3.1551/6152 Homework set 3 solution. fall 2003 3.155J/6,152J Microelectronic Processing Fall Term. 2003 Bob HAndley Martin schmidt Problem set 3 solutions Out Sept 29, 2003 Due Oct 08. 2003 (t) lTDt 4D d(2√ GDt Dt AD)jx4D/cs dc 2Dt dz 4D d-c 2 4 16D2t2 Clar)=C(z1\4Dt Thus (z, t)dcz,t) Problem 2 a) From Fig. 1. 16 in Plummer or 3. 4 in Campbell, n,=2x b)i) From table: D0=0.011, D.E=3.44eV, D.0=310, D.E=4.15 DE Using relation D= Do.Exp kl D ii)Use the equation D=D+D- to get D x(n/n )=9. x 10-5 cm/s thus: c) Diffusion length a=2/Dt for part i)a=0.062 um for part ii)a=0.13 um
3.155J/6.152J Homework set 3 solution, fall 2003 1 3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 3 Solutions Out Sept. 29, 2003 Due Oct.08, 2003 Problem 1 † C(z,t) = Q pDt exp - z 2 4Dt Ê Ë Á Á ˆ ¯ ˜ ˜ † dC dt = - Q 2t pDt + Q pDt z2 4Dt 2 Ê Ë Á Á ˆ ¯ ˜ ˜ exp - z 2 4Dt Ê Ë Á Á ˆ ¯ ˜ ˜ = C(z,t) z 2 2Dt -1 Ê Ë Á Á ˆ ¯ ˜ ˜ 1 2t † dC dz = -2z 4Dt C(z,t) † d2 C dz 2 = - 2 4Dt C(z,t) + 4z 2 16D2 t 2 C(z,t) = C(z,t) 2z 2 4Dt -1 Ê Ë Á Á ˆ ¯ ˜ ˜ 2 4Dt Thus, † dC(z,t) dt = D d2 C(z,t) dz2 Problem 2 a) From Fig. 1.16 in Plummer or 3.4 in Campbell, † ni = 2 ¥1019 b) i) From table: † D0 .O = 0.011, D0 .E = 3.44eV, D- .O = 31.0, D- .E = 4.15 Using relation † D = D0.0exp - D.E kT Ê Ë Á ˆ ¯ ˜ † Deff = D0 = 2.66 ¥10-15 cm2 s -1 ii) Use the equation † Deff = D0 + D- n ni Ê Ë Á ˆ ¯ ˜ to get D- ¥ (n/ni ) = 9.35 ¥ 10-15 cm2 /s thus: † Deff = 1.2 ¥10-14 cm 2 s -1 c) Diffusion length † a = 2 Dt for part i) a = 0.062 mm for part ii) a = 0.13 mm
3.1551/6152 Homework set 3 solution. fall 2003 Problem 3 a) Solubility limit for P at 1273 K is Co 2 1 x 10 cm". Intrinsic carrier concentration is 10 cm". Thus, before including higher order terms, Do=13x10-4 cm2/s. But with first order and second order correction terms (n/n ) for large P concentration, D=237 x 10+cm/s na2√DC Q=1.9 b) Find t o for zict=0.4 x 10-cm Solubility limit for phosphorous at 1373 K is Co z 6x 10 cm". Intrinsic carrier concentration at this temperature is 1.2 x 10cm D=110 X 10- cm?/s. Because the predep is at a higher concentration than the 3 With first order and second order correction terms(n/n )for large P concentratic solubility limit right inside the surface, the value of C(o, t)remains 6 x 10" throughout this drive-in process, and the solution must be an erfc. The depth dependence of the diffusion profile is: C(,)-Cer|- =N,=107cm 2√Dr N.102 1.67×10 2yDt)C. C Look up 1-1.67x10=0.99983 in the Plummers table for error functions and you find the argument must be 2.66=7/([2v(DD]. Solving for t gives t=5.4 sec c)As stated above, the concentration at the surface is continually fed from the predep and remains at the solubility limit of P, 6 x 10 cm Problem 4 Fig 8.3 gives the depth and standard deviation for boron implantation at 40 ke V. For ease of plotting I have expressed everything in nanometers: R, 140 nm and AR, 2 57 nm. Given Q=5x 10 cm2=50 nm, you can use Eq. 8.4 to get C=0.35nm (. x 10 cm), then plot Eq. 8. 1 as shown by the bold line below. Here the concentration is plotted as number per cubic nm Following the arguments around Eqs. 8.11 and 8.12-the profile evolves like V(2Dr) you can plot Eq,8.12 using d at950C≈1.94×106cm2s=19×102nm3/sto plot the fine line in the figure below To get the junction depth, set Eq Eq. 8.12 equal to 2 x 10 cm3=2 x 10nm, and solve it for x, the distance at what that concentration now occurs from the peak. If you're prone to math errors, you could plot it as shown in the second figure, to find that the junction occurs at about 464 nm
3.155J/6.152J Homework set 3 solution, fall 2003 2 Problem 3 a) Solubility limit for P at 1273 K is C0 ≈ 1 ¥ 1021 cm-3. Intrinsic carrier concentration is 1019 cm-3. Thus, before including higher order terms, † DO =1.3¥10-14 cm 2 /s. But with first order and second order correction terms (n/ni ) for large P concentration, D = 237 ¥ 10-14 cm2 /s. Using † Q = C0a p = 2 DtC0 p we get Q = 1.9 ¥ 1016 cm-2 b) Find t 0 for zjct = 0.4 ¥ 10-4 cm. Solubility limit for phosphorous at 1373 K is C0 ≈ 6 ¥ 1020 cm-3. Intrinsic carrier concentration at this temperature is 1.2 ¥ 1019 cm-3. With first order and second order correction terms (n/ni ) for large P concentration, D = 110 ¥ 10-13 cm2 /s. Because the predep is at a higher concentration than the solubility limit right inside the surface, the value of C(0,t) remains 6 ¥ 1020 cm-3 throughout this drive-in process, and the solution must be an erfc. The depth dependence of the diffusion profile is: † C(z,t) = Cs erfc - z 2 Dt Ê Ë Á ˆ ¯ ˜ = NA =1017 cm-3 † erfc - z 2 Dt Ê Ë Á ˆ ¯ ˜ = NA Cs = 1017 Cs =1.67 ¥10-4 Look up 1-1.67 ¥ 10-4 = 0.99983 in the Plummer’s table for error functions and you find the argument must be 2.66 = z/([2÷(Dt)]. Solving for t gives t = 5.4 sec. c) As stated above, the concentration at the surface is continually fed from the predep and remains at the solubility limit of P, 6 ¥ 1020 cm-3. Problem 4 Fig. 8.3 gives the depth and standard deviation for boron implantation at 40 keV. For ease of plotting I have expressed everything in nanometers: Rp ≈ 140 nm and DRp ≈ 57 nm. Given Q = 5 ¥ 1015 cm-2 = 50 nm-2, you can use Eq. 8.4 to get Cp = 0.35 nm-3 (3.5 ¥ 1020 cm-3), then plot Eq. 8.1 as shown by the bold line below. Here the concentration is plotted as number per cubic nm. Following the arguments around Eqs. 8.11 and 8.12 - the profile evolves like ÷(2Dt) - you can plot Eq. 8.12 using D at 950 C ≈ 1.94 ¥ 10-16 cm2 /s = 1.94 ¥ 10-2 nm2 /s to plot the fine line in the figure below. To get the junction depth, set Eq Eq. 8.12 equal to 2 ¥ 1015 cm-3 = 2 ¥ 10-6 nm3 , and solve it for x, the distance at what that concentration now occurs from the peak. If you’re prone to math errors, you could plot it as shown in the second figure, to find that the junction occurs at about 464 nm
3.1551/6152 Homework set 3 solution. fall 2003 0.3 0.2 0.1 x10-05 x10-06 10 4x106 2x106 Problem 5 At 30 keV, from Fig 8-3, R, 2 120 nm and AR= 50 nm. Given 0= 10cm=0.01 nm a) Peak occurs at depth where x=R, 120 nm b) From Eq.84,Cn=8×10°nm3=8×10°cm3 c)Atx=300nm,Eq.8.1 gives c=1.23×107nm3=1.23×104cm d) If the dose at 0.3 microns is 10 times larger than this even though the values of R, and C, are correct, then the wafer was not pure, but rather doped with boron to1.1×105cm3(1.11×103+1.23×104+=10
3.155J/6.152J Homework set 3 solution, fall 2003 3 Problem 5 At 30 keV, from Fig. 8-3, Rp ≈ 120 nm and DRp ≈ 50 nm. Given Q= 1012 cm-2 = 0.01 nm-2. a) Peak occurs at depth where x = Rp ≈ 120 nm. b) From Eq. 8.4, Cp = 8 ¥ 10-5 nm-3 = 8 ¥ 1016 cm-3. c) At x = 300 nm, Eq. 8.1 gives Cp = 1.23 ¥ 10-7 nm-3 = 1.23 ¥ 1014 cm-3. d) If the dose at 0.3 microns is 10 times larger than this even though the values of Rp and Cp are correct, then the wafer was not pure, but rather doped with boron to 1.1 ¥ 1015 cm-3 (1.11 ¥ 1015 + 1.23 ¥ 1014 = 10 ¥
3.1531/6152 Homework set 3 solution. fall 2003 1. 23 x 10). A later exposure to elevated temperature could not cause the increased concentration at 0.3 microns without shifting the standard deviation and peak value. Also, while the actual distribution may not be exactly Gaussian, for boron, the skew is toward increased concentration near the surface(due to enhanced backscatter of this light species)not what is asked here Problem 6 R=0. um(300 nm)demands an implant of boron at about 100 keV(Fig. 8-3). At this energy, AR,=85 nm, so the dose giving a peak concentration of 10cm is easily calculated to be 2.13 x 10cm". For a background doping of 1015 cm-3, the junction depth before diffusion is given by inverting Eq. 8. 1: x=558 nm or 0.558
3.155J/6.152J Homework set 3 solution, fall 2003 4 1.23 ¥ 1014). A later exposure to elevated temperature could not cause the increased concentration at 0.3 microns without shifting the standard deviation and peak value. Also, while the actual distribution may not be exactly Gaussian, for boron, the skew is toward increased concentration near the surface (due to enhanced backscatter of this light species) not what is asked here. Problem 6 Rp = 0.3 µm (300 nm) demands an implant of boron at about 100 keV (Fig. 8-3). At this energy, DRp ≈ 85 nm, so the dose giving a peak concentration of 1017 cm-3 is easily calculated to be 2.13 ¥ 1012 cm-2. For a background doping of 1015 cm-3, the junction depth before diffusion is given by inverting Eq. 8.1: x = 558 nm or 0.558 µm
