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

With R_{1} = R_{2}, from Equation (9.100), we have

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

v_{O1} = \left(\frac{δ}{4} \right) V^{+}
For V^{+} = 7.5  V and δ = 0.01, the maximum output of the bridge circuit is v_{O1} = 0.01875  V. 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

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

\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
We would like the ratios R_{4}^\prime/R_{3}^\prime and R_{2}^\prime/R_{1}^\prime to be the same order of magnitude. If we let R_{3}^\prime = 15.0  k \Omega and R_{4}^\prime = 187.0  k\Omega , then R_{4}^\prime/R_{3}^\prime = 12.467 and R_{2}^\prime/R_{1}^\prime = 10.195. If we set R_{2}^\prime = 200.0  k \Omega, then R_{1}^\prime = 19.62  k\Omega.
Resistance R_{1}^\prime 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.