Chapter 5, Problem 48P

The circuit in Fig. 5.80 is a differential amplifier driven by a bridge. Find v_o.

To determine

Calculate the output voltage $v_o$ of the differential amplifier.

Answer to Problem 48P

The output voltage $v_o$ is $\underset{\_}{-10.364\text{mV}}$.

Explanation of Solution

Given data:

Refer Figure 5.80 in the textbook for the differential amplifier that is driven by a bridge circuit.

Calculation:

Consider break this problem up into parts. The 15.5 mV voltage source splits the lower circuit from the upper circuit. In addition, there is no current flowing into the op amp input which means we now have the $40\text{kΩ}$ resistor in series with a parallel combination of the $60\text{kΩ}$ resistor and the equivalent $100\text{kΩ}$ resistor.

$\begin{array}{c}{R}_{\text{eq}}=40\text{k}\Omega +\left(\frac{60\text{k}\Omega \times 100\text{k}\Omega }{60\text{k}\Omega +100\text{k}\Omega }\right)\\ =40\text{k}\Omega +\left(\frac{6000}{160}\right)\text{k}\Omega \\ =40\text{k}\Omega +37.5\text{k}\Omega \\ =77.5\text{k}\Omega \end{array}$

Consider the expression for the current flows through the circuit.

$i=\frac{v_s}{R_{\text{eq}}}$

Substitute 15.5 mV for $v_s$ and $77.5\text{k}\Omega$ for $R_{\text{eq}}$.

$\begin{array}{c}i=\frac{15.5\text{mV}}{77.5\text{k}\Omega }\\ =200\times {10}^{-9}\text{A}\end{array}$

The voltage across the $60\text{kΩ}$ and equivalent $100\text{kΩ}$ is,

$v=i\left(60\text{kΩ}\parallel 100\text{kΩ}\right)$

Substitute $200\times {10}^{-9}\text{A}$ for i.

$\begin{array}{c}v=\left(200\times {10}^{-9}\text{A}\right)\left(60\text{kΩ}\parallel 100\text{kΩ}\right)\\ =\left(200\times {10}^{-9}\text{A}\right)\left(\frac{60\text{kΩ}\times 100\text{kΩ}}{60\text{kΩ}+100\text{kΩ}}\right)\\ =7.5\text{mV}\end{array}$

The voltage across the $80\text{kΩ}$  resistor is,

$\begin{array}{c}{v}_{80\text{k}\Omega }=0.8\times 7.5\text{mV}\\ =\text{6}\text{mV}\end{array}$

This voltage is also the voltage at both inputs of the op amp and the voltage between the $20\text{kΩ}$ and $80\text{kΩ}$ resistors in the upper circuit. Consider $v_1$ be the voltage to the left of the $20\text{kΩ}$ resistor of the upper circuit.

Modify Figure 5.80 by indicating the node voltages. The modified figure as shown in Figure 1.

Write a node equation at node $v_1$ using Kirchhoff’s current law.

$\begin{array}{l}\left(\frac{v_1-15.5\text{mV}}{10\text{kΩ}}\right)+\frac{v_1}{30\text{kΩ}}+\left(\frac{v_1-6\text{mV}}{20\text{kΩ}}\right)=0\\ 6v_1-93+2v_1+3v_1-18=0\\ v_1=10.091\text{mV}\end{array}$

Consider the expression for the current through the $20\text{kΩ}$ resistor, from left to right.

$i_{20\text{kΩ}}=\frac{v_1-v_{80\text{kΩ}}}{20\text{kΩ}}$

Substitute 10.091 mV for $v_1$ and 6 mV for $v_{80\text{k}\Omega }$.

$\begin{array}{c}{i}_{20\text{kΩ}}=\frac{10.091\text{mV}-6\text{mV}}{20\text{kΩ}}\\ =204.55\times {10}^{-9}\text{A}\end{array}$

Write the expression for the output voltage $v_o$ of the differential amplifier.

$v_o=6\text{mV}-i_{20\text{kΩ}}\left(80\text{kΩ}\right)$

Substitute $204.55\times {10}^{-9}\text{A}$ for $i_{20\text{kΩ}}$.

$\begin{array}{c}{v}_o=6\text{mV}-\left(204.55\times {10}^{-9}\text{A}\right)\left(80\text{kΩ}\right)\\ =6\text{mV}-16.364\text{mV}\\ =-10.364\text{mV}\end{array}$

Conclusion:

Thus, the output voltage $v_o$ is $\underset{\_}{-10.364\text{mV}}$.