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EECS 145L Final Examination Solutions (Fall 1996) EECS 145L Final Examination Solutions (Fall 1996)
UNIVERSITY OF CALIFORNIA, BERKELEY
College of Engineering, Electrical Engineering and Computer Sciences Department
1.a
The isolation amplifier isolates the input and output sections at low frequencies (e.g. d.c.
and 60 Hz) where dangerous currents exist and transmits signal and power at high frequency
(e.g. 500 kHz). The input section uses the input signal to amplitude modulate a high frequency
carrier and sends the result to the output section using a transformer that only transmits high
frequencies, The signal is recovered in the output section by demodulation. To avoid batteries,
a high frequency oscillator in the output section can transmit energy to the input section via a
second high frequency transformer.
The optical isolation amplifier uses a modulated light beam to transmit the signal to the output.
In this design, isolation is better, but batteries are required to power the input section.
1.b
The ground fault interrupter uses a differential transformer, rectification, and amplification
to detect any current mismatch between the power conductor (hot) and the current return
conductor (neutral). When the mismatch exceeds typically 5 mA, the amplified signal trips an
electromagnetic relay to shut off both conductors.
1.c
The thermistor is a solid-state temperature sensor whose electrical conductivity depends on
the electron population in the conduction band, which depends exponentially on temperature.
Resistance decreases with increasing temperature and increases with decreasing temperature.
[3 points off for not giving the dependence between temperature and resistance or for stating it
incorrectly- no description of electrical operation is complete without this information]
1.d
The bimetallic temperature switch consists of two metal strips with different thermal
expansion bonded together. As the temperature changes, the strips bend to make or break an
electrical contact.
[One common mistake students made in problem 1 was to describe the purpose of the devices
rather than the electrical operation that was asked for]
2.a
Battery
charger
circuit
Batteries
+ I
2b
Photon energy breaks the bond between a valence electron and a silicon atom, creating an
electron and a hole. The internal field of the diode separates the carriers and they drift to the n-
and p-type electrodes. If the circuit is completed, the electron can do electrical work on its way
to combining with the positive carrier.
2c
60 volts x 0.4 A = 24 watts (full credit for 30 watts)
2d
60 volts x 0.9 A = 54 watts (full credit for 60 watts)
December 18, 1996
page 1
S. Derenzo EECS 145L Final Examination Solutions (Fall 1996)
3a
Heat
sink
Fan or
skin
of aircraft
Power
amp
V
b
Instrumentation
amplifier
R
R
R
R
Thermally insulated box
Temperature-
sensitive circuit
Thermistor
T
Heat
pump
For maximum sensitivity and zero output at 20�C, all bridge resistors R are made equal to
resistance R
T
of the thermistor at 20�C. When thermistor resistance exceeds the bridge resistors
R (too cold), bridge imbalance causes the power amp to drive the heat pump so as to pump heat
into the insulated box. When the thermistor resistance is less than the bridge resistors (too hot),
the heat pump removes heat from the box. The instrumentation amplifier gain is set high so that
only a small bridge imbalance is needed to make the power amplifier control the heat pump.
[5 points off if no power amplifier to run heat pump]
[5 points off if there is no way described for heat sink to exchange heat with the surroundings-
a simple metal plate would heat up or cool excessively during an 18 hour flight
3b
At 0�C: The temperature inside the insulated box is slightly below 20�C, enough to imbalance
the bridge and cause the heat pump to heat the box. The heat comes from a combination of
Joule heating and heat transferred from the heat sink into the box. The heat sink cools, but this
process is limited by heat exchange with the surroundings (fan or skin of aircraft).
[2 points off if role of heat sink is not mentioned]
3c
At 40�C: The temperature inside the insulated box is slightly above 20�C, enough to imbalance
the bridge and cause the heat pump to cool the box. The heat pump transfers heat (both from
the box and Joule heating of the heat pump itself) to the heat sink. The heat sink warms, but
this process is limited by heat exchange with the surroundings (fan or skin of aircraft).
[2 points off if role of heat sink is not mentioned]
December 18, 1996
page 2
S. Derenzo EECS 145L Final Examination Solutions (Fall 1996)
4a
Leg
Arm
Arm
Ag(AgCl) skin
electrodes
1 mV
50 mV
Isolation amp
(gain 100)
Instr amp
(gain 50)
Band pass
filter
60 Hz
notch filter
Micro-
computer
5 V
5 V
To increase the signal from 1 mV to 5 V, need overall gain of 5,000. But 60 Hz differential
background of 10 mV will saturate, so notch filter is needed between the instrumentation
amplifier and the isolation amplifier. For subject safety, the instrumentation amplifier and the
notch filter must be powered by the isolation amplifier or by batteries.
The 60 Hz notch filter must drop the 10 mV differential to 1% of the 1 mV signal (10 �V). This
requires a gain of 10
3
at the notch, which is achievable.
The band pass filter requires a high pass stage that reduces the 1 mV x 5,000 = 5 volt drift by a
factor of 100 from 1 Hz to 100 Hz. A single stage filter would just barely work, two stages is a
better design. The figure below (part 4c) assumes two HPF stages, which reduces the drift by a
factor of 10,000.
The band pass filter also requires a low pass stage that drops the 300 kHz 100 mV to 1% of 5
V or 50 mV at 5,000 Hz. This is easily done with a single-stage filter, which reduces the 300
kHz background by a factor of 60.
4b
0.01
0.1
1
10
100
1,000
Voltage (peak-to-peak)
10
2
0.5
0.1
0.02
10
4
10
5
10
6
5
1
0.2
0.05
0.01
0.005
0.002
0.001
Electrode drift
ECG signal
300 kHz
60 Hz (saturated)
December 18, 1996
page 3
S. Derenzo EECS 145L Final Examination Solutions (Fall 1996)
4c
0.01
0.1
1
10
100
1,000
Frequency (Hz)
Voltage (peak-to-peak)
10
2
0.5
0.1
0.02
10
4
10
5
10
6
5
1
0.2
0.05
0.01
0.005
0.002
0.001
ECG signal
300 kHz
60 Hz
All backgrounds are reduced below 0.05 V (1% of the 5V ECG signal). The electrode drift was reduced
by a factor of 10,000, to 0.0005 V, which is below the bottom of the chart.
Final Examination score distribution:
131-140 1
141-150 1
151-160 4
161-170 0
171-180 3
181-190 2
191-200 2
undergraduate average = 164
graduate average = 184
145L Course Grade Distribution
Grade
Undergraduate
Graduate
Scores
Scores
A+
909
942, 961
A
887
882
A
852, 861
B +
825, 834
B
B
783, 797
C+
745, 766
Maximum
1000
1000
Average
826
928
December 18, 1996
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S. Derenzo