ECE-305: Spring 2015 Minority Carrier Diffusion Equation (MCDE) Professor Mark undstrom Electrical and Computer Engineering Purdue University, West afayette, IN USA lundstro@purdue.edu Pierret, Semiconductor Device Fundamentals (SDF) pp. 122-132 undstrom ECE 305 S15 2/11/15 outline Analyzing semiconductor problems often comes down to solving the minority carrier diffusion equation (MCDE). In this lecture, I will discuss HW5, which gives us practice solving the MCDE for several common situations. undstrom ECE 305 S15 2
the semiconductor equations p n! J p = i q + G p R p! J = i n q + G n R n 0 = i( ε E! ) + ρ!!!! J p = pqµ p E qd p p = pµ p ( Fp q)!!!! J n = nqµ n E + qd n n = nµn Fn q Three equations in three unknowns: p ( r! ), n ( r! ), V ( r! ) ρ = q p n + N + ( D N A )! E r! = V ( r! ) 3 minority carrier diffusion equation n = n 0 = N D! p J p = i q + G R p p p = d dx Δp J px q + G R p = d qd p dδp dx dx q + G Δp τ p (N-type semiconductor in low level injection) (hole continuity equation) (1D, generation by light) (low-level injection, no electric field) Δp d 2 Δp = D p Δp + G τ (D p p spatially uniform) 4
minority carrier diffusion equation Δp( x,t) = D d 2 Δp x,t Δp x,t p τ p + G To use the MCDE to solve a problem, first check to be sure that the simplifying assumptions needed to derive the MCDE from the continuity equation apply. Then further simplify the MCDE (if possible), specify the initial condition (if necessary) and the two boundary conditions (if necessary). 5 low level injection N-type semiconductor: n( x,t) = n 0 = N D p( x,t) Δp( x,t) >> p 0 = n 2 i n 0 Δp x,t = D d 2 Δp x,t Δp x,t + G p τ p P-type semiconductor: p( x,t) = p 0 = N A Δn x,t = D x,t Δn x,t + G n 2 n( x,t) Δn( x,t) >> n 0 = n i p 0 6
HW 5 P-type Si at T = 300 K N A = 10 17 cm 3 = p 0 µ n = 300 cm 2 /V-s D n = k B T q µ n = 7.8 cm 2 /s Δn( x,t) MCDE for electrons = D x,t Δn x,t n + G = 10 6 s n = D n = 27.9 µm 7 diffusion length HW5 #1: P-type sample in injection Steady-state, uniform generation (no spatial variation) Solve for Δn and for the QF s. Δn 0 = 0 Δn = D n Δn + G + G Δn = G 8
HW5 #1: solution Δn( x) Δn( x) = G Δn = G = 10 20 10 6 = 10 14 x = 0 x Steady-state, uniform generation, no spatial variation 9 HW5 #1: solution P-type / equilibrium E C n 0 = n 2 i = 10 3 cm -3 p 0 ( p 0 = n i e E i E F ) k B T 10 17 = 10 10 ( e E i E F ) k B T E i E V p 0 = 10 17 cm -3 E F E F = E i 0.41 ev Steady-state, uniform generation, no spatial variation 10
HW5 #1: solution P-type / out of equilibrium F p = E i 0.41 ev Δn = 10 14 cm -3 >> n 0 ( n Δn = n i e F n E i ) k B T E C E i E V p 0 = 10 17 cm -3 F n F p 10 14 = 10 10 ( e F n E i ) k B T F n = E i + 0.24 ev Steady-state, uniform generation, no spatial variation 11 HW5 #2 Now turn off the light. Transient, no generation, no spatial variation Solve for Δn and for the QF s. Δn x,t = D x,t Δn x,t + G n Δn Δn = 0 Δn + 0 = Δn τ p 12 Δn( t) = Δn( 0)e t/ = 1014 e t/
HW5 #2: Solution Δn( t) Δn( 0) = 10 14 Δn( x) = Δn( 0)e t/ t = 0 Δn = ( t ) = 0 t 13 HW5 #2: solution Δn( t = 0) = G Δn( x) How do the QF s vary with time? Δn( t) = Δn( t = 0)e t/ x = 0 x transient, no generation, no spatial variation 14
HW5 #2: solution F p = E i 0.41 ev n( t) Δn( t) = n i e F n t ( E i ) k B T 10 14 e t/ = ( 1010 e F n( t) E i ) k B T F n ( t) = E i + k B T ln 10 4 k B T t E C E i E V F n F p transient, no generation, no spatial variation 15 HW5 #4 Steady-state, sample long compared to the diffusion length. No generation. Δn( x = 0) = 10 12 cm -3 fixed Δn = D p Δn + G 0 = D p Δn + 0 Δn = 0 D p 16 Δn n 2 = 0 n D n
HW5 #4: solution Steady-state, sample long compared to the diffusion length. No generation. ( x) Δn( x = 0) = 10 12 cm -3 Δn( x) = 0 fixed 2 n D n n Δn( x) = Ae x/ n + Be+ x/ n Δn( x) = Ae x/ n Δn( x) = Δn( 0)e x/ n = 1012 e x/ n 17 HW5 #4: solution Δn( x) Δn( 0) Δn( x) = Δn( 0)e x/ n n = D n << x = 0 Δn( x ) = 0 x x = = 200 µm Steady-state, sample long compared to the diffusion length. 18
HW5 #4 Draw the energy band diagram with the QF s. Is there a hole current? 19 HW5 #5 Steady-state, sample is 5 micrometers long. No generation. Δn( x = 0) = 10 12 cm -3 Δn( x = 5 µm) = 0 fixed Δn = D p Δn + G 0 = D p Δn + 0 Δn = 0 2 n D n n 20
HW5 #5: solution Steady-state, sample is 5 micrometers long. No generation. Δn( x = 0) = 10 12 cm -3 Δn( x = 5 µm) = 0 fixed x ( x) = 0 Δn( x) = 0 2 n D n n Δn( x) = Ax + B Δn( x) = Δn ( 0 ) 1 x 21 HW5 #5: solution Δn( 0) p = D p τ p = 28 µm >> = 5 µm Δn( x) Δn( x) = Δn ( 0 ) 1 x Δn( x = ) = 0 x = 0 x = x Steady-state, sample short compared to the diffusion length. 22
HW5 #6 Steady-state, sample is 30 micrometers long. No generation. Δn( x = 0) = 10 12 cm -3 Δn( x = 30 µm) = 0 fixed Δn = D p Δn + G 0 = D p Δn + 0 Δn = 0 2 n D n n n = 28 µm = 30 µm 23 HW5 #6: solution Steady-state, sample is 30 micrometers long. No generation. Δn( x = 0) = 10 12 cm -3 Δn( x = 30 µm) = 0 fixed Δn = 0 2 n Δn( x) = Ae x/ n + Be+ x/ n Δn( 0) = A + B = 10 12 Δn( ) = Ae / n + Be+ / n = 0 24
HW5 #6 Δn( 0) n = D n Δn( x) Δn( x)= Δn 0 sinh ( x) / n sinh / n Δn( x = ) = 0 x x = 0 x = Steady-state, sample neither long nor short compared to the diffusion length. 25 ~HW5 #3 An infinitely long sample is uniformly illuminated with light for a long time. The optical generation rate G = 1 x 10 20 cm -3 sec -1. The minority carrier lifetime is 1 microsecond. The surface at x = 0 is highly defective, with a high density of R-G centers, so that Δn(0) = 0. Find the s.s. excess minority carrier concentration vs. position. D n d 2 Δp Δp + G = 0 Can we guess the solution? (This is like HW5 #3, but a little simpler.) 26
~ HW5 #3 Δn( x) G 1 e x p Δn( x ) = G τ p d 2 Δp D n Δp + G = 0 Δn( 0) = 0 x = 0 What does the energy band diagram look like? 27 x corresponding energy band diagram E E C F n ( x) E i F p ( x) x = 0 What does a gradient in the QF mean? E V x 28
summary ong region: decaying exponential solutions. Short region: linear Neither long nor short: hyperbolic functions 29