292 7.6 High-Throughput Techniques
objects, for example, from attaching smaller objects together in a modular fashion and util
izing origami methods to fold several smaller printed sheetlike structures together to gen
erate complex 3D shapes.
Worked Case Example 7.1: Microfabrication
A silicon substrate was spin-coated with an SU-8 photoresist by spinning at 3000 rpm in
order to ultimately generate a layer of silicon oxide of the same thickness as the sacrificial
photoresist material.
a
To generate a 0.5 μm thick layer of silicon oxide, how many minutes must the spin-
coating of the SU-8 proceed for?
In a subsequent step after the removal of the photoresist and deposition of the
silicon oxide, silicon oxide was coated with a 10 nm thick layer of gold for a surface
plasmon resonance application, employing evaporation deposition using a length of
gold wire evaporated at a distance of 5 cm away from the silicon substrate.
b
If the gold wire has a diameter of 50 μm and is wound tightly onto an electric heating
filament under a high vacuum, which melts and vaporizes the gold completely,
explain with reasoning how many centimeters of wire are needed to be used, stating
any assumptions you make.
(Assume that the density and dynamic viscosity of the SU-8 used are 1.219 g cm−3 and
0.0045 Pa · s.)
Answers
a
Assuming a high time approximation, we can rearrange Equation 7.7 to generate
the time t required for a given photoresist thickness h of such that
t
h
=
3
4
2
2
η
ρω
Thus,
t =
×
×
×
×
×
−
(3
0.0045) (4
(1.219
10 ) kg m
(3000/60
2 )
rad
3
3
2
2
/
π
s
(20
10
) m)
112 s
1.9 min
2
6 2
−
−
×
×
=
=
b
If a mass m of gold vaporizes isotropically, then the mass flux per unit area at a
distance d from the point of vaporization (here 5 cm) will be m/4πd2. Thus, over
a small area of the substrate δA, the mass of gold vapor deposited assuming it
solidifies soon after contact will be
δ
δ
π
ρ δ
δ
m
A m
d
A
Z
Au
=
⋅
=
⋅
/4
2
where the density of gold is ρAu, and the thickness of the deposited gold on the
silicon oxide substrate is δz, which is 10 nm here. Thus,
δ
π
ρ
z
m
d
Au
=
/4
2
But the mass of the gold is given by
m
rAU
Au
= /4
2
π
ρ
where l is the length of gold wire used of radius rAu.