Question

The circuit shown in Fig. 4.17a is a precision differential voltage-gain device. It is used to provide a single-ended input for an analog-to-digital converter. We wish to derive an expression for the output of the circuit in terms of the two inputs.

Answer 1

\( Solution: \) To accomplish this, we draw the equivalent circuit shown in Fig. 4.17b. Recall that the voltage across the input terminals of the op-amp is approximately zero and the currents into the op-amp input terminals are approximately zero. Note that we can write node equations for node voltages \( V_1 \) and  \( V_2 \) in terms of \( V_0 \) and \( V_a \). Since we are interested in an expression for \( V_0 \)

in terms of the voltages \( V_1 \) and \( V_2 \), we simply eliminate the \( V_a \) terms from the two-node equations. The node equations are

\( \frac{V_1-V_0}{R_2}+\frac{V_1-V_a}{R_1}+\frac{V_1-V_2}{R_G}=0 \)

\( \frac{V_2-V_a}{R_1}+\frac{V_2-V_1}{R_G}+\frac{V_2}{R_2}=0 \)

Combining the two equations to eliminate \( V_a \), and then writing \( V_0 \) in terms of \( V_1 \) and \( V_2 \), yields

\( V_0=(V_1-V_2) \bigg(1+\frac{R_2}{R_1}+\frac{2R_2}{R_G} \bigg) \)

 

Answer : Combining the two equations to eliminate \( V_a \), and then writing \( V_0 \) in terms of \( V_1 \) and \( V_2 \), yields

\( V_0=(V_1-V_2) \bigg(1+\frac{R_2}{R_1}+\frac{2R_2}{R_G} \bigg) \)