 0 1 ml n2 1 mt J y ,c = 6 e 0 J z ,c 0 0 0 0 1 mt E x E y Ez The factor of 6 is there because only 1/6th of the total conduction electron density in Silicon is in one valley ECE 407 Spring 2009 Farhan Rana Cornell University 7 electrical conductivity Example: Conduction Band of Silicon To find the conductivity tensor for Silicon one needs to sum over the current density contributions from all six valleys: J x ,c n2 J y ,c = 6 e J z ,c = 2 ml + 4 mt 0 0 0 2 ml + 4 mt 0 0 0 2 ml + 4 mt E x E y Ez Isotropic! E x r n e 2 E = E y me Ez 1 1 1 2 = Conductivity effective mass = + me 3 ml mt After adding the current density contributions from all six valleys, the resulting conductivity tensor in Silicon is isotropic and described by a conductivity effective mass ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity : Valence Band Consider a solid in which the energy dispersion for valence band near a band maximum is given by: Energy r r h2 r r Ev k = Ev ko + k ko 2 () () ( )T rr . M 1 . k ko ( ) Valence band r ko r k The velocity of electrons is: rr rr v v k = M 1 . h k ko r d 3k () ( ) The current density is (using the electron-hole transformation): r Jv = 2 e r near ko (2 ) 3 rr r fv k v v k = 2 e () () r d 3k r near ko (2 ) 3 rrr [1 fv (k )] vv (k ) In equilibrium, for every state with crystal momentum k ko that is unoccupied, rr the state k ko is also unoccupied and these two states have opposite velocities. ( ) ( r r ) Therefore in equilibrium: r Jv = 2 e r near ko r d 3k (2 ) 3 rrr [1 fv (k )] vv (k ) = 0 ECE 407 Spring 2009 Farhan Rana Cornell University 8 electrical conductivity : Valence Band Now assume that an electric field is present that shifts the crystal momentum of all electrons in the valence band: Energy r ko r k r E ky r 1 fv k () r ko r r e r k (t = ) = k E h e r E h ky Valence band kx r E = Ex x r ko kx Hole distribution in k-space when E-field is zero Distribution function: 1 fv k () r Hole distribution is shifted in k-space when E-field is not zero Distribution function: 1 fv k + r e r E h Since the wavevector of each electron is shifted by the same amount in the presence of the E-field, the net effect in k-space is that the entire electron distribution (and hole distribution) is shifted as shown ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity : Valence Band Current Density: r Jv = 2 e r dk r e r r r 1 fv k + E vv k r 3 h near ko (2 ) 3 () e r E h ky r E = Ex x r ko Do a shift in the integration variable: r Jv = 2 e r Jv = 2 e r near ko r near ko r d 3k (2 )3 r d 3k [ r r r e r 1 fv k v v k E h ( )] kx (2 ) r rr r [1 fv (k )] M 1 . h k ko eh E 3 r e r E 1 fv k + h r r v Jv = e 2 2 r 1 fv k M 1 . E 3 near k o (2 ) r 2 1 v Jv = p e M . E r = .E r d 3k [ ( )] Where the conductivity is now a tensor given by: = p e 2 M 1 ECE 407 Spring 2009 Farhan Rana Cornell University 9 electrical conductivity Example: Heavy-Hole Band of GaAs Consider the heavy-hole band of GaAs near the -point: 1 mhh M 1 = 0 0 This implies: 0 1 mhh 0 0 0 1 mhh Isotropic! r v J hh = phh e 2 M 1 . E J x ,hh 1 mhh 2 J y ,hh = phh e 0 J z ,hh 0 E x r p e 2 Ey = E = hh mhh Ez = phh e 2 mhh 0 1 mhh 0 0 0 1 mhh E x E y Ez ECE 407 Spring 2009 Farhan Rana Cornell University electrical conductivity Example: Light-Hole Band of GaAs Consider the light-hole band of GaAs near the -point: 1 mlh M 1 = 0 0 This implies: 0 1 mlh 0 0 0 1 mlh Isotropic! r r v J lh = plh e 2 M 1 . E = E = plh e 2 mlh The total valence band conductivity of GaAs can be written as the sum of the contributions from the heavy-hole and the light-hole bands: = phh e 2 plh e 2 + mhh mlh ECE 407 Spring 2009 Farhan Rana Cornell University 10 The Case of No Scattering: Bloch Oscillations Consider an electron in a 1D crystal subjected to a uniform electric field. The energy band dispersion and velocity are: E n (k x ) = E s 2 Vss cos(k x a ) v n (k x ) = Energy 1 dE n (k x ) = 2a Vss sin(k x a ) h dk x In the absence of scattering, the crystal momentum satisfies the dynamical equation: The time-dependent velocity of the electron is: d hk x (t ) = eEo dt eEo k x (t ) = t + k x (t = 0 ) h a FBZ kx a v n (t ) = 2a Vss sin(k x (t )a ) e a Eo = 2a Vss sin t + k x (t = 0 )a h Periodic! ECE 407 Spring 2009 Farhan Rana Cornell University r E = Eo x The Case of No Scattering: Bloch Oscillations A periodic velocity means that the electron motion in real space is also periodic: dx (t ) e a Eo t + k x (t = 0 )a = v n (t ) = 2a Vss sin dt h T dx o (t ) dt = x (t = T ) x (t = 0 ) = 0 where the period T is: dt T= 2 h e a Eo Reciprocal space: a kx t =0 a a T t= 4 kx a a T t= 2 kx a a 3T t= 4 kx a a kx t =T a Real space: t =0 t =T 0 T 4 3T t= 4 t= ECE 407 Spring 2009 Farhan Rana Cornell University t= T 2 x 11 The Phenomenology Of Transport: Basic Equations The presence of external fields, and scattering, the following relations work for electrons in any energy band near the band edge: r r r r r r r h k (t ) k d hk (t ) = e E ev n k B dt r r r r v n k (t ) = M 1 . h k (t ) ko r r r rr r rrr d 3k d 3k J n (t ) = 2 e fn k v n k (t ) = +2 e 1 fn k v n k (t ) 3 3 FBZ (2 ) FBZ (2 ) () [ ] ( ) ( ) () ( ) [ ( )] ( ) The first two can also be written as: M. rr rr rr r r M . v n k (t ) v n k rr d v n k (t ) = e E ev n k (t ) B dt ( ) ( ) [( ) ( )] Problem: One needs simple models for current transport so that non-specialists, like circuit designers, can understand devices and circuits without having to understand energy bands ECE 407 Spring 2009 Farhan Rana Cornell University The Phenomenology Of Transport: Electrons in Conduction Band We define the drift velocity for the electrons in conduction band as: r rr rr v e (t ) = v c k (t ) v c k ( ) () (1) The drift velocity is independent of wavevector and satisfies the equation: M. r r r r M . v e (t ) r d v e (t ) = e E ev e (t ) B dt Once the drift velocity is calculated, the electron current density is: r Je (t ) = 2 e r d 3k FBZ (2 ) f 3c r (k ) v c (k (t )) = n ( e ) ve (t ) r r (2) Electrons in conduction band are to be thought of as negatively charged particles with a mass replaced by a mass tensor and that satisfy (1) and (2). In case of multiple conduction band minima, as in silicon and germanium, individual contributions of each minima (or valley) to the current density are calculated separately and added in the end. ECE 407 Spring 2009 Farhan Rana Cornell University 12 The Phenomenology Of Transport: Holes in Valence Band We define the drift velocity for the holes in valence band as: r rr rr v h (t ) = v v k (t ) v v k ( ) () (1) The drift velocity is independent of wavevector and satisfies the equation: ( M ) . r r r r ( M ) . v h (t ) r d v h (t ) = + e E + ev h (t ) B dt Where realizing that the inverse effective mass tensor will have negative diagonal terms for valence band, I have multiplied throughout by a negative sign, with the result that the charge -e becomes +e Once the drift velocity is calculated, the hole current density is: r J h (t ) = +2 e r d 3k FBZ (2 ) 3 rrr [1 fv (k )] vv (k (t )) = p (+ e ) v h (t ) (2) Holes in valence band are to be thought of as positively charged particles with a mass replaced by a mass tensor and that satisfy (1) and (2). In case of degenerate valence band maxima, the heavy and light hole current density contributions are calculated separately and added in the end. ECE 407 Spring 2009 Farhan Rana Cornell University 13
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