Lecture 4, Sept. 6, 2007

Professor Ali Javey 9/06/2007

Semiconductor Fundamentals Lecture 4 Reading: Chapter 3 Announcements

• 1st HW due Tuesday. • New lecture room: 3106 Etcheverry, starting next Thursday (9/13) • TA OHs will be held in Cory 353 What to memorize?

• Learn the concepts • Understand the equations • Only memorize the most important and fundamental equations Drift

Electron and Hole Mobilities

• Drift is the motion caused by an electric field.

Effective Mass

In an electric field, , an electron or a hole accelerates.

electrons Remember : F=ma=-qE holes

Electron and hole effective masses Si Ge GaAs GaP

m n /m 0 0.26 0.12 0.068 0.82

m p /m 0 0.39 0.30 0.50 0.60 Remember : F=ma=mV/t = -qE Electron and Hole Mobilities

mpv = q τ mp

q τ v = mp mp

= μ v p v = −μn qτ mp qτmn μ p = μn = mp mn

• μp is the hole mobility and μn is the electron mobility Electron and Hole Mobilities

v =μ ; μ has the dimensions of v/ ⎡ cm/s cm2 ⎤ ⎢ = ⎥ . ⎣V/cm ⋅sV ⎦

Electron and hole mobilities of selected semiconductors Si Ge GaAs InAs 2 μ n (cm /V·s) 1400 3900 8500 30000 2 μ p (cm /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? Drift Velocity, Mean Free Time, Mean Free Path

2 EXAMPLE: Given μp = 470 cm /V·s, what is the hole drift velocity at 3 = 10 V/cm? What is τmp and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units. 2 3 5 Solution: ν = μp = 470 cm /V·s × 10 V/cm = 4.7× 10 cm/s 2 -31 -19 τmp = μp mp /q =470 cm /V ·s × 0.39 × 9.1×10 kg/1.6 ×10 C = 0.047 m2/V ·s × 2.2×10-12 kg/C = 1×10-13s = 0.1 ps

-13 7 mean free path = τmhνth ~ 1× 10 s × 2.2×10 cm/s = 2.2×10-6 cm = 220 Å = 22 nm This is smaller than the typical dimensions of devices, but getting close. Mechanisms of Carrier Scattering

There are two main causes of carrier scattering: 1. Phonon Scattering 2. Impurity (Dopant) Ion Scattering

Phonon scattering mobility decreases when temperature rises: 1 1 τμ ∝∝ ∝ ∝ T − 2/3 phonon phonon phonon density × carrier thermal ×TTvelocity 2/1

μ = qτ/m ∝ T v ∝ T1/2 th

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 2/3 vth T μ impurity ∝ ∝ + NN da + NN da Total Mobility 1600 111 1400 += τττ 1200 Electrons phonon impurity

) 111 -1 1000 += s

-1 phonon μμμimpurity

V 800 2

600

400 Holes Mobility (cm 200

1E14 1E15 1E16 1E17 1E18 1E19 1E20 -3 -3 Total ImpurityNa + Concenration Nd (cm ) (atoms cm ) Temperature Effect on Mobility

10 15

Question:

What Nd will make dμn /dT = 0 at room temperature? 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 and the velocity does not rise above a saturation velocity, vsat , no matter how large is.

Velocity saturation has a deleterious effect on device speed as we will see in the later chapters. Drift Current and Conductivity

unit area

2 2 Current density Jp = qpv A/cm or C/cm ·sec

EXAMPLE: If p = 1015cm-3 and v = 104 cm/s, then -19 15 -3 4 Jp = 1.6×10 C × 10 cm × 10 cm/s 2 2 = 6.1 =⋅ 1.6cmC/s A/cm Drift Current and Conductivity

unit area

+ -

Remember: • Holes travel in the direction of the Electric field • Electrons travel in the direction opposite to that of the E-field Drift Current and Conductivity

Jp,drift = qpv = qpμp

Jn,drift = –qnv = qnμn

Jdrift = Jn,drift + Jp,drift = (qnμn +qpμp) σ

∴ conductivity of a semiconductor is σ = qnμn + qpμp ∴ resistivity of a semiconductor is ρ = 1/σ Diffusion Current

Particles diffuse from a higher-concentration location to a lower-concentration location. Diffusion Current dn dp J = qD J −= qD ,diffusionn n dx ,diffusionp p dx D is called the diffusion constant. Signs explained: n p

x x Total Current – Review of Four Current Components

JTOTAL = Jn + Jp

dn J = J + J = qnμ + qD n n,drift n,diff n n dx

dp J = J + J = qpμ – qD p p,drift p,diff p p dx

JTOTAL = Jn,drift + Jn,diff + Jp,drift + Jp,diff Ereference V Ec and Ev vary in the opposite direction from the voltage. That x E is, Ec and Ev are higher where c the voltage is lower. E Ev E -E = -qV c reference x

Variation in Ec with position is called band bending

Einstein Relationship between D and μ Consider a piece of non-uniformly doped semiconductor.

−− fc kT/)EE( N-type semiconductor n-type semiconductor = ceNn

dn Nc −− fc kT/)EE( dEc Decreasing donor concentration −= e dx kT dx

E (x) n dE c −= c kT dx Ef n = − q kT Ev (x) Einstein Relationship between D and μ dn n = − q dx kT

dn J = qnμ + qD = 0 at equilibrium. n n n dx qD 0 = qnμ − qn n n kT kT kT D = μ Similarly, D = μ n q n p q p

These are known as the Einstein relationship. EXAMPLE: Diffusion Constant

What is the hole diffusion constant in a piece of 2 -1 -1 silicon with μp = 410 cm V s ?

Solution:

⎛ kT ⎞ −− 112 2 Dp = ⎜ ⎟μ p 26( ⋅= 410)mV = 11sVcm /scm ⎝ q ⎠

Remember: kT/q = 26 mV at room temperature.