2 Example: Energy-Band Diagram For Silicon at 300 K, where is EF if n = 1017 cm–3 ?Silicon at 300 K, ni = 1010 cm–3
3 Consider a Si sample at 300 K doped with 1016/cm3 Boron Consider a Si sample at 300 K doped with 1016/cm3 Boron. What is its resistivity?NA = 1016/cm3 , ND = 0(NA >> ND p-type)p 1016/cm3, n 104/cm3
4 Consider a Si sample doped with 1017cm–3 As Consider a Si sample doped with 1017cm–3 As. How will its resistivity change when the temperature is increased from T = 300 K to T = 400 K?The temperature dependent factor in (and therefore ) is n.From the mobility vs. temperature curve for 1017cm–3, we find that n decreases from 770 at 300 K to 400 at 400 K.As a result, increases by a factor of: 770/400 =1.93
5 1.a. (4.2)Calculate the equilibrium hole concentration in silicon at T = 400 K if the Fermi energy level is 0.27 eV above the valence band energy.1.b. (E4.3)Find the intrinsic carrier concentration in silicon at: (i) T = 200 K and (ii) T = 400 K.1.c. (4.13)Silicon at T = 300 K contains an acceptor impurity concentration of NA = 1016 cm–3. Determine the concentration of donor impurity atoms that must be added so that the silicon is n-type and the Fermi energy level is 0.20 eV below the conduction band edge.
6 Remark: kT/q = 25.86 mV at room temperature What is the hole diffusion coefficient in a sample of silicon at 300 K with p = 410 cm2 / V.s ?Remark: kT/q = mV at room temperature
7 Hint: In steady-state, generation rate equals recombination rate Consider a sample of Si doped with 1016 cm–3 Boron, with recombination lifetime 1 μs. It is exposed continuously to light, such that electron-hole pairs are generated throughout the sample at the rate of 1020 per cm3 per second, i.e. the generation rate GL = 1020/cm3/sa) What are p0 and n0?b) What are Δn and Δp?Hint: In steady-state, generation rate equals recombination rate
8 Consider a sample of Si at 300 K doped with 1016 cm–3 Boron, with recombination lifetime 1 μs. It is exposed continuously to light, such that electron-hole pairs are generated throughout the sample at the rate of 1020 per cm3 per second, i.e. the generation rate GL = 1020/cm3/s.c) What are p and n?d) What are np product?Note: The np product can be very different from ni2 in case of perturbed/agitated semiconductor
9 Photoconductor Photoconductor Photoconductivity is an optical and electrical phenomenon in which a material becomes more electrically conductive due to the absorption of electro-magnetic radiation such as visible light, ultraviolet light, infrared light, or gamma radiation.When light is absorbed by a material like semiconductor, the number of free electrons and holes changes and raises the electrical conductivity of the semiconductor.To cause excitation, the light that strikes the semiconductor must have enough energy to raise electrons across the band gap.
10 Net Recombination Rate (General Case) Chapter 3For arbitrary injection levels and both carrier types in a non-degenerate semiconductor, the net rate of carrier recombination is:Net Recombination Rate (General Case) Net Recombination Rate (General Case)whereET : energy level of R–G center
11 Continuity Equation JN(x) JN(x+dx) Area A, volume A.dx Flow of current Flow of electron
12 Continuity EquationTaylor’s Series ExpansionThe Continuity Equations
13 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 materialFor n-type materialEquilibrium minority carrier concentration n0 and p0 are independent of x (uniform doping).Low-level injection conditions prevail.
14 Minority Carrier Diffusion Equation Starting with the continuity equation for electrons:ThereforeSimilarly
15 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 materialnp is the electron concentration in p-type materialThus, the minority carrier diffusion equations are:
17 Minority Carrier Diffusion Length Consider the special case:Constant minority-carrier (hole) injection at x = 0Steady state, no light absorption for x > 0The hole diffusion length LP is defined to be:Similarly,
18 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.
19 Quasi-Fermi LevelsWhenever Δn = Δp ≠ 0 then 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
20 Example: Minority Carrier Diffusion Length Given ND=1016 cm–3, τp = 10–6 s. Calculate LP.From the plot,
21 Example: Quasi-Fermi Levels Consider a Si sample at 300 K with ND = 1017 cm–3 and Δn = Δp = 1014 cm–3.a) What are p and n?The sample is an n-typeb) 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.EcEvEiFN0.417 eVc) Find FN and FP?FP0.238 eV