Chapter 2 Motion and Recombination of Electrons and Holes 2.1 - PDF document

Chapter 2 Motion and Recombination of Electrons and Holes 2.1 Thermal Motion 3 1 = = 2 Average electron or hole kinetic energy kT mv th 2 2   - -  23 1 3 kT 3 1.38 10 JK 300K = = v th   - 31 m 0.26 9.1 10 kg n =  =  5 7 2 . 3 10 m/s 2 . 3 10 cm/s Slide 2-1 2.1 Thermal Motion • Zig-zag motion is due to collisions or scattering with imperfections in the crystal. • Net thermal velocity is zero. • Mean time between collisions is t m ~ 0.1ps Slide 2-2 1

Hot-point Probe can determine sample doing type Hot-point Probe distinguishes N and P type semiconductors. Thermoelectric Generator (from heat to electricity ) and Cooler (from electricity to refrigeration) Slide 2-3 2.2 Drift 2.2.1 Electron and Hole Mobilities • Drift is the motion caused by an electric field . Slide 2-4 2

2.2.1 Electron and Hole Mobilities = t m v q E mp p t q E = mp v m p = m = - m v E v E p n q t q t m = mp m = mn p n m m p n • m p is the hole mobility and m n is the electron mobility Slide 2-5 2.2.1 Electron and Hole Mobilities v = m E ; m has the dimensions of v/ E   2 cm/s cm = .    V/cm V s   Electron and hole mobilities of selected semiconductors Si Ge GaAs InAs m n (cm 2 /V∙s) 1400 3900 8500 30000 m p (cm 2 /V∙s) 470 1900 400 500 Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices? Slide 2-6 3

Drift Velocity, Mean Free Time, Mean Free Path EXAMPLE: Given m p = 470 cm 2 /V·s, what is the hole drift velocity at E = 10 3 V/cm? What is t mp and what is the distance traveled between collisions (called the mean free path )? Hint: When in doubt, use the MKS system of units. Solution: n = m p E = 470 cm 2 /V·s  10 3 V/cm = 4.7  10 5 cm/s t mp = m p m p /q = 470 cm 2 /V ·s  0.39  9.1  10 -31 kg/1.6  10 -19 C = 0.047 m 2 /V ·s  2.2  10 -12 kg/C = 1  10 -13 s = 0.1 ps mean free path = t mh n th ~ 1  10 -13 s  2.2  10 7 cm/s = 2.2  10 -6 cm = 220 Å = 22 nm This is smaller than the typical dimensions of devices, but getting close. Slide 2-7 2.2.2 Mechanisms of Carrier Scattering There are two main causes of carrier scattering: 1. Phonon Scattering 2. Ionized-Impurity (Coulombic) Scattering Phonon scattering mobility decreases when temperature rises : 1 1 - m  t    3 / 2 T phonon phonon   1 / 2 phonon density carrier thermal velocity T T m = q t /m  T v th  T 1/2 Slide 2-8 4

Impurity (Dopant)-Ion Scattering or Coulombic Scattering Boron Ion Electron _ - - + Electron Arsenic Ion There is less change in the direction of travel if the electron zips by the ion at a higher speed. 3 3 / 2 v T m   th impurity + + N N N N a d a d Slide 2-9 Total Mobility 1600 1 1 1 = + 1400 t t t phonon impurity 1200 Electrons 1 1 1 = + -1 ) 1000 -1 s m m m 2 V phonon impurity 800 Mobility (cm 600 400 Holes 200 0 1E14 1E15 1E16 1E17 1E18 1E19 1E20 N a + N d (cm -3 ) -3 ) Total Impurity Concenration (atoms cm Slide 2-10 5

Temperature Effect on Mobility 10 15 Question : What N d will make d m n / dT = 0 at room temperature? Slide 2-11 Velocity Saturation • When the kinetic energy of a carrier exceeds a critical value, it generates an optical phonon and loses the kinetic energy. • Therefore, the kinetic energy is capped at large E, and the velocity does not rise above a saturation velocity, v sat . • Velocity saturation has a deleterious effect on device speed as shown in Ch. 6. Slide 2-12 6

2.2.3 Drift Current and Conductivity E J p unit + area n + A/cm 2 or C/cm 2 ·sec J p = qpv Hole current density If p = 10 15 cm -3 and v = 10 4 cm/s, then EXAMPLE: J p = 1.6  10 -19 C  10 15 cm -3  10 4 cm/s =  2 = 2 1 . 6 C/s cm 1.6 A/cm Slide 2-13 2.2.3 Drift Current and Conductivity J p,drift = qpv = qp m p E J n,drift = –qnv = qn m n E J drift = J n,drift + J p,drift =  E =(qn m n +qp m p ) E  conductivity (1/ohm-cm) of a semiconductor is  = qn m n + qp m p 1/  = is resistivity (ohm-cm) Slide 2-14 7

Relationship between Resistivity and Dopant Density DOPANT DENSITY cm -3 P-type N-type RESISTIVITY (  cm)  = 1/  Slide 2-15 EXAMPLE: Temperature Dependence of Resistance (a) What is the resistivity (  ) of silicon doped with 10 17 cm -3 of arsenic? (b) What is the resistance (R) of a piece of this silicon material 1 m m long and 0.1 m m 2 in cross- sectional area? Solution: (a) Using the N-type curve in the previous figure, we find that  = 0.084  -cm. (b) R =  L/A = 0.084  -cm  1 m m / 0.1 m m 2 = 0.084  -cm  10 -4 cm/ 10 -10 cm 2 = 8.4  10 -4  Slide 2-16 8

EXAMPLE: Temperature Dependence of Resistance By what factor will R increase or decrease from T=300 K to T=400 K? Solution: The temperature dependent factor in  (and therefore  ) is m n . From the mobility vs. temperature curve for 10 17 cm -3 , we find that m n decreases from 770 at 300K to 400 at 400K. As a result, R increases by 770 = 1 . 93 400 Slide 2-17 2.3 Diffusion Current Particles diffuse from a higher-concentration location to a lower-concentration location. Slide 2-18 9

2.3 Diffusion Current dn dp = = - J qD J qD n , diffusion n p , diffusion p dx dx D is called the diffusion constant. Signs explained: n p x x Slide 2-19 Total Current – Review of Four Current Components J TOTAL = J n + J p dn J n = J n,drift + J n,diffusion = qn m n E + qD n dx dp J p = J p,drift + J p,diffusion = qp m p E – qD p dx Slide 2-20 10

2.4 Relation Between the Energy Diagram and V, E ( x ) V 0.7eV 0.7V x – + 0 N- N type Si E c and E v vary in the opposite E Ec (x) direction from the voltage. That - Ef (x) is, E c and E v are higher where the voltage is lower. Ev (x) dV 1 dE 1 dE c = 0.7V - = v E (x)= + dx q dx q dx x Slide 2-21 2.5 Einstein Relationship between D and m Consider a piece of non-uniformly doped semiconductor. - - ( E E ) / kT = n N e N-type semiconductor c f n-type semiconductor c dn N dE - c - ( E E ) / kT = - c e c f Decreasing donor concentration dx kT dx n dE E c (x) = - c kT dx E f n = - q E kT E v (x) Slide 2-22 11

2.5 Einstein Relationship between D and m dn n = - q E dx kT dn = + = m J qn qD 0 at equilibrium. E n n n dx qD = - m n 0 qn qn E E n kT kT kT = m = m D Similarly, D n n p p q q These are known as the Einstein relationship. Slide 2-23 EXAMPLE: Diffusion Constant What is the hole diffusion constant in a piece of silicon with m p = 410 cm 2 V -1 s -1 ? Solution:   kT =   m =  - - = 2 1 1 2 D ( 26 mV ) 410 cm V s 1 1 cm /s   p p q   Remember: kT/q = 26 mV at room temperature. Slide 2-24 12

2.6 Electron-Hole Recombination •The equilibrium carrier concentrations are denoted with n 0 and p 0 . •The total electron and hole concentrations can be different from n 0 and p 0 . • T he differences are called the excess carrier concentrations n’ and p’.  + n n n ' 0  + p p p ' 0 Slide 2-25 Charge Neutrality •Charge neutrality is satisfied at equilibrium ( n’= p’= 0) . • When a non-zero n’ is present, an equal p’ may be assumed to be present to maintain charge equality and vice-versa. •If charge neutrality is not satisfied, the net charge will attract or repel the (majority) carriers through the drift current until neutrality is restored. n = ' p ' Slide 2-26 13

Recombination Lifetime •Assume light generates n’ and p’. If the light is suddenly turned off , n’ and p’ decay with time until they become zero. •The process of decay is called recombination. •The time constant of decay is the recombination time or carrier lifetime, t . •Recombination is nature’s way of restoring equilibrium ( n’= p’= 0) . Slide 2-27 Recombination Lifetime • t ranges from 1ns to 1ms in Si and depends on the density of metal impurities (contaminants) such as Au and Pt. •These deep traps capture electrons and holes to facilitate recombination and are called recombination centers. E c Direct Recombination Recombination is unfavorable in centers silicon E v Slide 2-28 14

Direct and Indirect Band Gap Trap Indirect band gap Direct band gap Example: Si Example: GaAs Direct recombination is rare as k Direct recombination is efficient conservation is not satisfied as k conservation is satisfied. Rate of recombination (s -1 cm -3 )   d n n = - t dt   = n p     d n n p d p = - = - = t t dt dt Slide 2-30 15

EXAMPLE: Photoconductors A bar of Si is doped with boron at 10 15 cm -3 . It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of 10 20 /s·cm 3 . The recombination lifetime is 10 m s. What are (a) p 0 , (b) n 0 , (c) p’, (d) n’, (e) p , (f) n, and (g) the np product? Slide 2-31 EXAMPLE: Photoconductors Solution: (a) What is p 0 ? p 0 = N a = 10 15 cm -3 (b) What is n 0 ? n 0 = n i2 /p 0 = 10 5 cm -3 (c) What is p ’ ? In steady-state, the rate of generation is equal to the rate of recombination. 10 20 /s-cm 3 = p’/ t  p’= 10 20 /s-cm 3 · 10 -5 s = 10 15 cm -3 Slide 2-32 16