“Complementi Di Fisica” Lectures 25-26

“Complementi di Fisica” Lectures 25-26

Livio Lanceri Università di Trieste

Trieste, 14/15-12-2015

in these lectures

• Introduction – Non or quasi-equilibrium: “excess” carriers injection – Processes for “generation” and “recombination” of carriers – Continuity equations for carriers • Continuity equations: three important special cases – Steady-state injection from one side

• “diffusion length” Lp – Minority carriers recombination at the surface

• diffusion length and “surface recombination velocity” Slr – The Haynes-Shockley experiment • Evidence for simultaneous diffusion, drift and recombination • Are we describing the behaviour of minority carriers alone? What about majority carriers? – Why are “minorities” important? Some examples… – Built-in electric field (Gauss!) and “ambipolar” transport equations

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 2 In these lectures

• Reference textbooks – D.A. Neamen, Semiconductor Physics and Devices, McGraw- Hill, 3rd ed., 2003, p.189-230 (“6 Nonequilibrium excess carriers in semiconductors”) – R.Pierret, Advanced Semiconductor Fundamental, Prentice Hall, 2nd ed., p.134-174 (“5 Recombination-Generation Processes”) – J.Nelson, The Physics of Solar Cells, Imperial College Press, p. 79-117 (Ch.4, “Generation and Recombination”)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 3 Injection of “excess” carriers

Non-equilibrium!

(in some cases: quasi-equilibrium) Carrier injection - introduction

• “carrier injection” = process of introducing “excess” 2 carriers in a semiconductor, so that: np > ni – Optical excitation: • shine a light on a semiconductor crystal;

• if the energy of the photons is hν > Eg, then – Photons absorbed – “excess” electron-hole pairs are created: Δn = Δp – Other methods: • Forward-bias a pn junction • … – In an extrinsic semiconductor, the relative effect of Δn = Δp is very different for “majority” and “minority” carriers, since n ≠ p

– Let us work out an example (n-type Si, n0 > p0 at equilibrium)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 5 Carrier injection

thermal low high equilibrium injection injection

Example: n-type Si at 300K

majority carriers thermal equilibrium

intrinsic 2 n0 p0 = ni concentration 10 −3 ni =1.45×10 cm 15 −3 minority n0 ≈ N D =10 cm carriers 2 5 −3 p0 ≈ ni N D = 2.1×10 cm

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 6 Carrier injection

thermal low high equilibrium injection injection

Example: n-type Si at 300K

majority carriers “excess” carriers low-injection intrinsic 12 −3

Δn = Δp =10 cm << N D concentration 2 ⇒ np > ni minority n = n + Δn ≈ n 0 0 carriers =1015 +1012 ≈1015 cm−3

p = p0 + Δp ≈ Δp 5 12 12 −3 Large increase =10 +10 ≈10 cm in minority carriers

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 7 Carrier injection

thermal low high equilibrium injection injection

Example: n-type Si at 300K

majority carriers “excess” carriers high-injection intrinsic Δn = Δp > N D concentration 2 ⇒ np > ni n = n + Δn ≈ Δn minority 0 carriers

p = p0 + Δp ≈ Δp

Large increase for both carriers Similar concentrations

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 8 Carrier injection - summary

• “carrier injection” = process of introducing “excess” carriers in a semiconductor

• Several methods (optical, etc.)

• Low-level injection: relative effect on concentration – Negligible on majority carriers – Important for minority carriers (also called “minority carriers injection”)

• High-level injection – If very high, both concentrations become comparable – Sometimes encountered in device operation

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 9 Quasi-equilibrium

Example: “Injection” or “Generation” of an excess of electrons and holes by absorption of photons in a very short time, about 10-14 s

Excess electrons and holes relax separately to thermal equilibrium in about 10-12 s and remain for a much longer time in this quasi-equilibrium state

Recombination of electrons and holes via several possible mechanisms takes typically about 10-6 s; plenty of time to do something useful with “stable” electrons and holes!

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 10 Quasi-equilibrium and quasi-Fermi levels

FP FN • In quasi-equilibrium conditions:

– Two different “Quasi-Fermi Levels” FN and FP, describe the separate quasi-equilibrium concentrations of electrons and holes, – each population corresponding to a separate Fermi-Dirac pdf (one for electrons, another for holes)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 11 Quasi-Fermi levels: definition

• In quasi-equilibrium conditions:

– Two different “Quasi-Fermi Levels” FN and FP, describe the separate quasi-equilibrium concentrations of electrons and holes:

• Electrons:

(FN − Ei ) kT −(EC − FN ) kT n ≡ nie = NCe ⇔

FN ≡ Ei + kT ln(n ni ) = EC − kT ln(NC n)

• Holes:

(Ei − FP ) kT −(FN − EV ) kT p ≡ nie = NV e ⇔

FP ≡ Ei − kT ln( p ni ) = EV + kT ln(NV p)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 12 Modified mass action law

• Fermi level EF at equilibrium:

(EF − Ei ) kT (Ei − EF ) kT n0 = nie p0 = nie 2 n0 p0 = ni

• Quasi-Fermi levels at quasi-equilibrium:

(FN − Ei ) kT (Ei − FP ) kT n ≡ nie p ≡ nie

2 (FN − Fp ) kT 2 np = ni e > ni

Difference in total chemical potentials or quasi-Fermi levels

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 13 Quasi-Fermi levels: an example

Thermal equilibrium

Fermi level EF

n-type p-type

Non-equilibrium Quasi-Fermi levels

FN, FP equilibrium Non-equilibrium

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 14 Quasi-Fermi levels and currents

• At (quasi-)equilibrium, (quasi-)Fermi levels are constant – Why? Because from the thermodynamic point of view the Fermi level is the “total” chemical potential, including the “internal” chemical potential and “external” potential energy contributions, like for instance the electrostatic potential energy. • Off-equilibrium, the net movement of carriers is related to the changing total chemical potential or (quasi-)Fermi level by: ∂F ∂F J ≡ µ n N J ≡ µ p P n,x n ∂x p,x p ∂x

• From the definitions of FN, FP by substitution one obtains:

kBT ∂n kBT ∂p J = qµ nE + qµ J p,x = qµ p pEx − qµ p n,x n x n q ∂x q ∂x

drift diffusion Dn drift diffusion Dp NB: A quasi-Fermi level that varies with position in a band diagram immediately indicates that current is flowing in the semiconductor! (see exercise 1 for an application)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 15 Generation and Recombination

Charge carriers: electrons and holes Electrons and holes • Generation rate G : – G = number of free carriers generated (separating electrons from holes) per second and per unit volume – G is usually a function of the available energy (temperature, etc.)

• Recombination rate R: – R = number of free carriers “disappearing” due to recombination per second and per unit volume – R is usually proportional to the product of concentrations of “carriers” and “recombination centers” and to a “capture coefficient” defined as c = vthσ , where vth is the thermal velocity and σ is the recombination process “cross-section”

• Net recombination rate: U = R - G

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 17 Generation processes Generation processes

(for quantitative details: see bibliography)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 19 Radiative (light) generation Band structure for direct and indirect semiconductors

Ge: direct Si: indirect (+ indirect) (+ direct)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 20 Photon absorption: ingredients

direct transitions

indirect transitions

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 21 Photon absorption coefficient α

indirect direct

direct indirect

x I (0) x x + dx dI = −αI (x) ⇒ I (x) = I (0)e−α x dx

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 22 Recombination processes Recombination processes

Recombination via “traps” Auger recombination

often dominant in Silicon devices

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 24 Recombination via traps

• Recombination: often dominated by indirect processes through “recombination centers” or “traps” (direct recombination is negligible for Si)

• Example: in an n-type semicond., under low-injection conditions: – for the minority-carriers (holes !) excess-recombination, the bottleneck is “hole capture”, that

determines the hole “lifetime” τp – Once captured, the hole recombines quickly, since there are many U ≈ vthσ p Nt (pn − pn 0) electrons available 1 τ p ≡ vthσ p Nt

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 25

€ Recombination via traps: terminology

• “Capture”, “emission”: – From the point of view of the trap!

• In particular (figure, next slide): – (a) electron capture = (3) – (b) electron emission = (2) – (c) hole capture = (4) – (d) hole emission = (1)

• Detailed treatment: beyond our scope! – Shockley-Read-Hall model – See back-up slides and reference texts

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 26 Recombination via traps: “lifetime” approximation p-type p-type semiconductor: electron lifetime dominated by “electron capture” (3) “ ” mostly in empty RG centers empty U ≈ vthσ n Nt (n p − n p 0) 1 τ n ≡ ≈1.0µs (Si) vthσ n Nt

n-type n-type semiconductor: hole lifetime dominated € by “hole capture” (4) in “full” RG centers mostly full U ≈ vthσ p Nt (pn − pn 0) 1 τ p ≡ ≈ 0.3µs (Si) vthσ p Nt

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 27 € Continuity equations

Overall conservation of charge!

Detailed accounting of local carrier density as a function of time: Generation, recombination, drift, diffusion Summary of

(B)

(A)

From: The Feynman Lectures on Physics, vol.II 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 29 (A) Conservation of charge: continuity equations

• Any net flow of charge must come from some supply! ! ! ! d dQ ∫ J • nˆ dS = ∫ ∇ • J dV = − ∫ ρ dV = − S V dt V dt

! ! ∂J ∂Jy ∂J ∂ρ ∇ • J = x + + z = − ∂x ∂y ∂z ∂t

– The flux of a current from a closed surface is equal to the decrease of the charge inside the surface € – ρ is the net charge density (negative and positive, algebraic sum) • Let us consider electrons and holes, separately, in a semiconductor, in a simple one-dimensional case

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 30 Continuity for electrons

External voltage Volume element V Adx

How fast does the number of electrons change in A dx ? Net carriers per second “ ” 1 ∂ρ through the walls + generation −q ∂t - recombination $ ' ∂n Jn (x)A Jn (x + dx)A Adx = & − ) + (Gn − Rn )Adx ∂t % −q −q ( Substituting: ∂Jn € Jn (x + dx) = Jn (x) + dx + ... and dividing by A dx ∂x ⇒ see next page…

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 31 € Continuity for electrons and holes

One-dimensional Three-dimensional

! ! ∂n 1 ∂Jn ∂n 1 = + (Gn − Rn ) = ∇ • J n + (Gn − Rn ) ∂t q ∂x ∂t q

∂p 1 ∂Jp ∂p 1 ! ! = − + G − R = − ∇ • J + G − R t q x ( p p ) t q p ( p p ) ∂ ∂ ∂

€

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 32 Continuity for electrons and holes Minority carriers:

Electrons in p-type holes in n-type

∂np ∂p J = n q µ E + D J = p q µ E − D n n,x p n x n ∂x p,x n p x p ∂x One-dimensional, under low-injection conditions, for minority carriers: (electric field)

2 minority ∂np ∂E ∂np ∂ np np − np0 Electrons: = n µ x + µ E + D + G − carrier n in p-type p n n x n 2 n excess p ∂t ∂x ∂x ∂x τ n 2 holes: p ∂pn ∂Ex ∂pn ∂ pn pn − pn0 n = − p µ − µ E + D + G − minority in n-type n p p x p 2 p carrier ∂t ∂x ∂x ∂x τ p lifetime Recombination Simply substitute J=J(drift)+J(diffusion)… rate R

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 33 Continuity: generation, recombination, drift, diffusion

Summary and real-life applications Generation, Recombination, Continuity - 1 Net recombination rate (2) minority excess Δnp (Δnp); approximations: 1-dimensional, Electric field~0, U n = Rn − Gn For instance Uniform doping n0≠n0(x), p0 ≠ p0(x), Low-injection, photogeneration GL only U p = Rp − Gp Photogener. GL

General “minority diffusion” approx.

lifetime approx. (1)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 35 Generation, Recombination, Continuity - 2 Einstein relationships minority minority carriers (el.) carrier diffusion (h.) lifetimes lengths

“band bending”

“Quasi-Fermi”

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 36 Device simulations

• In real life, device designers use programs performing numerical integrations in discrete space and time steps, to obtain (*): Process simulator + boundary conditions, ! ! ρ ∇ • E = −∇2V = external fields ε “ ” doping N D , N A and excitations profiles ñ = −n + p − N A + N D ! ! ! J n = −qnµn∇V + qDn∇n ! ! ! J qn V qD n (*) carrier Device simulator p = − µn∇ − n∇ concentrations, 1 ! ! ∂n fields, currents ∇ • J n = (R − G)+ in: n, p q ∂t (1) equilibrium, el. field 1 ! ! ∂p (2) steady state, ! ! ∇ • J p = (R − G)+ q ∂t (3) transients J n , J p

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 37 “Equilibrium” vs “steady state”

“Equilibrium”: detailed balance, for each process

“steady state”: overall balance

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 38 Continuity equations: applications

Three examples

(1) Steady state injection from one side (2) Recombination at the surface (3) Haynes-Shockley experiment System of differential equations (A) Continuity (transport) equations for minority carriers, 1-d case (Sze notations): 2 ∂np ∂Ex ∂np ∂ np np − np0 = npµn + µnEx + Dn 2 + Gn − ∂t ∂x ∂x ∂x τ n 2 ∂pn ∂Ex ∂pn ∂ pn pn − pn0 = − pnµ p − µ pEx + Dp 2 + Gp − ∂t ∂x ∂x ∂x τ p (B) Gauss’ law, relating the divergence of the electric field with the local charge density, 1-d case: ∂E ρ x = ∂x ε Globally neutral, locally can be unbalanced! + − ρ = q(p − n + ND − NA ) ≈ q(p − n + ND − NA )

N = ND − NA To be solved with given boundary conditions!

€ 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 40 (ex.1) Steady-state injection from one side n-type semiconductor minority carriers: holes ∂pn p x ? = 0 steady state concentration n ( )= ∂t

Ex = 0 no applied field no generation Gp = 0 in the bulk pn0 at thermal equilibrium

pn (0)− pn0 excess injected at x = 0 (boundary condition)

2 ∂ (pn − pn0 ) 1 Continuity equation in this case: 2 = (pn − pn0 ) ∂x Dpτ p “Diffusion length”

−x Lp Solution: pn (x)= pn0 + (pn (0)− pn0 )e Lp = Dpτ p

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 41 (ex.1) Diffusion length - typical values

“Diffusion length” Lp = Dpτ p Ln = Dnτ n

2 2 2 2 µn [cm /Vs] Dn [cm /s] µp [cm /Vs] Dp [cm /s] Si € 1350 35€ 480 12.4 GaAs 8500 220 400 10.4 Ge 3900 101 1900 49.2

example −7 Lp = Dpτ p = (12.4)(5 ×10 ) = 25 µm

€

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 42 (ex.2) Recombination at the surface ∂p n = 0 steady state ∂t no applied field Ex = 0 Uniform generation GL ≠ 0 in the bulk !!! pn0 at thermal equilibrium

pn (0)− pn0 boundary condition

Here the boundary condition is fixed by the rate at which carriers disappear with “surface recombination velocity” Slr = vthσ p Nst depending on the “surface trap density” Nst cm s-1 cm s-1 cm2 cm-2 2 Equation to be solved ∂ Δpn Δpn GL 2 − + = 0 Δpn = pn (x)− pn0 (x > 0, bulk): ∂x Dpτ p Dp €

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 43 (ex.2) Solution with boundary conditions

x L −x L General solution: p p Δ pn (x) = Ae + Be + GLτ p

“complementary” “particular” (homegeneous) Lp = Dpτ p Boundary conditions: € Δpn (x) %x %→%+ ∞→ GLτ p ⇒ A = 0

Δpn (x) %x %→%0 → Δpn (0) ⇒ Δpn (0) = B + GLτ p

B = Δpn (0) − GLτ p Slr after some algebra, substituting A and B, our solution: % ( Δpn (0) − GLτ p −x L p € pn (x) = pn 0 + GLτ p'1 + e * Δpn (0) = ??? & GL τ p )

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 44

€ (ex.2) Surface boundary condition

surface bulk A

x − l + l

Consider a thin volume (A× 2l) enclosing the surface:

J x l A G v N p 0 Al v N p 0 A − x ( = ) = [ L − ( thσ p t )Δ n ( )] − ( thσ p st )Δ n ( )

diffusion Gener. – recomb. l → 0 Recombination current (bulk) (surface) € In the limit l → 0 : − J x (0)= −(vthσ p Nst )Δpn (0)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 45 (ex.2) Solution with surface recomb. velocity # dΔp & D n S p 0 −Jx (0) = −(vthσ p Nst ) Δpn (0) ⇒ p% ( = lr Δ n ( ) $ dx ' x= 0 cm2 s-1 cm-4 cm s-1 cm-3 from the general solution and boundary conditions: € $ B ' # dΔpn & B € D S G B % ( = − ⇒ p & − ) = lr ( Lτ p + ) Lp $ dx ' x= 0 Lp % (

−SlrGLτ p Δpn (0) = GLτ p + B ⇒ B = Dp Lp + Slr Solution expressed in terms of the surface recombination velocity: $ S τ ' € € lr p −x L p pn (x) = pn 0 + GLτ p&1 − e ) % Lp + Slrτ p (

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 46 € (ex.2) Minority carriers at the surface cm-3 s-1 2 p p G ∂ Δ n Δ n L S = v σ N 2 − + = 0 lr th p st ∂x Dpτ p Dp cm2 s-1 -1 -1 2 -2 cm2 s-1 s cm s cm s cm cm

€ €

cm Lp = Dpτ p cm S τ G τ lr p L p -3 Lp + Slrτ p cm cm €

€ 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 47 (ex.2) Limiting cases

Neglecting surface recombination:

Slrτ p << Lp ⇒ pn (x) = pn 0 + GLτ p

pn (0) = pn 0 + GLτ p as expected!

Large (“immediate”) surface recombination: S τ >> L ⇒ p x = p + G τ 1− e−x L p € lr p p n ( ) n 0 L p ( )

pn (0) = pn 0 as expected!

€

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 48 (ex.3) The Haynes-Shockley experiment localized applied light pulse Experimental set-up el. field

excess carrier distributions diffusion, at successive times t1 and t2, recombination no applied field

drift, excess carrier distributions diffusion, at successive times t1 and t2, recombination with a constant applied field

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 49 (ex.3) The Haynes-Shockley experiment

After a light pulse: GL = 0 no bulk generation ∂E x = 0 constant applied field ∂x Transport equation for excess minority carriers (n-type semiconductor): ∂Δp ∂Δp ∂2Δp Δp n € n n n = µp E x + Dp 2 − Δpn = pn − pn0 ∂t ∂x ∂x τ p Solution, no applied field: N & x 2 t ) diffusion, Δpn (x,t) = exp( − − + recombination 4πD t ( 4D t τ + € p ' p p * Solution, with applied field: & 2 ) x − µ E t drift, N ( ( p x ) t + Δpn (x,t) = exp − − diffusion, € 4πD t ( 4D t τ + p ' p p * recombination

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 50

€ The role of Gauss’ law

Dielectric relaxation

Ambipolar transport The role of Gauss’ law divergence-less fields Divergence of the electric field, 1-d case: can be non-zero !

∂Ex ρ (due to “external” charges) = ~ complete ionization ∂x ε ∂Ex + − = 0 ⇒ Ex = 0 ρ = q(p − n + ND − NA ) ≈ q(p − n + ND − NA ) ∂x

N = ND − NA Example: current in a semiconductor

Uniform resistivity: € uniform el.field, E x = const. no local charge

Non-uniform resistivity: € non-uniform field, local charge ≠ 0 !

E x1 ρ1 = Jx = E x2 ρ 2 ⇒ E x1 > E x2

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 52

€ Dielectric relaxation time constant

Example: in a short time inject an excess of holes Δp in a small region of a semiconductor crystal, that will experience a local unbalance of electric charge. How fast will be electrical neutrality restored? n-type ! ! ρ Poisson (Gauss) ∇ • E = ε ! ! Ohm J = σE ! ! ∂ρ Continuity ∇ • J = − ∂t ! ! ! ! σρ ∂ρ dρ ( σ + ∇ • J = σ∇ • E = = − ⇒ + * - ρ = 0 ε ∂t dt ) ε ,

−t τ d ε ρ(t) = ρ(0) e τ d = Dielectric relaxation time constant σ 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 53 Dielectric relaxation: Debye length

-5 “Debye length” LD (~ 10 cm): “dielectric relaxation time” τ % ε ( d L ≈ D ' * = D τ τ ≡ ε σ (~ 10-12 s) D p& σ ) p d d Expect no significant departures from electrical neutrality, over distances

greater than about 4 LD to 5 LD in uniformly doped extrinsic material, at thermal equilibrium (also true off-equilibrium!); € this process is much faster than the typical excess carrier lifetime (10-7 s)

16 -3 Numerical example for n-type Si, doped with donor concentration ND = 10 cm 11.7 8.85 ×10−14 ε εrε0 ( )( ) F ⋅ cm τ d = ≈ = −19 16 −1 σ qeµn ND (1.6 ×10 )(1200)(10 ) (Ω⋅ cm) = 5.4 ×10−13 s = 0.54 ps

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 54 € “ambipolar” transport

Two examples: (1) Bipolar diffusion (2) Shockley experiment

“Ambipolar transport” - equations

Combining the transport equations for electrons and holes with Gauss’ law, under some simplifying assumptions, (see back-up slides and reference texts):

p! ≡ p − p0 n! ≡ n − n0 n! ≈ p! Eint << Eapp

⇒ equations of coupled continuity for excess concentrations

∂ 2n" ∂n" ∂n" “ambipolar transport equation” D" + µ"E + g − R = ∂x2 x ∂x ∂t Non-linear!

With “ambipolar diffusion coefficient” and “ambipolar mobility”:

µnnDp + µ p pDn µnµ p (p − n) D" = µ" = µnn + µ p p µnn + µ p p

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 56 Example 1: “Ambipolar diffusion”

Excess electrons and holes produced by light close to the surface, in large concentrations compared to the equilibrium (dark) ones.

Electrons have larger mobility and move faster: electrons and holes partly separate (net charge positive close to the surface, negative inside) The resulting electric field is directed so as to compensate for the different mobilities (electrons slowed down, holes accelerated)

V < V VA B A

+ -

Delayed current pulse seen by probes at different distances: What is really moving ???

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 58 Example 2 - qualitative interpretation majority-carriers, locally slowed down

the minority-carrier drags along a majority-carrier concentration bump concentration bump

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 59 Lecture 32 - exercises

• Exercise 1: An intrinsic Si sample is doped with donors from one side such that ND=N0exp(-ax). (a) Find an expression for the built-in electric field E(x) at equilibrium over the range for which ND>>ni. (b) Evaluate E(x) when a = 1µm-1. • Exercise 2: An n-type Si slice of thickness L is inhomogeneusly doped

with phosphorous donor whose concentration profile is given by ND(x) = N0 +(NL – N0)(x/L). What is the formula for the electric potential difference between the front and the back surfaces when the sample is at thermal and electric equilibria regardless of how the mobility and diffusivity vary with position? What is the formula for the equilibrium electric field at a plane x from the front surface for a constant diffusivity and mobility?

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 60 Lecture 33 - exercises • Exercise 1: Calculate the electron and hole concentration under steady- 16 -3 -1 15 -3 state illumination in an n-type silicon with GL=10 cm s , ND=10 cm , and τn=τp=10 µs. • Exercise 2: An n-type silicon sample has 2x1016 arsenic atoms/cm3, 2x1015 bulk recombination centers/cm3, and 1010 surface recombination centers/ cm2. (a) Find the bulk minority carrier lifetime, the diffusion length, and the surface recombination velocity under low-injection conditions. The values -15 -16 2 of σp and σs are 5x10 and 2x10 cm , respectively. (b) If the sample is illuminated with uniformly absorbed light that creates 1017 electron-hole pairs/(cm2s), what is the hole concentration at the surface? • Exercise 3: The total current in a semiconductor is constant and is composed of electron drift current and hole diffusion current. The electron concentration is constant and equal to 1016 cm-3. The hole concentration is given by p(x)=1015 exp(-x/L) cm-3 (x>0), where L = 12µm. The hole diffusion 2 coefficient is Dp=12cm2/s and the electron mobility is µn=1000cm /(Vs). The total current density is J = 4.8 A/cm2. Calculate (a) the hole diffusion current density as a function of x, (b) the electron current density versus x, and (c) the electric field versus x.

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 61 Lecture 34 - exercises • Exercise 1: Excess electrons have been generated in a semiconductor so that at t = 0 the excess concentration is 15 -3 -6 Δn(0) = 10 cm . Assuming an excess-carrier lifetime τn = 10 s, calculate the excess electron concentration and the recombination rate for t = 4µs. • Exercise 2: Excess electrons and holes are generated at the end of a silicon bar (at x = 0); the silicon bar is doped 17 -3 with phosphorus atoms to a concentration ND = 10 cm . The minority lifetime is 10-6 s, the electron diffusion 2 coefficient is Dn = 25 cm /s, and the hole diffusion current is 2 Dp = 10 cm /s. Determine the steady-state electron and hole concentrations as a function of x (for x >0) and their diffusion currents at x = 10µm.

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 62 Back-up slides

(topics not included in the standard program!) Recombination via traps

Shockley-Read-Hall model “Low-injection minority lifetime” approximation p-type p-type semiconductor: electron lifetime dominated by “electron capture” (3) “ ” mostly in empty RG centers empty U ≈ vthσ n Nt (n p − n p 0) 1 τ n ≡ ≈1.0µs (Si) vthσ n Nt

n-type n-type semiconductor: hole lifetime dominated € by “hole capture” (4) in “full” RG centers mostly full U ≈ vthσ p Nt (pn − pn 0) 1 τ p ≡ ≈ 0.3µs (Si) vthσ p Nt

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 65 € “Shockley-Read-Hall” lifetimes - 1

• What happens if these approximations are not valid?

– n, p may be comparable (no longer true that n >> p or p >> n)

– carrier lifetime no longer dominated by availability of:

0 p-type: “empty” traps for “electron capture” (Nt = Nt(1-F) ≈ Nt) - n-type: “full” or “ionized” traps for “hole capture” (Nt = NtF ≈ Nt)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 66 “Shockley-Read-Hall” lifetimes - 2

All four “indirect” processes must be taken into account

(see also SZE 2.4.2, “indirect recombination”, or Neamen 6.5.1)

R = e N 1− F 1=d “hole emission” (from a trap) d p t ( ) 2=b “electron emission” (from a trap) Rb = en Nt F 3=a “electron capture” (in a trap) Ra = cn Nt (1− F) 4=c “hole capture” (in a trap) Rc = c p Nt F Net recombination rates for electrons U = R − R and holes separately: n a b U = R − R € p c d 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 67

€ “Shockley-Read-Hall” lifetimes - 3

From equilibrium conditions

GL =0; detailed balance: Ra - Rb = Rc - Rd = 0) – emission coefficients (en, ep) in terms of: – capture coeff. (cn = vthσn, cp = vthσp)

(Ei −Et ) kT en = cnn1 n1 = nie

(Et −Ei ) kT ep = c p p1 p1 = nie

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 68 “Shockley-Read-Hall” lifetimes - 4

non-equilibrium steady-state

(GL = constant ≠ 0, and: Un = Ra - Rb = Up = Rc - Rd ≠ 0 ): see SZE eq. (63)

np − n 2 np − n 2 U = U = U = i = i n p 1 1 τ n (n + n1) + τ p (p + p1) (n + n1) + (p + p1) cn Nt c p Nt

€

This is a general result, usually implemented in device simulations – A special case: the previous “Low-injection minority lifetime” result – For very high doping concentrations, direct transitions become likely: this can be modeled by making τn and τp concentration-dependent

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 69 “ambipolar” transport

Two examples: (1) Bipolar diffusion (2) Shockley experiment

“Ambipolar transport” - equations

Special case: homogeneous semiconductor ⇒ Thermal equilibrium concentrations n0, p0 constant (time and space)

2 ∂ p" $ ∂p" ∂Ex ' p ∂n" Dp 2 − µ p & Ex + p ) + gp − = ∂x % ∂x ∂x ( τ p ∂x 2 ∂ n" $ ∂n" ∂Ex ' n ∂n" Dn 2 + µn & Ex + n ) + gn − = ∂x % ∂x ∂x ( τ n ∂x ! ! ∂Ex q ∇ • E = = ( p" − n") p" ≡ p − p0 n" ≡ n − n0 ∂x ε Assume: - Small internal electric field, Eint << Eapp with respect to the applied field - Almost complete balance n! ≈ p! of electron and hole concentrations - Generation, recombination n p gn = g p ≡ g = ≡ R τ nt τ pt

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 71 “Ambipolar transport” - equations

We get then :

2 ∂ n" $ ∂n" ∂Ex ' ∂n" Dp 2 − µ p & Ex + p ) + g − R = × µ p p ∂x % ∂x ∂x ( ∂t 2 ∂ n" $ ∂n" ∂Ex ' ∂n" Dn 2 + µn & Ex + n ) + g − R = × µnn ∂x % ∂x ∂x ( ∂t

Multiply (see above), add and divide by µnn + µ p p ∂ 2n" ∂n" ∂n" “ambipolar transport equation” D" + µ"E + g − R = ∂x2 x ∂x ∂t Non-linear!

With “ambipolar diffusion coefficient” and “ambipolar mobility”:

µnnDp + µ p pDn µnµ p (p − n) D" = µ" = µnn + µ p p µnn + µ p p

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 72 “Ambipolar transport”

In an extrinsic semiconductor under low injection, the ambipolar mobility coefficients reduce to the minority-carrier parameter values, that are constant

∂ 2n" ∂n" n" ∂n" p-type D + µ E + g" − = minority: electrons n 2 n x ∂x ∂x τ n0 ∂t

2 n-type ∂ p" ∂p" p" ∂p" minority: holes Dp 2 − µ p Ex + g" − = ∂x ∂x τ p0 ∂t

The behaviour of excess majority carriers follows that of minority!!!

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 73 Example 1: “Ambipolar diffusion”

Excess electrons and holes produced by light close to the surface, in large concentrations compared to the equilibrium (dark) ones.

The coupled motion is called + ambipolar diffusion. Since electrons and holes move with the - same velocity in the same direction, there is no net charge current associated with this motion ! 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 74 Example 1: “Ambipolar diffusion” ∂n Charge currents jx,n = q Dn + q nµn E x for electrons and holes: ∂x ∂p j = −q D + q pµ E x,p p ∂x p x The net current density vanishes! Associated electric field:

Dn ∂n ∂x − Dp ∂p ∂x jx = jx,n + jx,p = 0 ⇒ E x = € nµn + pµp

The particle currents are therefore equal for electrons and holes:

Dn nµn + Dp pµp ∂n ∂n € je = jh = = Damb nµn + pµp ∂x ∂x D nµ + D pµ n n p p “ ” Damb = is the ambipolar diffusion coefficient nµn + pµp

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 75

€ A “steady-state” example: locally illuminated semiconductor bar Ingredients and qualitative expectations n-type; non-equilibrium; open-circuit; Local steady illumination Diffusion of excess carriers (p’, n’ )

p$ ≡ p − p0 = Δp n$ ≡ n − n0 = Δn

Diffusion currents, but also

drift currents due to the electric field Ex dp! dn! J = qµ pE − qD J = qµ nE + qD h h x h dx e e x e dx

J = J e + J h = 0

Electric field Ex (charge unbalance!) dE q x = (p" − n")≠ 0 dx ε The local charge unbalance is small! p! − n! p! − n! ≈ <<1 p! n!

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 77 Ingredients and qualitative expectations holes (h): minority n-type; non-equilibrium; open-circuit; electrons (e): majority Local steady illumination drift (e,h): n >> p Diffusion of excess carriers (p’, n’ )

qµenEx >> qµh pEx p$ ≡ p − p0 = Δp n$ ≡ n − n0 = Δn

dp! Diffusion currents, but also J h = qµh pEx − qDh dx drift currents due to the electric field Ex dn! J e = qµenEx + qDe Diffusion (e,h): opposite currents, dx comparable sizes

J = J e + J h = 0 Electric field Ex (charge unbalance!) dE q dp! x = (p" − n")≠ 0 ⇒ qµh pEx << qDh dx dx ε The local charge unbalance is small! in this case dp! p! − n! p! − n! ⇒ J ≈ −qD minority carriers flow ≈ <<1 h h p n dx mainly by diffusion ! ! 14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 78 the minority-carrier current Under conditions of: will be comparable to - comparable mobilities the majority-carrier current - small injection only if in uniform extrinsic material minority carriers flow mainly by diffusion

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 79 d 2 p p g Qualitative results $ $ L 2 − = − ; 0 < x < δ 2 dx Dhτ h Dh = 0 ; x > δ 2

The continuity equations above can be solved analytically to obtain

p’(x) and Jh(x) ⇒ Je(x) = - Jh(x)

J e + J h = 0 ⇒ J e = −J h

If De = Dh ⇒ Ex = 0, p" = n"

If De > Dh ⇒ Majority diffusion larger ⇒ Majority drift current: same direction as Jh(x)

Electric field Ex J e(drift) = qµe (n0 + n")Ex ≈ qµen0 Ex

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 80 the minority-carrier current Under conditions of: will be comparable to - comparable mobilities the majority-carrier current - small injection only if in uniform extrinsic material minority carriers flow mainly by diffusion

The large supply of majority carriers effectively “shields” the minority ones from producing any significant space charge.

The small fields that are generated by slight departures from neutrality serve to adjust the majority-carrier current to the general conditions of the problem, without producing significant effects on minority carriers.

An approximate calculation of Jh, Je, Ex, p’, n’ in the “quasi-neutral” n’ ≈ p’ approximation (without enforcing Gauss’ law with p’ - n’ = 0) will be quite satisfactory; of course, the small p’ - n’ will not be very accurately

determined from Ex found in this way

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 81 Approximate quantitative solution

Assuming “quasi-neutral” behaviour: dp! dn! p! ≈ n! ≈ dx dx (well justified in most cases)

J e = J e(drift) + J e(diffusion) = −J h

dn" dp" De J e(diffusion) = qDe ≈ qDe = − J h dx dx Dh & D # J J J J $ e 1! e(drift) = − e(diffusion) − h ≈ h $ − ! % Dh "

J e(drift) J e(drift) J h (De Dh −1) Ex = ≈ ≈ qµen qµen0 qµen0

Approximate charge unbalance (dEx/dx)

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 82 Nearly exact solution

A more accurate solution, not using the p’ ≈ n’ approx. to evaluate Je (diffusion)

In this example:

δ ≈ Lh >> LD light beam width δ hole diffusion length Lh Debye length LD

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 83 Nearly exact solution

A more accurate solution, not using the p’ ≈ n’ approx. to evaluate Je (diffusion)

In this example:

δ ≈ Lh >> LD light beam width δ hole diffusion length Lh Debye length LD

14/15-12-2015 L.Lanceri - Complementi di Fisica - Lectures 25-26 84