the working principle IC transducers

Experiment 1 : Control and monitor temperature of oven using IC transducer
1) Understand the working principle IC transducers
UNIT OBJECTIVE
At the completion of this unit, you will be able to explain the operation of the IC temperature
transducer and its function as a temperature measurement and control device.
UNIT FUNDAMENTALS
The IC temperature transducer used on your circuit board is an epoxyencapsulated integrated circuit.
Although it is packaged in a 3-terminal TO-92 case, the transducer itself is a 2-
terminal device. The center lead is not internally connected.
The IC functions as a current source whose output current is a function of temperature.
At a reference point of 0°C (the freezing point of water), the output current (IREF) is
273.2 µA.
Every temperature transducer has a temperature coefficient that describes the way
the transducer’s characteristics change as temperature changes.
The temperature coefficient α (the Greek letter alpha) of the IC transducer on your
circuit board is one microamp per degree Celsius (α = 1 µA/°C).
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A positive or negative temperature change from the 0°C reference point causes a positive
or Negative current change of 1 µA/°C.
For any temperature T, the current at that temperature (IT) may be expressed as follows:
IT = (α x T) + 273.2 µA (where IT is in µA, T is in °C, and α = 1 µA/°C).
What is the IC transducer’s output current at –46.8°C?
I-46.8°C = 2.7*10^-4 µA (Recall Value 1)
This figure shows the output characteristics, schematic symbol, and a list of advantages
and disadvantages for the IC temperature transducer.
The advantages and disadvantages are relative to the other types of temperature
transducers on your circuit board. For example, the IC transducer has the highest
output level as compared to the thermocouple, thermistor, and RTD.
For which feature would you choose the IC transducer over the other types?
a. Linearity
b. Operating speed
c. Wide temperature range
The IC temperature transducer on your circuit board has a positive temperature
coefficient. This means that the transducer’s temperature-dependent parameter (current)
increases as temperature increases.
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Other types of IC transducers can have a negative temperature coefficient. With a
negative Temperature coefficient, increasing temperature causes
a. Increasing current.
b. Decreasing current.
This is a simplified block diagram of the IC TRANSDUCER circuit block.
The transducer’s current output [I(T)] drives an op amp that is configured as a currentto-Voltage converter.
The resulting output is a voltage [V (T)] that is a function of the transducer’s temperature.
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The remaining circuitry allows the block to operate as a temperature controller
that regulates the temperature inside the oven.
Resistor RSP is used to select a set point, or the temperature at which the oven is to
be regulated.
A second op amp, configured as a comparator, determines whether the oven
temperature is above or below the set point.
The comparator’s output drives a transistor that switches a heater resistor on if the
temperature is below the set point, or off if the temperature is above the set point.
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This circuit controls oven temperature by
a. Alternately turning power to the heater on and off.
b. Applying more power to the heater for a larger temperature change and less
power for a small temperature change.
Experimental Procedure
1. Connect the minus (black) lead to TP2 and the plus (red) probe to TP1. Set
the multimeter for volts dc.
2. Place the shunt on the TEMP header in the 35°C position
3. Insert a two-post connector in the OVEN ENABLE position
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9. Connect the + meter lead to OUT and the – lead to a circuit ground point
10. Measure the circuit output voltage at 35°C
VOUT 2.9518 V
11. What is your finding? The calculated is similar to the measured result.
12. Move the TEMP shunt to the 40°C position. The oven temperature is now
rising toward 40°C
13. Calculate the transducer current at 40°C
IT T 273.2 A
where T = set point temperature and α = 1 A
IT = 313.2 A
/°C
14. Calculate the op amp output voltage at 40°C
VOUTT 30 0.5V /0 C
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Experiment. 2: Control and monitor temperature of oven using Thermocouple
EXERCISE OBJECTIVE
At the completion of this unit, you will be able to describe and demonstrate
the operation of a thermocouple.
EXERCISE FUNDAMENTALS
A thermocouple is a temperature transducer consisting of two wires made
of different metals soldered or welded together.
The junction of the two wires exhibits a phenomenon called the Seebeck Effect,
whereby the junction generates a voltage that is a function of its temperature.
Different combinations of metals result in different voltage-temperature
characteristics. In industry, thermocouples are usually classified by a one letter
type designation that describes their response.
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This figure shows the voltage-temperature curves and wire compositions
for thermocouple types R, K, J, and E.
What metals are used in a type E thermocouple?
a. Iron and constantan
b. Chromel and constantan
c. Chromel and alumel
d. Platinum and rhodium
The THERMOCOUPLE circuit block on your circuit board uses a J type device, which
is composed of iron and constantan.
Compared to the other types, the type J has the
a. Highest temperature range.
b. Lowest temperature range.
c. Highest output voltage range.
If you attempt to measure a thermocouple’s output with a voltmeter, the copper
meter leads contact the thermocouple leads and form two additional junctions.
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Since the metals are dissimilar, each of these measuring junctions also produces
a Seebeck Effect voltage that is dependent on temperature.
The voltages at these junctions must be subtracted from the meter reading
to obtain the true voltage at the sensing junction.
This figure shows how you can use a thermistor bridge circuit, also known as a
Wheatstone bridge, to cancel the effects of the Seebeck voltages at the
measuring junctions. The temperature of the measuring junctions must be known
in order to determine their voltages and subtract them from the meter reading.
For this reason, the thermistor is placed in the same thermal environment as
the measuring junctions so that it is at the same temperature.
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The sensitivity and polarity of the bridge are designed to cancel the Seebeck
voltages at the measuring junctions. The meter then reads only the desired
voltage at the sensing junction.
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Thermocouple Characteristics
EXERCISE OBJECTIVE
When you have completed this exercise, you will be able to describe and
demonstrate the characteristics of a thermocouple. You will verify your results
with a multimeter.
DISCUSSION
The schematic symbol of a thermocouple reflects its construction – two wires
joined at their ends.
The thermocouple’s response is nearly linear and is considered linear over
small temperature ranges.
What type of temperature coefficient is indicated by the curve?
a. Positive
b. Negative
Since a thermocouple generates a voltage, it is considered a self-powered device.
Thermocouples are simple, rugged, low-cost transducers that operate over a very
wide temperature range.
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Disadvantages include a low output level, stability, and sensitivity, and the need
for a reference for accurate temperature measurement.
Manufacturers supply tables of voltage versus temperature for their thermocouples.
Since the thermocouple on your circuit board has a linear response over the
trainer’s operating temperature range, you can simply use its temperature
coefficient (α = 51 µV/°C) to calculate voltage at a given temperature.
This schematic shows how the thermocouple is connected in a thermistor bridge
on your circuit board.
The thermistor and the measuring junctions are located on the underside of
the circuit board so that they are in the same temperature environment.
The bridge is adjusted to offset the effects of the measuring junction voltages.
This configuration is called a reference junction circuit.
When the bridge is balanced, its output voltage equals the thermocouple’s voltage
at 0°C, which is the freezing point of water. For this reason, the circuit is also
called the electronic ice point reference.
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The thermocouple on your circuit board has an output of 0 V at 0°C. What is
the output voltage of the calibrated bridge at 0°C?
a. 0 V
b. +0.5 V
c. Cannot be determined
You can connect the bridge outputs to the instrumentation amplifier set to a gain
of 100 (AV = 100) to boost the voltage to more practical levels.
Since the temperature coefficient (α) equals 51 µV/°C, you can use the
following equation to determine the circuit output at any temperature T:
IA OUT = α x T x AV
What voltage would the circuit output at 28.6°C?
IA OUT = _________ mV (Recall Value 1)
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PROCEDURE
In this PROCEDURE, you will calibrate the electronic ice point reference to read
the thermocouple voltage output. You will verify your results by calculating and
measuring the output voltage at different set points and comparing the values.
1. Enable the oven and select the 40°C set point. Complete the following steps
as you allow the oven to reach the set point.
2. Set your multimeter to measure Vdc and connect the leads across the output
of the instrumentation amplifier as shown.
3. In the THERMOCOUPLE circuit block, connect +OUT and –OUT from the
reference circuit to +IN and –IN respectively, in the
INSTRUMENTATION AMPLIFIER circuit block.
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4. Set the instrumentation amplifier’s gain to 100.
5. Calculate the voltage to which you should calibrate the circuit output at 40°C:
IA OUT = α x T x AV
α = 51 µV/°C
AV = 100
IA OUT = 204 mV (Recall Value 1)
6. Observe the OVEN ON LED and allow it to complete several cycles to make
sure the oven has reached its set point.
7. Adjust the REF potentiometer for a voltmeter reading of 204 mV at the IA
OUT terminal.
You have calibrated the circuit for a bridge output of 51 µV/°C. Do not
disturb the REF pot setting for the remainder of this PROCEDURE.
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8. Move the TEMP shunt to the 35°C position. Complete the following steps as
you allow the oven to reach the new set point.
9. Calculate the circuit output voltage at 35°C:
IA OUT = α x T x AV
α = 51 µV/°C
AV = 100
IA OUT = 178.5 mV (Recall Value 2)
10. Observe the OVEN ON LED an allow it to complete several cycles to make
sure the oven has reached its set point.
11. Measure the output voltage at 35°C.
IA OUT = 188 mV (Recall Value 3)
12. Move the TEMP shunt to the 45°C position. Complete the following steps as
you allow the oven to reach the new set point.
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13. Calculate the circuit output voltage at 45°C:
IA OUT = α x T x AV
α = 51 µV/°C
AV = 100
IA OUT = 229.5 mV (Recall Value 4)
14. Observe the OVEN ON LED and allow it to complete several cycles to make
sure the oven has reached its set point.
15. Measure the output voltage at 45°C.
IA OUT = 231 mV (Recall Value 5)
16. Move the TEMP shunt to the 50°C position. Complete the following steps as
you allow the oven to reach the new set point.
17. Calculate the circuit output voltage at 50°C:
IA OUT = α x T x AV
α = 51 µV/°C
AV = 100
IA OUT = 255 mV (Recall Value 6)
18. Observe the OVEN ON LED and allow it to complete several cycles to make
sure the oven has reached its set point.
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