UNIVERSITY OF CALIFORNIA College of Engineering Deartment of Electrical Engineering and Comuter Sciences Fall 2000 EE143 Midterm Exam #1 Family Name First name Signature Make sure the exam aer has 8 ages lus an aendix age at the end. Instructions: DO ALL WORK ON EXAM PAGES This is a 90-minute exam (3 sheets of notes allowed) Grading: To obtain full credit, show correct units and algebraic sign in answers. Numerical answers which are orders of magnitude off will receive no artial credit. Problem 1 (20 oints) Problem 2 (25 oints) Problem 3 (30 oints) Problem 4 (25 oints) TOTAL (100 oints)
Problem 1 Process Sequence Descrition (20 oints total) We would like to imlement a simle RC low-ass filter in integrated form. The to view, the equivalent circuit, and the cross-section along line AB are shown below. The resistor body is lightly doed (n - ) oly- Si and the to caacitor late of is heavily doed (n + ) oly-si. Cross-sections along AB after major rocessing stes are sketched in the right column of the table shown below. Fill in the sequence of rocess stes used in the left column. Al Al Si (bottom late of caacitor) A n- oly Si B contact hole Al oly Si (to late of caacitor) Al n- oly Si (lightly doed) oly Si (heavily doed) Gate oxide CVD -substrate (lightly doed) Process Descrition 1) Starting wafer, - Si Cross Section along AB -substrate (lightly doed) Boron Channel Sto Imlant ad oxide hotoresist Si3N4 -substrate (lightly doed) -substrate (lightly doed)
-substrate (lightly doed) n- oly Si (lightly doed) As imlant (low dose) -substrate (lightly doed) n- oly Si (lightly doed) hotoresist As imlant (high dose) oly Si (heavily doed) -substrate (lightly doed) n- oly Si (lightly doed) oly Si (heavily doed) CVD -substrate (lightly doed) n- oly Si (lightly doed) oly Si (heavily doed) CVD -substrate (lightly doed) Al n- oly Si (lightly doed) oly Si (heavily doed) CVD -substrate (lightly doed)
Problem 2 Thermal oxidation ( 25 oints total) (a) (10 oints) For a articular oxidation rocess, it is known that the oxidation rate (dx ox /dt) is 0.48µm /hour when the oxide thickness is 0.5 µm and it slows down to 0.266 µm /hour when the oxide thickness is 1 µm. Find the linear oxidation constant (B/A) and the arabolic oxidation constant B. Give answers in roer units. (b) (15 oints) A lightly doed Si wafer has been rocessed by some unknown IC rocessing stes. You then erform a thermal oxidation exeriment [fixed temerature and fixed oxidizing ambient] with this wafer and observed the following oxide growth results: Oxidation Time Thickness 0 hour 0 1 hour 2000 Å 4 hours 2500 Å Are the following conjectures TRUE or FALSE? You have to give brief exlanations to justify your answers. Conjecture 1 : The rocessed Si wafer was oxidized first to an oxide thicknes of 100 Å and then have the oxide dissolved in HF. Conjecture 2 : The rocessed Si wafer has a highly doed surface layer ( doing > 10 19 /cm 3 ) which is less than 1000 Å thick. Conjecture 3: The rocessed Si wafer has a thin layer of oly-si layer on to surface.
Problem 3 Ion Imlantation (30 oints total) (a) Phoshorus ions are imlanted through 50nm of SiO 2 into an underlying silicon substrate. Design an imlantation ste which will give the following hoshorus rofile which has a eak concentration of 10 17 /cm 3 and a concentration of 10 15 /cm 3 at the Si/SiO 2 interface. Assume the stoing ower of SiO 2 is the same as silicon. C(x) [ log scale] R =-7.14745 +12.33417 E +0.00323 E 2-8.086e-6 E 3 +3.766e-9 E 4 R =24.39576+4.93641 E -0.00697 E 2 +5.858e-6 E 3-2.024e-9 E 4 SiO 2 10 15 /cm 3 x= 0 x =50nm R Si 10 17 /cm 3 deth x Projected Range & Straggle in Angstrom 10000 1000 100 P 31 into Si 10 10 100 1000 R Ion Energy E in kev R (i) Exress the distance in terms of the straggle R of the imlant rofile. (ii) Choose the required aroximate energy for the Phoshorus ions. (iii) Calculate the dose required.
Problem 3 continued (b) (15 oints) To form the source/drain regions of a state-of-the-art MOSFET by in ion imlantation followed by an annealing ste, it is required to have (1) very shallow junction deths x j ; AND (2) very low sheet resistance R S. (i) Discuss two major hysical mechanisms which limit the formation of ultra-shallow junction deths. (ii) Discuss one major hysical mechanism which limits the formation of low sheet resistance source/drain layers. (iii) Discuss the difficulty to satisfy both the shallow junction deths x j AND low sheet resistance R S requirements.
Problem 4 Diffusion (25 oints total) (a) (10 oints) A boron redeosition ste is erformed into an n-tye Si substrate with a background concentration C B of 1 10 15 /cm 3. The redeosition thermal cycle is 975 o C for 60 minutes. Given: Boron solid solubility at 975 o C = 3.5 10 20 /cm 3 Boron diffusion constant at 975 o C = 1.5 10-14 cm 2 /sec (i) Calculate the junction deth x j. (ii) Calculate the incororated boron dose Q. (b) (10 oints) You are faced with the following three choices for forming the source and drain of a NMOS transistor. oly-si gate L x j -Si ( N a =1E15/cm3 ) (Process i) Shallow diffusion redeosition dose of Q hoshorus atoms /unit area, followed by a drive-in at 1100 C for 60 minutes. (Process ii) Shallow diffusion redeosition dose of Q hoshorus atoms /unit area, followed by a drive-in at 1150 C for 30 minutes. (Process iii) Shallow imlantation dose of Q hoshorus atoms / unit area, followed by an anneal at 950 C for 10 minutes Use the following diffusivity values: TEMPERATURE D(Phoshorus) 950 C 5 10-5 µm 2 / min 1100 C 2 10-3 µm 2 / min 1150 C 5 10-3 µm 2 / min
Problem 4 continued (i) Which rocess will give the shortest MOSFET channel length L? Justify your answer. (iii) If the substrate doing is increased from 1 10 15 to 1 10 16 boron atoms /cm 3, which of the three rocesses will exhibit the greatest change in channel length? Exlain your reasoning. (c) (5 oints) For high-concentration Arsenic drive-in diffusion, the diffusion deth rofile is aroximately rectangular with a eak concentration equals to the solid solubility C S. Qualitatively lot the Irvin Curve [i.e., log C S versus log (R S x j )] for low background concentrations N B. What is the vlaue of the sloe for this lot? Show your derivations. Log C(x) C S N B x j Deth x
erf (z) = 2 π e -y2 dy 0 z APPENDIX for Midterm #1 erfc (z) 1 - erf (z) erf (0) = 0 erf( ) = 1 erf(- ) = - 1 erf (z) 2π z for z <<1 erfc (z) 1 e -z2 for z >>1 π z d erf(z) dz = - d erfc(z) dz = 2π e-z2 z erfc(y)dy = z erfc(z) + 1 π (1-e-z2 ) erfc(z)dz = 1 π 0 0 The value of erf(z) can be found in mathematical tables, as build-in functions in calculators and sread sheets. If you have a rogrammable calculator, you may find the following aroximation useful (it is accurate to 1 art in 10 7 ): erf(z) = 1 - (a 1 T + a 2 T 2 +a 3 T 3 +a 4 T 4 +a 5 T 5 ) e -z2 1 where T = and P = 0.3275911 1+P z a 1 = 0.254829592 a 2 = -0.284496736 a 3 = 1.421413741 a 4 = -1.453152027 a 5 = 1.061405429 1 0.1 0.01 0.001 ex(-z^2) 0.0001 erfc(z) 0.00001 0.000001 0.0000001 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 z
EE143 Midterm #1 Solutions Fall 2000 Problem 1 Process Descrition Cross Section along AB 1) Starting wafer, - Si -substrate (lightly doed) 2) Pad oxide growth Si 3 N 4 deosition Active area atterning (Mask #1) Channel sto Boron imlantation Remove hotoresist Boron Channel Sto Imlant ad oxide hotoresist Si3N4 -substrate (lightly doed) 3) Local oxidation to grow field oxide Stri Si 3 N 4 Stri ad oxide (chemical di) -substrate (lightly doed) 4) As imlantation to form n + in active region Gate oxide oxidation 5) Undoed oly-si deosition Blanket As imlant (low dose) to form n - oly-si -substrate (lightly doed) n- oly Si (lightly doed) As imlant (low dose) 6) Masking resistor region with hotoresist (Mask #2) As imlantation (high dose) to form n + oly-si n- oly Si (lightly doed) hotoresist -substrate (lightly doed) As imlant (high dose) oly Si (heavily doed) 7) Pattern oly-si (Mask #3) Deosit CVD SiO 2 Diffuse and activate doants in oly-si with short time annealing ( 900 C) -substrate (lightly doed) n- oly Si (lightly doed) oly Si (heavily doed) CVD 8) Metal contact oening (Mask #4) n- oly Si (lightly doed) -substrate (lightly doed) oly Si (heavily doed) CVD -substrate (lightly doed)
9) Al deosition Al interconnect atterning (Mask #5) Al n- oly Si (lightly doed) oly Si (heavily doed) CVD -substrate (lightly doed) Problem 2 (a) From the Grove model, we have : x ox 2 + Ax ox = B(t+τ) Therefore, dx ox 2x ox dt + A dx ox = B dt or dx ox B = dt A+2x ox B B From 0.48 = and 0.266 =, we get A+0.5 2 A+1 2 A=0.243 µm, B= 0.597 µm 2 /hour and B/A = 2.46 µm/hour (b) The observed indicates the growth rate is slower than the Deal-Grove model after 2000 Å of oxide is grown. Even if we take the limit that growth rate is roortional (time) 1/2, the oxide will be thicker than 4000 Å after 4 hours of oxidation. The Si surface layer must have a faster oxide rate than the bulk of the wafer. Conjecture 1 : The rocessed Si wafer was oxidized first to an oxide thickness of 100 Å and then have the oxide dissolved in HF. FALSE Doant segregation can enhance oxidation rate but the small oxide growth cannot change the surface layer doant concentration due to doant segregation. Conjecture 2 : The rocessed Si wafer has a highly doed surface layer ( doing > 10 19 /cm 3 ) which is less than 1000 Å thick. TRUE The Si wafer has a highly doed surface layer ( N > 10 19 /cm 3 ) which is less than 1000 Å thick ( i.e., 0.46 2000 Å 1000 Å). The underneath substrate is lightly doed. The Si wafer has a highly doed region underneath a lightly doed surface region. Oxidation rate is higher when the doing concentration is higher than 10 19 /cm 3, mainly through the linear term B/A. After this layer of highly doed Si is consumed, the growth rate slows down. Conjecture 3 : The rocessed Si wafer has a thin layer of oly-si layer on to surface. TRUE Initial oxidation of oly-si is very fast. After oly-si is all consumed, the oxidation rates slows down. Problem 3 (a) (i) Let z = N ex[ - ( )2 2( R ) 2] = N B, ex[-z 2 ] = 10-2 imlies z = 2.15 from gaussian chart. 2 R Therefore, = 2.15 2 R = 3.04 R (ii) The required hoshorus energy has to satisfy R = 500 Å+ 3.04 R From the R and R curves, an energy of ~300keV will satisfy this condition with R = 3600 Å and R = 1000 Å
(iii) N = 0.4 φ R Dose φ = 1017 10-5 = 2.5 10 0.4 12 /cm 2. (b) (i) (A) Ion channeling--- Although a tilt-and-rotate geometry can minimize the rimary channeling effect, we can still have a small fraction of the scattered ions bouncing into various axis or lanes, giving a channeling tail of the doing rofile. The only way to eliminate this lucky ion channeling effect is to reamorhise the surface region with Si imlantation. However, the risk is the formation of residual defects after annealing. (B) Transient Enhanced Diffusion--- The excess oint defects created by ion imlantation will enhance doant diffusion, creating deeer junction deths than desired. There is no established technology yet to eliminate these effect. One ossibility is to co-imlant Ge or C into the S/D regions which sink the excess oint defects. (ii) Doants have a finite solid solubility in Si. The excess concentration above solid solubility is not electrically active. This sets the lower limit for sheet resistance for a given junction deth. (iii ) The sheet resistance is inversely roortional to the area under the concentration rofile curve. A shallow junction deth requires a small Dt roduct (i.e., a lower diffusion temerature or a very short annealing time). However, too low a diffusion temerature may not activate all the imlanted doants or comletely restore the crystal damage. Imlanting a very high dose of doant will increase the height of the doant rofile (i.e., lowers R S ) but we are limited by the solid solubility. Problem 4 (a) x (i) C(x,t)=C S erfc( 2 Dt ) 3.5 10 20 x j erfc[ 2 Dt ] = x j 1015 2 Dt = erfc-1 (2.9 10-6 ) = 3.3 Since 2 Dt = 1.47 10-5 cm, x j = 1.47 10-5 3.3 = 0.49 µm (ii ) Q = 2C S π Dt = 1.47 10-5 3.5 10 20 1.77 = 2.9 10 15 /cm 2 (b) The lateral junction deth is roortional to the vertical junction deends on Dt roduct. ( i ) Dt of (ii) > Dt of (i) > Dt (iii) Process (ii) has the largest lateral diffusion Process(ii) gives shortest MOSFET channel. (ii)
conc. (log scale) a smaller sloe imlies a larger change in x j N B2 N B1 distance x j2 x j1 Since the dose Q is the same, a lager Dt imlies a smaller sloe (i.e., more doant sread-out) Process (ii) gives the greatest change in channel length (c) With dose Q T = C s x j and the mobility constant, the sheet resistance log Cs R s = 1/ (qµq T ) = 1/ (qµc s x j ). sloe = -1 C s = 1/(qµR s x j ). The log (C s ) versus log (R s x j ) lot will be a straight line with a sloe of -1 log R X s j