1.
a. 1 pt. for a correct cross-section
With perfect selectivity
PIT mask:
DIMPLE is 30um on a side
P1_P2_VIA is 16um on a side
ANCHOR2 is 6um on a side
b.
i. 1 pt. for a correct cross-section
ii. 1 pt. for a correct cross-section
Assuming a 175% overetch, we etch the full 1.5um of
deposited POLY2. The nitride at the center of the pit gets
(1.5um)(1.75)/5=0.525um etched away during the overetch.
This does not fully breach the 0.6um of nitride.
iii. 1 pt. for stating that the nitride is not etched
The nitride is not breached. Assuming a more aggressive
overetch or a lower selectivity, the nitride could potentially be
etched away. It should noted that in Fouladi et al. the nitride
breach actually etches 3um into the silicon substrate! (Figure
2).
If you made assumptions about the other overetches that gave you different
answers (like P1P2via etching into ox1, and ANCHOR2 etching into nitride), give
yourself full credit as long as your results are consistent with your
assumptions. When I wrote this problem I thought that nitride was 0.5um thick (-1
for me!) and chose the selectivity such that nitride would be breached. -ksjp
2. 1 pt showing that the polysilicon in the structure will get etched away
3. 1 pt. for using oxide as the structural material
1 pt. for using POLY0 at the base of the cantilever as the piezoresistive
sensing element
1 pt. for etching the substrate to release the cantilever
1 pt. for relating force to strain at the base of a cantilever
1 pt. for estimating the minimum detectable force using RMS noise
The layout and cross-section of the basic structure are shown below. The
only two masks are POLY0 to define the piezoresistor and PIT to define
where the silicon substrate is exposed and etched away by the TMAH etch.
Some simplifying assumptions:
 The poly0 is at bottom surface at the base of the cantilever
 The entire structure has the Young’s Modulus of oxide
The maximum strain will occur at the surface of the beam where z=a/2
𝜖=
𝑧
𝑎
=
𝜌 2𝜌
Radius of curvature can be related to the moment on the beam by
1 𝑀
=
𝜌 𝐸𝐼
The moment along the beam is given by
𝑀(𝑥) = 𝑀0 + 𝐹(𝐿 − 𝑥)
For a pure point force at the tip of the beam, 𝑀0 = 0. At the base of the beam,
x=0
1 𝑀(0) 𝐹𝐿
=
=
𝜌
𝐸𝐼
𝐸𝐼
2
𝐹𝐿
𝑎𝐹𝐿
𝜖=
→𝜖=
𝑎
𝐸𝐼
2𝐸𝐼
Assuming we have the piezoresistor in a wheatstone bridge configuration
with excitation voltage 𝑉𝑥
𝑉𝑜𝑢𝑡 =
𝐺𝑉𝑥
𝐺𝑉𝑥 𝑎𝐹𝐿
𝜖=
4
4 2𝐸𝐼
For 𝐺 = 30, 𝑉𝑥 = 1.5V, 𝑎 = 3.5um (thickness of the oxides), 𝐿 = 500um,
𝑉
𝐸 = 70GPa, and beam width 𝑏 = 5um, 𝑉𝑜𝑢𝑡 ≅ 30000 [𝑁] 𝐹
For the minimum detectable signal, we will have noise from the damping of
the cantilever and the voltage across the resistor
𝑃𝑛,1
𝑃𝑛,2
2
𝐹̅𝑛,1
=
= 4𝑘𝐵 𝑇∆𝑓
𝑏
𝑉 2
𝑣̅𝑛2 30000 [𝑁] 𝐹̅𝑛,2
=
=
= 4𝑘𝐵 𝑇∆𝑓
𝑅
𝑅
For 𝑏 = 10−5 , 𝐹𝑛,1 = (4 × 10−13 )
For 𝑅 = 10000𝛺, 𝐹𝑛,2 = (7.5 ×
N
√∆𝑓
√Hz
N
10−11 )
√∆𝑓
√Hz
The voltage noise dominates for large resistor values. We can bring the
resistor value down so that our smallest source of noise is the damping of the
cantilever. This happens when 𝑅 < 1𝛺
4. 1 pt. for finding the variance
𝑘 𝑇
Going from -40C to +85C, 𝐵2 increases by over 54%
5. 1 pt. for correct reasoning
𝑘𝐵 𝑇
is greatly reduced close to absolute zero. This lowers the noise floor of
2
the experiment, allowing detection of very small signals.
6. 1 pt. for an
The power spectral density is given by
𝑃𝑛 = 4𝑘𝐵 𝑇∆𝑓
The power dissipated through the resistor will be
𝑣̅𝑛2
𝑃𝑛 =
= 4𝑘𝐵 𝑇∆𝑓
𝑅
𝑣𝑛 = √4𝑘𝐵 𝑇𝑅∆𝑓
We know that vn= 4nV/rt-Hz for R=1k, so for f=10kHz, vn=400nV. The noise
equivalent acceleration will be the noise voltage divided by the sensitivity, so
400nV/(1mv/’g’) = 0.4 m’g’ or
𝑣𝑛 √4𝑘𝐵 𝑇𝑅∆𝑓
𝑎𝑛 =
=
𝑆
𝑆
√4 (1.38 × 10−23 J ) (300K)(1000Ω)(10000Hz)
K
𝑎𝑛 =
mV
1 g
𝑎𝑛 = 4 × 10−4 g
7. 1 pt.
W
Power spectral density is always 4𝑘𝐵 𝑇 [Hz]. There is no dependence on any
other physical parameter besides temperature.
8. 1 pt. for each part (5pts. total)
a.
1
1
𝑘𝐵 𝑇
𝑘𝑥̅𝑛 2 = 2 𝑘𝐵 𝑇 → 𝑥𝑛 = √
2
𝑘
=√
4×10−21 𝑁𝑚
0.2 𝑁/𝑚
= 1.4 × 10−10 m
W
b. 𝑃𝑛 = 4𝑘𝐵 𝑇𝛥𝑓 = 1.7 × 10−20 Hz 𝛥𝑓
W
c. 𝐹𝑛 = √4𝑘𝐵 𝑇𝑏𝛥𝑓 = √(1.7 × 10−20 Hz) (10−7
Ns
m
) (1Hz) = 4 × 10−14 N
d.
i. 𝑥𝑛,𝐷𝐶 =
𝐹𝑛
𝑘
=
4×10−14 N
0.2 N/m
= 2 × 10−13 m
ii. 𝑥𝑛,𝜔𝑛 = 𝑄𝑥𝑛,𝐷𝐶 = 30(2nm) = 60 × 10−13 m
𝑥𝑛,𝐷𝐶
iii. 𝑥𝑛,10𝜔𝑛 = 10
= 0.1 × 10−13 m
2
e. They are consistent. Equipartition, used in part a, tells you the noise
on every energy storage element such as a capacitor or a spring.
Fluctuation-dissipation tells you that there is constant noise power
across all frequencies and this presents itself in elements like
resistors and dampers. Integrating the noise density that you get from
fluctuation-dissipation across all frequencies will give you
equipartition.
9. 1 pt. for each correct answer (5 pts. total)
a. Release, the drying is performed at the critical point of CO2, so that
the structure can dry without crossing a phase boundary.
b. Both, dimples reduce device-substrate overlapping surface area and
effective length of suspended structures
c. Release, the XeF2 etch is gaseous. There is no liquid
d. Both, SAM’s make surfaces hydrophobic, reducing stiction
e. Release, same reason as XeF2. It is a gas phase etch.