热平衡及载流子浓度

载流子就是带有电荷、并可运动而输运电流的粒子,包括电子、离子等。半导体中的载流子有两种,即带负电的自由电子和带正电的自由空穴。实际上,空穴也就半导体中的价键空位,一个空位的运动就相当于一大群价电子的运动;只不过采用数量较少的空穴这个概念来描述数量很多的价电子的运动要方便得多。
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1. Lecture #3 OUTLINE • Thermal equilibrium • Fermi-Dirac distribution – Boltzmann approximation • Relationship between EF and n, p • Temperature dependence of EF, n, p Finish reading Chapter 2 Spring 2003 EE130 Lecture 3, Slide 1 Review: Energy Band Model and Doping Spring 2003 EE130 Lecture 3, Slide 2 1

2.Spring 2003 EE130 Lecture 3, Slide 3 Important Constants • Electronic charge, q • Permittivity of free space, εo • Boltzmann constant, k • Planck constant, h • Free electron mass, mo • Thermal voltage kT/q Spring 2003 EE130 Lecture 3, Slide 4 2

3. Thermal Equilibrium • No external forces applied: – electric field = 0 – magnetic field = 0 – mechanical stress = 0 • Thermal agitation –> electrons and holes exchange energy with the crystal lattice and each other ⇒ Every energy state in the conduction and valence bands has a certain probability of being occupied by an electron Spring 2003 EE130 Lecture 3, Slide 5 Analogy for Thermal Equilibrium Sand particles Dish Vibrating table • There is a certain probability for the electrons in the conduction band to occupy high-energy states under the agitation of thermal energy (vibrating atoms, etc.) Spring 2003 EE130 Lecture 3, Slide 6 3

4. Fermi Function Probability that an available state at energy E is occupied: 1 f (E) = 1 + e( E − EF ) / kT EF is called the Fermi energy or the Fermi level There is only one Fermi level in a system at equilibrium. Spring 2003 EE130 Lecture 3, Slide 7 Effect of Temperature on f(E) Spring 2003 EE130 Lecture 3, Slide 8 4

5. Boltzmann Approximation If E − E F > 3kT , f ( E ) ≈ e − ( E − EF ) kT If E F − E > 3kT , f ( E ) ≈ 1 − e E − EF kT Spring 2003 EE130 Lecture 3, Slide 9 Density of States E gc(E) ∆Ε Ec Ec Ev Ev gv(E) g(E)dE = number of states per cm3 in the energy range between E and E+dE Near the band edges: mn* 2mn* (E − Ec ) gc ( E ) = E ≥ Ec π 2h 3 m*p 2m*p (Ev − E ) gv ( E ) = E ≤ Ev π 2h 3 Spring 2003 EE130 Lecture 3, Slide 10 5

6. Equilibrium Distribution of Carriers • Obtain n(E) by multiplying gc(E) and f(E) • Obtain p(E) by multiplying gv(E) and 1-f(E) Spring 2003 EE130 Lecture 3, Slide 11 Equilibrium Carrier Concentrations • Integrate n(E) over all the energies in the conduction band to obtain n top of conduction band n=∫ g c ( E ) f ( E )dE Ec • By using the Boltzmann approximation, and extending the integration limit to ∞, we obtain 3/ 2  2πmn*kT  n = N c e −( Ec − EF ) / kT where N c = 2 2   h  Spring 2003 EE130 Lecture 3, Slide 12 6

7.• Integrate p(E) over all the energies in the valence band to obtain p g v ( E )[1 − f ( E )]dE Ev p=∫ bottom of valence band • By using the Boltzmann approximation, and extending the integration limit to -∞, we obtain 3/ 2  2πm*p kT  p = N ve − ( E F − Ev ) / kT where N v = 2 2    h  Spring 2003 EE130 Lecture 3, Slide 13 Intrinsic Carrier Concentration ni Multiply n = N c e − ( Ec − E F ) / kT and p = N v e − ( EF − Ev ) / kT − E g / kT np = N c N v e − ( Ec − Ev ) / kT = N c N v e 2 Recall that np = ni − E g / 2 kT ni = N c N v e Spring 2003 EE130 Lecture 3, Slide 14 7

8. Intrinsic Fermi Level • To find EF for an intrinsic semiconductor (n = p = ni): n = N c e( Ei − EC ) / kT = N v e( EV − Ei ) / kT = p EC + EV kT  NV  Ei = + ln  2 2  N C  EC + EV 3kT  m p  * Ei = + ln *  2 4  mn  Spring 2003 EE130 Lecture 3, Slide 15 n(ni, EF) and p(ni, EF) Spring 2003 EE130 Lecture 3, Slide 16 8

9. Shifting the Fermi Level Spring 2003 EE130 Lecture 3, Slide 17 Example: Energy-band diagram Question: Where is EF for n = 1x1017 cm-3? Spring 2003 EE130 Lecture 3, Slide 18 9

10.Carrier Concentration vs. Temperature intrinsic regime n = ND ln n “freeze-out” regime 1/T high room cryogenic temp. temperature temperature Spring 2003 EE130 Lecture 3, Slide 19 Dependence of EF on Temperature Ec 300K 400K ed nor-dop Ef, Do Ef , Acce ptor-dop ed 400K 300K Ev 1013 1014 1015 1016 1017 1018 1019 1020 N A or N D (cm-3) Spring 2003 EE130 Lecture 3, Slide 20 10

11. Dopant Ionization Q: Nd = 1017 cm-3. What fraction of the donors are not ionized? Solution: First assume that all the donors are ionized. n = N D = 1017 cm −3 ⇒ E f = Ec − 146meV 45meV 146 meV Ed Ec EF Ev 1 1 Probability of non-ionization ≈ = = 0.02 1 + e( Ed − EF ) / kT 1 + e((146−45) meV ) / 26 meV Therefore, it is reasonable to assume complete ionization, i.e., n = ND Spring 2003 EE130 Lecture 3, Slide 21 Summary Thermal equilibrium: Balance between internal processes with no external stimulus (no electric field, no light, etc.) => Electron-hole pair (EHP) generation rate = EHP recombination rate • Fermi function: probability that a state at energy E is filled with an electron under equilibrium conditions: 1 f (E) = ( E − E F ) / kT 1+ e – Boltzmann approximation: For high E, i.e. E - EF > 3kT: f ( E ) ≅ e − ( E − EF ) / kT For low E, i.e. EF – E > 3kT: 1 − f ( E ) ≅ e − ( E F − E ) / kT Spring 2003 EE130 Lecture 3, Slide 22 11

12. n = N c e − ( Ec − EF ) / kT = ni e( EF − Ei ) / kT p = N v e − ( EF − Ev ) / kT = ni e( Ei − EF ) / kT Spring 2003 EE130 Lecture 3, Slide 23 12