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Semiconductor Device Physics Lecture 5 Dr. Gaurav Trivedi, EEE Department, IIT Guwahati 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 n EF Ei kT ln ni 17 10 5 0.56 8.62 10 300 ln 10 eV 10 0.56 0.417 eV 0.977 eV Consider a Si sample at 300 K doped with 1016/cm3 Boron. What is its resistivity? (NA >> ND p-type) NA = 1016/cm3 , ND = 0 p 1016/cm3, n 104/cm3 1 qn n qp p 1 qp p (1.6 10 )(470)(10 ) 1.330 cm 19 16 1 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 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. What is the hole diffusion coefficient in a sample of silicon at 300 K with p = 410 cm2 / V.s ? kT DP p q 25.86 meV 410 cm 2 V 1s 1 1e cm 2 25.86 mV 410 V s 10.603 cm2 /s 1 eV 1 V 1e 1 eV 1.602 1019 J • Remark: kT/q = 25.86 mV at room temperature 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/s a) What are p0 and n0? p0 1016 cm3 10 2 2 10 ni 4 3 10 cm n0 1016 p0 b) What are Δn and Δp? p n GL 1020 106 1014 cm3 • Hint: In steady-state, generation rate equals recombination rate 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? p p0 p 1016 1014 1016 cm 3 n n0 n 104 1014 1014 cm 3 d) What are np product? 30 3 np 1016 1014 10 cm ni2 • Note: The np product can be very different from ni2 in case of perturbed/agitated semiconductor 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. Net Recombination Rate (General Case) Chapter 3 For arbitrary injection levels and both carrier types in a nondegenerate semiconductor, the net rate of carrier recombination is: ni2Rate np (General Case) Net p Net Recombination n t R G t RRecombination p (n n1 ) n (Rate p p1(General ) G Case) where n1 ni e( ET Ei ) kT ( Ei ET ) kT p1 ni e • ET : energy level of R–G center Continuity Equation Area A, volume A.dx Flow of current JN(x) JN(x+dx) Flow of electron dx n 1 Adx J N ( x dx) J N ( x) A t q Continuity Equation J N ( x) J N ( x dx) J N ( x) dx x n 1 J N ( x) t q x • Taylor’s Series Expansion The Continuity Equations n 1 J N ( x) n n t q x t thermal t other R G p 1 J P ( x) p t q x t processes p thermal t other R G processes 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: n n J N q n nE qDN qDN x x p p J P qp pE qDP qDP x x • 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. Minority Carrier Diffusion Equation Starting with the continuity equation for electrons: n 1 J N ( x) n GL t q x n (n0 n) 1 (n0 n) n qDN GL t q x x n Therefore n 2 n n DN GL 2 t x n Similarly p 2 p p DP GL 2 t x p n t thermal R G n t other processes n n GL 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: np t DN 2 np x 2 np n GL pn 2 pn pn DP GL 2 t x p Simplifications (Special Cases) Steady state: np pn 0, 0 t t 2 np x 2 2 pn 0, DP 0 2 x 0, pn No diffusion current: DN No thermal R–G: np No other processes: GL 0 n p 0 Minority Carrier Diffusion Length Consider the special case: Constant minority-carrier (hole) injection at x = 0 Steady state, no light absorption for x > 0 2 pn pn 0 DP 2 x p pn (0) pn0 GL 0 for x 0 2 pn pn pn 2 2 x DP p LP The hole diffusion length LP is defined to be: LP DP p Similarly, LN DN n Minority Carrier Diffusion Length 2 pn pn 2 is: The general solution to the equation 2 x LP pn ( x) Ae x LP Bex LP A and B are constants determined by boundary conditions: pn () 0 B0 pn (0) pn0 A pn0 Therefore, the solution is: pn ( x) pn0e x LP • Physically, LP and LN 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 ≠ ni2 and we are at nonequilibrium conditions. In this situation, now we would like to preserve and use the relations: n ni e( EF Ei ) kT , p nie( Ei EF ) 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 Ei ) kT n FN Ei kT ln ni p nie( Ei FP ) kT p FP Ei kT ln ni • The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions Example: Minority Carrier Diffusion Length Given ND=1016 cm–3, τp = 10–6 s. Calculate LP. From the plot, p 437 cm2 V s kT DP p q 25.86 mV 437 cm2 V s 11.3cm2 s LP DP p 11.3cm 2 s 106 s 3.361103 cm = 33.61 m 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-type 2 n n0 ND 1017 cm3 , p0 i 103 cm3 n0 n n0 n 1017 +1014 1017 cm3 p p0 p 103 +1014 1014cm3 b) What is the np product? np 1017 1014 =1031cm3 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 FN c) Find FN and FP? FN Ei kT ln n ni Ei FN Ei 8.62 105 300 ln 1017 1010 0.417 eV FP Ei kT ln p ni 5 FP Ev 0.238 eV Ei FP 8.62 10 300 ln 10 10 0.238 eV Ec 14 10 np ni e FN Ei kT 1010 e 0.417 0.02586 ni e Ei FP kT 1010 e 1.000257 1031 1031cm3 0.238 0.02586