Summing amplifier V0 = 3V1 + 2V2 + V3

A summing amplifier gives an output equal to the weighted sum of its inputs. A single inverting summer gives a negative sum, so the required positive output V0 = 3V1 + 2V2 + V3 is obtained by following an inverting summer with a unity-gain inverter.

Stage 1: inverting summing amplifier

 V1 o--[ R1 ]--+
               |
 V2 o--[ R2 ]--+-----[ Rf ]-----+
               |                |
 V3 o--[ R3 ]--+     |\         |
               +-----|- \       |
                     |   >------+----o V01
               +-----|+ /
               |     |/
              GND

For an ideal op-amp (virtual ground at the inverting input):

V01 = −Rf (V1/R1 + V2/R2 + V3/R3)

We need the weights 3, 2 and 1, so Rf/R1 = 3, Rf/R2 = 2 and Rf/R3 = 1.

Choose Rf = 30 kΩ (assumed value):

• R1 = Rf/3 = 10 kΩ
• R2 = Rf/2 = 15 kΩ
• R3 = Rf/1 = 30 kΩ

Then V01 = −(3V1 + 2V2 + V3).

Stage 2: unity-gain inverting amplifier

               +-----[ R ]-----+
               |               |
               |     |\        |
 V01 o--[ R ]--+-----|- \      |
                     |   >-----+----o V0
               +-----|+ /
               |     |/
              GND

With both resistors equal (R = 10 kΩ), V0 = −(R/R) V01 = −V01.

Result

V0 = −V01 = 3V1 + 2V2 + V3

Answer: R1 = 10 kΩ, R2 = 15 kΩ, R3 = 30 kΩ and Rf = 30 kΩ in the first stage, followed by an inverter with two equal resistors (10 kΩ each).

The two resistors of the second stage can be any equal pair. A 741 op-amp with a ±12 V supply can be used, as long as the output stays within the supply limits.