Microelectronic Fabrication Technology Course VLSI June05 Batch Answers to Assignment #6 (Due by Oct 03, 2005)
Q.1. Q.1.
Side Sidewa wall ll slo slope pe of of con conta tact ct hol holes es duri during ng RIE: RIE: Taking advantage of photoresist erosion during reactive ion etching of SiO2, we can vary the slope angle of SiO2 contact holes. The following schematic shows the resist and SiO2 cross-sections cross-sections before (solid lines) and after (dash lines) reactive ion etching. Given: θ = 75° VRV = vertical etching rate of resist = 100 nm/min AR = degree of anisotropy of etching rate of resist = 0.5 VOV = vertical etching rate of oxide = 100 nm/min AO = degree of anisotropy of etching rate of oxide = 1.0
(i) Use algebra to prove that the slope of SiO2 after etching is a straight line with constant Φ. (ii) If all etch rates have a variation of ±10% from run to run, find the m aximum and minimum values of Φ.
Before RIE
resist
θ oxide silicon
Φ
After RIE when SiO2 is just cleared
t=0
Before RIE
t = t1
t = to
Intermediate position at time ‘t1’ after start of RIE
resist yox
oxide
is just cleared at ‘to’
xo
(x, y)
Φ silicon
After RIE when SiO2
θ y
x
Ans: Oxide etch is completely anisotropic. Resist etch is partly anisotropic. (i)
Let ‘yox’ be thickness of oxide. Let ‘to’ be time required to clear the SiO2, to = yox / VOv and and ‘xo’ be amount of PR laterally etched at foot of the resist in time ‘to’. Consider a point (x,y) on the oxide profile. The oxide will etch at ‘x’ when the foot of the photoresist is eroded to a distance ‘x’. Let ‘t1’ be the time that the foot of resist is eroded to a distance, x, from original position (i.e, x=0), t1 = x / (VRv cotθ + VRl) Therefore, the vertical thickness, y, of SiO2 etched at the position x’ is y = VOv × (to-t1) = VOv × [(yox / VOv ) - x / (VRv cotθ + VRl)] So, y is linear with x with constant slope tanΦ w.r.t. ti me with tanΦ = VOv / (VRv cotθ + VRl).
(ii)
tan Φ = yox / xo yox = VOv to xo =(VRv cotθ + VRl) to VRv = 100 nm/min, VRi = VRv x AR = 100 x 0.5 = 50 nm/min So, tan Φ = VOv / (VRv cotθ + VRl) Maximum Φ, Φmax, will be obtained, when VOv will be 10% higher and VR’s will be 10% lower.
tan Φmax = 1.1VOv / [0.9(VRv cotθ + VRl)] = 1.1 ×100 / [0.9(100 x cot75° +50)] = 1.59 So, Φmax = tan-1(1.59) = 57.8° Minimum Φ, Φmax, will be obtained, when VOv will be 10% lower and VR’s will be 10% higher. tan Φmin = 0.9VOv / [1.1(VRv cotθ + VRl)] = 0.9 ×100 / [1.1(100 x cot75° +50)] = 1.065 So, Φmin = tan-1 (1.065) = 46.8 °
Q.2.
a) In a certain process, it is desired that the pitch of metal lines be equal to or less than 1.0 µm (the pitch equals one metal line width plus one spacing between metal lines, measured at top of features). Assume that the metal line width and spacing are equal (that is, 0.5µm each). The height of such structures is also 0.5µm, and the minimum lithographic dimension possible is 0.25µm. What is the minimum degree of anisotropy needed in the etch process in order to produce such a structure? Ans: After anisotropic etching (with minimum degree of anisotropy), the profile will be like follows to obtain the required line width and pitch at top of the m etal line:
0.5µm
0.5µm
0.5µm PR
0.5µm
metal
metal
0.25µm So, here hf = 0.5 µm, df = 0.5 µm and dm = 1 – 0.25 = 0.75 µm B = df – dm = 0.5 – 0.75 = - 0.25 Degree of anisotropy = Af = 1- B/(2 x hf ) = 1 – 0.25/(2 x 0.5) = 0.75
b) The cross-section of a Si trench with Etching mask
vertical sidewalls is shown at the right. Assume the etching mask is not attacked by the following etching processes.
1 μm
(i) The structure is first subjected to a completely isotropic wet-etch process. The 0.5 μm
etching rate is 0.5 μm/min and etching time is 1 minute. Sketch proportionally the
Silicon
etched Si cross-section with dashed lines in the same figure. (ii) The structure as fabricated in part (i) is followed by a completely anisotropic plasma -etch process. The etching rate is 1 μm/min and etching time is 1 minute. Sketch the etched Si cross-section with dotted lines in the same figure and show dimensions of all changes in the sketch. Ans:
Etching mask
1µm
1.5 μm
Dotted line after step (i)
Etching mask
1µm
2.5 μm
0.5µm
Dotted line after step (ii)
Q.3.
a) 500 nm of poly-Si over SiO2 is to be patterned using RIE with an etching rate of 50 nm/min. • poly-Si thickness has a variation of ±10% • poly-Si etching rate has a variation of ±5%. • poly-Si is running over steps. Over-etch time fraction = 10%. (i) Use worst-case consideration to find the difference between maximum time to clear poly-Si and minimum time to clear poly-Si. (ii) If process requirement allows less than 10 nm of SiO2 being removed during the poly etching. What is the minimum selectivity required for poly-Si : SiO2 ( i.e., vertical poly etching rate / vertical SiO2 etching rate) ? Ans:
(i) Maximum poly-Si etch rate Poly-Si
= 50 + 5% = 52.5 nm/min Minimum poly-Si etch rate
500± 10%
10%
= 50 - 5% = 47.5 nm/min
silicon
Maximum poly-Si thickness = 500x1.1 = 550 nm Minimum poly-Si thickness = 500x0.9 = 450 nm Over etch time fraction at step = 10% Maximum thickness to be etched at step = 550 x1.1 nm Maximum time to clear poly-Si at the foot of a step = 550x1.1/47.5 = 12.74 min Minimum time to clear poly-Si on flat surface = 450/52.5 = 8.57 min Difference between maximum time to clear poly-Si and minimum time to clear polySi = 12.74 – 8.57 = 4.17 min (ii) SiO2 will be maximum etched for 4.17 min. Maximum thickness of SiO2 that can be etched = 10 nm Maximum etch rate allowed for SiO2 = 10/4.17 = 2.4 nm/min Minimum selectivity required for poly-Si : SiO2 = 50/2.4 = 20.85
b) To pattern Si substrates using reactive ion etching, one can use the CF4+O2
mixture as the plasma gas. How will the degree of anisotropy vary if one changes the following RIE conditions? (= increase, = decrease, 0 = no change)
Process Parameter
Plasma Ion bombardment energy O2 percentage in gas mixture
For a given (CF4+O2) mixture, add H2
Separation between the two electrodes in the parallel plate etcher Chamber Pressure
Degree of Anisotrop y
Brief explanation
Higher energy will clean up the horizontal surface causing (increase) horizontal etch rate to increase. O2 will reduce selectivity and remove CF3 thereby reducing the sidewall protection. This will (decrease) reduce the degree of anisotropy. Adding H2 to (CF4 + O2) plasma reduces F* radicals through the H+F*=HF reaction. Less (increase) chemical reactions and less chemical etching Takes place (less isotropic) This will reduce plasma density thereby reducing the availability (decrease) of ions for physical etching. Higher chamber pressure will scatter ions and will result in (decrease) decrease of the number of ions traveling vertically.
