Problem 6.6-8

Let’s call a the inverting input node, and b the non-inverting input node.

In this problem, we shall call V_a as the inverting input node voltage and V_b as the non-inverting input node voltage, as V1 and V2 are already used.

We thus have V_a = V_b

Question a.

At b, the KCL gives, as the current entering the op amp is zero:\

(V2 – V_b)/R3 = V_b /R4

which gives

Moreover, we have V₀ – V_a + V_a – V1 = – R2*io – R1*io

V₀ –V1 = – io*(R2 + R1)

So V₀ = – io*(R2 + R1) + V1

But from:

V_a – V1 = – R1*io we can get i0 = V1/R1 – V_b /R1

Then we get V₀ = – io*(R2 + R1) + V1

By substituting io above in this equation we get the result:

We need to solve

That corresponds to R2 = 11*R1 and also 1 + R2/R1 = 4*(1 + R3/R4), which is 2*R4 = R2

All the different combinations that satisfy those two equations fit here.