Match List I with List II :List IList II(a) ion implantation(1) interconnect between devices(b) etching(2) insulating dielectric layer(c) oxidation(3) removal of unwanted material(d) metallization(4) low temperature impurity insertionCodes : (a) (b) (c) (d)

The correct code is (4) (3) (2) (1).(a) Ion implantation inserts dopant impurities at low temperature, unlike diffusion.(b) Etching removes unwanted material.(c) Oxidation grows the insulating dielectric layer of SiO2.(d) Metallization forms the interconnects between devices.Option A reverses every pair, and options C and D misplace etching, oxidation or ion implantation.

Match List I with List II and select the correct answer using the codes given below :List IList II(a) Chemical analysis(i) Probe station(b) Structural characteristics(ii) SEM(c) Topology(iii) XRD(d) Electrical characteristics(iv) EDAX

The correct matching is a-iv, b-iii, c-ii, d-i.Chemical analysis: EDAX (energy-dispersive X-ray analysis) identifies elements from their characteristic X-rays (iv).Structural characteristics: XRD (X-ray diffraction) reveals crystal structure and lattice spacing (iii).Topology: SEM (scanning electron microscope) images surface topography (ii).Electrical characteristics: a probe station makes electrical contact with devices on a wafer for I-V and C-V measurements (i).The other codes swap XRD and SEM, or give chemical analysis to the probe station.

CMOS Fabrication Process Questions and Answers

Q: ................... cannot be fabricated on an IC.

The components that cannot be fabricated on an IC are large inductors and transformers.Integrated circuits are built in thin layers on a silicon chip. Transistors, diodes and resistors (diffused or thin-film) are made easily in those layers, and small capacitors too. A useful inductance, and especially a transformer, needs many turns of wire and usually a magnetic core, which cannot be built in the tiny planar area and thickness of a chip; only very small spiral inductors of a few nanohenries are possible at radio frequencies.So large inductors and transformers are connected externally.

Q: Distribution of implanted impurity ions by Ion implantation method is :

The implanted ions are distributed in a Gaussian profile.Each ion loses energy through random collisions, so the depths at which they stop are spread around a mean, the projected range Rp, with a standard deviation called the straggle \(\Delta R_{p}\). To a first approximation the concentration is\[ \begin{aligned} N(x) &= N_{p}\, e^{-(x - R_{p})^{2}/2\Delta R_{p}^{2}} \\ N_{p} &= \frac{Q}{\sqrt{2\pi}\,\Delta R_{p}} \end{aligned} \]where Q is the implanted dose. It is not constant, because the peak is below the surface; not parabolic; and, although each stopping event is random, the resulting distribution is not described as random but as Gaussian.

Q: Which is the final step in wafer processing sequence ?

The final step is metallization.Wafer processing runs through oxidation, photolithography, etching, diffusion or ion implantation and deposition, repeated for each layer that builds the transistors. Only when the devices are complete is a metal layer, usually aluminium or copper, deposited and patterned to form the contacts and interconnections between them, followed by passivation.Oxidation and photolithography come early and are repeated throughout, and chemical vapour deposition is used for intermediate layers, so none of them is the last step.

Q: With help of Scanning Electron Microscope (SEM) one can resolve objects as small as :

The keyed resolution of an SEM is 1 nm.A scanning electron microscope builds an image by scanning a finely focused electron beam over the sample. The short wavelength of the electrons and a probe only a few nanometres across let modern field-emission SEMs resolve features of about 1 nm, far below the limit of light microscopes.10 mm is visible to the naked eye. About 1 micron (more exactly about 0.2 μm) is the limit of an optical microscope, which is what the SEM improves on, and 10 nm is the resolution of older thermionic instruments rather than the best an SEM can do.

Q: In the process of photolithography, an optical stepper is used for the exposure of the wafers. Write the correct order of various processes starting from the light source to the die on wafer in an optical stepper.(1) Illumination diffuser(2) Condenser lens(3) Reficle(4) Interferometer

The order is (1), (2), (3), (4): illumination diffuser, condenser lens, reticle, interferometer.In an optical stepper, light from the source first passes through the illumination diffuser, which spreads it uniformly. The condenser lens then collects the light and directs it evenly onto the reticle (printed 'Reficle' in the paper), the mask carrying the pattern of one or a few dies. Projection optics image the reticle onto the wafer, whose stage position is measured by a laser interferometer so that each die is exposed in exactly the right place.The other options put the reticle before the condenser, or start with the interferometer, which belongs to the wafer stage at the end of the path, not at the light source.

Q: Wet oxidation is given by :

Wet oxidation is Si + \(\mathrm{2H_2O}\).In wet (steam) oxidation, silicon reacts with water vapour at about 900-1100 °C to form silicon dioxide and release hydrogen:\[ \mathrm{Si + 2H_2O \rightarrow SiO_2 + 2H_2} \]It grows oxide much faster than dry oxidation and is used for thick field oxides. Si + \(\mathrm{O_2}\) is dry oxidation, which grows a slower, denser oxide used for gate oxides. HF does not oxidise silicon; it etches silicon dioxide. Hydrogen does not form an oxide at all.

Q: Positive photo resists are used more than negative photo resists because :

The key gives the option stating that negative photoresists do not have as high a resolution as positive photoresists.A negative resist becomes insoluble where it is exposed. During development the unexposed regions dissolve, but the exposed, cross-linked regions absorb the developer and swell, which distorts fine lines. A positive resist does not swell this way, so it reproduces much smaller features. That higher resolution is why positive resists are preferred for modern VLSI, even though negative resists react to light more readily.The option claiming positive resists have the lower resolution reverses the decisive fact. Negative resists are not less sensitive to light, and the lower sensitivity of positive resists is a drawback, not the reason they are used more.

11.

................... cannot be fabricated on an IC.

A.Transistors
B.Diodes
C.Resistors
D.Large inductors and transformers

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The components that cannot be fabricated on an IC are large inductors and transformers.

Integrated circuits are built in thin layers on a silicon chip. Transistors, diodes and resistors (diffused or thin-film) are made easily in those layers, and small capacitors too. A useful inductance, and especially a transformer, needs many turns of wire and usually a magnetic core, which cannot be built in the tiny planar area and thickness of a chip; only very small spiral inductors of a few nanohenries are possible at radio frequencies.

So large inductors and transformers are connected externally.

Discuss in Forum

12.

Match List I with List II :

List I	List II
(a) ion implantation	(1) interconnect between devices
(b) etching	(2) insulating dielectric layer
(c) oxidation	(3) removal of unwanted material
(d) metallization	(4) low temperature impurity insertion

Codes : (a) (b) (c) (d)

A.(1) (2) (3) (4)
B.(4) (3) (2) (1)
C.(4) (2) (1) (3)
D.(2) (3) (4) (1)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The correct code is (4) (3) (2) (1).

(a) Ion implantation inserts dopant impurities at low temperature, unlike diffusion.
(b) Etching removes unwanted material.
(c) Oxidation grows the insulating dielectric layer of SiO₂.
(d) Metallization forms the interconnects between devices.

Option A reverses every pair, and options C and D misplace etching, oxidation or ion implantation.

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13.

Distribution of implanted impurity ions by Ion implantation method is :

A.Parabolic
B.Gaussian
C.Random
D.Constant

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The implanted ions are distributed in a Gaussian profile.

Each ion loses energy through random collisions, so the depths at which they stop are spread around a mean, the projected range R_p, with a standard deviation called the straggle \(\Delta R_{p}\). To a first approximation the concentration is

\[ \begin{aligned} N(x) &= N_{p}\, e^{-(x - R_{p})^{2}/2\Delta R_{p}^{2}} \\ N_{p} &= \frac{Q}{\sqrt{2\pi}\,\Delta R_{p}} \end{aligned} \]

where Q is the implanted dose. It is not constant, because the peak is below the surface; not parabolic; and, although each stopping event is random, the resulting distribution is not described as random but as Gaussian.

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14.

Which is the final step in wafer processing sequence ?

A.Photolithography
B.Chemical vapor
C.Metallization
D.Oxidation

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The final step is metallization.

Wafer processing runs through oxidation, photolithography, etching, diffusion or ion implantation and deposition, repeated for each layer that builds the transistors. Only when the devices are complete is a metal layer, usually aluminium or copper, deposited and patterned to form the contacts and interconnections between them, followed by passivation.

Oxidation and photolithography come early and are repeated throughout, and chemical vapour deposition is used for intermediate layers, so none of them is the last step.

Discuss in Forum

15.

With help of Scanning Electron Microscope (SEM) one can resolve objects as small as :

A.10 mm
B.1 micron
C.10 nm
D.1 nm

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The keyed resolution of an SEM is 1 nm.

A scanning electron microscope builds an image by scanning a finely focused electron beam over the sample. The short wavelength of the electrons and a probe only a few nanometres across let modern field-emission SEMs resolve features of about 1 nm, far below the limit of light microscopes.

10 mm is visible to the naked eye. About 1 micron (more exactly about 0.2 μm) is the limit of an optical microscope, which is what the SEM improves on, and 10 nm is the resolution of older thermionic instruments rather than the best an SEM can do.

Discuss in Forum

16.

In the process of photolithography, an optical stepper is used for the exposure of the wafers. Write the correct order of various processes starting from the light source to the die on wafer in an optical stepper.

(1) Illumination diffuser
(2) Condenser lens
(3) Reficle
(4) Interferometer

A.(1), (3), (2), (4)
B.(1), (2), (3), (4)
C.(3), (1), (2), (4)
D.(4), (1), (2), (3)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The order is (1), (2), (3), (4): illumination diffuser, condenser lens, reticle, interferometer.

In an optical stepper, light from the source first passes through the illumination diffuser, which spreads it uniformly. The condenser lens then collects the light and directs it evenly onto the reticle (printed 'Reficle' in the paper), the mask carrying the pattern of one or a few dies. Projection optics image the reticle onto the wafer, whose stage position is measured by a laser interferometer so that each die is exposed in exactly the right place.

The other options put the reticle before the condenser, or start with the interferometer, which belongs to the wafer stage at the end of the path, not at the light source.

Discuss in Forum

17.

Wet oxidation is given by :

A.Si + \(\mathrm{O_2}\)
B.Si + HF
C.Si + \(\mathrm{2H_2O}\)
D.Si + \(\mathrm{H_2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Wet oxidation is Si + \(\mathrm{2H_2O}\).

In wet (steam) oxidation, silicon reacts with water vapour at about 900-1100 °C to form silicon dioxide and release hydrogen:

\[ \mathrm{Si + 2H_2O \rightarrow SiO_2 + 2H_2} \]

It grows oxide much faster than dry oxidation and is used for thick field oxides. Si + \(\mathrm{O_2}\) is dry oxidation, which grows a slower, denser oxide used for gate oxides. HF does not oxidise silicon; it etches silicon dioxide. Hydrogen does not form an oxide at all.

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18.

After the photoresist pattern is formed, which of the process is followed :

A.Oxidation
B.Masking
C.Etching
D.Bonding

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

After the photoresist pattern is formed, the next process is etching.

In photolithography the resist is coated, exposed through a mask and developed, leaving a pattern of resist on the wafer. That pattern then serves as a mask: etching removes the exposed underlying layer, such as silicon dioxide or metal, while the resist protects the covered areas. The resist is stripped afterwards.

Oxidation is done before the resist is applied, to grow the layer that will be patterned. Masking is what the exposure step already did with the photomask. Bonding comes at the packaging stage, long after the wafer has been patterned.

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19.

Match List I with List II and select the correct answer using the codes given below :

List I	List II
(a) Chemical analysis	(i) Probe station
(b) Structural characteristics	(ii) SEM
(c) Topology	(iii) XRD
(d) Electrical characteristics	(iv) EDAX

A.a-i, b-ii, c-iv, d-iii
B.a-iv, b-ii, c-i, d-iii
C.a-iv, b-iii, c-ii, d-i
D.a-iii, b-i, c-ii, d-iv

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The correct matching is a-iv, b-iii, c-ii, d-i.

Chemical analysis: EDAX (energy-dispersive X-ray analysis) identifies elements from their characteristic X-rays (iv).
Structural characteristics: XRD (X-ray diffraction) reveals crystal structure and lattice spacing (iii).
Topology: SEM (scanning electron microscope) images surface topography (ii).
Electrical characteristics: a probe station makes electrical contact with devices on a wafer for I-V and C-V measurements (i).

The other codes swap XRD and SEM, or give chemical analysis to the probe station.

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20.

Positive photo resists are used more than negative photo resists because :

A.Negative photo resists are more resistive to light, but their photolithographic resolution is not as high as that of the positive photo resists
B.Positive photo resists are more sensitive to light, but their photolithographic resolution is not as high as that of negative photo resists
C.Negative photo resists are less sensitive to light
D.Positive photo resists are less sensitive to light

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The key gives the option stating that negative photoresists do not have as high a resolution as positive photoresists.

A negative resist becomes insoluble where it is exposed. During development the unexposed regions dissolve, but the exposed, cross-linked regions absorb the developer and swell, which distorts fine lines. A positive resist does not swell this way, so it reproduces much smaller features. That higher resolution is why positive resists are preferred for modern VLSI, even though negative resists react to light more readily.

The option claiming positive resists have the lower resolution reverses the decisive fact. Negative resists are not less sensitive to light, and the lower sensitivity of positive resists is a drawback, not the reason they are used more.

Discuss in Forum

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