An **inverting amplifier** is an operational amplifier circuit which produces amplified output that is 180° out of phase to the applied input. Inverting amplifier is also known as **inverting operational amplifier** or **inverting op-amp**.

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## Inverting operational amplifier

Figure 1 shows the basic inverting operational amplifier. The input signal V_{i} is applied to the inverting terminal through the R_{1} resistor. The non-inverting terminal of the op-amp is grounded.

### Analysis

**Case-1: When op-amp is ideal i.e. open-loop gain is infinite.**

The analysis of the inverting amplifier circuit is shown in figure 2. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V_{−}) is equal to the voltage of the non-inverting terminal (V_{+} = 0), according to the **virtual ground concept**.

V_{− }= V_{+} = 0 Volt

The currents entering both terminals of the op-amp are zero since the op-amp is ideal.

\begin{equation} \label{eq:poly}

I_1=\frac{V_i-0}{R_1}=\frac{V_i}{R_1}

\end{equation}

\begin{equation} \label{eq:poly}

I_2=\frac{0-V_o}{R_2}=-\frac{V_o}{R_2}

\end{equation}

Apply KCL at node **P**

$I_{1} = 0 + I_2$

\begin{equation} \label{eq:poly}

I_1= I_{2}

\end{equation}

From equations (1), (2) and (3), we have

$\Rightarrow \frac{V_i}{R_1}=-\frac{V_o}{R_2}$

$\Rightarrow \frac{V_0}{V_i}=-\frac{R_2}{R_1}$

The closed-loop voltage gain A_{v} is given by

\[

\quicklatex{color=”#000000″ size=20}

\boxed{A_{v}=\frac{V_0}{V_i}=-\frac{R_2}{R_1}}

\]

From voltage gain A_{v}, we can see that the output is 180° out of phase with the input.

**Case-2: When op-amp has a finite open-loop gain, A _{OL} (Practical case)**

The analysis of the inverting amplifier circuit is shown in figure 3. Since the op-amp has a finite open-loop gain, Hence the voltage of the inverting terminal (V_{−}) is not equal to the voltage of the non-inverting terminal (V_{+} = 0) i.e.

V_{− }≠ V_{+}

Assuming that the currents entering both terminals of the op-amp are zero.

\begin{equation} \label{eq:poly}

I_1=\frac{V_i-V_-}{R_1}

\end{equation}

\begin{equation} \label{eq:poly}

I_2=\frac{V_–V_o}{R_2}

\end{equation}

Apply KCL at node **P**

$I_{1} = 0 + I_2$

\begin{equation} \label{eq:poly}

I_1= I_{2}

\end{equation}

From equations (4), (5) and (6), we have

$\Rightarrow \frac{V_i-V_-}{R_1}=\frac{V_–V_o}{R_2}$

\begin{equation} \label{eq:poly}

$\Rightarrow \frac{V_i}{R_1}=-\frac{V_o}{R_2}+V_-(\frac{1}{R_1}+\frac{1}{R_2})$

\end{equation}

As we know that,

$V_o=A_{OL}(V_+-V_-)$

$V_o=A_{OL}(0-V_-)=-A_{OL}V_-$

\begin{equation} \label{eq:poly}

V_-=-\frac{V_0}{A_{OL}}

\end{equation}

From equations (7) and (8), we have

$\Rightarrow \frac{V_i}{R_1}=-\frac{V_o}{R_2}-\frac{V_0}{A_{OL}}(\frac{1}{R_1}+\frac{1}{R_2})$

$\Rightarrow \frac{V_i}{R_1}=-V_o(\frac{1}{R_2}+\frac{1}{A_{OL}}(\frac{1}{R_1}+\frac{1}{R_2}))$

The closed-loop voltage gain A_{v} is given by

$\Rightarrow A_v=\frac{V_o}{V_i}=\frac{-\frac{1}{R_1}}{\frac{1}{R_2}+\frac{1}{A_{OL}}(\frac{1}{R_1}+\frac{1}{R_2})}$

$\Rightarrow A_v=\frac{V_o}{V_i}=\frac{-\frac{R_2}{R_1}}{1+\frac{1}{A_{OL}}(1+\frac{R_2}{R_1})}$

Therefore

\[

\quicklatex{color=”#000000″ size=20}

\boxed{A_v=\frac{V_o}{V_i}=\frac{-\frac{R_2}{R_1}}{1+\frac{(1+\frac{R_2}{R_1})}{A_{OL}}}}

\]

From voltage gain A_{v}, we can see that the output is 180° out of phase with the input.