Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

6.3. Reactive Sputtering Deposition Consider the reactive sputtering model of Section 16.3 with equal target and substrate areas. Let a=m/c and Y= 2rsr/ic be the

image text in transcribedimage text in transcribedimage text in transcribed

6.3. Reactive Sputtering Deposition Consider the reactive sputtering model of Section 16.3 with equal target and substrate areas. Let a=m/c and Y= 2rsr/ic be the normalized sputtering coefficient and flux, respectively. DEPOSITION AND IMPLANTATION (a) Show using (16.3.3) and (16.3.4) that the surface coverages of the compound on the target and the substrate are given respectively by ts=1+YY=Y2+2Y+aY2+2Y (b) Show using (16.3.5) and (16.3.6) that the reactive gas flow and the sputtering flux are given respectively by dtdNrsputY(1+Y1+Y2+2Y+aa)1+Ya+Y c) For the limiting case a1, graph dNr/dt versus Y and sput versus Y. From these graphs, sketch sput versus dNr/dt and show that the curve exhibits hysteresis similar to that shown in Figure 16.5. Is there hysteresis for the case a1 ? Prove your answer. ntdtdt=irsr(1t)ict=0 where i is the number of atoms per molecule of reactive gas (e.g., i=2 for O2 gas). Sputtered compound molecules and metal atoms are assumed to be evenly deposited over the substrate surface. The coverage s of compound on the substrate increases because reactive gas molecules are incident on the metallic part (1S), and because a fraction (1s) of the compound flux sputtered from the target is deposited on the metallic part of the substrate. Similarly, s decreases because a fraction s of the metal-atom flux sputtered from the target is deposited on the compound part of the substrate. Hence, accounting for the ratio of target and substrate areas, we obtain nsdtds=irsr(1s)+ict(AsAt)(1s)im(1t)(AsAt)s=0 6.3. Reactive Sputtering Deposition Consider the reactive sputtering model of Section 16.3 with equal target and substrate areas. Let a=m/c and Y= 2rsr/ic be the normalized sputtering coefficient and flux, respectively. DEPOSITION AND IMPLANTATION (a) Show using (16.3.3) and (16.3.4) that the surface coverages of the compound on the target and the substrate are given respectively by ts=1+YY=Y2+2Y+aY2+2Y (b) Show using (16.3.5) and (16.3.6) that the reactive gas flow and the sputtering flux are given respectively by dtdNrsputY(1+Y1+Y2+2Y+aa)1+Ya+Y c) For the limiting case a1, graph dNr/dt versus Y and sput versus Y. From these graphs, sketch sput versus dNr/dt and show that the curve exhibits hysteresis similar to that shown in Figure 16.5. Is there hysteresis for the case a1 ? Prove your answer. ntdtdt=irsr(1t)ict=0 where i is the number of atoms per molecule of reactive gas (e.g., i=2 for O2 gas). Sputtered compound molecules and metal atoms are assumed to be evenly deposited over the substrate surface. The coverage s of compound on the substrate increases because reactive gas molecules are incident on the metallic part (1S), and because a fraction (1s) of the compound flux sputtered from the target is deposited on the metallic part of the substrate. Similarly, s decreases because a fraction s of the metal-atom flux sputtered from the target is deposited on the compound part of the substrate. Hence, accounting for the ratio of target and substrate areas, we obtain nsdtds=irsr(1s)+ict(AsAt)(1s)im(1t)(AsAt)s=0

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Ceramic Matrix Composites Fiber Reinforced Ceramics And Their Applications

Authors: Walter Krenkel

1st Edition

3527313613, 978-3527313617

More Books

Students also viewed these Chemical Engineering questions