Summing Amplifier or Op-amp Adder | Applications | Electricalvoice

Summing Amplifier or Op-amp Adder | Applications

Summing Amplifier is an electronic circuit that produces output as a weighted sum of the applied inputs. It is also known as Op-amp adder. Basically it performs mathematical operation of addition. In this article, we will see the summing amplifier circuit, its working and its applications.

Summing Amplifier Circuit

Summing Amplifier
Fig. 1 Summing Amplifier

The summing amplifier circuit is shown in figure 1. The input V1 , V2, and V3 are applied. Vo is the output voltage. The non-inverting terminal of the op-amp is connected to the ground. This means that the voltage of the non-inverting terminal is zero volts.

Analysis

The analysis of the summing 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+ = 0V), according to the virtual short concept.

V= V+ = 0V

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

Fig. 2 Summing amplifier analysis

The currents I1, I2 and I3 flowing through resistances R1, R2 and R3 respectively.

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

(2)   \begin{equation*}  I_2=\frac{V_{2}-0}{R_2}=\frac{V_{2}}{R_2} \end{equation*}

(3)   \begin{equation*}  I_3=\frac{V_{3}-0}{R_3}=\frac{V_{3}}{R_3} \end{equation*}

Apply KCL at node Q

(4)   \begin{equation*}  $I_1+I_2+I_3 = I$ \end{equation*}

Apply KCL at node P

I = 0+I_f

(5)   \begin{equation*}  $I = I_f$ \end{equation*}

(6)   \begin{equation*}  I_f=\frac{0-V_{o}}{R_f}=-\frac{V_{o}}{R_f} \end{equation*}

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

\Rightarrow I_1+I_2+I_3=-\frac{V_{o}}{R_f}

putting values of I1, I2 and I3 from equations (1), (2) and (3), we have

\Rightarrow \frac{V_{1}}{R_1}+\frac{V_{2}}{R_2}+\frac{V_{3}}{R_3}=-\frac{V_{o}}{R_f}

\Rightarrow V_o=-(\frac{R_f}{R_1}V_1+\frac{R_f}{R_2}V_2+\frac{R_f}{R_3}V_3)

Therefore,

    \[ \boxed{V_o=-(\frac{R_f}{R_1}V_1+\frac{R_f}{R_2}V_2+\frac{R_f}{R_3}V_3)} \]

If R_1=R_2=R_3=R_f=R

Then, we have

V_o=-(\frac{R}{R}V_1+\frac{R}{R}V_2+\frac{R}{R}V_3)

    \[ \boxed{V_o=-(V_1+V_2+V_3)} \]

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