Inverting Amplifier – Applications

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.

Inverting operational amplifier

Figure 1 shows the basic inverting operational amplifier. The input signal Vi is applied to the inverting terminal through the R1 resistor. The non-inverting terminal of the op-amp is grounded.

Inverting amplifier
Fig. 1 Inverting operational amplifier

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.

Fig. 2 Inverting amplifier circuit analysis

(1)   \begin{equation*}  I_1=\frac{V_i-0}{R_1}=\frac{V_i}{R_1} \end{equation*}

(2)   \begin{equation*}  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

(3)   \begin{equation*}  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 Av is given by

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

From voltage gain Av, 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, AOL (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.

Fig. 3 Inverting amplifier circuit analysis

(4)   \begin{equation*}  I_1=\frac{V_i-V_-}{R_1} \end{equation*}

(5)   \begin{equation*}  I_2=\frac{V_--V_o}{R_2} \end{equation*}

Apply KCL at node P

I_{1} = 0 + I_2

(6)   \begin{equation*}  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}

(7)   \begin{equation*}  $\Rightarrow \frac{V_i}{R_1}=-\frac{V_o}{R_2}+V_-(\frac{1}{R_1}+\frac{1}{R_2})$ \end{equation*}

As we know that,

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

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

(8)   \begin{equation*}  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 Av 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

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

From voltage gain Av, we can see that the output is 180° out of phase with the input.

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.

error: Content is protected !!

Adblocker detected! Please consider reading this notice.

We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading.

We don't have any banner, Flash, animation, obnoxious sound, or popup ad. We do not implement these annoying types of ads!

We need fund to operate the site, and almost all of it comes from our online advertising.

Please add electricalvoice.com to your ad blocking whitelist or disable your adblocking software.

×