Electrical Resistance I + V _ W Material with resistivity ρ t L Resistance R V I = L ρ Wt (Unit: ohms) where ρ is the electrical resistivity 1
Adding parts/billion to parts/thousand of dopants to pure Si can change resistivity by 8 orders of magnitude! Resistivity Range of Materials Si with dopants SiO2, Si3N4 1 Ω-m = 100 Ω-cm 2
The Si Atom The Si Crystal diamond structure High-performance semiconductor devices require defect-free crystals 3
Carrier Concentrations of Intrinsic (undoped) Si electron - Bottom of conduction band Energy gap =1.12 ev hole + Top of valence band n (electron conc) = p (hole conc) = n i 4
Purity of Device-Grade Si wafer 99.999999999 % (so-called eleven nines )!! Maximum impurity allowed is equivalent to 1 mg of sugar dissolved in an Olympic-size swimming pool. 5
Dopants in Si By substituting a Si atom with a special impurity atom (Column V or Column III element), a conduction electron or hole is created. Donors: P, As, Sb Acceptors: B, Al, Ga, In 6
Energy Band Description of Electrons and Holes Contributed by Donors and Acceptors E C = bottom of conduction band E V = top of valence band E D = Donor energy level E A = Acceptor energy level At room temperature, the dopants of interest are essentially fully ionized Donors Acceptors 7
Semiconductor with both acceptors and donors has 4 kinds of charge carriers Hole Electron Mobile; they contribute to current flow with electric field is applied. Ionized Donor Ionized Acceptor Immobile ; they DO NOT contribute to current flow with electric field is applied. However, they affect the local electric field 8
Charge Neutrality Condition Even N A is not equal to N D, microscopic volume surrounding any position x has zero net charge Valid for homogeneously doped semiconductor at thermal equilibrium Si atom Ionized Donor Ionized Acceptor Hole Electron electron-hole pair due to transition from valence band to conduction band 9
How to Calculate Electron and Hole Concentrations for homogeneous Semiconductor n: electron concentration (cm -3 ) p : hole concentration (cm -3 ) N D : donor concentration (cm -3 ) N A : acceptor concentration (cm -3 ) Assume completely ionized 1) Charge neutrality condition: N D + p = N A + n 2) At thermal equilibrium, np = n i 2 ( Law of Mass Action ) Note: Carrier concentrations depend on NET dopant concentration (N D - N A )! 10
N-type and P-type Material If N D >> N A (so that N D N A >> n i ): n N D N A and p N D n 2 i n >> p material is n-type N A If N A >> N D (so that N A N D >> n i ): p N A N D and n N A n 2 i p >> n material is p-type N D 11
Carrier Drift When an electric field is applied to a semiconductor, mobile carriers will be accelerated by the electrostatic force. This force superimposes on the random thermal motion of carriers: E.g. Electrons drift in the direction opposite to the E-field Current flows 3 2 2 1 3 1 4 electron 4 electron 5 5 E=0 E Average drift velocity = v = µ E Carrier mobility 12
Carrier Mobility Mobile carriers are always in random thermal motion. If no electric field is applied, the average current in any direction is zero. Mobility is reduced by Si collisions with the vibrating atoms - phonon scattering - deflection by ionized impurity atoms - B- - As+ 13
Carrier Mobility µ Mobile charge-carrier drift velocity is proportional to applied E-field: v = µ E µ n Mobility depends on (N D + N A )! (Unit: cm 2 /V s) µ p 14
Electrical Conductivity σ When an electric field is applied, current flows due to drift of mobile electrons and holes: electron current J n = ( q) nvn = qnµ ne density: hole current density: total current density: conductivity J J J σ = ( + q) pv = qp µ p p p n = σe qnµ n p p n E = J + J = ( qnµ + qpµ ) E + qpµ p 15
Electrical Resistivity ρ ρ 1 σ = 1 qn µ + qpµ n p ρ 1 qnµ n for n-type ρ 1 qpµ p for p-type (Unit: ohm-cm) Note: This plot does not apply for compensated material! 16
Consider a Si sample doped with 10 16 /cm 3 Boron. What is its electrical resistivity? Answer: Example Calculation N A = 10 16 /cm 3, N D = 0 (N A >> N D p-type) p 10 16 /cm 3 and n 10 4 /cm 3 ρ = = 1 1 qn µ + qpµ qpµ n p [ ] 19 16 1 (1.6 10 )(10 )(450) = 1.4 Ω cm p From µ vs. ( N A + N D ) plot 17
Example: Dopant Compensation Consider the same Si sample (with 10 16 /cm 3 Boron), doped additionally with 10 17 /cm 3 Arsenic. What is the new resistivity? Answer: N A = 10 16 /cm 3, N D = 10 17 /cm 3 (N D >>N A n-type) = * The sample is converted to n-type material by adding more donors than acceptors, and is said to be compensated. n 9x10 16 /cm 3 and p 1.1x10 3 /cm 3 ρ = 1 1 qn µ + qpµ qnµ n p [ ] 19 16 1 (1.6 10 )(9 10 )(600) = 0.12 Ω cm n 18
Summary of Doping Terminology intrinsic semiconductor: undoped semiconductor extrinsic semiconductor: doped semiconductor donor: impurity atom that increases the electron concentration group V elements (P, As)in Si acceptor: impurity atom that increases the hole concentration group III elements (B, In) in Si n-type material: semiconductor containing more electrons than holes p-type material: semiconductor containing more holes than electrons majority carrier: the most abundant mobile carrier in a semiconductor minority carrier: the least abundant mobile carrier in a semiconductor mobile carriers: Charge carriers that contribute to current flow when electric field is applied. 19
Sheet Resistance R S R L = ρ Wt R s is the resistance when W = L R s ρ t The R s value for a given layer (e.g. doped Si, metals) in an IC or MEMS technology is used for design and layout of resistors for estimating values of parasitic resistance in a device or circuit = R s L W (unit in ohms/square) if ρ is independent of depth x 20
R S when ρ(x) is function of depth x I + V _ ρ 1, dx ρ 2, dx ρ 3, dx. ρ n, dx 1 R S = W dx ρ 1 + dx ρ 2 + dx ρ 3 + L... + dx ρ n = ( σ 1 + σ 2 + t.. σ n depth x )dx For a continuous σ(x) function: R S = = t 0 1 σ ( x) dx 1 t [ qµ n( x) n( x) + qµ p( x) p( x) ] 0 dx 21
Electrical Resistance of Layout Patterns (Unit of R S : ohms/square) Metal contact Top View W = 1µm 1m R = R s L=1µm R = R s 1m R 2.6R s R = 3R s R = R s /2 R = 2R s 22
How to measure R S? (Typically, s 1 mm >>t) The Four-Point Probe is used to measure R s 4 probes are arranged in-line with equal spacing s 2 outer probes used to flow current I through the sample 2 inner probes are used to sense the resultant voltage drop V with a voltmeter For a thin layer (t s/2), 4. 532V R s = I For derivation, see EE143 Lab Manual http://www-inst.eecs.berkeley.edu/~ee143/fa05/lab/four_point_probe.pdf If ρ is known, then R s measurement can be used to determine t 23
Electron mobility vs. T For reference only Hole mobility vs. T 24