Semiconductor Device Physics Lecture 6 Dr. Gaurav Trivedi, EEE Department, IIT Guwahati Metallurgical Junction Doping profile Poisson’s Equation Poisson’s equation is a well-known relationship in electricity and magnetism. It is now used because it often contains the starting point in obtaining quantitative solutions for the electrostatic variables. E K S 0 D v D E K S 0 In one-dimensional problems, Poisson’s equation simplifies to: E x K S 0 Equilibrium Energy Band Diagram pn-Junction diode Qualitative Electrostatics Equilibrium condition Band diagram Electrostatic potential V 1 q ( E c E ref ) V ( x ) E dx Qualitative Electrostatics Equilibrium condition Electric field E dV dx E ( x) K S 0 x Charge density E x K S 0 Formation of pn Junction and Formation of pn Junction and Charge Distribution Charge Distribution q( p n N D N A ) qNA– qND+ Formation of pn Junction and Charge Distribution Built-In Potential Vbi • Vbi for several materials: Ge ≤ 0.66 V Si ≤ 1.12 V GeAs ≤ 1.42 V qV bi ( E F E i ) n side ( E i E F ) p side For non-degenerately doped material, ( E F E i ) n-side ( E i E F ) p-side n ND kT ln kT ln n n i i p NA kT ln kT ln n n i i qV bi NAND kT ln 2 n i The Depletion Approximation The Depletion Approximation On the p-side, ρ = –qNA dE dx E ( x) E ( x) qN A S qN A S qN A S x c1 ( x xp ) with boundary E(–xp) 0 On the n-side, ρ = qND E ( x) qN D S ( xn x ) with boundary E(xn) 0 Solution for ρ Step Junction with VA 0 qN A , x p x 0 qN D , 0 x xn 0, o th e rw ise Solution for E E ( x) Solution for V qN A S qN D S ( x p x ), ( x n x ), xp x 0 0 x xn qN A 2 ( x x ) , xp x 0 p 2 S V ( x) qN D 2 V bi ( xn x ) , 0 x xn 2 S Step Junction with VA 0 At x = 0, expressions for p-side and n-side for the solutions of E and V must be equal: N A xp N D xn qN A 2 S ( x p ) V bi 2 qN D 2 S ( xn ) 2 Relation between ρ(x), E(x), and V(x) 1.Find the profile of the built-in potential Vbi 2.Use the depletion approximation ρ(x) With depletion-layer widths xp, xn unknown 3.Integrate ρ(x) to find E(x) Boundary conditions E(–xp) 0, E(xn)0 4.Integrate E(x) to obtain V(x) Boundary conditions V(–xp) 0, V(xn) Vbi 5.For E(x) to be continuous at x 0, NAxp NDxn Solve for xp, xn Depletion Layer Width Eliminating xp, Eliminating xn, Summing xn xp 2 S NA q ND (NA ND ) 2 S ND q NA (NA ND ) xn xp W V bi V bi 2 S 1 1 q NA ND Exact solution, try to derive V bi ND NA xn One-Sided Junctions If NA >> ND as in a p+n junction, W xn 2 S V bi q ND , xp xn ND 0 NA If ND >> NA as in a n+p junction, W xp 2 S V bi q NA , xn xp NA 0 ND Simplifying, W 2 S V bi q N where N denotes the lighter dopant density Step Junction with VA 0 • To ensure low-level injection conditions, reasonable current levels must be maintained VA should be small Step Junction with VA 0 In the quasineutral, regions extending from the contacts to the edges of the depletion region, minority carrier diffusion equations can be applied since E ≈ 0. In the depletion region, the continuity equations are applied. Step Junction with VA 0 Built-in potential Vbi (non-degenerate doping): V bi NAND N A kT ND kT ln ln ln 2 q q q ni ni ni kT Depletion width W : W xp xn xp 2 S 1 1 V bi V A q NA ND 2 S ND q NA NA ND xp V bi V A , ND NA ND xn W , xn 2 S NA q ND NA ND NA NA ND W V bi V A Effect of Bias on Electrostatics • If voltage drop , then depletion width • If voltage drop , then depletion width Quasi-Fermi Levels Whenever Δn = Δp ≠ 0 then np ≠ ni2 and we are at nonequilibrium conditions. In this situation, now we would like to preserve and use the relations: n ni e ( E F E i ) kT , p ni e ( E i E F ) kT On the other hand, both equations imply np = ni2, which does not apply anymore. The solution is to introduce to quasi-Fermi levels FN and FP such that: n ni e ( FN E i ) kT n F N E i kT ln ni p ni e ( E i FP ) kT p FP E i kT ln ni • The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions Example: Quasi-Fermi Levels a) What are p and n? 2 n 0 N D 10 cm 17 3 , p0 n n 0 n 10 +10 3 31 n0 14 10 cm np 10 10 =10 cm 14 3 10 cm 17 b) What is the np product? 17 10 cm 14 17 p p 0 p 10 +10 ni 3 14 3 3 3 Example: Quasi-Fermi Levels Consider a Si sample at 300 K with ND = 1017 cm–3 and Δn = Δp = 1014 cm–3. 0.417 eV c) Find FN and FP? F N E i kT ln n n i F N E i 8.62 10 5 300 ln 10 17 10 10 Ec FN Ei FP Ev 0.417 eV 0.238 eV FP E i kT ln p n i E i FP 8.62 10 5 0.238 eV np n i e 300 ln 10 14 10 10 FN E i kT ni e E i FP 0.417 kT 0.238 10 e 0.02586 10 e 0.02586 10 10 1.000257 10 10 cm 31 3 31 Linearly-Graded Junction E 1 S dx V E dx Linearly-Graded Junction Homework Assignment Derive the relations of electrostatic variables, charge density and builtin voltage Vbi for the linearly graded junction.