Summing Amplifier or Op-amp Adder – Applications

April 15, 2021September 5, 2020 by Electricalvoice

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.

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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. V_o 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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