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UNIVERSITY OF CALIFORNIA
College of Engineering Department of Electrical Engineering and Computer Sciences
EECS 145L: Electronic Transducer Laboratory
FINAL EXAMINATION December 18, 1995 5:00 - 8:00 PM
You have three hours to work on the exam, which is to be taken closed book. Calculators are OK, but not needed. You will not receive full credit if you do not show your work. Total points = 210 out of 1000 for the course.
1 ______________ (48 max) 2 ______________ (60 max)
3 ______________ (42 max) 4 ______________ (60 max)
TOTAL _____________ (210 max)
COURSE GRADE SUMMARY
LAB REPORTS (5 required, 100 points each):
4 __________ 5 __________ 6 __________ 7 __________ 11 __________
12 __________ 13 __________ 14 __________ 15 __________ 16 __________
17 __________ 18 __________ 19 __________ 25 __________
LAB TOTAL
LAB PARTICIPATION
MID-TERM
MID-TERM
FINAL EXAM
TOTAL COURSE GRADE
____________
____________
____________
____________
____________
____________
(500 max)
(100 max)
(100 max)
(90 max)
(210 max)
(1000 max)
COURSE LETTER GRADE
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Problem 1 (48 points)
In 50 words or less, describe the following:
1.1 (8 points) The differences between the electrical functions of (a) the ground fault interrupter circuit and (b) the circuit breaker.
1.2 (8 points) The differences between the electronic properties of (a) the operational amplifier and (b) the instrumentation amplifier.
1.3 (8 points) The differences between (a) common mode gain and (b) differential gain.
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Problem 2 (60 points)
Design a system for the amplification and analog filtering of EEG (brain-wave) data, given that
- The electrical signals ( V 1 and V 2 ) are taken from two skin electrodes placed on the head (a third “ground” electrode is placed on the neck).
- The wires from the skin electrodes to your system are approximately 1 m long. The 60 Hz electromagnetic interference received by one wire is 100 mV (peak-to-peak) and by the other wire is 110 mV (peak-to-peak).
- The desired differential EEG signal has an amplitude of 50 μV peak-to-peak and is in the 0.5 to 30-Hz frequency band.
- Electrode drift produces a differential voltage V ED2 – V ED1 of 1 mV in the 0 Hz to 0.1 Hz frequency range, and can be ignored at frequencies above this range.
- The EMG background amplitude V EMG2 – V EMG1 from the head muscles is 100 μV and is in the 100- Hz to 3-kHz band.
- To summarize the voltages present on the two wires: V 1 = V ED1 + V EEG1 + V EMG1 + (0.050 volts) sin(2π tf 0 ) ( f 0 = 60 Hz) V 2 = V ED2 + V EEG2 + V EMG2 + (0.055 volts) sin(2π tf 0 )
- You wish to see the differential EEG signal V EEG2 – V EEG1 undistorted (variations in gain less than 10% from 0.5 to 30 Hz) and reduce all other backgrounds to below 2% of the EEG signal
- You decide to use an instrumentation amplifier followed by analog filtering.
- Your system should amplify the EEG signal to 5 volts peak-to-peak for input to a microcomputer analog input circuit with an input impedance of 10 kΩ.
2.1 (10 points) Using the grid below, show the magnitude of |V 2 – V 1 | as a function of frequency before amplification and filtering. Label all signals and backgrounds.
Frequency (Hz)
| V
V
| (mV) 2
1
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2.2 (15 points) Sketch a block diagram of your system, showing all essential components and signal lines
2.3 (15 points) Plot the differential voltage gain | V out/( V 2 – V 1 )| of your system after amplification and filtering, using the grid below. (You may use the voltage ratio or dB for the vertical axis.)
Frequency (Hz)
Voltage gain
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Problem 3 (42 points)
After considering how sensitive strain gauges are to the thermal expansion of the element to which they are bonded, you invent a new temperature sensor that consists of two resistive strain gauges cemented to a small aluminum plate.
Assume the following:
- You use the two strain gauges (unstrained resistance 100Ω, gauge factor = 2) in a bridge circuit
- The thermal expansion coefficient of aluminum is 23 ppm/°C (ppm = parts per million)
- The maximum power that the strain gauges can dissipate is 250 mW
- You use an instrumentation amplifier with a noise level of 10 nV/Hz1/2^ (relative to the input)
3.1 (21 points) Sketch your circuit design, including all components and wires.
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3.2 (7 points) What bridge bias voltage gives maximum bridge sensitivity?
3.3 (7 points) What is the bridge output sensitivity in mV/C°?
3.4 (7 points) What is the noise level in terms of C° at 1M Hz and 1 Hz?
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4.2 (25 points) List the steps that the system must perform to measure the output offset voltage at the nine temperatures.
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Equations, some of which you may need:
R ( T ) = R ( T 0 )exp β ( 1 T − T^10 ) V rms = B [( D 1 G )^2 + ( D 0 )^2 ] ∆ x = ∆ V
dV / dx V ( t ) = V 0 sin(ω t ) ω= 2 π f V 0 = A ( V + − V − )
G = 1
1 + ( f / fc )^2 n^
tan φ n
^
=^
f fc
G = (^ f^ /^ fc )
n
1 + ( f / fc )^2 n^
tan φ n
^
− f fc
N ( x ) = N (0) e −^ x μ^ I = I 0 e −^ kLC^ T = T 2 − ( T 2 − T 1 ) e − t^ /^ τ
x = e −α^ t^ [ A cos(ω t ) + B sin(ω t )] = Re −α^ t^ cos(ω t +δ ) V = q / C
v = v 0 + at x = x 0 + v 0 t + 0.5 at^2 (constant a ) g = 10 m s– I rms = 2 qI ( F 2 − F 1 ) q = 1.60 x 10–19^ Coulombs
V rms = 4 kTR ( F 2 − F 1 ) k = 1.38 x 10–23^ Volt^2 sec ohm–1^ °K–
RT = R 3 Vb^ R^1 −^ V^0 (^ R^1 +^ R^2 ) Vb R 2 + V 0 ( R 1 + R 2 ) V^0 =^ G ±^ ( V +^ −^ V −^ )^ +^ Gc^ ( V +^ +^ V −^ )
fc = 1 2 π RC
“CMRR” = G ±
Gc
“CMR” = 20log 10 G ± Gc
^
R =ρ A / L ∆ R R
= Gs ∆ L L
V 0 = Vb Gs ∆ L L
^
VT = V BE2 − V BE1 = kT q
ln I^1 I 2
^
^
k / q = 86.17 μV / K
PR = σ AT^4 σ = 5.6696 × 10 −^8 Wm−^2 K^4
E = hc / λ hc = 1240 eV ⋅ nm λmax = (2.8978 × 106 nm K)/ T
η= Tn^ +^2 −^ Tn^ +^1 Tn + 1 − Tn
T equ = Tn + 1 + Tn +^2 −^ Tn +^1 1 −η
Q = π I + I^2 R /2 + Kp ( Ts − T 0 )+ Ka ( Ta − T 0 ) T equ =
π I + I^2 R /2 + Kp Ts + KaTa Kp + Ka
μ ≈ a = (^) m^1 ai i = 1
m
∑ σ a^2 =^ m^1 − 1 (^ ai −^ a )
2 i = 1
m
∑ σ^ a =
σ a m
