Problem 6.5-8

Let’s call a the node on the upper part of the bridge and b the one on the lower.

Following the example on page 215 of the textbook, we can write an equivalent Thevenin equivalent for the bridge. We have Rt = (R1//R2) in series with (R3//R4)

Thus we get Rt = (10k//30k) in series with (30k//10k)

So Rt = 75k + 75k = 150 kΩ

Then the Thevenin’s voltage is such that

Voc = (R2/(R1 + R2) – R4/(R3 + R4)) *Vs

Here we get Voc = 1*Vs = 6 V

The equivalent circuit is the same as figure 6.5-2 (d) on page 215. We can apply the KVL for Va. Let V1 be the voltage for the non-inverting input node.

We get Va = V1 + Voc + Rt*i1 but we have i1 = 0 because of the op amp and V1 = 0 also. Thus Va = Voc

Now we can write the node equation at node a:

We have i1 = (V0 – Va)/30 k – Va/10 k

 

(V0 – Voc)/30 k = Voc/10 k

So V0 – Voc = 3*Voc

V0 = 4*Voc = 24 V