# UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences. Professor Ali Javey. Spring 2009.

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1 UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences EE143 Professor Ali Javey Spring 2009 Exam 1 Name: SID: Closed book. One sheet of notes is allowed. There are a total of 10 pages on this exam, including the cover page. Problem 1 20 Problem 2 30 Problem 3 30 Problem 4 20 Total 100 1

2 Physical Constants Electronic charge q C Permittivity of vacuum ε F cm -1 Relative permittivity of silicon ε Si /ε Boltzmann s constant k x 10-5 ev/ K or J K -1 Thermal voltage at T = 300K kt/q V Effective density of states N c 2.8 x cm -3 Effective density of states N v 1.04 x cm -3 Intrinsic carrier concentration of n i cm -3 Silicon at T=300K 2

3 Problem 1. True/False (20 pts) Circle either (T)rue or (F)alse (2 pts each) 1) We cannot mix organic wastes with acids, because it can cause fire or even explosion. 2) In the clean room, the light is always yellow to prevent bacteria from growing in the room. 3) We always use plastic (e.g., Teflon) beakers for etchants containing HF. 4) We add sulfuric acid (H 2 SO 4 ) to hydrogen peroxide (H 2 O 2 ) when preparing Piranha. 5) The bandgap of a semiconductor material decreases at higher temperatures. 6) Piranha cleaning removes native oxide layer. 7) Increased spinning speed will give you thinner photoresist coating. 8) Aluminum is an acceptor in silicon. 9) At thermal equilibrium, semiconductors always have equal number of free electrons as free holes. 10) Silicon can be either a donor or acceptor in GaAs, depending on the processing conditions. 3

4 Problem 2. Introduction to Materials (30 pts) A semiconductor is doped with an impurity concentration N such that N>>n i and all the impurities are ionized. Also, n=n and p=ni 2 /N. a) Is the impurity a donor or an acceptor? (2 pts) donor b) Specify the majority carrier and minority carrier in this semiconductor material. (4 pts) majority: electrons minority: holes c) Assume the material is Si, and N=10 17 cm -3. Find the electron and hole concentration at T=300K. (6 pts) electrons: cm -3 holes: 10 3 cm -3 d) Find the Fermi level (with respect to either E c or E v ) for this semiconductor at room temperature, and draw the energy band diagram. Make sure that you clearly label E f, E c, E v. (4 pts) n = n e E E E E F F c F i E E ( EF Ei ) / kt i i = = : 0.419eV above E = 1.12eV / eV below E, 0.98eV i c above E v 0.14eV 0.98eV E c E F E i E v e) Briefly describe the meaning of Fermi level (one sentence max). (4 pts) The energy level at which the probability of electron presence is 1/2. 4

5 f) Fermi level typically lies somewhere between E c and E v. However, we know that the energy regime between E c and E v corresponds to the forbidden regime as it represents the band gap. Briefly, explain how the Fermi level could lie in the band gap without violating the laws of physics (2 sentences max). (4 pts) E F is an arbitrary energy level. Assuming there is an energy state, the probability of the state being occupied by an electron is 1/2. The density of state at Fermi level can be zero which is the case here. g) Briefly explain the difference between a metal, semiconductor, and insulator in terms of the band diagram (3 sentences max). Please draw the band diagram for each material, clearly labeling the valence and conduction bands, and the Fermi level. (4 pts) Semiconductors have a bandgap of about <~ 4eV between their valence and conduction bands with the conduction band being mostly empty of electrons (depending on the doping level). Insulators have a bandgap exceeding ~ 4eV. For metals, the Fermi level is in the middle of the conduction band, so there is no bandgap near the highest occupied and lowest unoccupied states. E F E F E F h) Do amorphous and crystalline silicon have the same band gap? Briefly explain (2 sentences max). (4 pts) No. In amorphous Si, the atomic arrangement is random. So the atomic orbital interactions and coupling are different. As a result, the bandgap is different (it s actually larger) 5

6 Problem 3. Photolithography (30 pts) (a) The figure below shows the image patterns of three basic kinds of photolithography methods: contact, proximity, and typical projection litho. Please identify what photolithography method each of the labeled imaging patterns (A, B, C) corresponds to? (3 pts) B A C A: Projection B: Proximity C: Contact (b) Briefly explain what phenomenon causes the shape of the pattern of image B shown above. (2 sentences max) (4 pts) Because of diffraction effects, the light bends away from the aperture edges, producing some resist exposure outside the aperture edges. Also, there is ringing in the intensity distribution within the aperture dimension, because of the constructive and destructive interference due to the wave nature of light. 6

7 (c) List the pros and cons of contact photolitho compared to proximity photolitho? (4 pts) pros: better resolution cons: damage possibility on wafer/mask (d) List the pros and cons of contact photolitho compared to typical projection photolitho? (4 pts) pros: less expensive cons: damage (projection cons: feature reduction possible better resolution) (e) List three possible routes for improving the resolution limit of a projection photolithography tool, without changing the light wavelength or the optics of the system. (3 pts) 3 among these 4; - Immersion lithography - Phase shifting mask - Partial coherence light - Proximity correction (f) The resolution of a projection lithography system can be readily improved by using a lower wavelength light. List three limitations for how low in the wavelength we can go. (3 pts) 3 among these 4 - absorption in optic system, therefore, requiring the use of complex mirrors instead - poor DOF - light source - appropriate PR 7

8 (g) With g-line (436nm), what is the necessary numerical aperture, NA, needed to produce a minimal feature size of 400 nm? Assume k 1 =0.6. (4 pts) l m = k 1 (λ/na) 400 = 0.6 (436/NA) NA=0.654 (h) Assume that projection lithography is being conducted on the surface of a wafer with maximum height variation of 1 µm. The minimal feature size is 400 nm. Suggest wavelength and NA to satisfy both the surface height variation and the feature size. Assume k1=0.6. (4 pts) DOF= k 2 λ/(na) 2 = 500 nm -- (1) l m = k 1 (λ/na) = 0.6 (λ/na) = 400 nm -- (2) (2)/(1) NA = 400/1000/0.6 = λ = 400/0.9 = 444 nm 8

9 Problem 4. Photoresists (20 pts) Consider the following mask and wafer cross-section: a) Draw the cross section of the wafer after exposing and developing if positive PR is used. (4 pts) b) Redraw the cross-section in part a) if a negative PR is used instead. (4 pts) c) Briefly explain the exposure/developing mechanism for a positive PR (3 sentences max). (4 pts) Inhibitor in PR (PAC, Photo Active Compound) is not soluble in the developer. When the PR is exposed, the weak N 2 bond is broken by photochemical reaction. After several chain reactions, the molecule finally transforms into carboxylic acid, which is soluble in the developer. (No change for polymer itself) 9

10 d) Briefly explain the exposure/developing mechanism for a negative PR (3 sentences max). (4 pts) Light breaks N-N in light sensitive additive. Then the reactive intermediate produces a cross-linked network of polymers in the exposed part and thus increases the molecular weight. Cross-linked chains have significantly low solubility in developer. (no change for polymer itself) e) List 2 advantages and 2 disadvantages of positive vs. negative PR. (4 pts) Positive PR advantages - Higher resolution - Aqueous solvent Disadvantages - Less sensitive - Less chemical resistance - Less tolerant of developing conditions 10

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