Download soln_1-99

EE 452 and EE 830
Assignment 1, 1999
Solutions
4. Choose a thermocouple:
Type K is most linear
Type E has the greatest sensitivity
Types B, C, J, R and S do not cover the necessary range
I will choose Type K because of its better linearity. [Note: when making an arbitrary
choice, it is a good idea to explain why that choice has been made. It will make your past
work a much better reference for future work.]
From the thermocouple reference tables for Type K, the output voltage at -100 ºC is
-3.554 mV, and at +500 ºC, it is +20.644mV. Our desired output is -1000 mV and +5000
mV, respectively. For a single gain to apply over the entire range, we have to solve the
following simultaneous equations:
-1000 mV = G * (-3.554 mV) + Vo and
+5000 mV = G * (20.644 mV) + Vo,
where G is the gain and Vo is the offset.
This results in a gain of 248 and an offset of -121 mV. Note some of you chose to have
zero offset, but then your output wasn't -1.00 V and +5.00 V at -100 and + 500 ºC,
respectively. That is, without other techniques for correcting the nonlinearity, a choice
must be made as to how the error will be incorporated into the overall response. If you
are adding an offset voltage, then you should show how it is to be implemented. That is,
you can't just put in a power supply as you might in the lab.
Temper
ature
degrees
celsius
-100
-50
0
50
100
150
200
250
300
350
400
450
500
Thermo
couple Desired Expected
Desired output
Error in temperature
Output output
output referred to the input due to nonlinearity
mV
V
V
mV
degrees celsius
-3.544
-1.889
0
2.023
4.069
6.138
8.138
10.153
12.209
14.293
16.397
18.516
20.644
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
-1.00
-0.59
-0.12
0.38
0.89
1.40
1.90
2.40
2.91
3.42
3.95
4.47
5.00
-3.544
-1.528
0.488
2.504
4.520
6.536
8.552
10.569
12.585
14.601
16.617
18.633
20.649
0
10
12
11
10
10
10
10
9
7
5
2
0
You were asked to use an instrumentation amplifier. Usually the second (differential
stage) is set for unity gain to improve the common-mode rejection performance. In this
application, that is not a major consideration. A gain of 248 is too high to accomplish in a
single stage. Therefore, the differential stage can be designed for a gain that is easy to
achieve with fixed value resistors, such as a gain of ten. Any combination of resistors of
R/10R will then be possible over a range of about 1 k to 10 M.
From a noise performance perspective, it is better to keep the resistance values small, say
1 k and 10 k. Note: one of the resistors in the differential stage (usually the one
shown in the figure) must be adjustable. This is done so that the resistor ratios of the two
halves of the differential stage can be set to be equal. The common-mode rejection
performance degrades substantially for small differences in the resistor ratios.
The first stage of the instrumentation amplifier can be used to obtain a gain of 24.8 by
having fixed feedback resistors and a fixed/variable gain-controlling resistor. Therefore, I
arbitrarily chose a value of 22 k for the feedback resistors, which means the gain
control resistor must be 1.849 k. This can be achieved with a 1.6-k, fixed, 5% resistor
in series with a 500- potentiometer, or a 1.78-k, fixed, 1-% resistor in series with a
100- potentiometer.
Note: an online table of standard resistor values can be found at:
http://www.action-electronics.com/resist1p.htm
Note also: a comparable table of standard capacitor values can be found at:
http://www.action-electronics.com/capacito.htm In this case, note that the capacitor
values are not all available over the full range of decades as is the case for resistors.
Figure 1 in the accompanying pdf link shows the instrumentation amplifier circuit.
5.
Using the formula provided in class, the shunt resistor for the range 0 to 40 ºC for this
thermistor is 9.136 k. I used an actual value of 9.09 k as that is the closest 1-%
standard value. The table below shows the resulting linearized resistance versus
temperature and the calculated input deviation with and without the shunt resistor.
Temperature Resistance Rsh//Rt "linearized"
input
"linearized"
input
degrees
ohms
ohms temperature deviation temperature deviation
celsius
without
without with shunt with shunt
shunt
shunt
resistor
resistor
-10
42,000
7473
-28.3
-18.3
-8.3
1.7
-5
34,000
7172
-13.2
-8.2
-4.6
0.4
0
27,000
6800
0.0
0.0
0.0
0.0
5
22,000
6432
9.4
4.4
4.5
-0.5
10
18,000
6040
17.0
7.0
9.3
-0.7
15
15,000
5660
22.6
7.6
14.0
-1.0
20
12,000
5172
28.3
8.3
20.0
0.0
25
10,000
4762
32.1
7.1
25.0
0.0
30
8,300
4339
35.3
5.3
30.2
0.2
35
6,900
3923
37.9
2.9
35.3
0.3
40
5,800
3541
40.0
0.0
40.0
0.0
45
4,900
3184
41.7
-3.3
44.4
-0.6
50
4,200
2873
43.0
-7.0
48.2
-1.8
From this table, it can be seen that the maximum input deviation without the shunt
resistor is 8.3 ºC, whereas with the shunt resistor it is 0.7 ºC. The nonlinearity is 21%
and 1.75% respectively. Thus, the linearization technique allows us to achieve 1-degree
accuracy. This level of accuracy holds to at least -5 and +45 ºC.
9.
Many of you developed a circuit in which the current from the AD590 was sensed by a
series resistor, and the voltage across the resistor was used directly as the input to a
comparator. While these circuits would work in principle, the change in voltage is so
small that it makes the setting of the hysteresis on the comparator very sensitive to actual
component values. The other point of interest is that many comparators can sink much
more current (when their output is at 0 V) than they can source (when their output is
+Vcc). Therefore, the preferred design would be to have the drive current to the
transistor (or directly to the relay) occur when the comparator output is low.
The circuit of Figure 2 attached shows one possible configuration based on a comparator
with hysteresis. The first stage is a current to voltage converter for the AD590.
However, because of the large offset current (290 A), the summing node (at the
inverting terminal) is used to subtract 300 A. This means the output of the first
operational amplifier is positive for temperatures below 27 ºC. (If the comparator cannot
withstand negative input voltages and if higher temperatures are expected, then the
subtracted current would have to be increased.) The comparator with hysteresis is set up
with a simple reference voltage, basically a voltage divider. One of the three resistor
values must be chosen arbitrarily (R1 = 100 k). Then two simultaneous equations can
be set up at the inverting node of the comparator. One equation is for the output voltage
of +15 V and a desired node voltage of 3.0 V; the other equation for the output voltage of
0 V and a desired node voltage of 1.5 V. Solution of the two simultaneous equations
yields R2 = 80.6 k and R3 = 12.7 k. All these resistor values are the closest standard
1% values.
12.
The main point of this question was to get you thinking about using self-heating for a
functional purpose rather than as a side-effect to be minimized. Once an appropriate
voltage has been generated, a comparator can be used to generate a control signal as in
the previous question. The first decision to be made is the degree of self-heating to be
used. To maximize the voltage difference between air versus liquid, a relatively large
self-heating temperature should be used when the sensor is in air, say T = 50 ºC. This
means that 400 mW must be dissipated. Since there is an 8:1 ratio in dissipation
constant, T should be about 6 ºC, when the sensor is in the liquid. The sensor has a
resistance of 19 ohms at 50 ºC. For 400-mW of heat to be generated with this resistance,
a current of 145 mA is required. If we assume a 15-V supply is available, then we need a
total resistance of 103 . We can use a 100- (5%) resistor for this. When the sensor is
in the liquid, the resistance will be 83 . Thus the voltage across the sensor will go from
0.24 V when it is in air to 6.8 V when it is in the liquid.