Question 23.3: The RTD shown in Fig. 23-7 has a resistance ranging from 1 k...

The buffer amplifiers provide very high input impedance for the differential amplifier. Using Eq. (17-8), the closed loop input impedance is calculated as shown here:

Z_{in(CL) }= (1 + A_{VOL}B)R_{in} ∥ R_{CM}

In a voltage follower, the feedback fraction “B” is 1. This reduces the previous equation to:

Z_{in(CL) }= (1 + A_{VOL})R_{in} ∥ R_{CM}

Substituting in the values for A_{VOL},R_{in}, and R_{CM} for a 741 operational amplifier,A_{VOL}=  100,000, R_{in} = 2 \ MΩ, and R_{CM}= 200 MΩ:

Z_{in(CL)} = (1 + 100,000) 2 MΩ ∥ 200 MΩ

Z_{in(CL)}=\frac{1}{\frac{1}{200 \ G\Omega }+\frac{1}{200 \ M\Omega } } =199.8 \ M\Omega

Z_{in(CL)} ≅ R_{CM}

The transducer has a ΔR of 100 Ω and a Bridge R value of 1 kΩ.

4 × R = 4 kΩ
2 × ΔR = 200 Ω

\frac{4R}{2\Delta R}=\frac{4(1 \ k\Omega )}{2(100 \ \Omega )} =20

The value of 4R is only 20 times larger than 2ΔR; therefore, Eq. (18-12) is used to determine the input voltage:

v_{in}=\frac{\Delta R }{4R+2\Delta R} V_{CC}

Substituting in the resistance values shown in Fig. 23-7:

v_{in}=\frac{100 \ \Omega }{(4 × 1 \ kΩ) + (2 × 100 \ Ω)} (10 \ V)= 238.1 \ mV

The differential voltage gain for the amplifier is calculated using Eq. (18-4):

A_V=\frac{-R_2}{R_1}

A_V=\frac{−10 \ kΩ}{1 \ k\Omega } =-10

The differential amplifier has a voltage gain of −10. The differential output voltage can be calculated by multiplying the voltage gain of the amplifier by the differential input voltage:

v_{out} = A_V × v_{in}
v_{out} = −10 (238.1 mV) = −2.381 V