Question 18.4: In Fig. 18-18, R1 = 1 kΩ, R2 = 100 kΩ, and R = 10 kΩ. What i...
In Fig. 18-18, R_1 = 1 \ kΩ, R_2 = 100 \ kΩ, and R = 10 kΩ. What is the differential voltage gain of the instrumentation amplifier? What is the common-mode voltage gain if the resistor tolerances in the second stage are ±0.01 percent? If v_{in} = 10 mV and v_{in(CM)} = 10 \ V, what are the values of the differential and common-mode output signals?
With the equations given in Fig. 18-18, the voltage gain of the preamp is
A_V=\frac{100 \ k\Omega }{ 1 \ kΩ } +1=101Since the voltage gain of the second stage is −1, the voltage gain of the instrumentation amplifier is −101.
The common-mode voltage gain of the second stage is:
A_{V(CM)} = ±2(0.01 \%) = ±2(0.0001) = ±0.0002
Since the first stage has a common-mode voltage gain of 1, the common-mode voltage gain of the instrumentation amplifier is ±0.0002.
A differential input signal of 10 mV will produce an output signal of:
A common-mode signal of 10 V will produce an output signal of:
v_{out(CM)} = ±0.0002(10 V) = ±2 mV
Even though the common-mode input signal is 1000 times greater than the differential input, the CMRR of the instrumentation amplifier produces a common-mode output signal that is approximately 500 times smaller than the differential output signal.