1.5 [= 2R2/R1f + 100 = 249Rf/R1f + 100 The resulting value of R 1f is R1 f = 0.6

Question

1.5 [= 2R2/R1f + 100 = 249Rf/R1f + 100 The resulting value of R 1f is R1 f = 0.606 k omega, which yields R2 = 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. EXERCISE PROBLEM Ex 9.8: For the instrumentation amplifier in Figure 9.26, the parameters are R4 = 90k omega, R3 = 30 omega, and R2 = 50 k omega. Resistance R1 is a series combination of a fixed 2k omega resistor and a 100 k omega potentiometer. (a) Determine the range of the differential voltage gain. (b) Determine the maximum current in R1 for input voltages in the range -25 mV to + 25 mV. (Ans. (a) 5.94

Explanation / Answer

1.5 = 249R1f/(R1f + 100)

=> 1.5R1f + 150 = 249R1f

=>247.5R1f = 150

=>R1f = 0.606 Kohm

Now, 1.5 = 2R2 / (R1f + 100)

Substituitng value of R1f

1.5 = 2R2 / 0.606 + 100

=>1.5 = 2R2 / 100.606

=> 2R2 = 150.909

=>R2 = 75.45 Kohm

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