EE 471: Transport Phenomena in Solid State Devices Spring 2018 Lecture 3 Transport in Semiconductors Bryan Ackland Department of Electrical and Computer Engineering Stevens Institute of Technology Hoboken, NJ 07030 Adapted from Modern Semiconductor Devices for Integrated Circuits, Chenming Hu, 2010 1
Thermal Motion Carriers (conduction electrons, holes) have a finite kinetic energy which is a function of temperature For conduction electrons, kinetic energy = (E E c ) aaaaaaaaaaaaaa eeeeeeeeeeeeeeee kkkkkkkkkkkkkk eeeeeeeeeeee = tttttttttt kkkkkkkkkkkkkk eeeeeeeeeeee nnnnnnnnnnnn oooo eeeeeeeeeeeeeeeeee = ff EE. DD EE. EE EE cc dddd ff EE. DD EE dddd Substituting and using Boltzmann approximation: aaaaaaaaaaaaaa kkkkkkkkkkkkkk eeeeeeeeeeee = 3 2. kkkk Same result for holes Statistical mechanics: kt/2 per degree of freedom 2
Thermal Velocity 1 2 mm nn. vv ttt 2 = 3 2. kkkk where m n is effective electron mass, v th is thermal velocity vv ttt = 3kkkk mm nn At 300 K, and using m n = 0.26 m 0 vv ttt = 3 1.38 10 23 J/K 300K 0.26 9.1 10 31 kg = 2.3 10 5 m/sec = 2.3 10 7 cm/sec 670 times speed of sound in air or 1/1000 speed of light 3
Scattering Electrons and holes move in a zig-zag pattern due to collisions or scattering of atoms in the crystal At thermal equilibrium, net thermal velocity is zero does not generate DC current, but does generate noise mean free time between collisions 0.1 ps at 300 K mean distance between collisions 25 nm about 50x lattice constant 4
Non-equilibrium conditions Hot-point probe: distinguishes N and P-type semiconductor Thermo-electric generator : converts heat to electric power Or a thermo-electric cooler: solid-state refrigerator 5
Drift EE Drift is the motion of charged carriers in presence of an electric field Drift is superimposed on thermal motion Drift velocity vv dddddddddd is average net velocity due to electric field 6
Drift Velocity Suppose we have a block of semiconductor that contains NN pp holes under the influence of an electric field EE. each hole is accelerated by electric field until it collides with the lattice assume each hole loses its entire drift momentum at each collision Total drift momentum being added to all the holes by the electric field in time tt is given by: pp EE = NN pp. EE. qq. tt If ττ mmmm is the mean free time between collisions, then there will be: ( NN pp. tt) ττ mmmm collisions in time tt Total drift momentum lost to collisions is given by: pp CC = NN pp. tt. (mmmmmmmm hoooooo dddddddddd mmmmmmmmmmmmmmmm) ττ mmmm = NN pp. tt. mm pp. vv pp,dddddddddd ττ mmmm where vv pp,dddddddddd is the mean hole drift velocity 7
Drift Velocity and Mobility In steady state, pp = pp EE + pp CC =0 which means: NN pp. tt. mm pp. vv pp,dddddddddd = NN ττ pp. qq. EE. tt vv pp,dddddddddd = qq. EE. ττ mmmm mmmm mm pp vv pp,dddddddddd = μμ pp. EE μμ pp = qq. ττ mmmm mm pp μμ pp is the hole mobility vv nn,dddddddddd = μμ nn. EE μμ nn = qq. ττ mmnn mm nn μμ nn is the electron mobility 8
Electron & Hole Mobilities vv = μμ. EE: µ has the dimensions of vv/ee cccc ss VV/cccc cccc2 VV. ss Electron & hole mobilities, room temperature, lightly doped: Si Ge GaAs InAs µ n (cm 2 /V s) 1400 3900 8500 30000 µ p (cm 2 /V s) 470 1900 400 500 Mobility describes ability of carrier to respond to EE Higher mobility leads to higher speed devices 9
Example: Drift Velocity Given µ p = 470 cm 2 /V.s for Si, what is the hole drift velocity at EE = 10 3 V/cm? What is the mean free path? 10
Effective Mass of Electron (Hole)??? We have used the term effective mass of electrons and holes (mm nn, mm pp ) in both calculation of density of quantum states and in calculation of mobility. Effective mass has different value in these two formulae because particle is moving through 3-D lattice and effective mass is single term that tries to capture effects of lattice in all dimensions 1/3 in density of states: mm dd = KK 1. mm xx. mm yy. mm zz 1 in mobility: mm bb = KK 2. 1 mm xx + 1 mm yy + 1 mm zz Density of States Mobility mm nn mm 0 1.08 0.26 mm pp mm 0 0.56 0.386 where mm 0 = 9.11 10 31 kkkk 11
Phonon Scattering Mobility depends on τ mn and τ mp What determines mean time between scattering (collisions)? Electrons & holes are scattered by non-uniformities in crystal lattice. Two main causes: Phonon scattering Ionized impurity scattering Phonons are (quantum mechanical) particle representation of thermal vibration in the crystal lattice 1 ττ ppp ppppppppppp dddddddddddddd cccccccccccccc ttttttttt. vvvvvvvvvvvvvvvv 1 TT TT 1/2 so μμ ppp ττ ppp TT 3/2 Phonon-limited mobility decreases with temperature 12
Ionized Impurity Scattering Electrons and holes can be scattered by donor (positive) ion or acceptor (negative) ion: Less change in direction if electron is travelling at higher speed μμ iiiiii ττ iiiiii (cccccccccccccc ttttttttt vvvvvvvvvvvvvvvv)3 iiiiiiiiiiiiiiii dddddddddddddd TT3/2 NN aa + NN dd Impurity-limited mobility increases with temperature 13
Electron & Hole Mobility electrons 1 μμ = 1 + 1 μμ ppp μμ iiiiii holes Holes have about 1/3 mobility of electrons At low dopant concentration, mobility dominated by phonon scattering At high concentration, mobility further reduced by impurity scattering 14
Mobility vs. Temperature At high dopant concentrations and low temperature, mobility is dominated by impurity ion scattering 15
Velocity Saturation Under low electric fields, drift velocity << thermal velocity Under high field conditions, drift velocity adds appreciably to overall kinetic energy of electron (hole) This reduces mean free time between collisions which increases phonon induced scattering (which reduces µ) Drift velocity saturates: For Si at 300 K: v sat 10 7 cm/s at EE 10 5 VV/cccc 16
Carrier Velocity and Drift Current Current density J is charge per second crossing unit area plane normal to direction of current flow (units are A/cm 2 ) JJ pp,dddddddddd = qq. pp. vv pp,dddddddddd Example: if pp = 10 15 /cccc 3 and vv = 10 4 cccc/ss: JJ pp,dddddddddd = 1.6 10 19 10 15 10 4 = 1.6 AA/cccc 2 17
Drift Current and Conductivity JJ pp,dddddddddd = qq. pp. vv pp,dddddddddd = qq. pp. μμ pp. EE JJ nn,dddddddddd = qq. nn. vv nn,dddddddddd = qq. nn. μμ nn. EE JJ dddddddddd = (JJ nn,dddddddddd + JJ pp,dddddddddd ) = (qq. nn. μμ nn + qq. pp. μμ pp ). EE Define conductivity σσ as: JJ dddddddddd = σσ. EE Units of σσ are (ohm.cm) -1 then σσ = 1 ρρ = qq. nn. μμ nn + qq. pp. μμ pp (ρ is resistivity) 18
Ohms Law If we have a bar of semiconductor: JJ dddddddddd = II AA EE = VV LL II AA = σ VV LL i.e. V= II. RR where RR = ρρ. LL AA 19
Resistivity vs. Dopant Density Dopant Density (cm -3 ) N-type P-type ρρ = 1 σσ Resistivity ρ (ohm.cm) 20
Example: Silicon Resistivity a) What is the room temperature resistivity ρ of silicon doped with 10 17 cm -3 of arsenic? b) What is the resistance of a piece of this material that is 1µm long and 0.2 µm 2 in area? c) By what factor will resistance change as we go from 27 C to 127 C? 21
Diffusion x=a All particles are in constant thermal motion Because (density) x<a is greater than (density) x>a, more particles cross x=a from left-to-right than from right-to-left Net particle flow from high to low concentration If particles are charged (carriers) net current 22
Diffusion Current Rate of particle flow proportional to concentration gradient ffffffff pppppp unit area = D. dd(cccccccccccccccccccccccccc) dddd where D is the diffusion constant ffffffff pppppp unit area = cccccccccccccccccccccccccc vvvvvvvvvvvvvvvv vv nn,dddddddd = 1 nn. DD nn. dddd dddd leads to a diffusion current density: JJ nn,dddddddd = qq. DD nn. dddd dddd JJ nn,dddddddd = nn. qq. vv nn.dddddddd JJ pp,dddddddd = qq. DD pp. dddd dddd 23
Diffusion Current JJ nn,dddddddd = qq. DD nn. dddd dddd JJ pp,dddddddd = qq. DD pp. dddd dddd Units of Diffusion Constant (DD nn, DD pp ) are cccc 2 ss 24
Example: Diffusion Current Assume that in a sample of N-type silicon the electron concentration varies linearly from 1 10 18 to 7 10 17 over a distance of 0.1mmmm. Calculate the diffusion current density if DD nn = 7.8 cccc 2 /ss. 25
Total Current in Semiconductor JJ = JJ nn + JJ pp JJ nn = JJ nn,dddddddddd + JJ nn,dddddddd = qq. nn. μμ nn. EE + qq. DD nn. dddd dddd JJ pp = JJ pp,dddddddddd + JJ pp,dddddddd = qq. pp. μμ pp. EE qq. DD pp. dddd dddd Four terms in total Frequently only need to consider one term at any one time at any one point in semiconductor 26
Applied Voltage and Energy Band Diagrams When a voltage is applied to a semiconductor, it tilts the band diagram Positive voltage raises potential of holes, decreases potential of electrons lowers energy bands E c and E v always separated by E g EE cc xx = cccccccccccccccc qq. VV(xx) Remember: Ec and Ev move in opposite direction to applied V EE xx = dddd dddd = 1 qq ddee cc dddd = 1 qq ddee vv dddd 27
Graded Impurity Distribution Consider a bar of N-type silicon in thermal equilibrium, more heavily doped on left side: Electrons will diffuse from left to right (diffusion current right to left) This moves negative charge (electrons) to RHS Creates electric field from left to right which induces drift current that draws electrons back to LHS Electrons diffuse until drift current balances diffusion current N decreasing donor concentration E c (x) E F E v (x) Note that band diagram is consistent with: difference in carrier density (as seen in E c -E F ) presence of electric field EE = 1 ddee cc qq dddd 28
If bar is in thermal equilibrium: remember so Einstein Relationship JJ nn = 0 = qq. nn. μμ nn. EE + qq. DD nn. dddd dddd (EE nn = NN cc. ee cc EE FF ) kkkk dddd dddd = NN cc kkkk ee (EE cc EE FF kkkk). ddee cc dddd = nn kkkk. ddee cc = nn. qq. EE dddd kkkk qq. nn 0 = qq. nn. μμ nn. EE qq. DD nn. kkkk. EE DD nn = kkkk qq. μμ nn DD pp = kkkk qq. μμ pp 29
Example: Diffusion Constant A piece of silicon is doped with 3 10 15 cccc 3 of donors and 7 10 15 cccc 3 of acceptors. What are the electron and hole diffusion constants at 300 K? 30
Example: Potential due to density gradient Consider a n-type semiconductor at T=300 K in thermal equilibrium. Assume that the donor concentration varies as xx NN dd xx = NN ddd. ee LL over the range 0 xx 5LL where NN ddd = 10 16 cccc 3 and LL = 10μμμμ. a) Determine the electric field as a function of x for 0 xx LL. b) Calculate the potential difference between x=0 and x= 25 µm. 31
Example: Electron collision frequency An electron is moving in a piece of very lightly doped silicon at room temperature under an applied field such that its drift velocity is one-tenth of its thermal velocity. Calculate the average number of collisions it will experience in traversing by drift a region 1µm long. 32