Carrier Action: Motion, Recombination and Generation. What happens after we figure out how many electrons and holes are in the semiconductor?
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1 Carrier Action: Motion, Recombination and Generation. What happens after we figure out how many electrons and holes are in the semiconductor? 1
2 Carrier Motion I Described by 2 concepts: Conductivity: σ (or resistivity: ρ = 1/σ) Mobility: µ Zero Field movement: Random over all e - Thermal Energy Distribution. Motion Electrons are scattered by impurities, defects etc. What happens when you apply a force? 2
3 Apply a force: Electrons accelerate: Carrier Motion II F r = qe r = qexˆ x -n 0 qe x =dp x /dt {from F=ma=d(mv)/dt} Electrons decelerate too. Approximated as a viscous damping force (much like wind on your hand when driving) dp x = -p x dt/τ {dt = time since last randomizing collision and τ = mean free time between randomizing collisions.} Net result: deceration = dp x /dt = -p x /τ 3
4 Carrier Motion III Acceleration=Deceleration in steady state. dp x /dt(accel) + dp x /dt(decel) = 0 -n 0 qe x - p x /τ = 0. Algebra: p x /n 0 = -qτe x = <p x > But <p x > = m n *<v x > Therefore: qτ < vx >= m * n qτ, qτ µ m * p * n m p µ = = n E x µ E n x 4
5 Currents Current density (J) is just the amount of charge passing through a unit area per unit time. J x = (-q)(n 0 )<v x > in C/(s m 2 ) or A/m 2 = +(qn 0 µ n )E x for e - s acting alone. = σ n E x (defining e - conductivity) If both electrons and holes are present: r J r µ µ σ E o n + po p E = E = ρ = q( n ) r r 5
6 Current, Resistance How do we find: current (I)? We integrate J. I = J y = = tow z 0 0 tot dydz resistance (R)? R = L ρ( x) dx ρl 1 L = = w( x) t( x) wt σ wt 0 t w x L V Provided ρ, w, t are all constants along the x-axis. 6
7 Mobility changes Although it is far too simplistic we use: µ n = qτ/m n * τ depends upon: τ is the mean free time. m n * is the effective mass. (depends on material) # of scatter centers (impurities, defects etc.) More doping => lower mobility (see Fig. in books) More defects (worse crystal) => smaller mobility too. The lattice temperature (vibrations) Increased temp => more lattice movement => more scattering => µ smaller τ and smaller µ. Increasing Doping 7
8 Mobility Changes II Mobility is also a function of the electric field strength (E x ) when E x becomes large. (This leads to an effect called velocity saturation. ) <v x > 10 7 cm/s 10 6 cm/s 10 5 cm/s Here µ is constant (low fields). Note constant µ => linear plot. V sat At ~10 7 cm/s, the carrier KE becomes the same order of magnitude as k B T. Therefore: added energy tends to warm up the lattice rather than speed up the carrier from here on out. The velocity becomes constant, it saturates. E x (V/cm)
9 What does E x do to our Energy Band Diagram? Drift currents depend upon the electric field. What does an electric field do to our energy band diagrams? It bends them or causes slope in E C, E V and E i. We can show this. Note: E electron = Total E = PE + KE How much is PE vs. KE??? E electron e - h + E g E C E V 9
10 Energy Band Diagrams in electric fields E C is the lower edge for potential energy (the energy required to break an electron out of a bonding state.) Everything above E C is KE then. PE always has to have a reference! We ll choose one arbitrarily for the moment. (E REF = Constant) Then PE = E C -E REF We also know: PE=-qV E electron KE PE h + e - KE E g E C = PE E V = PE E REF 10
11 Energy Band Diagrams in electric fields II Electric fields and voltages are related by: E = - V (or in 1-D E=-dV/dx) So: PE = E C -E REF = -qv or V = -(E C -E REF )/q E x = -dv/dx = -d/dx{-(e C -E REF )/q} or E x = +(1/q) de C /dx r r E = 1 E q C or Ex = r r E = 1 E q V or Ex = r r E = 1 E or E = 1 q i x 1dE q dx 1dE q dx dei q dx C V 11
12 Energy Band Diagrams in electric fields III The Electric Field always points into the rise in the Conduction Band, E C. E electron E x E C E i E g E V E REF What about the Fermi level? What happens to it due to the Electric Field? 12
13 Another Fermi-Level Definition The Fermi level is a measure of the average energy or electro-chemical potential energy of the particles in the semiconductor. THEREFORE: The FERMI ENERGY has to be a constant value at equilibrium. It can not have any slope (gradients) or discontinuities at all. The Fermi level is our real-life E REF! 13
14 Let s examine this constant E F Note: If current flows => it is not equilibrium and E F must be changing. + V - Semiconductor In this picture, we have no connections. Therefore I=0 and it is still equilibrium! E electron E x E x E C Brings us to a good question: If electrons and holes are moved by E x, how can there be NO CURRENT here??? Won t E x move the electrons => current? E i E F E V The answer lies in the concept of Diffusion. Next Looks N-type Looks P-type 14
15 Examples: Perfume, Diffusion I Heater in the corner (neglecting convection), blue dye in the toilet bowl. What causes the motion of these particles? Random thermal motion coupled with a density gradient. ( Slope in concentration.) 15
16 Green dye in a fishbowl If you placed green dye in a fishbowl, right in the center, then let it diffuse, you would see it spread out in time until it was evenly spread throughout the whole bowl. This can be modeled using the simple-minded motion described in the figure below. L-bar is the mean (average) free path between collisions and τ the mean free time. Each time a particle collides, it s new direction is randomly determined. Consequently, half continue going forward and half go backwards. Dye Concentration l x 16
17 Diffusion II Over a large scale, this would look more like: t=0 t 1 t 2 Let s look more in depth at this section of the curve. t 3 t equilibrium 17
18 Diffusion III What kind of a particle movement does Random Thermal motion (and a concentration gradient) cause? n(x) n b0 n b1 n b2 It causes net motion from large concentration regions to small concentration regions. Bin (0) x l Bin (1) Half of e - go left half go right. Bin (2) x x + l Line with slope: n b2 n b 1 dn l dxx0 x-axis 18
19 Diffusion IV Net number of electrons crossing x 0 is: Number going right: 0.5*n b1 *l*a Minus Number going left: 0.5*n b2 *l*a Net is = 0.5*l*A*(n b1 -n b2 ) (note l*a=volume of a bin.) Flux = # of particles crossing a plane per unit time and unit area. Symbol is: φ φ = 0.5*l*A*(n b1 -n b2 ) τ*a Or φ = 0.5*l (n b1 -n b2 ) τ (τ = mean free time.) 19
20 Diffusion V Using the fact that slope (dn/dx) = -(n b1 -n b2 )/l gives: φ = - 0.5*l 2 dn or φ = -D n *dn/dx (electrons) τ dx or φ = -D p *dp/dx (holes) Now: When charges move we get current. Consequently, the current density is directly related to the particle flux. The equations are: (electrons) J n = qφ n = + qd = + qd n n dn dx n (1 (3 D) D) J (holes) p = + qφ = qd = qd p p p dp dx p (1 (3 D) D) 20
21 Diffusion VI Let s look at an example: n(x) dn/dx = 0 here x J(x) x The electrons are diffusing out of the center and toward the edges. 21
22 Currents round-up So now we know that our total currents have 2 components: DRIFT due to any electric field we apply DIFFUSION due to any (dp/dx, dn/dx) we apply and thermal motion. J J n p r r = qnµ E + qd n n Drift Diffusion r r = qpµ E qd p p n p J = J + total p J n 22
23 Answering that old question How can we have an electric + V - Field and still have no current? (Still have J = 0?) Semiconductor Diffusion must E electron E x balance Drift! E x E C Example: J p = r qpµ E p qd p r p = 0 E i E F E V or r E = Dp µ p 1 p dp dx Looks N-type Looks P-type 23
24 Einstein Relationship We next remember: p=n i exp((e i -E F )/k B T) Plugging this into our equation for the electric field and noting that de F /dx = 0 we get The Einstein Relationships. D µ p p kbt D = q µ n and = n kbt q These are very useful. You will never find a table for both D p and µ p as a result of these. Once you have µ, you have D too, by this relationship. 24
25 A sanity check Pretend we have: What will be the fluxes and currents? E x n(x) p(x) x Holes Mechanism Electrons Diffusion Flux (φ) Current Density (J) Drift Flux (φ) Current Density (J) 25
26 Recombination Generation I Generation (G): How e - and h + are produced or created. Recombination (R): How e - and h + are destroyed or removed The concepts are visually seen in the energy band diagram below. E e G R At equilibrium: r = g and since the generation rate is set by the temperature, we write it as: r = g thermal hv E C E V hv x 26
27 Recombination Generation II Recombination must depend upon the # of electrons: n o the # of holes: p o (If no e - or h +, nothing can recombine!) From the chemical reaction e - + h + Nothing we can know that r = α r n o p o = α r n i2 = g thermal The recombination rate coefficient When the temperature is raised g thermal increases Therefore n i must increase too! 27
28 Recombination Generation III A variety of recombination mechanisms exist: E e Direct, Band to Band G R Auger E C E e hv hv G R E V x E C E e Indirect via R-G centers G R E V x E C R-G Center Energy Level E V x 28
29 GaAs is a Direct Band Gap Semiconductor Direct recombination of electrons with holes occurs. The electrons fall from the bottom of the CB to the VB by giving off a photon! E g The Band Gap Energy GaAs band structure produced by J. R. Chelikowsky and M. L. Cohen, Phys. Rev. B 14, 556 (1976) using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966). 29
30 GaAs band structure produced by W. R. Frensley, Professor of UTD using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966). 30
31 Si is an Indirect Band Gap Semiconductor E g The Band Gap Energy Only indirect recombination of electrons with holes occurs. The electrons fall from the bottom of the CB into an R-G center and from the R-G center to the VB. No photon! Silicon band structure produced by J. R. Chelikowsky and M. L. Cohen, Phys. Rev. B 14, 556 (1976) using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966). 31
32 Silicon band structure produced by W. R. Frensley, Professor of UTD using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966). 32
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