Question

Design a 3-channel op-amp based circuit to give following output relation.
7V1-5V2+dV3/dt

Answer 1

See the diagram.

    The three channels are the V1, V2, V3 branches at the inputs. For the differentiation, we use a capacitor. We use Kirchoff's laws for voltage and current at input ports "-" and "+" of the Opamp.


   V₊ = V₂ R₄ /(R₄+R₂) = V₂ / [ 1 + R₂/R₄ ]
    V₋ ≈ V₊
   Q = (V₃ - V₋) C
   Current through capacitor = i₃ = dQ/dt = C dV₃/dt  as V₋ ≈ V₊ and it is constant.

   i = (V₋ - V₀)/R₃ = C dV₃/dt + (V₁ - V₋) / R₁

=> - V₀ = (R₃ C) dV₃/dt + V₁ *R₃/R₁ - V₋ * [ 1 + R₃/R₁] 
=> V₀ = - { V₁ * R₃/R₁ + (R₃ C  dV₃/dt  - V₂ * [1 + R₃/R₁] / [1 + R₂/R₄]  }

Given function 7 V₁ - 5 V₂ + dV₃/dt

So comparing the expressions: 
     R₃ / R₁ = 7                  R₃ C = 1
     1 + R₂/R₄ = (7+1) / 5
     R₂ / R₄ = 0.60

We get the function of the output, the sign is negative.

For example:   R₁ = 1 MΩ    then  R₂ = 600 Ω      R₃ = 7 MΩ    R₄ = 1 kΩ
                        C = 0.14 μF