ECE611 / CHE611: Electronic Materials Processing Fall 017 John Labram Solutions to Homework Due at the beginning of class Thursday October 19 th Question 1 [3 marks]: a) Piranha solution consists of a mixture of sulfuric acid and hydrogen peroxide. State the two main reactions that take place when these compounds are mixed. [1 marks] These are found in Lecture 4: H SO 4 + H O H SO 5 + H O H SO 4 + H O H 3 O + + HSO 4 + O b) Describe why elemental oxygen is important in cleaning wafers? [ marks] Organic (C-containing molecules) such as dust from humans are one of the biggest contaminants for wafers. Elemental oxygen is highly reactive with many organic compounds, in particular the following reactions are important for turning solid carbon-based molecules to gas or liquid: O + O + C CO H + H + O H O Question [6 marks]: We are using a galvanic cell to deposit copper onto a 300mm (diameter) silicon wafer, at an elevated temperature of 60 C. The anode is solid zinc, in an aqueous solution of SO 4 - and Zn +. The cathode is our wafer, in an aqueous solution of SO 4 - and Cu +. a) We wish to establish a voltage of 1V in our cell. The concentration of zinc ions in the H O:SO 4 - :Zn + solution is [10 M], what concentration of Cu + is required in the H O:SO 4 - :Cu + solution is required to establish this voltage? [3 marks] This is from Lecture 5. Here we need to use the Nernst Equation to get the reaction quotient in terms of the modified cell potential: E = E 0 RT ln Q nf E 0 E = RT ln Q nf nf RT (E0 E) = ln Q Q = exp [ nf RT (E0 E)] The reaction quotient, Q, is the ratio of the product to the reactant: Q = [Zn+ ] [Cu + ] 1
So we can say: [Zn + ] [Cu + = exp [nf ] RT (E0 E)] Inverting: [Cu + ] [Zn + = exp [nf ] RT (E E0 )] So: The standard cell potential is given in the notes: [Cu + ] = [Zn + ] exp [ nf RT (E E0 )] E 0 0 = E reduction 0 E oxidation E 0 = 0.34 ( 0.76) = 1.104V We are told the target potential: E = 1V, and the concentration of Zn + ions is [Zn + ] = 10M. The number of electrons involved in each half reaction is n =. We are told the temperature in C, T = 60 C, therefore T = 333 K. The other values are constants. Putting the numbers in: [Cu + 9.65 104 ] = 10 exp [ (1 1.104)] 8.314 333 [Cu + ] = 0.0071 M b) With a voltage of 1V established in our cell, we want to control the deposition rate with a variable resistor between the electrodes in our galvanic cell. What resistance (in ohms) do we require to deposit a 100 nm thick layer of copper on our wafer over a period of 1 hour? [3 marks] Here we need to use Ohm s law: R = V I We are told the potential is 1V, so we need to determine the current required for this deposition. We use the equation relating current to the number of moles of atoms deposited, N, as: We can then substitute in the resistance: I = nfn t R = tv nfn
We are told t and V,we know n, so we need to determine the number of moles deposited. For a mass, m, deposited, the number of moles deposited is: Where m Cu is the mass of one mol of copper. N = The mass we can determine from the geometry of the wafer and the thickness deposited: m m Cu 300mm 100nm We can work in g and cm, since none of our constants have length dimensions or mass. The thickness of the layer deposited is: L = 1 10-5 cm. The surface area of the wafer is: Where d = 30 cm. So the volume is: A = πd 4 LA = Lπd 4 With knowledge of the density of copper, ρ = 8.96 g/cm 3, we can then determine the mass deposited: m = ρla So the number of moles of copper deposited is: N = ρlπd 4m Cu We can finally substitute this back into our equation for resistance: R = tv4m Cu nfρlπd Putting in the numbers: 3
Giving us: R = 60 60 1 4 63.546 9.65 10 4 8.96 10 5 π 30 R = 18.7 Ω Question 3 [ marks]: We are using a plasma to etch a SiO layer under a patterned photoresist, as shown in the figure below. The photoresist layer is 00 nm thick and the feature width is 100nm. What is the minimum selectivity we require to achieve a feature with an aspect ratio of 10, without etching any undesired regions? Assume anisotropic etching is taking place. This is from Lecture 6. The aspect ratio is just the ratio of the depth to width of the feature. AR = D W We are told the feature width is 100nm and we are etching anisotropically, so the feature width will not change during the etch process. So, to achieve an aspect ratio of 10 we therefore need to etch 1 μm of the SiO, before the 00nm of photoresist is completely removed. The ratio of these thicknesses is: 1μm 00nm = 5 This means we need to etch the SiO 5 times as fast as we etch the photoresist. I.e. the selectivity is 5. Question 4 [4 marks]: We have a plasma chamber that has plates 10 cm apart, and we are using pure nitrogen (N ) as our process gas. If the number of secondary electrons produced per ionization event is 0.5, use Paschen s Law to determine the pressure at which the minimum ignition voltage occurs for this chamber, at what this voltage is. For N the parameters in the Paschen Equation are as follows: = 413 VPa -1 m -1. A = 13 VPa -1 m -1. From Lecture 6, the Paschen Equation is as follows: 4
V = pd ln(apd) ln [ln (1 + 1 )] We are given the secondary electron emission coefficient: = 0.5, and the distance between plates, d = 0.1m. Voltage is V, pressure is p, and we are given A and in the question. To find the minimum voltage we have to differentiate voltage with respect to pressure: We will use the quotient rule: For: The first derivative is: dv dp = 0 f(x) = g(x) h(x) f (x) = g (x)h(x) g(x)h (x) [h(x)] Where we are using Newton s notation for derivatives: f (x) = df dx, g (x) = dg dx, h (x) = dh dx If V(p) = f(p), then: g(p) = pd h(p) = ln(apd) ln [ln (1 + 1 )] We need to determine the derivatives of these functions with respect to p. Start with g(p): g (p) = dg dp = d Now we need to differentiate, h(p), which is a little harder. Since is a constant, we will just define the following variable for now: So we can just say: Now take the derivative: α = ln [ln (1 + 1 )] h(p) = ln(apd) α 5
h (p) = dh dp = 1 p Now substitute g(p), g (p), h(p), and h (p) into the quotient rule formula: dv d[ln(apd) α] d = dp [ln(apd) α] dv d[ln(apd) α 1] = dp [ln(apd) α] We could multiply out the denominator, but since we are only interested in the minimum we can just set: And just multiply both sides by the denominator: Divide both sides by d: Take exponentials: Recall from earlier we defined: So now we can put this back in: dv d[ln(apd) α 1] = dp [ln(apd) α] = 0 d[ln(apd) α 1] = 0 ln(apd) α 1 = 0 ln(apd) = α + 1 Apd = e α+1 Apd = e α e 1 α = ln [ln (1 + 1 )] e α = ln (1 + 1 ) Apd = e 1 ln (1 + 1 ) p = e1 1 ln (1 + ) Ad So this is the pressure at the minimum ignition voltages, as a function of d, A, and. Note this is not a function of Putting numbers in: p =.718 1 ln (1 + 13 0.1 0.5 ) p =.97 Pa 6
To get the voltage, we then just enter this pressure, along with all the other variables, into the Paschen Equation: V = pd ln(apd) ln [ln (1 + 1 )] Giving us: V = 413.97 0.1 ln(13.97 0.1) ln [ln (1 + 1 0.5 )] V = 94.874V You can check this by plotting the function if you want: 1000 Voltage (V) 100 10 1 10 100 Pressure (Pa) Question 5 [4 marks]: The table below shows some published values of the growth parameters of SiO on silicon. For the dry growth process, the parameter τ is already evaluated. 7
a) At a temperature of 1100 C, what starting oxide thickness does this data for the dry oxide growth assume? [ marks] This is from Lecture 7. Here we are given τ for dry oxide growth, as well as A and. So we can just use our equation for τ: We need to solve this for x 0 : τ = (x 0 + Ax 0 ) τ = x 0 + Ax 0 Writing this in standard quadratic form: x 0 + Ax 0 τ = 0 We can then use the quadratic equation: Now we just enter values from the table: x 0 = A ± A + 4τ x 0 = A ± A + 4τ (Alternatively, you could have just used the equation from the lectures and said x 0 = x ox when t=0). For the dry oxide growth at 1100 C, we can see these values are: A = 0.090 μm. = 0.07 μm /hr. τ=0.076 hr. x 0 = 0.090 ± 0.090 + 4 0.07 0.076 x 0 = x 0 = 0.090 ± 0.0163 0.090 ± 0.18 Since (-0.09-0.18) is negative, only the positive root is physically meaningful: x 0 = 0.090 + 0.18 x 0 = 0.0377 x 0 = 0.0189μm b) Using the parameters in the above table, determine total thickness if a dry oxidation was carried out for 5 hours at 1100 C. [ marks] 8
This is a simple exercise of putting the numbers into the equation: x ox = A ± A + 4(t + τ) Giving: x ox = 0.09 + 0.09 + 4 0.07 (5 + 0.076) x ox = 0.330μm Question 6 [5 marks]: The rate constants associated with silicon growth are often described with an Arrhenius relationship: = C 1 e E 1 k T A = C e E k T Some example parameters for these relationships are given in the below table. a) Using the parameters in the above table, determine the time required to grow a 100nm total thickness layer of SiO on a Si (111) surface which already has a 5nm native oxide, using a dry reactor at 100 C. [ marks] Again, here we are basically just re-arranging and putting numbers into the equations. For Si (111) growth in a dry reactor, the parameters are: 9
C 1 = 7.7 10 μm /hr C = 6.3 10 6 μm/hr E 1 = 1.3 ev E =.0 ev. First evaluate : = C 1 e E 1 k T = 7.7 10 1.3 1.60 10 19 exp ( 1.38 10 3 (100 + 73) ) (Remembering to convert to Joules and Kelvin) This gives you: Now evaluate /A: = 4.76 10 μm /hr A = C e E k T A = 6.3.0 1.60 10 19 106 exp ( 1.38 10 3 (100 + 73) ) Giving: To get A, just divide by /A: A = 8.90 10 1 μm/hr Now we need to evaluate τ: ( 4.76 10 ) = A = A 8.90 10 1 = 0.0535 μm τ = (x 0 + Ax 0 ) We have just worked out A and, and x 0 is given in the question as: 0.005 μm. Put in the numbers: Giving: τ = (0.005 + 0.0535 0.005) 4.76 10 τ = 0.069 hr We can then take our quadratic equation from the notes: x ox + x ox A = (t + τ) 10
Re-arrange in terms of t: Finally, put in the numbers: t = t = x ox + x ox A τ 0.1 0.1 + 0.1 0.139 1.04 10 0.069 t = 0.316 hr b) If growing a wet oxide on a (100) silicon surface, again use the values in this table to determine the temperature you would need to hold your reactor at, for growth to transition from linear to parabolic when the total oxide thickness is 100nm.[3 marks] We know the transition from linear to parabolic growth occurs at: x ox = A We need to now get an expression for A in terms of T. From before we have: = C 1 e E 1 k T So: A = C e E k T A = ( A ) = C 1e E 1 k T C e E k T So: A = C 1 C e E 1 k T e E k T We are looking for the situation when x ox = A, so: Take logs: = C 1 C exp [ E 1 k T ] exp [ E k T ] A = C 1 exp [ (E 1 E ) ] C k T x ox = C 1 exp [ (E 1 E ) ] C k T x ox C = exp [ (E 1 E ) ] C 1 k T (E 1 E ) k T (E E 1 ) k T = ln ( x oxc C 1 ) = ln ( x oxc C 1 ) 11
k T = (E E 1 ) ln ( x oxc C 1 ) T = (E E 1 ) k ln ( x oxc C 1 ) For the wet oxidation on (100) silicon, the values are: C 1 = 3.86 10 μm /hr C = 0.97 10 8 μm/hr E 1 = 0.78 ev E =.05 ev. Entering the values: T = (.05 0.78) 1.60 10 19 1.38 10 3 0.1 0.97 108 ln ( 3.86 10 ) T = 1455K = 118 C 1