Question

Show that when the four resistances are equal, the following circuit becomes a subtracting circuit.

Short answer

The input and output of the given circuit is related by, $v_o=v_2-v_1$.

Step-by-step solution

Step 1, Given Information

Given an amplifier circuit with an Op-Amp and four resistors.

We are asked to prove that it is a subtracting circuit

Step 2. Explanation

The Op-Amp has an inverting and a non-inverting terminal.

Let the voltages at the inverting and a non-inverting terminal be $v_{-}$and $v_{+}$respectively.

Also, the current through $R_1,R_{2}and{R}_{k1}$is $i_1,i_{2}and{i}_{k1}$respectively.

Therefore, applying Kirchhoff's law in the loop including the inverting terminal, we get;

$i_1=0+i_{k1}$

Op-Amp has a large input impedance, so the input current will only pass through the feedback resistor.

Therefore, we can write;

$\frac{v_1-v_{-}}{R_1}=\frac{v_{-}-v_o}{R_{k1}}$

$\frac{v_1}{R_1}-\frac{v_{-}}{R_1}=\frac{v_{-}}{R_{k1}}-\frac{v_o}{R_{k1}}$

$\frac{v_1}{R_1}+\frac{v_o}{R_{k1}}=\frac{v_{-}}{R_{k1}}+\frac{v_{-}}{R_1}$

$\Rightarrow v_1{R}_{k1}+v_o{R}_1=v_{-}(R_{k1}+R_1)$

$\Rightarrow v_{-}=\frac{v_1{R}_{k1}+v_o{R}_1}{(R_{k1}+R_1)}$

But given that all resistances are equal.

$\Rightarrow v_{-}=\frac{v_{1}R+v_{o}R}{2R}$

Now, based on voltage division we get;

$v_{+}=\left(\frac{R_{k2}}{R_{k2}+R_2}\right)v_2$

But as all resistances are equal.

localid="1645523564252" $v_{+}=\frac{R{v}_2}{2R}=\frac{v_2}{2}$

Step 3. The input and output of the given circuit is related

Based on the virtual ground concept;

$v_{-}=v_{+}$

Now substituting the values of voltages at the inverting and non-inverting terminals, we get;

$$\begin{gathered} \frac{v_{1}R+v_{o}R}{2R}=\frac{v_2}{2} \\ \frac{v_1+v_o}{2}=\frac{v_2}{2} \\ \Rightarrow v_o=v_2-v_1 \end{gathered}$$

This implies that the output is the difference of inputs and hence the circuit acts as a subtracting circuit when the resistances in the difference amplifier configuration given are made equal.