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Key Questions 2/28/14 Things you should know when you leave… Key Questions ECE 340 • Why do carriers diffuse? Lecture 16 and 17: Diffusion • What happens when we add an electric of Carriers field to our carrier gradient? • How can I visualize this from a band Class Outline: diagram? • What is the general effect of including recombination in our considerations? •Diffusion Processes • What is the relationship between •Diffusion and Drift of Carriers diffusion and mobility? M.J. Gilbert ECE 340 – Lecture 16 and 17 Diffusion Processes Diffusion Processes What happens when we have a concentration discontinuity?? Let’s shine light on a localized part of a semiconductor… Now let’s monitor the system… Consider a situation where we spray perfume in the corner of a room… •Assume thermal motion . • If there is no convection or motion of air, then the scent spreads by diffusion. •Carriers move by interacting with the lattice or impurities. • This is due to the random motion of particles. •Thermal motion causes particles to jump to an •Particles move randomly until they collide with an air molecule which changes adjacent compartment. it’s direction. •After the mean-free time (τ ), half of • c If the motion is truly random, then a particle sitting in some volume has equal particles will leave and half will remain a probabilities of moving into or out of the volume at some time interval. certain volume. 1024 T = 0 512 512 512 256 256 384 384 Shouldn’t the same thing happen T1 = 0 in a semiconductor if we have 1024 t = 3τ t = 0 t =τ c t = 2τ c 128 128 c 256 spatial gradients of carriers? T2 = 0 • 320 256 Process continues until uniform concentration. 192 •We must have a concentration gradient for T3 = 0 t = 6τ c diffusion to start. M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 1 2/28/14 Diffusion Processes Diffusion Processes How do we describe this physical process?? So we have a flux of particles… We want to calculate the rate at which electrons diffuse in a simple one- The rate of electron flow in the +x direction dimensional example. Consider an arbitrary electron distribution… (per unit area): λ φ = n − n •Divide the distribution into n ( 1 2 ) 2τ c incremental distances of the mean-free path (λ). Since the mean-free path is a small differential •Evaluate n(x) in the center length, we can write the electron difference as: of the segments. φn = In the limit of small Δx, or small •Electrons on the left of x n(x)− n(x + Δx) 0 n − n = λ mean-free path between collisions… have a 50% chance of 1 2 Δx moving left or right in a 2 time, τc. λ n(x)− n(x + Δx) λ •Same is true for electrons φn (x) = lim Δx→0 τ c Δx λ λ to the right of x0. λ2 dn(x) = − Net # of electrons moving from left 1 1 τ dx Diffusion coefficient (n1λA)− (n2λA) c 2 to right in one τc. 2 2 (cm /sec) λ λ M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 Diffusion Processes Diffusion and Drift of Carriers But we already expected this… How do we handle a concentration gradient and an electric field? Define the carrier flux for E electrons and holes: e- The total current must be the sum of the electron and hole currents resulting dn x J (x) = J n + J p ( ) from the drift and diffusion processes φn (x)= −Dn dx h+ n(x) dn x ( ) Electrons dp(x) J n (x)= qµnn(x)E(x)+ qDn φ p (x)= −Dp dx dx p(x) dp(x) And the corresponding current densities associated with diffusion… J n (x)= qµ p p(x)E(x)− qD p Holes x dx n dn(x) J diff = qDn Drift Diffusion dx Where are the particles and currents flowing? φn (diff) φ (drift) p dp(x) φp (diff and drift) n J = −qD diff n dx Jp (diff and drift) Jp (diff) Jn (drift) Carriers move together, currents Dashed Arrows = Particle Flow ! !Solid Arrows = Resulting Currents! opposite directions. !! M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 2 2/28/14 Diffusion and Drift of Carriers Diffusion and Drift of Carriers A few extra observations… Can we relate the diffusion coefficient to the mobility? φn (diff) φ (drift) We can by using what we know about drift, diffusion, and band bending… φp (diff and drift) n Jp (diff and drift) Jp (diff) Jn (drift) •In equilibrium, no current flows. •Any fluctuation that would begin a diffusion current also Dashed Arrows = Particle Flow ! !Solid Arrows = Resulting Currents! sets up an electric field which redistributes the carriers !! by drift. dn x •Diffusion currents are in opposite directions. ( ) J n (x)= qµnn(x)E(x)+ qDn = 0 •Drift currents are in the same direction. dx •Currents depend on: Dn 1 dn(x) •Relative electron and hole concentrations. Solve for the electric field E(x): E(x) = •Magnitude and directions of electric field. µn n(x) dx •Carrier gradients. (EF −Ei ) k T b DN kbT dn x It’s equilibrium, so we know n(x): n(x) = nie ( ) •Diffusion currents can be large even = J n (x)= qµnn(x)E(x)+ qDn E E q dx if the carriers are in the minority by ( F − i ) q µ N dn(x) ni k T dEi q b (qnE)µn − (qnE) DN = 0 several orders of magnitude. = − e = − nE k T dp x dx kbT dx kbT b ( ) •Not true for drift currents. DP kbT J n (x)= qµ p p(x)E(x)− qD p = dx Assuming E is non-zero µP q M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 Diffusion and Drift of Carriers Diffusion and Drift of Carriers These relations are called the Einstein relations… Recall the previous example… Ec D k T Assume that: N b = Ei •It is silicon maintained at 300 K. µ N q Ef •E – E = E /4 at ± L and E – E = DP kbT f i g f i Ev = Eg/4 at x = 0. µP q •Choose the Fermi level as the reference energy. •The balance of drift and diffusion currents creates a built-in x electric field to accompany any gradient in the bands. -L 0 L E = −∇V E •Gradients in the bands can occur at equilibrium when: 1 V V = − (Ec − Eref ) 1 dE 1 dE 1 dE q = c = v = i q dx q dx q dx E V • the band gap varies. = −∇ • 1 dEc 1 dEv 1 dEi x alloy concentration varies. = = = x • q dx q dx q dx dopant concentrations vary. -L 0 L -L 0 L M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 3 2/28/14 Diffusion and Drift of Carriers Diffusion and Drift of Carriers Question: Is it in equilibrium? Ec What are the electron and hole Ec current densities at ± L/2: Energy Ei Ei Ef Ef Ef It is in equilibrium, so JP and JN = Ev 0. Ev Material 1 Material 2 DOS – N1(E) DOS – N2(E) FD – f1(E) FD – f2(E) x Roughly sketch n and p inside the x •Assume two materials in sample: intimate contact. -L 0 L -L 0 L •In thermal equilibrium. p • Rate1-2 N E f E • N E 1− f E No current. 1( ) 1( ) 2 ( )[ 2 ( )] n •No net energy transfer. i •Carriers moving from 2 to 1 Rate2-1 N2 (E)f2 (E)• N1(E)[1− f1(E)] must be balanced by carriers n moving from 1 to 2. Rate1-2 = Rate2-1 x dEF = 0 YES -L 0 L Therefore… f1(E) = f2(E) Ef1 = Ef2 dx M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 Diffusion and Drift of Carriers Diffusion and Recombination What are the electron diffusion Ec So what does this mean? current at ± L/2? If so, in what direction? Ei There is a diffusion current at Ef both L/2 and –L/2. Ev dn n At –L/2: > 0 J dx diff dn n x At L/2: < 0 J diff dx What are the electron drift current -L 0 L Consider this semiconductor: at ± L/2? If so, in what direction? drift •The hole current density leaving the differential area may be Jn = −qn vdn = qnµnE n larger or smaller than the current density that enters the area. At –L/2: J drift •This is a result of recombination and generation. n •Net increase in hole concentration per unit time, dp/dt, is At L/2: J drift difference between hole flux per unit volume entering and DP kbT What is the diffusion = Use Einstein 11.9 cm2/sec leaving, minus the recombination rate. coefficient? relation µP q M.J. Gilbert ECE 340 – Lecture 16 and 17 M.J. Gilbert ECE 340 – Lecture 16 and 17 4 2/28/14 Diffusion and Recombination Diffusion and Recombination How can we explain this? Are there any simplifications? The net increase in hole concentration per unit If the current is carried mainly by diffusion (small drift) we can replace the time is the difference between the hole flux currents in the continuity equation… entering and leaving minus the recombination rate… n ∂n dn(x) J diff = qDN J x = qµ n x E x + qD ∂x n ( ) n ( ) ( ) n ∂p 1 J P (x)− J P (x + Δx) Δp dx = − p ∂p ∂t x x x q Δx τ p dp x → +Δ J diff = −qDP ( ) J n (x)= qµ p p(x)E(x)− qD p ∂x dx Rate of hole Increase in hole Recombination We put this back into the continuity equations… buildup.
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