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Unformatted text preview: MatSE 361 4 RCSize Material NOTES Old Chapter B: Resistance and Capacitance 1
MatSE 361 Chapter B Resistance and Capacitance OUTLINE 2
MatSE 361 Chapter B Resistance and Capacitance Introduction: Current Flow and Capacitance
Current flow from point A to point B: Seems fairly straightforward: Electrical measurements are critical for the testing performance and material characterization of electronic devices. ISSUES: Contacts: MetalSemiconductor Junction, Size effects: Width, thickness 3
MatSE 361 Chapter B Resistance and Capacitance Electric Field, Voltages and Energies Electrical Force "F"on a charge "q" by Electric field "E" F = q*E External electrical field is induced by an external voltage "V" E = dV/dx More comprehensive: Gradients in electrochemical potential or fermi level provides the net force for movement of the charges.
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MatSE 361 Chapter B Resistance and Capacitance Mobility and Conductivity Energy and velocity of electrons (gas, not a metal) 1/2mv2th = 3/2 kT (kT at RT =0.025 eV) Under an electric field "E" the electrons accelerate "a" and acquire a "drift" velocity "vd": vd a vd = (e/m*) E Materials characteristic: = (e/m*) = n e = 1/ Materials characteristic: "n" is the number of carriers
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MatSE 361 Chapter B Resistance and Capacitance Drift velocity calculations calculations Derivation on Board
Comparison between values of drift velocity, thermal velocity, and Fermi velocity for metal and Si. 6
MatSE 361 Chapter B Resistance and Capacitance Drift Vd, Mean Free Path (o), Mean Free Time () Derivation on Board
Calculation of mean path and mean time between collisions. 7
MatSE 361 Chapter B Resistance and Capacitance Wafer Characteristics
Characteristics sent by wafer manufacturer. One side polished Gettering on back surface during annealing Cost of polishing 8
MatSE 361 Chapter B Resistance and Capacitance Resistance, Resistivity, and Ohm's Law = collision time, mfp = mean free path 9
MatSE 361 Chapter B Resistance and Capacitance Sheet Resistance = resistivity(cm) L = length(cm) R = resistance () R = ( /t) * (L/W) t = thickness(cm) W = width (cm) For device performance calculations we typically need the value of the resistance "R" between two points in the circuit. We could calculate R by measuring material. However, this is not easy since thickness is small (typically 100's of ) is often unknown. [Can't use book/bulk value] To circumvent this problem a useful semimaterials property parameter called R (Sheet Resistance) is used. R = /t = units of If we have the value of R we can easily obtain R since the lateral/length dimensions are readily available. R is easy to measure (mapping of entire wafer is automatic).
MatSE 361 Chapter B Resistance and Capacitance 10 2pt and 4pt Resistivity Measurements
There several ways to measure the sheet rho. In the final device circuit special patterns are made. However, in asdeposited films where there are no patterns the system is a "blanket layer" semiinfinite sheet of material. We apply probes to this sheet and measure a resistance. 4pt. Method 2pt. Method
+ I V  + I V  11
MatSE 361 Chapter B Resistance and Capacitance 2Point Probe Resistivity Measurement
I V RC1 Sample RS I V RC2 Rmeasured = V/I = RS + RC1 + RC2 No problem if RS RC1 + RC2 then Rmeasured ~ RS Problem: We don't know the values for RC1 + RC2 ~ may be very difficult to obtain values. Source: Oxidation, spreading resistance, metal/semiconductor interface ... General Solution: Use four probes instead of two.
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MatSE 361 Chapter B Resistance and Capacitance 4Point Resistivity Measurement
I V V RC1 Sample RC2 RS1
4pt method RMeasured = V/I = RS2 I RC3 RS2 RS3 RC4 Compare with 2pt method RMeasured = V/I = RS + RC1 + RC2
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MatSE 361 Chapter B Resistance and Capacitance 4point inline sheet rho method
4pt. Method
+ I V  s For the 4Point Probe geometry: R = K * (V/I) where "K" is a geometric factor If probes are (a) inline (b) equally spaced (c) distance "s" between the probes thickness of the sheet then K = ( / ln2) and if the thickness is known then = R * t Advantages: contact resistance are almost eliminated.
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MatSE 361 Chapter B Resistance and Capacitance Contact Resistance
We can obtaining good values R for even though the contact resistance Rc may be high. However, when we operate the device the there can not high values for Rc. Example contacts to source, drain and gate. RAluminum RAl/Silicide contact
Contact resistance can be much larger than the other resistance in the circuit. And since these are series resistances the largest resistance dominates! RSilicide RSilicide/Si contact Causes of high contact resistance? Oxide/insulating interface layers "improper" doping in the silicide/Si junction (Schottky barrier)
MatSE 361 Chapter B Resistance and Capacitance 15 Rc Contact Resistance vs. Doping Source/Drain
Schottky Barrier I
Rc(V=0) = dV/dI V "Ohmic" Contact 16
MatSE 361 Chapter Ion Implantation Contact Resistance 1: TLM method
material (Wolf II page 84) [Reeves & Harrison IEEE Elect. Dev. Lett. Vol. EDL3, (1982)] Contact Resistance  Transmission Line Model (Shockley 1964) metal R
L2 L1 L = distance between pads (cm) Lt = transmission length (cm) Rc = contact resistance () Rsh = sheet rho of semiconductor () c = contact resistivity (cm2) L3 R L1 semicinductor R() w1*Rsh (slope) 2*Rc 2*Lt
L (m)
17 c = Rsh * Lt2 MatSE 361 Chapter B Resistance and Capacitance Contact Resistance 2: TLM (Case Study)
"Low Resistance Ohmic Contacts on GaN,", Appl. Phys. Lett. 64, 1003 (1994) "Ohmic contacts to nGaAs using In/Pd metallization," et al., "Appl. Phys. Lett. 51, 326 (1987) Typical processing of "Ohmic" contact include metallization and thermal annealing of the metal/semiconductor conatcts. 18
MatSE 361 Chapter B Resistance and Capacitance Contact Resistance 2: Kelvin Method RAl/W contact
2 RW
0 RW/Al contact Using this 3dimension 4point method we can measure resistance problems at vias in a structure very close to a realistic circuit. Current is applied to one of the top leads and one bottom leads. Voltage is measured on the remaining top leads and one bottom leads. Rmeas = V/I = Ral/W contact + RW + RW/Al contact 19
MatSE 361 Chapter B Resistance and Capacitance Contact Resistance 3: W vias (IBM case study)
Kelvin contact measurements showed Rmeas 1,000,000 . Origin of problem? W ? Interace? Which interface? SEM shows that W growth was not complete or uniform. Materials Analysis Growth not uniform: starts at pinholes Nucleation limited growth "Bad" interface
Resistance was dominated by an interface made of insulating "ceramic" type compound C, O ,F. Origin of interface? "Dirty" interface before W deposition CVD initial "bad" products
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MatSE 361 Chapter B Resistance and Capacitance Contact Resistance 4: IBM vias (case study)
"Current Crowding at Vias ...", J. Appl. Phys. 70 253 (1991) In estimating the resistance of a 3D circuit it is important from device aspects to include issues associated with actual nature of interconnect. This includes contact resistance sheet resistance and the geometry of the system, in this case current crowding. 21
MatSE 361 Chapter B Resistance and Capacitance Contact Resistance 4: IBM vias (case study)
In industry (IBM) finiteelement methods are often used to give precise estimates of the effective "resistance" of a connection at a via. 22
MatSE 361 Chapter B Resistance and Capacitance Resistance: Size effects (1)
Values in the literature for transport characteristics such as resistivity are generally expressed in terms of their large "bulk" size properties. However, the "characteristic" size of the phenomenon (e.g. MPF) must be interpreted in terms of other "size" constraints: grain boundary, size of the sample (thinfilms) .... Film thickness Mean Free Path Fuchs Theory (1952)
(Tung PRL, 54, 1840 (1985) 3 1 (u  u (1  p)[1  exp( K / u)] =1 du 2 0 1  p exp(k / u)
3 = resistivity = d/MFP p = scattering factor 0 < p < 1 Diffuse Scattering: p=0 inelastic scattering carrier loses coherence Specular Scattering p=1 elastic scattering carrier remains coherent
23 MatSE 361 Chapter B Resistance and Capacitance Resistivity: Size Effects Case Study(2) (Tung) 3 1 (u  u (1  p)[1  exp( K / u)] =1 du 2 0 1  p exp(k / u)
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MatSE 361 Chapter B Resistance and Capacitance Case Study: Jackson Organic LED
H. Klauk, J. R. Huang, J. A. Nichols, T. N. Jackson Abstract Using ionbeam sputtering we have prepared ultrathin transparent metal contacts with large broadband optical transmittance and low electrical sheet resistance. Metal films deposited by ionbeam sputtering have exceptionally small surface roughness, and films as thin as about 20 are continuous and conductive and provide optical transmittance as large as 80 %. Fig 5. Schematic cross section of an Alq / TPD organic light emitting diode with ultrathin ionbeam deposited transparent metal contacts. Fig 6. Photograph of an array of Alq / TPD organic light emitting diodes fabricated on a flexible Mylar PET substrate. 25 Resistivity: Size Effects Case Study(3)  Jackson)
"Ion Beam Deposited ..." Klauk, Jackson, et al, Thin Solid Films 366 272 (2000) Scattering Effects: Surface & Grain Boundary Fig 2. Electrical resistivity of ionbeam deposited nickel as a function of film thickness. Fitted to the experimental data is the curve predicted by Fuchs' theory (below) where o is the bulk resistivity of the metal, lambda is the electron mean free path,and theta is one of the polar coordinates. (1) Surface Flatness & (2) Islanding Discontinuous Films (percolation) 26
MatSE 361 Chapter B Resistance and Capacitance Resistivity: Size Effects Case Study(3) (Reynolds Fig) 27
MatSE 361 Chapter B Resistance and Capacitance Resistivity: Size Effects Case Study(3) (Reynolds Ref) 28
MatSE 361 Chapter B Resistance and Capacitance Nanowires ErSi
Contact Resistance Mean free path Size dependence Wire diameter ~ 3nm Sheet R () Line R (nm)
L 29 Electrical Properties of metals Al  Interconnect metallization  old Si contact material Cu  New interconnect metallization (12 years) GOOD: Conductivity, Electromigration, Thermal POOR: Reactivity High CVD Chemistry W  "plug" via metallization
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MatSE 361 Chapter B Resistance and Capacitance Hall Effect 1
Uniform Configuration Vh Vr I Hall Coefficient qE + q vd B = 0 E =  vd B E = vd B E = Vh/w vd = j/nq j = i / area = i / wt Vh /w = vdB = ( j / n q) B = ( iB / w t n q) Rh = 1 / nq Rh =  Vh t / i B Resitivity R = (Vr / I) (w/L) = t R = (Vr / I) (t w/L) B
w t L Mobility and Carrier concentration h = Rh / =  Vh t / i B n = iB / Vh t q 31
MatSE 361 Chapter B Resistance and Capacitance Hall Effect 2 Drift velocity Lorentz force vx = Ex FB = qvxBy FB = q Ex By Restoring force by EH is established then q Ex By = q EH Mobility By = EH / Ex =VHLx / LZVx type and concentration of carrier Hall Coefficient RH = 1/nq Comments: Size dependence, Strength of field (Al)
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MatSE 361 Chapter B Resistance and Capacitance Mean Free Path (Kittel)
Results from following pages using vF (Fermivelocity  assume the electrons do not act as ideal gas of electrons but as a Fermi gas) " " is obtained by just by conductivity measurements "" and assuming the carrier
concentration "n", and mass "m": MFP = vF * = ne*e /m = ne * e/m 1/RH H MFP of Cu is 420 MFP of Ag is 570 Kittel pg.208 EF = (!2/2m)(32N/V) ! ! mv2F = EF vF = (!2/m)(32N/V) ! MatSE 362 Lectures/Allen 33 Hall Effect: Size Effects Chopra (pg 369) Table: Mean Free path 34 Thermal Heat and Chips 35
MatSE 361 Chapter B Resistance and Capacitance Thermal 1: Thermal Management
The thermal management of microprocessors is becoming increasingly important in design and material selection. All the heat which is input to the chip must be extracted otherwise the temperature of the chip decreases efficiency of switching, reliability (electromigration), stress, ... 20 years ago mainframes (IBM) used BJT Bipolar Junction Transitors  PNP emitter/base/collector. BJTCurrent controlled device needs large amounts of current for switching. Eventually need watercooled packages. Then the MOSFET technology matured all PC's run on this type of transistor. MOSFET voltage controlled devices requiring much less current. PC transistor density has increased. Before (1985)  no/small heat sink, Now (2000)  with fans Soon (2005?)  liquid cooled processors
W/cm2 > 100 W/cm2 [From Principals of Electronic Packaging, Seraphim (1990)] 36
MatSE 361 Chapter B Resistance and Capacitance Thermal 1a Heat Power Consumption 37
MatSE 361 Chapter 0 Syllabus, Intro Thermal 2: Source of Heat
It takes power for a transistor to switch from one state to another. It takes power to get the signal to/from the transistor (interconnects). Electrical Energy Heat via Joule Heating. Power input Temperature Q/ t = I2 R = V I Q/m = Cp T (Black box) ... Q, t, I, R, V ... Q, m, Cp, T EXAMPLE: Suppose we apply 100 A through an Al wire (resistor) with dimensions 100m * 1m * 1 m. Calculate the change in temperature for the wire in 1 sec. Resistance: R = L / (h*w) = 2.5 = 2.5 x 108 J Thermal energy: Q = I2 R t Temperature T = Q / M*Cp = 92 C Current density: J ~ 10,000 A/cm2. Power density h 2.5 W (side) 0.025 W (bottom) L
MatSE 361 Chapter B Resistance and Capacitance w
38 Thermal3: Heat transfer
In order to remove this heat generated by the transistor or interconnect line the heat must be transferred away (convection, conduction, or radiation). EXAMPLE: What is temperature difference between top/bottom Si wafer if a constant heat flux occurs at the top and cooled at the bottom with a Cu heatsink. 100 W Al
250 m 20C Heat sink Si Cu Heat Flux F = Kth*T/ x Kth : 1.48 W/cmK for Si 0.014 W/cmK  SiO2 T =  F *x/ Kth = 2 C If additional 10 m of SiO2 inserted between Al/Si then T = 10 C The 300 mm wafers the thickness is 800 m .
39 MatSE 361 Chapter B Resistance and Capacitance Thermal4: Thermal Conductivity of Materials There is a wide range of values for Kth in materials used for chip making. The ratio of Kth in for Al/SiO2 is large~ 200. Heat transfer in metals occurs via electrons Heat transfer in nonconductive material is by phonons (lattice vibrations).
Au 3.18 Ag 4.29 Graphite* 0.06/19.6 Diamond** 9.923.2 * / $$ basal plane ** (various phases) . The WiedemannFranz Law Kth/ T = L where L = Lorentz number ~2.4 /X 108 W/K2 Kth = Thermal conductivity = Electgrical conductivity
MatSE 361 Chapter B Resistance and Capacitance 40 Thermal 3 (Example)
Cooling  switch from Al to Cu metallization  helps cool from the top! INTEL Fellow (Greene's group) T1 T1 Cu ~ 10 m SiO2 ~ 1 m Si ~ 800 m T1 41
MatSE 361 Chapter B Resistance and Capacitance Thermal 4 (Example) Problem: Suppose there is 1 mA of current in a 1m *1 m crosssection interconnect wire. (a) What is the temperature of the wire after 1 minute of operation without a heat sink? (b)What is the temperature in the middle of the wire after a steady state has been established if the wire is heatsunk by an infinite thick slab of Si? 42
MatSE 361 Chapter B Resistance and Capacitance Thermal 5 43
MatSE 361 Chapter B Resistance and Capacitance Dielectrics 1 Vacuum Parallel Plate Capacitor Q = oE Q = oV/L C = QA/V = oA/L Dielectric C = QA/V = roA/L Memory Design : Keep "C" constant but decrease size A Options Increase r (high "K") or decrease L Low power applications (laptops) would like to reduce "V" Interconnects: Reduce "C" decrease r (low "K") or increase L
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MatSE 361 Chapter B Resistance and Capacitance Dielectrics 2
Vacuum Dielectric Constant = 1.0 Askeland pg 632 45
MatSE 361 Chapter B Resistance and Capacitance ...
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 Fall '08
 Rockett

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