Question 9.8: Objective: Determine the range required for resistor R1, to ...
Objective: Determine the range required for resistor R_{1}, to realize a differential gain adjustable from 5 to 500.
The instrumentation amplifier circuit is shown in Figure 9.26. Assume that R_{4} = 2R_{3}, so that the difference amplifier gain is 2.
Assume that resistance R_{1} is a combination of a fixed resistance R_{1f} and a variable resistance R_{1v} , as shown in Figure 9.28. The fixed resistance ensures that the gain is limited to a maximum value, even if the variable resistance is set equal to zero. Assume the variable resistance is a 100 kΩ potentiometer.
From Equation (9.67), the maximum differential gain is
v_{O} = \frac{R_{4}}{R_{3}} \left(1 + \frac{2R_{2}}{R_{1}} \right) (v_{I 2} − v_{I 1}) (9.67)
500 = 2 \left( 1 + \frac{2R_{2}}{R_{1 f}} \right)and the minimum differential gain is
5 = 2 \left( 1 + \frac{2R_{2}}{R_{1 f} + 100} \right)
From the maximum gain expression, we find that
2R_{2} = 249 R_{1 f}
Substituting this R_{2} value into the minimum gain expression, we have
1.5 = \frac{2R_{2}}{R_{1 f} + 100} = \frac{249 R_{1 f}}{R_{1 f} + 100}
The resulting value of R_{1 f} is R_{1 f} = 0.606 k \Omega, which yields R_{2} = 75.5 k \Omega.
Comment: We can select standard resistance values that are close to the values calculated, and the range of the gain will be approximately in the desired range.
Design Pointer: An amplifier with a wide range of gain and designed with a potentiometer would normally not be used with standard integrated circuits in electronic systems. However, such a circuit might be very useful in specialized test equipment.
