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Semiconductor Device Physics

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.

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Semiconductor Device Physics

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  1. Semiconductor Device Physics Lecture 6 Dr. GauravTrivedi, EEE Department, IIT Guwahati

  2. Metallurgical Junction • Doping profile

  3. Poisson’s Equation • Poisson’s equation is a well-known relationship in electricity and magnetism. • It is now used because it often containsthe starting point in obtaining quantitative solutions for the electrostatic variables. • In one-dimensional problems, Poisson’s equation simplifies to:

  4. Equilibrium Energy Band Diagram • pn-Junction diode

  5. Qualitative Electrostatics • Equilibrium condition Band diagram Electrostatic potential

  6. Qualitative Electrostatics • Equilibrium condition Electric field Charge density

  7. Formation of pn Junction and Charge Distribution Formation of pn Junction and Charge Distribution – + qNA qND

  8. Formation of pn Junction and Charge Distribution

  9. Built-In Potential Vbi • Vbi for several materials: • Ge ≤ 0.66 V • Si ≤ 1.12 V • GeAs ≤ 1.42 V • For non-degenerately doped material,

  10. The Depletion Approximation The Depletion Approximation • On the p-side,ρ = –qNA with boundary E(–xp) = 0 • On the n-side,ρ = qND with boundary E(xn) = 0

  11. Step Junction with VA=0 • Solution for ρ • Solution for E • Solution for V

  12. 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:

  13. Relation between ρ(x),E(x),and V(x) • Find the profile of the built-in potential Vbi • Use the depletion approximation ρ(x) • With depletion-layer widths xp, xn unknown • Integrate ρ(x) to find E(x) • Boundary conditionsE(–xp) = 0, E(xn)=0 • Integrate E(x) to obtain V(x) • Boundary conditions V(–xp) = 0, V(xn) =Vbi • For E(x) to be continuous at x = 0, NAxp =NDxn • Solve for xp, xn

  14. Depletion Layer Width • Eliminating xp, • Eliminating xn, Exact solution, try to derive • Summing

  15. One-Sided Junctions • If NA>>NDas in a p+njunction, • If ND>> NAas in a n+pjunction, • Simplifying, • where N denotes the lighter dopant density

  16. Step Junction with VA 0 • To ensure low-level injection conditions, reasonable current levels must be maintained  VA should be small

  17. 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.

  18. Step Junction with VA 0 • Built-in potential Vbi (non-degenerate doping): • Depletion width W :

  19. Effect of Bias on Electrostatics • If voltage dropâ,thendepletion width â • If voltage drop á, then depletion width á

  20. Quasi-Fermi Levels • Whenever Δn =Δp ≠0then np ≠ ni2 and we are at non-equilibrium conditions. • In this situation, now we would like to preserve and use the relations: • 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: • The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions

  21. Example: Quasi-Fermi Levels • a) What are p and n? • b) What is the np product?

  22. 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 Ec FN • c) Find FN and FP? Ei FP Ev 0.238 eV

  23. Linearly-Graded Junction

  24. Linearly-Graded Junction Homework Assignment • Derive the relations of electrostatic variables, charge density and builtin voltage Vbi for the linearly graded junction.

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