ELECTRONIC THERMOMETER WITH AN INSTRUMENTATION AMPLIFIER Objective: Design an electronic thermometer with an instrumentation amplifier to provide the necessary amplification. Specifications: The temperature range to be measured is 0 to 100 °F. The output voltage is to be in the range of 0 to 5 V

Question

ELECTRONIC THERMOMETER WITH AN INSTRUMENTATION AMPLIFIER

Objective: Design an electronic thermometer with an instrumentation amplifier to provide the necessary amplification.

Specifications: The temperature range to be measured is 0 to 100 °F. The output voltage is to be in the range of 0 to 5 V corresponding to 100 °F.
Design Approach: In Chapter 1, we began a design of an electronic thermometer
using the temperature characteristics of a pn junction diode. Here, we expand on that design.
Figure 9.48(a) shows a circuit with two diodes, each biased with a constant current source. Figure 9.48(b) shows the same circuit, but with the constant current sources implemented with transistor circuits. The current source circuits were briefly described and analyzed in Chapter 5. The two diodes, [latex]D_{1}[/latex] and [latex]D_{2}[/latex], are assumed to be matched or identical devices. We also assume that all transistors are matched. Ne-glecting base currents, we have [latex]I_{1} = I_{REF1}[/latex] and [latex]I_{2} = I_{REF2}[/latex].
Choices: Ideal matched silicon diodes and bipolar transistors are available. In addition, ideal op-amps are available

Accepted Answer

(Diodes): From Chapter 1, we can write the voltage drops across each diode as
[latex]V_{D1} = V_{T} \ln \left( \frac{I_{1}}{I_{S}} \right) [/latex] (9.101(a))
and
[latex]V_{D2} = V_{T} \ln \left( \frac{I_{2}}{I_{S}} \right) [/latex] (9.101(b))
We may note that, since the two diodes are matched, the reverse-saturation current, [latex]I_{S}[/latex] , is the same in the two expressions.
The output voltage is defined as the difference between the voltages across the two diodes, or
[latex]V_{AT} = V_{D1} − V_{D2} = V_{T} \left[ \ln \left(\frac{I_{1}}{I_{S}} \right) − \ln \left(\frac{I_{2}}{I_{S}} \right) \right] [/latex] (9.102(a))
or
[latex]V_{AT} = V_{T} \ln \left( \frac{I_{1}}{I_{2}} \right) = \frac{kT}{e} \ln \left( \frac{I_{REF 1}}{I_{REF 2}} \right) [/latex] (9.102(b))
The output voltage, [latex]V_{AT}[/latex] , is now directly proportional to absolute temperature T, hence the subscript AT.
If we let [latex]I_{REF1}/I_{REF2} = 5[/latex], then Equation (9.102(b)) can be written as
[latex]V_{AT} = (0.0259) \left( \frac{T}{300} \right) \ln(5) = (1.3895 × 10^{−4})T[/latex] (9.103)

Letting [latex]I_{REF1}/I_{REF2} \gt 0[/latex] provides a small amount of gain. Converting absolute temperature to degrees Celsius and then to degrees Fahrenheit, we find
[latex]T = T_{C} + 273.15 [/latex] (9.104)
and
[latex]T_{F} = 32 + \frac{9}{5} T_{C} ⇒ T_{C} = (T_{F} − 32) \left( \frac{5}{9} \right)[/latex] (9.105)
where [latex]T_{C}[/latex] and [latex]T_{F}[/latex] are temperatures in degrees Celsius and degrees Fahrenheit, respectively.
Combining Equations (9.104) and (9.105), we obtain
[latex]T = (T_{F} − 32) \left( \frac{5}{9} \right) + 273.15 = \frac{5}{9} T_{F} + 255.37 [/latex] (9.106)
The output voltage from Equation (9.103) can now be written as
[latex]V_{AT} = (1.3895 × 10^{−4}) \left(\frac{5}{9} T_{F} + 255.37 \right)[/latex]

[latex]= (7.719 × 10^{−5})T_{F} + 3.5484 × 10^{−2}[/latex] (9.107)
(Instrumentation Amplifier): Since neither terminal of the output voltage is at ground potential, we can apply this voltage to an instrumentation amplifier to obtain a voltage gain. The output of the instrumentation amplifier will be applied to a summing amplifier in addition to an offset voltage. The objective of the design is to obtain an output voltage of zero volts at [latex]T_{F} = 0[/latex] and an output voltage of 5 V at [latex]T_{F} = 100 °F.[/latex]
If the gain of the instrumentation amplifier is A = −129.55, then the output of
the instrumentation amplifier is as follows:

[latex]T_{F}[/latex]	[latex]V_{AT}[/latex]	[latex]V_{O1}[/latex]
0	0.035484	−4.5970
100	0.043203	−5.5970

(Output Stage): The offset voltage can be generated by using the nonin-
verting op-amp circuit with a Zener diode, as shown in Figure 9.49. If we use a Zener diode with a breakdown voltage of 3.60 V and if we set [latex]R_{3}/R_{4} = 0.277[/latex], then the output voltage is [latex]V_{O2} = +4.597 V[/latex]. Applying the output voltage of the instrumentation amplifier, [latex]V_{O1}[/latex], and the offset voltage, [latex]V_{O2}[/latex], to a summing amplifier with a gain of −5 as shown in Figure 9.49, we achieve the desired specifications. That is [latex]V_{O} = 0[/latex] at [latex]T_{F} = 0[/latex] and [latex]V_{O} = 5 V[/latex] at [latex]T_{F} = 100 °F[/latex].
Comment: The primary advantage of this system is that the output voltage is a linear function of temperature.
In Chapter 16, we can apply the analog output voltage [latex]V_{O}[/latex] to an A/D converter and use a seven-segment display so that the output signal is actually displayed in terms of degrees Fahrenheit

Question 9.DA.8: ELECTRONIC THERMOMETER WITH AN INSTRUMENTATION AMPLIFIER Obj...

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