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Microsoft Word - 145L_2005_FS EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 1
S. Derenzo
UNIVERSITY OF CALIFORNIA, BERKELEY
College of Engineering, Electrical Engineering and Computer Sciences Department
1.1 A PIN photodiode is a diode with a transparent contact and works by converting incident light into
electron-hole pairs that are separated by an electric field and collected at the output leads to produce
a current that is proportional to the light intensity
[4 points off for stating how it is used but not how it works]
[4 points off if photons or light not mentioned]
[4 points off if electrons not mentioned- this is essential in describing how light is converted into a
current]
1.2 A light emitting diode is a doped wide-band-gap semiconductor that works by using electrical power
to raise electrons to an excited state. Light is emitted when the electrons return to the ground state.
[4 points off if electrons not mentioned- this is essential in describing how current is converted into
light]
1.3 The metal foil strain gauge consists of a thin strip of metal bonded to an insulating layer and works
by changing its resistance when strain changes its dimensions.
[4 points off if not stated that the change in resistance is caused by a change in dimension]
1.4 The Peltier thermoelectric heat pump works by using electrical power to drive electrons through
alternating p-type and n-type semiconductor elements. Cooling occurs when the electrons enter the
n-type material. Heating occurs when the electrons are forced into the p-type material.
[2 points off for not mentioning the role of electrons]
[2 points off for not mentioning alternating p-type and n-type semiconducting elements]
1.5 The ground fault interrupter uses a differential transformer, rectification, and amplification to detect
the current difference between two current carrying conductors. If the difference exceeds 5 mA, the
amplified signal trips an electromagnetic relay that disconnects both conductors. The relay is reset
with a manual push button.
[3 points off not for describing that the GFI detects the difference between power and ground (hot
and neutral)]
[2 points off for not giving a specific current difference that causes a trip]
[2 points off for not describing the reset function]
1.6 The bimetal switch consists of two strips with different thermal expansion coefficients bonded
together and works by bending to make an electrical contact at a selected temperature.
[4 points off if two strips are not mentioned]
1.7 The Ag(AgCl) electrode consists of a silver element bonded to AgCl. It can convert electrons in the
silver into Cl ions in solution by the reaction AgCl + e Ag + Cl and convert Cl ions in
solution into electrons in the silver by the reverse reaction.
[2 points off for describing only one charge transfer reaction]
[4 points off for not describing either reaction] EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 2
S. Derenzo
2.1
Instrumentation
amplifier
User's
amplifier
gain G
R
T
V
b
V
1
V
2
V
0
V
1
V
2
R


V
OO
= G V
RTI
+ V
RTO



exp 1T 1
293













Note: Increases with increasing T


R
T
= 1 k exp +
1T 1293













Note: Decreases with increasing T


V
2
= V
b
R
R
T
V
b
R
1 k




exp 1T 1
293













Note: R
T
in denominator so V
2
has proper T dependence


V
0
= G(V
1
V
2
)
+ V
OO
To eliminate the offset voltage, we want

V
0
= GV
1
which requires

GV
2
= V
OO


V
b
GR
1 k




exp 1T 1
293











= GV
RTI
+ V
RTO



exp 1T 1
293













V
b
R
= (1 k)(1 µV + 100 µV/G)
[3 points off if compensation decreases with increasing temperature- a common mistake was to use
the thermistor as the feedback resistor of an inverting amplifier]
[2 points off if input to the compensation circuit does not have a high input impedance]
[5 points off if it is not clear how the output offset compensating circuit depends on the values of
V
RTI
, V
RTO
, and G.
[6 points off if the thermistor is used in a bridge with 1 k (or unspecified) resistors- 1/(R
T
+ 1 k)
has a very different temperature dependence than the desired 1/R
T
] EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 3
S. Derenzo
2.2 For G = 100 at 20°C, R
T
= 1 k, V
b
= 2 mV, R = 1
3.1 Sensor: thermistor or solid-state temperature sensor
[3 points off for thermocouple, which would require an electronic ice point- if you had an electronic
ice point, you would not need the thermocouple]
Actuator: thermoelectric heat pump
[3 points off for heater which is not a specific actuator]
3.2 Sensor: thermocouple or platinum resistance thermometer
[3 points off for using a thermistor or solid-state temperature sensor- they are semiconductors that
would be destroyed at 500 °C]
Actuator: resistive heater
[3 points off for using a thermoelectric heat pump as an actuator- it is a semiconductor that would
be destroyed at 500 °C]
Another correct answer is a bimetallic switch, which acts both as a sensor and an actuator
3.3 Sensor: piezoelectric crystal or embedded piezoresistors in a silicon disk
Actuator: motor plus compresssor pump or solinoid plus piston
3.4 Sensor: Circular resistor or digital encoder
Actuator: stepping motor
4.1 Green is reflected, so all other wavelengths are absorbed.
4.2 Fluorescent
4.3 Assuming 100% quantum efficiency, the power received by a 1 cm
2
photodetector under full
sunlight is (10
4
m
2
) (1000 W m
2
) = 0.1 W.
0.1 W of 2 eV photons produces 50 mA of closed-loop photovoltaic current.
Open-circuit conditions produces 0 mA and 0.6 V max (voltage saturation).
Load resistor must be < 0.6 V/0.05 A = 12 or photodiode will saturate.
A 10 load resistor will produce 0.5 V when the photodiode receives full sunlight. This determines
the set point voltage. EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 4
S. Derenzo
0.5 V
10
50 mA
Eror signal
Controller
integrator or
rectifier
Power
amplifiers
High power,
efficient lamps
[5 points off if photodiode current not calculated]
[2 points off if set point not calculated or load resistor is > 12 (which would require a
photovoltaic output > 0.6 volts)]
[3 points off for omitting a control circuit that uses the error signal to power the artificial lights]
4.4
Sensor
Controller,
amplifiers
Greenhouse
Natural light
Artificial
light EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 5
S. Derenzo
5.1
Light
source
Bread
Reflectance
signal R
Set point =
desired D
Error
signal
Heater
control
P h o t o -
diode
Toast
mode
Difference
amplifier
D =
Rc R
Rc
R
Difference
amplifier
Calibrate
mode
C
V
b
R
5.2

The photodiode constantly records R, the light reflected from the surface of the toast. Use optical
filtering so that the photodiode does not see the red glow of the heater elements. In calibrate mode, the heater is off and the reflected signal Rc is stored on capacitor C In toast mode, the heater is on, capacitor C is disconnected from the photodiode, and the first
difference amplifier generates D = Rc R, which is the amount the reflected light intensity has been
reduced by toasting. The second difference amplifier generates an error signal, which is the
difference between D and the desired darkening D. The heater control turns off the heaters when the
error signal goes negative.
[5 points off for not describing a light source for illuminating the bread]
[5 points off for not describing the light sensing circuit]
[5 points off for not saving the initial reflected signal]
[5 points off for not generating a set point that corresponds to the desired level of toasting]
[5 points off for not turning off the toaster when D = RcR > Desired D.
An alternative accepted solution used a matched second light sensing system that used a piece of toast as
a reference and the error signal was derived from the difference between the two light sensing systems.
The disadvantage of using a reference toast is that it will become moldy in a few days and this will
change its color. This is readily fixed by using a painted reference surface, which will have much better
color stability. EECS 145L Final Examination Solutions (Fall 2005)
December 17, 2005
page 6
S. Derenzo
145L FINAL EXAM GRADE STATISTICS
Problem
1
2
3
4
5
Total
Average
51.8
10.4
32.1
41.2
30.4
166.0
rms
9.2
4.1
7.0
7.4
3.5
26.7
Maximum
63
20
36
48
33
200
Total score distribution:
70-79 0
80-89 1
90-99 0
100-109 0
110-119 0
120-129 1
130-139 1
140-149 0
150-159 2
160-169 3
170-179 5
180-189 3
190-199 3
200 0
145L COURSE GRADE STATISTICS
Grade
Undergraduate
Graduate
Scores
Scores
A+
960.5
A
955.4, 951.3
A
939.3, 926.4
B+
915.9, 910.4, 908.8, 908.0, 896.3
B
873.5, 869.8, 865.3
B
C+
812.4, 810.0, 788.5
C
750.3
C
D+
D
D
F
Maximum
1000
1000
Average
884.8
rms
62.4
Note: the average grade for the lab report 4, 6, 12, 14 series was 90.1 and the average grade for the
lab report 5,