Objective: Design an amplifier system that will produce an output voltage of ±5 V when the resistance R3 deviates by ±1% from the value of R2. This would occur, for example, in a system where R3 is a thermistor whose resistance is given by R3 = 200 [1 + (0.040)(T − 300)/300] kΩ where T is the

Question

Objective: Design an amplifier system that will produce an output voltage of ±5 V when the resistance [latex]R_{3}[/latex] deviates by ±1% from the value of [latex]R_{2}[/latex]. This would occur, for example, in a system where [latex]R_{3}[/latex] is a thermistor whose resistance is given by

[latex]R_{3} = 200 \left[1 + \frac{(0.040)(T − 300)}{300} \right] kΩ [/latex]

where T is the absolute temperature. For [latex]R_{3}[/latex] to vary by ±1% means the temperature is in the range 225 ≤ T ≤ 375 K.
Consider biasing the bridge circuit at [latex]V^{+} = 7.5 V[/latex] using a 5.6 V Zener diode.
Assume ±10 V is available for biasing the op-amp and reference voltage source, and that [latex]R_{1} = R_{2} = 200 kΩ[/latex] .

Accepted Answer

With [latex]R_{1} = R_{2}[/latex], from Equation (9.100), we have

[latex] v_{O1} = δ \left(\frac{R_{1}||R_{2}}{R_{1} + R_{2}} \right) V^{+} [/latex] (9.100)

[latex]v_{O1} = \left(\frac{δ}{4} \right) V^{+}[/latex]
For [latex]V^{+} = 7.5 V[/latex] and δ = 0.01, the maximum output of the bridge circuit is [latex]v_{O1} = 0.01875 V[/latex]. If the output of the amplifier system is to be +5 V, the gain of the instrumentation amplifier must be 5/0.01875 = 266.7. Consider the instrumentation amplifier shown in Figure 9.26. The output voltage is given by Equation (9.67), which can be written

[latex]v_{O} = \frac{R_{4}}{R_{3}} \left(1 + \frac{2 R_{2}}{R_{1}} \right) (v_{I2} − v_{I1}) [/latex] (9.67)

[latex]\frac{v_{O}}{v_{O1}} = \frac{R_{4}^\prime}{R_{3}^\prime} \left(1 + \frac{2 R_{2}^\prime}{R_{1}^\prime} \right) = 266.7 [/latex]
We would like the ratios [latex]R_{4}^\prime/R_{3}^\prime [/latex] and [latex]R_{2}^\prime/R_{1}^\prime [/latex] to be the same order of magnitude. If we let [latex]R_{3}^\prime = 15.0 k \Omega[/latex] and [latex]R_{4}^\prime = 187.0 k\Omega[/latex] , then [latex]R_{4}^\prime/R_{3}^\prime = 12.467 [/latex] and [latex]R_{2}^\prime/R_{1}^\prime = 10.195[/latex]. If we set [latex]R_{2}^\prime = 200.0 k \Omega[/latex], then [latex]R_{1}^\prime = 19.62 k\Omega[/latex].
Resistance [latex]R_{1}^\prime[/latex] can be a combination of a fixed resistance in series with a potentiometer, to permit adjustment of the gain.
Comment: The complete design of this instrumentation amplifier is shown in Figure 9.45. Correlation of the reference voltage source design is left as an exercise.
Design Pointer: The design of fairly sophisticated op-amp circuits is quite straightforward when the ideal op-amp parameters are used.

Question 9.12: Objective: Design an amplifier system that will produce an o...

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