Understanding Semiconductor Carrier Action and Diffusion Equations

2. Chapter 3 Carrier Action Area A, volume A.dx JN(x) JN(x+dx) dx Continuity Equation • Consider carrier-flux into / out of an infinitesimal volume:

3. Chapter 3 Carrier Action Continuity Equation • Taylor’s Series Expansion • The Continuity Equations

4. Chapter 3 Carrier Action Minority Carrier Diffusion Equation • The minority carrier diffusion equations are derived from the general continuity equations, and are applicable only for minority carriers. • Simplifying assumptions: • The electric field is small, such that: • For p-type material • For n-type material • Equilibrium minority carrier concentration n0 and p0 are independent of x (uniform doping). • Low-level injection conditions prevail.

5. Chapter 3 Carrier Action Minority Carrier Diffusion Equation • Starting with the continuity equation for electrons: • Therefore • Similarly

6. Chapter 3 Carrier Action Carrier Concentration Notation • The subscript “n” or “p” is now used to explicitly denote n-type or p-type material. • pn is the hole concentration in n-type material • np is the electron concentration in p-type material • Thus, the minority carrier diffusion equations are:

7. Chapter 3 Carrier Action Simplifications (Special Cases) • Steady state: • No diffusion current: • No thermal R–G: • No other processes:

8. Chapter 3 Carrier Action Minority Carrier Diffusion Length • Consider the special case: • Constant minority-carrier (hole) injection at x = 0 • Steady state, no light absorption for x > 0 • The hole diffusion length LP is defined to be: Similarly,

9. Chapter 3 Carrier Action Minority Carrier Diffusion Length • The general solution to the equation is: • A and B are constants determined by boundary conditions: • Therefore, the solution is: • Physically, LP and LN represent the average distance that a minority carrier can diffuse before it recombines with majority a carrier.

10. Chapter 3 Carrier Action Example: Minority Carrier Diffusion Length • Given ND=1016 cm–3, τp = 10–6 s. Calculate LP. • From the plot,

11. Chapter 3 Carrier Action 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

12. Chapter 3 Carrier Action Example: Quasi-Fermi Levels • Consider a Si sample at 300 K with ND = 1017 cm–3 and Δn = Δp = 1014 cm–3. • The sample is an n-type • a) What are p and n? • b) What is the np product?

13. Chapter 3 Carrier Action 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

14. Chapter 2 Carrier Action Homework 4 • 1. (6.17) • A semiconductor has the following properties:DN = 25 cm2/s τn0 = 10–6 s • DP = 10 cm2/s τp0 = 10–7 s • The semiconductor is a homogeneous, p-type (NA = 1017 cm–3) material in thermal equilibrium for t ≤ 0. At t = 0, an external source is turned on which produces excess carriers uniformly at the rate GL = 1020 cm–3 s–1. At t = 2×10–6 s, the external source is turned off. (a) Derive the expression for the excess-electron concentration as a function of time for 0 ≤ t ≤ ∞ (b) Determine the value of the excess-electron concentration at (i) t = 0, (ii) t = 2×10–6 s, and (iii) t = ∞ (c) Plot the excess electron concentration as a function of time. • 2. (4.38) • Problem 3.24Pierret’s “Semiconductor Device Fundamentals”. • Deadline: 17 February 2011, at 07:30.