Chapter 2 Motion and Recombination of Electrons and

Chapter 2 Motion and Recombination of Electrons and

Chapter 2 Motion and Recombination of Electrons and Holes 2.1 Thermal Motion 3 2 1 2 Average electron or hole kinetic energy kT mvth2 3kT 3 1.38 10 23 JK 1 300K vth meff 0.26 9.110 31 kg 2.3 105 m/s 2.3 107 cm/s Modern Semiconductor Devices for Integrated Circuits (C. Hu) 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 m ~ 0.1ps Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-2 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) Modern Semiconductor Devices for Integrated Circuits (C. Hu)

Slide 2-3 2.2 Drift 2.2.1 Electron and Hole Mobilities Drift is the motion caused by an electric field. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-4 2.2.1 Electron and Hole Mobilities m p v qE mp v v pE p q mp mp qE mp mp

v nE q mn n mn p is the hole mobility and n is the electron mobility Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-5 2.2.1 Electron and Hole Mobilities 2 v = E ; has the dimensions of v/E cm/s cm . V/cm V s Electron and hole mobilities of selected semiconductors

Si Ge GaAs InAs n (cm2/Vs) 1400 3900 8500 30000 p (cm2/Vs) 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? Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-6 Drift Velocity, Mean Free Time, Mean Free Path EXAMPLE: Given p = 470 cm2/Vs, what is the hole drift velocity at E= 103 V/cm? What is 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: = pE = 470 cm2/Vs 103 V/cm = 4.7 105 cm/s mp = pmp/q =470 cm2/V s 0.39 9.110-31 kg/1.610-19 C = 0.047 m2/V s 2.210-12 kg/C = 110-13s = 0.1 ps mean free path = mhth ~ 1 10-13 s 2.2107 cm/s = 2.210-6 cm = 220 = 22 nm

This is smaller than the typical dimensions of devices, but getting close. Modern Semiconductor Devices for Integrated Circuits (C. Hu) 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: phonon phonon 1 1 3/ 2 T phonon density carrier thermal velocity T T 1 / 2 = q/m T

Modern Semiconductor Devices for Integrated Circuits (C. Hu) vth T1/2 Slide 2-8 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. impurity 3 th 3/ 2 v T Na Nd Na Nd Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-9 Total Mobility

1600 1 1 1 phonon impurity 1400 Electrons 1 1 1 phonon impurity 1000 2

-1 -1 M obility (cmV s ) 1200 800 600 400 Holes 200 0 1E14 1E15 1E16

1E17 1E18 1E19 1E20 Na +Concenration Nd (cm-3) Total Impurity (atoms cm ) -3 Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-10 Temperature Effect on Mobility 10 1 5 Question:

What Nd will make dn/dT = 0 at room temperature? Modern Semiconductor Devices for Integrated Circuits (C. Hu) 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, vsat . Velocity saturation has a deleterious effect on device speed as shown in Ch. 6. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-12 2.2.3 Drift Current and Conductivity E

unit area + + Hole current density Jp Jp = qpv A/cm2 or C/cm2sec EXAMPLE: If p = 1015cm-3 and v = 104 cm/s, then Jp= 1.610-19C 1015cm-3 104cm/s 2 2 1 . 6

C/s cm 1.6 A/cm = Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-13 2.2.3 Drift Current and Conductivity Jp,drift = qpv = qppE Jn,drift = qnv = qnnE Jdrift = Jn,drift + Jp,drift = E =(qnn+qpp)E conductivity (1/ohm-cm) of a semiconductor is = qnn + qpp 1/ = is resistivity (ohm-cm) Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-14

DOPANT DENSITY cm-3 Relationship between Resistivity and Dopant Density P-type N-type RESISTIVITY ( cm) = 1/ Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-15 EXAMPLE: Temperature Dependence of Resistance (a) What is the resistivity () of silicon doped with 1017cm-3 of arsenic? (b) What is the resistance (R) of a piece of this silicon material 1m long and 0.1 m2 in crosssectional 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 / 0.1 m2 = 0.084 -cm 10-4 cm/ 10-10 cm2 = 8.4 10-4 Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-16 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 n. From the mobility vs. temperature curve for 1017cm-3, we find that n decreases from 770 at 300K to 400 at 400K. As a result, R increases by 770 1.93 400 Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-17

2.3 Diffusion Current Particles diffuse from a higher-concentration location to a lower-concentration location. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-18 2.3 Diffusion Current J n ,diffusion dn qDn dx J p ,diffusion dp qD p dx D is called the diffusion constant. Signs explained:

p n x Modern Semiconductor Devices for Integrated Circuits (C. Hu) x Slide 2-19 Total Current Review of Four Current Components JTOTAL = Jn + Jp Jn = Jn,drift + Jn,diffusion = qnnE Jp = Jp,drift + Jp,diffusion = qppE + dn qDn dx

dp qD p dx Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-20 E 2.4 Relation Between the Energy + Diagram and V, E 0.7eV Si (a )

V (x ) 0 .7 V + N- N type Si Ec and Ev vary in the opposite direction from the voltage. That is, Ec and Ev are higher where 1 dEc 1 dEv the voltagedV is lower. E(x)= dx q dx q dx

x 0 (b ) E - E - E c(x ) E cE (x) ( ) f x E f (x) E v(x ) E v (x) 0 .7 V

+ x x Modern Semiconductor Devices for Integrated Circuits (C. Hu) + 0.7V (c ) Slide 2-21 E 2.5 Einstein Relationship between D and f E c( x ) Consider a piece of non-uniformly doped semiconductor.

n N ce N n- -ttyy pp ee s es me mi c o inc d ou cn t do ru c t o r D e c r e a s in g d o n o r c o n c e n t r a ti o n ( Ec E f ) / kT dn N ( E E ) / kT dEc c e c f dx kT dx n dEc Ec(x) kT dx Ef Ev(x)

n qE kT Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-22 2.5 Einstein Relationship between D and dn n qE dx kT dn 0 at equilibrium. dx qD n E 0 qn E qn

n kT Jn qn nE qD n kT Dn n q kT p Similarly, Dp q These are known as the Einstein relationship. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-23 EXAMPLE: Diffusion Constant What is the hole diffusion constant in a piece of silicon with p = 410 cm2 V-1s-1 ? Solution:

kT p (26 mV) 410 cm 2 V 1s 1 11 cm 2 /s D p q Remember: kT/q = 26 mV at room temperature. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-24 2.6 Electron-Hole Recombination The equilibrium carrier concentrations are denoted with n0 and p0. The total electron and hole concentrations can be different from n0 and p0 . The differences are called the excess carrier concentrations n and p. n n0 n' p p0 p ' Modern Semiconductor Devices for Integrated Circuits (C. Hu)

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' Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-26 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, . Recombination is natures way of restoring equilibrium (n= p= 0). Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-27 Recombination Lifetime 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. Ec Direct Recombination is unfavorable in silicon Recombination

centers Ev Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-28 Direct and Indirect Band Gap Trap Direct band gap Example: GaAs Indirect band gap Example: Si Direct recombination is efficient as k conservation is satisfied. Direct recombination is rare as k conservation is not satisfied

Modern Semiconductor Devices for Integrated Circuits (C. Hu) Rate of recombination (s-1cm-3) dn n dt n p dn n p dp dt dt Modern Semiconductor Devices for Integrated Circuits (C. Hu)

Slide 2-30 EXAMPLE: Photoconductors A bar of Si is doped with boron at 1015cm-3. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of 1020/scm3. The recombination lifetime is 10s. What are (a) p0 , (b) n0 , (c) p, (d) n, (e) p , (f) n, and (g) the np product? Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-31 EXAMPLE: Photoconductors Solution: (a) What is p0? p0 = Na = 1015 cm-3 (b) What is n0 ? n0 = ni2/p0 = 105 cm-3 (c) What is p? In steady-state, the rate of generation is equal to the

rate of recombination. 1020/s-cm3 = p/ p= 1020/s-cm3 10-5s = 1015 cm-3 Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-32 EXAMPLE: Photoconductors (d) What is n? n= p= 1015 cm-3 (e) What is p? p = p0 + p= 1015cm-3 + 1015cm-3 = 21015cm-3 (f) What is n? n = n0 + n= 105cm-3 + 1015cm-3 ~ 1015cm-3 since n0 << n (g) What is np? np ~ 21015cm-3 1015cm-3 = 21030 cm-6 >> ni2 = 1020 cm-6. The np product can be very different from ni2. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-33 2.7 Thermal Generation If n is negative, there are fewer

electrons than the equilibrium value. As a result, there is a net rate of thermal generation at the rate of |n|/ . Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-34 2.8 Quasi-equilibrium and Quasi-Fermi Levels Whenever n = p 0, np ni2. We would like to preserve and use the simple relations: n N c e ( Ec E f ) / kT p N v e ( E f Ev ) / kT But these equations lead to np = ni2. The solution is to introduce two quasi-Fermi levels Efn and Efp such that ( Ec E fn ) / kT n N c e

p N v e ( E fp Ev ) / kT Even when electrons and holes are not at equilibrium, within each group the carriers can be at equilibrium. Electrons are closely linked to other electrons but only loosely to holes. Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-35 EXAMPLE: Quasi-Fermi Levels and Low-Level Injection Consider a Si sample with Nd=1017cm-3 and n=p=1015cm-3. (a) Find Ef . n = Nd = 1017 cm-3 = Ncexp[(Ec Ef)/kT] Ec Ef = 0.15 eV. (Ef is below Ec by 0.15 eV.) Note: n and p are much less than the majority carrier concentration. This condition is called low-level injection. Modern Semiconductor Devices for Integrated Circuits (C. Hu)

Slide 2-36 EXAMPLE: Quasi-Fermi Levels and Low-Level Injection Now assume n = p = 1015 cm-3. (b) Find Efn and Efp . n = 1.0110 cm = 17 -3 Nce ( Ec E fn ) / kT EcEfn = kT ln(Nc/1.011017cm-3) = 26 meV ln(2.81019cm-3/1.011017cm-3) = 0.15 eV Efn is nearly identical to Ef because n n0 . Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-37

EXAMPLE: Quasi-Fermi Levels p = 10 cm = 15 -3 Nve ( E fp Ev ) / kT EfpEv = kT ln(Nv/1015cm-3) = 26 meV ln(1.041019cm-3/1015cm-3) = 0.24 eV Ec Ef Efn Efp Ev Modern Semiconductor Devices for Integrated Circuits (C. Hu)

Slide 2-38 2.9 Chapter Summary vp pE vn - nE J p ,drift qp pE Jn ,drift qn nE dn dx dp qD p dx J n ,diffusion qDn J p ,diffusion kT Dn n q kT Dp p

q Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-39 2.9 Chapter Summary is the recombination lifetime. n and p are the excess carrier concentrations. n = n0+ n p = p0+ p Charge neutrality requires n= p. rate of recombination = n/ = p/ Efn and Efp are the quasi-Fermi levels of electrons and ( E E ) / kT holes. n N c e c p N v e fn

( E fp Ev ) / kT Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-40

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