Problem + Hard 47-2.
Source: Exam Question for Electronic Engineer
- Petrobras - 2012.
Resistive transducers are used in many industrial control systems. One
way to measure the electrical resistance of a resistive transducer RS
is illustrated in the circuit shown in the Figure 47-02.1, where we consider the amplifiers
ideal operations. Determine the value of electrical resistance RS of the transducer through
from a mathematical expression that is a function of the values of the resistors that make up the circuit and the
output voltage Vo.
Figure 47-02.1
Solution of the Problem 47-2
In the circuit shown in the Figure 47-02.1, names were given to the nodes and in the input circuit we calculated the
Thevenin equivalent. Knowing that the input current of the ideal operational amplifier is
zero and that the input voltages of OP 1 are virtually equal, we can find a relationship between
Vb and VS. Using the nodal voltage method we have:
Vb/R1 + Vb/R1 - VS/R1 = 0
And from there, we get the desired relation, or:
Va = Vb = VS /2
eq. 47-2a
Let's save this relationship because we will need it later. On the other hand, we can find
a relation between Va and Vx making the input mesh of OP 1.
Note that the current passing through the feedback resistor (2 R1) is the same
current flowing through R1 and the input source. Then, we can write:
(Va + 5)/ R1 = (Vx - Va) / 2 R1
We can eliminate R1 and after some algebraic manipulation, we find:
Vx = 3 Va + 10 = (3/2) VS + 10
eq. 47-2b
Note that we used the eq. 47-2a to arrive at eq. 47-2b. Since we have two variables, we need one more
equation that relates Vx and VS to solve the system.
From the circuit we can easily see that ix = ib + iS. Thus,
We can relate the two variables of interest as follows:
ix = ( Vx - VS) / R1 = VS / 2 R1 + VS / RS
Where ib = VS / 2 R1 and iS = VS / RS.
And after some algebraic manipulation we arrive at the desired partial result, or:
Vx = VS [ ( 3/2) + ( R1 / RS) ]
eq. 47-2c
Note that we now have the eq. 47-2b and the eq. 47-2c relating the two variables of interest. So, equating the
two equations we arrive at:
VS [ ( 3/2) + ( R1 / RS) ] = (3/2) VS + 10
And once again doing proper algebraic work we find the relation:
RS = ( R1 / 10) VS
eq. 47-2d
Note that so far we have only worked with OP 1. At this point, how do we need to relate
RS depending on Vo and other components, let's direct
our attention to OP 2. Note that the same is in the non-inverting configuration. Therefore,
the following relation between Vo and VS holds:
Vo = [( R2 + R3) / R3] VS
And since we need the value of VS we can transform the above equation into:
VS = [ R3 / ( R2 + R3)] Vo
Substituting this value in eq. 47-2d we will obtain the relation requested by the problem statement.