Chapter 9, Problem 9.67P

Assume the instrumentation amplifier in Figure 9.26 has ideal op-amps. Thecircuit parameters are $R_1=10k\Omega ,R_2=40k\Omega ,R_3=40k\Omega ,$ and $R_4=120k\Omega$ . Determine $v_{O1},v_{O2},v_O$ , and the current in $R_1$ for(a) $v_{I2}=1.2+0.08\sin \omega t\left(V\right),v_{I1}=1.2-0.08\sin \omega t\left(V\right)$ ; and (b $v_{I2}=-0.60-0.05\sin \omega t\left(V\right),v_{I1}=-0.65+0.05\sin \omega t\left(V\right)$

To determine

The values of the voltages $v_{O1}$ , $v_{O2}$ and $v_O$ .

Also, the current through the resistance $R_1$ .

Answer to Problem 9.67P

The expression for the voltage $v_{O2}$ is $1.2+0.72\sin \omega t\text{V}$ , voltage $v_{O1}$ is $1.2+0.72\sin \omega t\text{V}$ , $v_O$ is $4.32\sin \omega t\text{V}$ and the value of the current through the resistance $R_1$ is $-16\times {10}^{-3}\sin \omega t\text{A}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

  

The expression for the voltage $v_b$ by voltage division rule is given by,

  $v_b=\frac{R_4{v}_{O2}}{R_3+R_4}$

The expression for the current at the node $v_a$ is given by,

  $v_a=v_b$

Substitute $\frac{R_4{v}_{O2}}{R_3+R_4}$ for $v_b$ in the above equation.

  $v_a=\frac{R_4{v}_{O2}}{R_3+R_4}$

Apply KCL at the node $v_a$ .

  $\begin{array}{l}\frac{v_{O1}-v_a}{R_3}=\frac{v_a-v_O}{R_4}\\ \frac{R_4}{R_3}{v}_{O1}-v_a\left[\frac{R_4+R_3}{R_3}\right]=-v_O\end{array}$

Substitute $v_b$ for $v_a$ in the above equation.

  $\frac{R_4}{R_3}{v}_{O1}-v_b\left[\frac{R_4+R_3}{R_3}\right]=-v_O$

Substitute $\frac{R_4{v}_{O2}}{R_3+R_4}$ for $v_b$ in the above equation.

  $\begin{array}{l}\frac{R_4}{R_3}{v}_{O1}-\left(\frac{R_4{v}_{O2}}{R_3+R_4}\right)\left[\frac{R_4+R_3}{R_3}\right]=-v_O\\ v_O=\frac{R_4}{R_3}\left(v_{O2}-v_{O1}\right)\end{array}$ …… (1)

The expression for the current $i_1$ is given by,

  $i_1=\frac{v_{I1}-v_{I2}}{R_1}$ …… (2)

Apply KVL in the loop 1

  $\begin{array}{c}{v}_{O1}-v_{O2}=i_1{R}_2+i_1{R}_1+i_1{R}_2\\ =i_1\left(R_1+2R_2\right)\end{array}$

Substitute $\frac{v_{I1}-v_{I2}}{R_1}$ for $i_1$ in the above equation.

  $v_{O1}-v_{O2}=\left(\frac{v_{I2}-v_{I1}}{R_1}\right)\left(R_1+2R_2\right)$

Substitute $\left(\frac{v_{I2}-v_{I1}}{R_1}\right)\left(R_1+2R_2\right)$ for $v_{O1}-v_{O2}$ in equation (1).

  $v_O=\frac{R_4}{R_3}\left(\frac{v_{I2}-v_{I1}}{R_1}\right)\left(R_1+2R_2\right)$

Substitute $10\text{k}\Omega$ for $R_1$ , $40\text{k}\Omega$ for $R_2$ , $40\text{k}\Omega$ for $R_3$ and $120\text{k}\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{v}_O=\frac{120\text{k}\Omega }{\left(40\text{k}\Omega \right)}\left(\frac{v_{I2}-v_{I1}}{10\text{k}\Omega }\right)\left(10\text{k}\Omega +2\left(40\text{k}\Omega \right)\right)\\ =27\left(v_{I2}-v_{I1}\right)\end{array}$ …… (3)

Substitute $1.2+0.08\sin \omega t\text{V}$ for $v_{I1}$ and $1.2-0.08\sin \omega t\text{V}$ for $v_{I2}$ in the above equation.

  $\begin{array}{c}{v}_O=27\left(1.2+0.08\sin \omega t\text{V}-1.2+0.08\sin \omega t\text{V}\right)\\ =4.32\sin \omega t\text{V}\end{array}$

Substitute $10\text{k}\Omega$ for $R_1$ , $1.2+0.08\sin \omega t\text{V}$ for $v_{I1}$ and $1.2-0.08\sin \omega t\text{V}$ for $v_{I2}$ in equation (2).

  $\begin{array}{c}{i}_1=\frac{1.2-0.08\sin \omega t\text{V}-\left(1.2+0.08\sin \omega t\text{V}\right)}{10\text{k}\Omega }\\ =-16\times {10}^{-3}\sin \omega t\text{mA}\end{array}$

The expression for the current $i_1$ is given by,

  $\begin{array}{l}{i}_1=\frac{v_{O1}-v_{I1}}{R_2}\\ v_{O1}=i_1{R}_2+v_{I1}\end{array}$ …… (4)

Substitute $-0.08\sin \omega t$ for $v_{I1}$ , $-16\times {10}^{-3}\sin \omega t\text{mA}$ for $i_1$ , $40\text{k}\Omega$ for $R_2$ in the above equation.

  $\begin{array}{c}{v}_{O1}=\left(-16\times {10}^{-3}\sin \omega t\text{mA}\right)\left(40\text{k}\Omega \right)-0.08\sin \omega t\\ =1.2-0.72\sin \omega t\text{V}\end{array}$

The expression for the voltage $v_{O2}$ is given by,

  $v_{O2}=v_{I2}-i_1{R}_2$ …… (5)

Substitute $0.08\sin \omega t$ for $v_{I2}$ , $-16\times {10}^{-3}\sin \omega t\text{mA}$ for $i_1$ , $40\text{k}\Omega$ for $R_2$ in the above equation.

  $\begin{array}{c}{v}_{O2}=1.2+0.08\sin \omega t-\left(-16\times {10}^{-3}\sin \omega t\text{mA}\right)\left(40\text{k}\Omega \right)\\ =1.2+0.72\sin \omega t\text{V}\end{array}$

Conclusion:

Therefore, theexpression for the voltage $v_{O2}$ is $1.2+0.72\sin \omega t\text{V}$ , voltage $v_{O1}$ is $1.2+0.72\sin \omega t\text{V}$ , $v_O$ is $4.32\sin \omega t\text{V}$ and the value of the current through the resistance $R_1$ is $-16\times {10}^{-3}\sin \omega t\text{A}$ .

To determine

The values of the voltages $v_{O1}$ , $v_{O2}$ and $v_O$ .

Also the current through the resistance $R_1$ .

Answer to Problem 9.67P

The expression for the voltage $v_{O2}$ is $-0.4-0.45\sin \omega t\text{V}$ , voltage $v_{O1}$ is $-0.85+0.45in\omega t\text{V}$ , $v_O$ is $1.35-2.7\sin \omega t\text{V}$ and the value of the current through the resistance $R_1$ is $\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}$ .

Explanation of Solution

Calculation:

Substitute $-0.60-0.05\sin \omega t\text{V}$ for $v_{I2}$ and $-0.65+0.05\sin \omega t\text{V}$ for $v_{I1}$ equation (3).

  $\begin{array}{c}{v}_O=27\left(-0.60-0.05\sin \omega t\text{V}-0.60+0.05\sin \omega t\text{V}\right)\\ =1.35-2.7\sin \omega t\text{V}\end{array}$

Substitute $10\text{k}\Omega$ for $R_1$ , $-0.65+0.05\sin \omega t\text{V}$ for $v_{I1}$ and $-0.60-0.06\sin \omega t\text{V}$ for $v_{I2}$ in equation (2).

  $\begin{array}{c}{i}_1=\frac{-0.65+0.05\sin \omega t\text{V}-\left(-0.60-0.06\sin \omega t\text{V}\right)}{10\text{k}\Omega }\\ =\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}\end{array}$

Substitute $-0.65+0.05\sin \omega t\text{V}$ for $v_{I1}$ , $\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}$ for $i_1$ , $40\text{k}\Omega$ for $R_2$ in the above equation.

  $\begin{array}{c}{v}_{O1}=\left(\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}\right)\left(40\text{k}\Omega \right)-0.65+0.05\sin \omega t\text{V}\\ =-0.85+0.45in\omega t\text{V}\end{array}$

Substitute $-0.60-0.05\sin \omega t\text{V}$ for $v_{I2}$ , $\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}$ for $i_1$ , $40\text{k}\Omega$ for $R_2$ equation (5)

  $\begin{array}{c}{v}_{O2}=-0.60-0.05\sin \omega t\text{V}-\left(\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}\right)\left(40\text{k}\Omega \right)\\ =-0.4-0.45\sin \omega t\text{V}\end{array}$

Conclusion:

Therefore, the expression for the voltage $v_{O2}$ is $-0.4-0.45\sin \omega t\text{V}$ , voltage $v_{O1}$ is $-0.85+0.45in\omega t\text{V}$ , $v_O$ is $1.35-2.7\sin \omega t\text{V}$ and the value of the current through the resistance $R_1$ is $\left(-5+10\right){10}^{-3}\sin \omega t\text{mA}$ .