Electric field ex dv x v x and e x q dx choosing

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Unformatted text preview: 18 Band Diagram Illustra\on: Carriers Moving in an Electric Field E From Previous Lecture •  Ÿ༉ •  •  The electron (hole) kine\c energy is zero at the conduc\on (valence) band edge Total energy = Kine\c Energy + Poten\al Energy The Poten\al Energy is given by the rela\ve posi\on of the band edge 19 Band Diagrams With Electric Fields •  As the electron is accelerated in the field, it loses poten\al energy and gains kine\c energy •  When an electric field is present, the bands are \lted to reflect the loss in poten\al energy as the electron moves in the direc\on opposite to the field V ( x ) ≡ Electrostatic Potential E ( x ) ≡ Electron Potential Energy E ( x ) ≡ Electric Field E(x) dV ( x ) V (x) = and E ( x ) = − (− q ) dx Choosing E i as a spatial reference: E (x) = − dV ( x ) d ⎡ E ⎤ 1 dEi =− ⎢ i ⎥= dx dx ⎣ (− q ) ⎦ q dx Ÿ༉ Ÿ༉ Ÿ༉ Ÿ༉ 20 Deriva\on of Einstein Rela\onship •  In equilibrium, there is no net current, and therefore no net mo\on of charge •  Also in equilibrium, the Fermi level is invariant J p ( x ) = qp( x )µ pE ( x ) − qD p Ÿ༉ Ÿ༉ dp( x ) =0 dx EF D 1 dp( x ) so E ( x ) = p µ p p( x ) dx since po = ni e( Ei − EF )/kT D 1 d ⎡ Ei − EF ⎤ E (x) = p ni e( Ei − EF )/kT µ p ni e( Ei − EF )/kT dx ⎢ kT ⎥ ⎣ ⎦ Ÿ༉ E (x) = D p 1 ⎛ dEi dEF ⎞ D p 1 1 dEi − = E (x) ( qE ( x ) − 0 ) since ⎜ ⎟= µ p kT ⎝ dx dx ⎠ µ p kT q dx so 1 = Dp q D kT and p = µ p kT µp q Ÿ༉ We can also have tilted bands with no externally applied potential! 21 Key Points •  The Einstein rela\on allows us to calculate either D or μ from a measurement of the other •  The Fermi level EF is constant in equilibrium •  The intrinsic energy Ei is fixed rela\ve to the band edges •  Material proper\es that cause Ei to move rela\ve to EF will create built- in fields that balance dris and diffusion –  Doping concentra\on, doping type, bandgap (alloy composi\on), etc. 22 Example: Internal Electric Field With Graded Doping N n(x) n( x ) = N d ( x ) = N o e− ax As the electrons diffuse away, the Φ n (diff ) o ni Φ n (drift ) Nd+ e- N d+ ee ee Ec Ei Ev pull from the ionized donors is no longer screened and field increases to compensate Ef h h dn( x ) J n ( x ) = 0 = qn( x )µnE ( x ) + qDn dx dn( x ) qn( x )µnE ( x ) = − qDn dx Dn 1 dn( x ) Dn kT E (x) = − = a= a µn n( x ) dx µn q = (0.0259 eV )(10 4 cm −1 ) E ( x ) = 259V / cm Graded Doping: 1/a = 1 micron D kT = µ q 23 Bonus Material Haynes- Shockley Experiment Light pulse or current pulse t d = tl − t o = tl L vd = td vd vd µp = = E V /L Hall Effect: Majority carrier mobility Haynes-Shockley: Minority carrier mobility 25 Haynes- Shockley Experiment ⎛ δ p⎞ With Negligible Recombination: ⎜ =0 τp ⎟ ⎝ ⎠ ∂δ p( x, t ) ∂ 2 δ p ( x, t ) = Dp ∂t ∂x2 Solution: ⎡ ΔP ⎤ − 4 x t D δ p ( x, t ) = ⎢ ⎥e p ⎢ 2 π D pt ⎥ ⎣ ⎦ Δp δp= ← Amplitude of Peak 2 π D pt 2 − e x2 4 D pt ← Spread of Pulse in +/-x Directions At 1 Point: e ˆ ˆ δ p ⋅ e−1 = δ p ⋅ e − ( 2) Δx Δx = Δt ⋅ vd = Δt ⋅ 2 4 D pt d Δx = 4 D pt d = 4 D p Dp L L =4 vd µE Dp (Δx )2 = 16td L td Δt 2 L2 = 3 16td Δt is more easily measured 26 Haynes- Shockley Experiment Δx = Δt ⋅ vd = Δt ⋅ Dp (Δx )2 = 16td L td Δt 2 L2 = 3 16td Since vd = µ pE = µ p µp = V L = Lo t d L ⋅ Lo V ⋅ td 27 Haynes- Shockley Experiment Example: Ge, Sample length 1cm Probes separated by 0.95 cm Voltage of 2V Pulse arrives at td=0.25 ms Pulse width Δt=117µs Dp ( Δx )2 = 16t d −6 Δt 2 L2 (117 x10 s ) ⋅ ( 0.95 cm ) = = = 49.4 cm 2 / s 3 −3 3 16t d 16 ⋅ (0.25 x10 s ) 2 2 L 0.95 cm vd t d 0.25 x10 −3 s = 1900 cm 2 / V − s µp = = = Eo 2V E Lo 1cm Dp µp = 0.026V = kT q 28 Assignments •  Homework assigned every Friday, due following Friday •  Reading: –  Fri 2/21: §'s 4.3.1, 4.3.3 –  Mon 2/24: §'s 4.3.3, 4.3.4 –  Wed 2/26: §'s 4.4, 4.4.1, 4.4.2 –  Fri 2/28: §'s 4.4.2, 4.4.3 •  Chapter 3 in Pierret covers similar material 30 Outline, 2/28/14 •  The Con\nuity Equa\on •  Steady- State Carrier Injec\on 32...
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