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16A
Subject
Electrical Engineering
Date
Dec 6, 2023
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Last Updated: 2023-10-28 12:46
1
EECS 16A
Designing Information Devices and Systems I
Fall 2023
Homework 9
This homework is due November 3, 2023, at 23:00
Self-grades are due November 6, 2023, at 23:00.
Submission Format
Your homework submission should consist of
one
fi
le.
•
hw9.pdf
: A single PDF
fi
le that contains all of your answers (any handwritten answers should be
scanned).
Submit the
fi
le to the appropriate assignment on Gradescope.
1. Reading Assignment
For this homework, please read Notes 16 and 17. Note 16 will provide an introduction to capacitors (a cir-
cuit element which stores charge), capacitive equivalence, and the underlying physics behind them. Sections
17.1 - 17.2 in Note 17 will provide an overview of the capacitive touchscreen and how to measure capac-
itance. 17.3 will introduce you to the op-amp and how it can be used as a comparator. 17.4 will discuss
implementing a capacitive touchscreen with an op-amp comparator.
(a) Describe the key ideas behind how a capacitor works. How are capacitor equivalences calculated?
Contrast this with how we calculate resistor equivalences.
(b) Consider the capacitive touchscreen. Describe how it works, and compare and contrast it to the resistive
touchscreens we have seen in previous lectures and homeworks.
2. Equivalent Capacitance
(a) Find the equivalent capacitance between terminals
a
and
b
of the following circuit in terms of the given
capacitors
C
1
,
C
2
,
and
C
3
. Leave your answer in terms of the addition, subtraction, multiplication, and
division operators
only
.
a
C
1
b
C
2
C
3
(b) Find and draw a capacitive circuit using three capacitors,
C
1
,
C
2
, and
C
3
, that has equivalent capacitance
of
C
1
(
C
2
+
C
3
)
C
1
+
C
2
+
C
3
UCB EECS 16A, Fall 2023, Homework 9, All Rights Reserved. This may not be publicly shared without explicit permission.
1
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2
3. Dynamic Random Access Memory (DRAM)
Nearly all devices that include some form of computational capability (phones, tablets, gaming consoles,
laptops, ...) use a type of memory known as Dynamic Random Access Memory (DRAM). DRAM is where
the “working set” of instructions and data for a processor is typically stored, and the ability to pack an
ever increasing number of bits on to a DRAM chip at low cost has been critical to the continued growth in
computational capability of our systems. For example, a single DRAM chip today can store
>
8 billion bits
and is sold for
≈
$3-$5.
At the most basic level and as shown below, every bit of information that DRAM can store is associated
with a capacitor. The amount of charge stored on that capacitor (and correspondingly, the voltage across the
capacitor) determines whether a “1” or a “0” is stored in that location.
C
bit
+
−
V
bit
Access Switch
Single DRAM Bit Cell
In any real capacitor, there is always a path for charge to “leak” off the capacitor and cause it to eventually
discharge. In DRAMs, the dominant path for this leakage to happen is through the access switch, which we
will
model as a leakage to ground
. The
fi
gure below shows
a model of this leakage
:
C
bit
I
leak
Fun Fact: This leakage is actually responsible for the “D” in “DRAM” – the memory is “dynamic” because
after a cell is “written” by storing some charge onto its capacitor, if you leave the cell alone for too long,
the value you wrote in will disappear because the charge on the capacitor leaked away.
Let’s now try to use some representative numbers to compute how long a DRAM cell can hold its value
before the information leaks away.
Let
C
bit
=
28fF (note that 1fF
=
1
×
10
−
15
F) and the capacitor be
initially charged to 1
.
2V to store a “1.”
V
bit
must be
>
0
.
9V in order for the circuits outside of the column
to properly read the bit stored in the cell as a “1.”
What is the maximum value of
I
leak
that would allow the DRAM cell retain its value for
>
1ms
?
Hint: Start by writing out the equations you know about charge and current related to the capacitor. Note
here that the current source is discharging the capacitor.
4. It’s
fi
nally raining!
UCB EECS 16A, Fall 2023, Homework 9, All Rights Reserved. This may not be publicly shared without explicit permission.
2
Last Updated: 2023-10-28 12:46
3
A lettuce farmer in Salinas Valley has grown tired of weather.com’s imprecise rain measurements. Therefore,
they decided to take matters into their own hands by building a rain sensor.
They placed a rectangular
tank outside and attached two metal plates to two opposite sides in an effort to make a capacitor whose
capacitance varies with the amount of water inside.
a
C
air
b
C
H
2
O
h
H
2
O
w
h
tot
The width and length of the tank are both
w
(i.e., the base is square) and the height of the tank is
h
tot
.
(a) What is the capacitance between terminals
a
and
b
when the tank is full? What about when it is empty?
Note:
the permittivity of air is
ε
, and the permittivity of rainwater is 81
ε
.
(b) Suppose the height of the water in the tank is
h
H
2
O
. Modeling the tank as a pair of capacitors in parallel,
fi
nd the total capacitance between the two plates. Call this capacitance
C
tank
.
(c) After building this capacitor, the farmer consults the internet to assist them with a capacitance-measuring
circuit. A fellow internet user recommends the following:
I
s
C
tank
I
c
+
−
V
C
(
t
)
,
V
C
(
0
) =
0V
In this circuit,
C
tank
is the total tank capacitance that you calculated earlier.
I
s
is a known current
supplied by a current source.
The suggestion is to measure
V
C
for a brief interval of time, and then use the difference to determine
C
tank
.
Determine
V
C
(
t
)
, where
t
is the number of seconds elapsed since the start of the measurement. You
should assume that before any measurements are taken, the voltage across
C
tank
, i.e.
V
C
, is initialized
to 0V, i.e.
V
C
(
0
) =
0.
(d) Using the equation you derived for
V
C
(
t
)
, describe how you can use this circuit to determine
C
tank
and
h
H
2
O
.
(e) However, after spending some time thinking about different ways of measuring this capacitance you
came up with a better idea. You decided to use the circuit proposed in part (c) along with a comparator,
as show in the
fi
gure below.
What you are basically interested in, is the time
T
1
needed for
V
c
to
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4
reach
V
re f
. In order to measure time you use a timer. When voltage
V
c
becomes larger than
V
re f
, the
comparator
fl
ips its value and you stop the timer. How would you measure in that case the value of the
capacitance?
5. Capacitive Touchscreen
(Contributors: Deepshika Dhanasekar, Panos Zarkos, Richard Liou, Wahid Rahman, Urmita Sikder)
The model for a capacitive touchscreen can be seen in Figure 1. See Table 1 for values of the dimensions.
The green area represents the contact area of the
fi
nger with the top insulator. It has dimensions
w
2
x
d
1
,
where
w
2
is the horizontal width of the
fi
nger contact area and
d
1
is the depth (into the page) of the
fi
nger
contact area. The top metal (red area) has dimensions
w
1
×
d
1
. The bottom metal (grey area) has dimensions
w
×
d
2
, where
w
is larger than both
w
1
and
w
2
.
Table 1: Touchscreen Dimension Values
d
1
10 mm
d
2
1 mm
t
1
2 mm
t
2
4 mm
w
1
1 mm
w
2
2 mm
(a) Draw the equivalent circuit of the touchscreen that contains the nodes
F
,
E
1
, and
E
2
when: (i) there no
fi
nger present; and (ii) when there is a
fi
nger present. Express the capacitance values in terms of
C
0
,
C
F
−
E
1
, and
C
F
−
E
2
.
Hint: Note that node F represents the
fi
nger. When there is no touch node F would be non-existent.
Hint: Treat E
1
as the "top node", E
2
as the "bottom node", and the
fi
nger F as an intermediate node
when present.
(b) What are the values of
C
0
,
C
E
1
−
F
, and
C
F
−
E
2
? Assume that the insulating material has a permittivity
of
ε
=
4
.
43
×
10
−
11
F
/
m
and that the thickness of the metal layers is small compared to
t
1
(so you can
ignore the thickness of the metal layers).
(c) What is the difference in effective capacitance between the two metal plates (nodes
E
1
and
E
2
) when a
fi
nger is present?
UCB EECS 16A, Fall 2023, Homework 9, All Rights Reserved. This may not be publicly shared without explicit permission.
4
Last Updated: 2023-10-28 12:46
5
Top view
Side view
Side view
d
2
d
1
t
1
w
1
Figure 1: Model of capacitive touchscreen.
6. Super-Capacitors
In order to enable small devices for the “Internet of Things” (IoT), many companies and researchers are cur-
rently exploring alternative means of storing and delivering electrical power to the electronics within these
devices. One example of these are “super-capacitors” - the devices generally behave just like a “normal”
capacitor but have been engineered to have extremely high values of capacitance relative to other devices
that
fi
t in to the same physical volume. They can function as a power supply for low power applications such
as IoT devices and have the advantage that they can be charged and discharged many times without losing
maximum charge capacity. That property makes super-capacitors suitable to store energy from intermittent
power sources such as those from energy harvesting. Suppose you are tasked with designing a power supply
with a super-capacitor in an IoT device.
(a) Assuming that your electronic device (load) can be modeled as a constant current source with a value of
i
load
, draw circuit models for your device using super-capacitors as the power supply with the following
con
fi
gurations:
• Con
fi
g 1: a single super-capacitor as the power supply
• Con
fi
g 2: two super-capacitors stacked in series as the power supply
• Con
fi
g 3: two super-capacitors connected in parallel as the power supply
(b) If each super-capacitor is charged to an initial voltage
v
init
and has a capacitance of
C
sc
, for each of the
three con
fi
gurations above, write an expression for the voltage supplied to your electronic device as a
function of time after the device has been activated (i.e. connected to the super-capacitor(s)).
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5
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6
(c) Now let’s assume that your electronic device requires some minimum voltage
v
min
in order to function
properly. For each of the three super-capacitor con
fi
gurations, write an expression for the lifetime of
the device.
(d) Assume that a single super-capacitor doesn’t provide you suf
fi
cient lifetime and so you have to spend
the extra money (and device volume) for another super-capacitor.
You consider the two following
con
fi
gurations:
• Con
fi
g 2: two super-capacitors stacked in series
• Con
fi
g 3: two super-capacitors connected in parallel
When is Con
fi
g 3 (parallel) better than Con
fi
g 2 (series)? Your answer should involve conditions on
v
init
and
v
min
.
(e) Calculate the amount of energy delivered by the super-capacitors in Con
fi
g 3 (parallel) over the de-
vice’s lifetime.
7. Homework Process and Study Group
Who did you work with on this homework? List names and student ID’s. (In case you met people at home-
work party or in of
fi
ce hours, you can also just describe the group.) How did you work on this homework?
If you worked in your study group, explain what role each student played for the meetings this week.
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Q
bit
=
V
bit
∗
C
bit
Write down equations that are known: I
leak
= (1
.
2
V
−
0
.
9
V
)
∗
28
fF/
1
∗
10
−
3
S
I
leak
=
dV
bit
/dt
∗
C
bit
Differentiate with respect to time: Plug in Values: Final Answer: 8
.
4
pA
C
empty
=
ε
air
h
tot
w
w
=
ε
h
tot
C
full
=
ε
H
20
h
tot
w
w
= 81
ε
h
tot
C
air
=
ε
air
(
h
tot
−
h
H
2
O
)
w
w
=
ε
(
h
tot
−
h
H
2
O
)
C
tank
=
C
water
+
C
air
=
ε
(
h
tot
+ 80
h
H
2
O
)
I
C
=
C
tank
dV
C
dt
, I
C
=
I
s
dV
C
dt
=
I
s
C
tank
V
C
(
t
) =
I
s
C
tank
t
It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, positive charge accumulates on one plate, and negative charge accumulates on the other plate, creating an electric field between them. The capacitance (C) of a capacitor measures its ability to store charge per unit voltage. It is calculated using the formula:
C=Q/V, where Q is the charge stored on the plates and V is the voltage across the capacitor.
Series Capacitors: When capacitors are connected in series, their total capacitance (C
_
total) is calculated using the formula
Parallel Capacitors: When capacitors are connected in parallel, their total capacitance is the sum of the individual capacitances:
C
0
=
ε
d
2
w
1
t
1
= 2
.
215
∗
10
−
14
F
C
F
−
E
1
=
ε
d
1
w
1
t
2
−
t
1
= 2
.
215
∗
10
−
13
F
C
F
−
E
2
=
ε
d
2
(
w
2
−
v
1
)
t
2
= 1
.
108
∗
10
−
14
F
V
out
(
t
) =
V
init
×
e
−
t
R
eq
C
sc
(b) Expression for Voltage Supplied to the Electronic Device:
Config 1: Single Super-Capacitor
Config 2: Two Super-Capacitors in Series
V
out
(
t
) = 2
V
init
×
e
−
t
R
eq
C
sc
Config 3: Two Super-Capacitors in Parallel
V
out
(
t
) =
V
init
×
e
−
2
t
R
eq
C
sc
(c) Expression for Device Lifetime:
For each configuration, the lifetime of the device can be calculated
using the time it takes for Vout (t) to decrease to Vmin.
−
R
eq
C
sc
ln
±
V
sa
V
sa
²
−
R
e
C
sc
2
ln
±
V
sa
V
sa
²
−
R
eq
C
sc
2
ln
±
V
min
V
int
²
Config 3 (parallel) is better than Config 2 (series) when the following condition is met:
The total energy delivered by the super-capacitors in Config 3 over the device's lifetime can be calculated using the following formula:
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ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON

Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,