Example: Energy-Band Diagram For Silicon at 300 K, where is E F if n = 10 17 cm –3 ? Silicon at 300 K, n i = 10 10 cm –3
Consider a Si sample at 300 K doped with 10 16 /cm 3 Boron. What is its resistivity? N A = 10 16 /cm 3, N D = 0(N A >> N D p-type) p 10 16 /cm 3, n 10 4 /cm 3
Consider a Si sample doped with 10 17 cm –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 10 17 cm –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
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 N A = 10 16 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.
What is the hole diffusion coefficient in a sample of silicon at 300 K with p = 410 cm 2 / V.s ? Remark: kT/q = 25.86 mV at room temperature
Consider a sample of Si doped with 10 16 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 10 20 per cm 3 per second, i.e. the generation rate G L = 10 20 /cm 3 /s a)What are p 0 and n 0 ? b) What are Δn and Δp? Hint: In steady-state, generation rate equals recombination rate
Consider a sample of Si at 300 K doped with 10 16 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 10 20 per cm 3 per second, i.e. the generation rate G L = 10 20 /cm 3 /s. c)What are p and n? d) What are np product? Note: The np product can be very different from n i 2 in case of perturbed/agitated semiconductor
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. Photoconductor
Net Recombination Rate (General Case) E T : energy level of R–G center Chapter 3 For arbitrary injection levels and both carrier types in a non- degenerate semiconductor, the net rate of carrier recombination is: where Net Recombination Rate (General Case)
Continuity Equation JN(x)JN(x)J N (x+dx) dx Area A, volume A.dx Flow of current Flow of electron
Taylor’s Series Expansion The Continuity Equations Continuity Equation
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 n 0 and p 0 are independent of x (uniform doping). Low-level injection conditions prevail.
Minority Carrier Diffusion Equation Starting with the continuity equation for electrons: Therefore Similarly
Carrier Concentration Notation The subscript “n” or “p” is now used to explicitly denote n-type or p-type material. p n is the hole concentration in n-type material n p is the electron concentration in p-type material Thus, the minority carrier diffusion equations are:
Simplifications (Special Cases) Steady state: No diffusion current: No thermal R–G: No other processes:
Minority Carrier Diffusion Length Similarly, Consider the special case: Constant minority-carrier (hole) injection at x = 0 Steady state, no light absorption for x > 0 The hole diffusion length L P is defined to be:
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, L P and L N represent the average distance that a minority carrier can diffuse before it recombines with majority a carrier.
Quasi-Fermi Levels Whenever Δn = Δp ≠ 0 then np ≠ n i 2 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 = n i 2, which does not apply anymore. The solution is to introduce to quasi-Fermi levels F N and F P such that: The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions
Example: Minority Carrier Diffusion Length Given N D =10 16 cm –3, τ p = 10 –6 s. Calculate L P. From the plot,
Example: Quasi-Fermi Levels Consider a Si sample at 300 K with N D = 10 17 cm –3 and Δn = Δp = 10 14 cm –3. The sample is an n-type a)What are p and n? b)What is the np product?
Example: Quasi-Fermi Levels Consider a Si sample at 300 K with N D = 10 17 cm –3 and Δn = Δp = 10 14 cm –3. c)Find F N and F P ? FPFP 0.238 eV FNFN 0.417 eV EcEc EvEv EiEi