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Published byEmmeline Riley Modified over 9 years ago
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CZ processing Ingot diameter varies inversely with pull rate: L = latent heat of fusion N = density = Stephan-Boltzman constant m = thermal conductivity at T m T m = melt temperture (1417 oC for Si) o = C s /C o increases as ingot grows The dopant concentration is given by: I L =I o (1-V s /V o ) ko and C s = -dI L /dV s = C o k o (1-f) ( o-1) C, I and V are concentration, number of impurities and volume when o: initial L : liquid and s: solid
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Float Zone Processing In Float Zone refining, solid concentration varies with initial concentration as follows: Concentration Liquid Temperature Solid CoCo CsCs CLCL
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MSE-630 4-point probe Outer probe forces current through wafer; inner probes measure voltage drop Used to measure sheet resistivity = 1/q( n n+ p p) -cm us. n>>p or p>>n, so only one term is of interest = 2 s V/I Typically, 0.5-mm< s <1.5-mm When s = 1.588 mm, 2 s = 1 cm If t is not >>s, use correction factor: r = t/ln(2)*V/I If x j is the dopant depth, we measure square resistivity: s = /x j
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MSE-630 2-point probe Useful to determine material type (n- or p-type) Apply two probes, one 25 – 100 o C hotter than other Thermally excited electrons flow away from hot probe, leaving holes and build up around cold electrode Measure Seebeck voltage using high impedance volt meter If material is p-type polarity will be reversed We can measure either short circuit current or open circuit voltage. Current for an n-type material is: J n = q n nP n dT/dx P n is thermoelectric power, either (-) for e - or (+) for h +
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MSE-630 Hall effect Using Hall effect, we can determine material: Type Carrier concentration Carrier mobility 1.Current, Ix, is forced through sample 2.Results in measurable voltage drop, V x = wt/s V x /I x 3.Applying a magnetic field, B, deflects electrons: F = q( + v x B) electrons will be forced in –y direction Where v = electron velocity
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MSE-630 Hall effect Since no current flows in the y-direction, an electric field e must build to offset magnetic force: F y = q( y + v x B z ) = 0 → y = -v x B z Since v x = I x /qwtn, y = B z I x /qwtn, or V y = B z I x /qtn Define the Hall Coefficient: R H = tV y /B z I x = 1/qn → n = ± 1/qR H The “Hall mobility”, H is H = ׀ R H ׀ / = ׀ R H ׀ Hall mobility is typically ~2 x e - or h + mobility Consistent units for calculating Hall effect: V = voltsA = Amps length = meters B = Tesla (1T = 10 4 Gauss = 1 V-s/m 2 )R H = m 3 /C
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Typical defects in crystals Typical defects are: Point defects – vacancies & interstitials Line defects – dislocations Volume defects – stacking faults, precipitates The equilibrium number of vacancies varies with temperature: n v = n o exp(-E v /kT) Thermal stresses cause dislocations. Thermal stress is: = E T = stress, E = Young’s modulus, = thermal expansion coefficient ( m/m/ o C) O and C are also defects with concentrations of 10 17 -10 18 cm -3 and 10 15- 10 16 cm -3 Other impurities are in the ppb range
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Where do defects come from? Naturally occurring vacancies and interstitials Thermal stresses inducing dislocations and stacking faults Precipitation of second phases inducing dislocations and stacking faults Impurities Process damage (e.g, ion implantation)
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The oval shaped area with a lighter contrast is the emitter of a bipolar transistor. In preferential etching this would look similar to what was shown as an illustration for etching.illustration for etching Some of these small stacking faults have a peculiar, "sailing-boat" like shape (marked by "S" in the picture above). Below, a detailed view of a "sailing boat stacking fault":
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Oxidation of Silicon produces interstitials in supersaturation. These surplus interstitials tend to agglomerate in discs - i.e. stacking fault loops. The difficult part is the nucleation; it determines what will happen. We have to consider two ways of oxidizing Si, we first consider Surface oxidation: The surface oxidizes homogeneously by exposing it to an oxidizing atmosphere at high temperatures. This is the normal oxidation process. The emission of interstitials occurs at the interface; the interstitials diffuse into the bulk; the supersaturation decreases with the distance from the surface. There is no easy nucleation for an interstitial type dislocation loop as long as the interface is defect free. If defects are present, most prominent small precipitates of metal impurities (Fe, Ni, Cu) may serve as nucleation centers for the interstitials; a stacking fault penetrating in a semicircular fashion into the bulk is formed. If many precipitates are available, a large density of small stacking faults may be observed: precipitates of metal impurities
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The name "swirl" comes from the spiral "swirl-like" pattern observed in many cases by preferential etching as shown on the right. Close inspection revealed two types of etch features which must have been caused by different kinds of defects. Lacking any information about the precise nature of the defects (which etching can not give), they were termed "A-" and "B-swirl defects".
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MSE-630 Imaging Defects Dislocations and stacking faults introduced during processing and nucleated by oxygen, thermal stress and oxidation processes Chemical etchants reveal density, size and location of defects. Etchants attack areas with high chemical or strain energy and are visible with a microscope EtchComposition SirtlCr 2 O 3 (5M): HF 1:1 SeccoK 2 Cr 2 O 7 (1.5M):HF or Cr 2 O 3 (.15M):HF 1:2 DashHF:HNO 3 :acetic acid 1:3:10
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MSE-630 FTIR: Fourier Transform Infrared Spectroscopy Used to measure concentrations of O and C (down to ~10 15 /cm 3 ) Molecules absorb energy at characteristic wavelengths E = h = hc/ Si-O-Si absorbs at wave number 1106/cm C absorbs at wave number 607/cm
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MSE-630 1.IR beam split & follows two separate paths to sample and detector 2.Moving mirror causes two beams to interfere constructively or destructively in a sinusoidal manner 3.The Fourier transform of the signal will be a delta function proportional to incident beam intensity 4.If the frequency of the source is swept, the FT of the resulting transform will produce an intensity spectrum 5.If we insert a sample, the intensity spectrum will change because of absorption of specific wavelengths 6.Scan of sample is compared to a baseline scan to identify absorbed frequencies
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