EE 147/247A Prof. Pister Fall 2020
Homework Assignment #2 Rubric
1. [7 pts total] You have a solid cube of aluminum of side length S sitting a distance d beneath a large flat conductive plate, creating a parallel plate gap between them. You apply a voltage V between the cube and the plate, which generates an upwards force on the cube. If d=S/10
a. [1 pt] Does the magnitude of the force depend upon S? 1 point for โnoโ. It would be nice to see the derivation, but not required.
๐น =12 ๐&๐
( ๐ด๐( =
12 ๐&๐
( ๐ โ ๐
- ๐10/( =
12 ๐&๐
((100)
b. [3 pts] Using the approximation that ยฝ e0(15V)2=1nN, what is the force on the cube if
V=1.5V, 15V, and 150V? If you used a calculator instead of the approximation, take 1 point off! There should be no more than one significant figure in your answer.
1.5V: 1003(๐&(1.5๐)( = 1๐๐
15V: 1003(๐&(15๐)( = 100๐๐
150V: 1003(๐&(150๐)( = 10๐ข๐
c. [3 pts] For the three voltages above, calculate the range of values of S for which the mass
of the cube is less than the electrostatic force lifting it up (note: -1 point for me for comparing โthe mass of the cubeโ to a force. I should have said the weight of the cube - ksjp) If you just solved for the value at which the forces are equal, but didnโt give the range of values, subtract 1 point total even if you did it three times. Itโs fine if you wrote โor lessโ or anything that indicates a range.
๐น9: = ๐ โ ๐ = ๐ โ ๐=> โ ๐ = ๐?(2710) โ 10=27100๐?๐ โ 25000๐? N
1.5Vโ 1๐๐:๐น9: < 1๐๐ ๐? < 4 โ 10F3G๐?
S<3.5 โ 10FI๐
15Vโ 100๐๐:๐น9: < 100๐๐ ๐? < 4 โ 10F3(๐?
S<1.5 โ 10FG๐
150Vโ 10๐ข๐:๐น9: < 10๐ข๐ ๐? < 4 โ 10F3&๐?
S<7 โ 10FG๐
2. [14 pts. total] In a single-mask SOI process in which your minimum lithography line and space is
l, you wish to make an accelerometer with a variable capacitor readout, like our simplest accelerometer (e.g. in lecture W2L2 at around 34 minutes). Two of the many capacitor finger pairs are shown below, showing the desired front gap gf, and the undesired back gap gb.
a. [4 pts] For a displacement of the proof mass over the range y = [-gf, gb], sketch the front gap capacitance, back gap capacitance, and total capacitance assuming gb =10 gf. You
should have some asymptotes! Show clearly the relative front and back capacitance values when y=0.
1 point each for getting the asymptotes right on each curve (3 pts total)
1 point for showing that the back gap is 10 times lower capacitance than front gap at y=0
b. [1 pt] What is the impact of the back gap on dC/dy near y=0? 1 point for stating that the back gap reduces the overall dC/dy, or saying that it has a negative impact.
Donโt need a numerical answer for full credit, but it would be nice.
๐ถ(๐ฆ) = ๐&๐ก๐ฟ(1
๐N โ ๐ฆ+
1๐Q + ๐ฆ
)
๐ถR =๐๐ถ(๐ฆ)๐๐ฆ = ๐&๐ก๐ฟ(
1(๐N โ ๐ฆ)(
โ1
(๐Q + ๐ฆ)()
๐๐ถ(๐ฆ)๐๐ฆ
TUV&
= ๐&๐ก๐ฟ(1
(๐N)(โ
1(๐Q)(
)
So when gb=10gf, the magnitude of dC/dy is reduced by only 1%. Clearly we can get a bigger dC/dy signal if we put the fingers closer than this.
c. [5 pts] Repeat parts a and b, assuming gb = gf.
๐ถR =๐๐ถ(๐ฆ)๐๐ฆ = ๐&๐ก๐ฟ(
1(๐N โ ๐ฆ)(
โ1
(๐Q + ๐ฆ)()
๐๐๐N = ๐Q๐๐๐๐ฆ = 0, ๐กโ๐๐๐ถR(0) =๐๐ถ(0)๐๐ฆ = 0
Must state that dC/dy=0 to get the last point. Note that dC/dy=0 means that there is no net electrostatic force, and no capacitive signal to detect. So now weโve definitely put the fingers too close!
Making gb too big means we donโt get very many fingers, so we donโt have much capacitance per layout area, and there is an upper limit on how much area we can use. Making gb too small means that we can put in a lot of fingers, but the change in capacitance per finger gets smaller. Somewhere there is an optimum.
d. [4 pts] Assuming that the width of the capacitor fingers and the front gap is l, what is the value of the back gap that maximizes dC/dy at y=0 per unit layout area? Iโve drawn a representative tile-able layout area in red below.
2 points for writing an equation for dC/dy per layout area (1 point if you wrote an equation for capacitance per area instead)
1 point for taking a derivative with respect to the back gap gb 1 point for getting the right answer
As in part b: ]^(U)
]U= ๐&๐ก๐ฟ(
3(:_FU)`
โ 3(:abU)`
)
๐๐ถ(๐ฆ)๐๐ฆ
TUV&
= ๐&๐ก๐ฟ(1
(๐N)(โ
1(๐Q)(
)
The Area denoted by the dotted red line is A=L(๐N + ๐Q + 2๐ค);๐คโ๐๐๐๐ค๐๐ ๐กโ๐๐๐๐๐๐๐๐ค๐๐๐กโ
Area Density AD =gh(i)gi j
ikl
=
AD = mlno(
p
qr_sit`F
p(raui)`
)
v(:_b:ab(w)
But ๐Q = ๐ค = ๐ so AD = mlno(
p
qr_t`F
py`)
v(:_b?z)
And we can cancel and L before deriving ]={]:_
= mln(:_|F?:_z`F}z|):_|z|(:_b?)`
= 0 ๐&๐ก(๐N? โ 3๐N๐( โ 6๐?) = 0
The only real root is ๐N = 2.35๐ As an example, if you have 2 um line and space lithography and you are making a gap-closing actuator or capacitive sensor, you maximize your force or capacitance change per unit area by making the front gap and fingers 2 um wide, and the back gap 4.7 um wide. Note that if the fingers and the front gap are not the same size (sometimes we make the fingers wider to avoid electrostatic pull-in) then the optimal back gap will vary somewhat.
3. [4 pts] For this problem, assume that the fracture strain of silicon is 1%. You may use 50 kg as
your mass if youโd rather not disclose personal information. a. [2pts] What is the minimum cross-sectional area of a silicon wire that will hold your
weight without breaking? 2 points if you get the right answer. 1 point for some explanation and the wrong answer
๐ = 50๐๐ ๐น = ๐๐ = ๐๐ฅ=๏ฟฝ=
o๐ฅ = ๐ธ๐ด๐ โ ๐ด = ๏ฟฝ
๏ฟฝm =9:
๏ฟฝm
๐ด =I&๏ฟฝ:โ3&๏ฟฝ
๏ฟฝ`
3I&๏ฟฝ๏ฟฝ๏ฟฝโ&.&3 = 3&
?โ3&๏ฟฝs` = 3 โ 10F๏ฟฝ ๐(
1
y gf
gb