Resolved Problems about Multiple Operational Amplifier

Problem 47-4 Source: Problem 5.87 - page 187 - ALEXANDER, Charles K., SADIKU, Matthew N. O. - Book: Fundamentos de Circuitos Elétricos - 5ª Edição - Ed. McGraw Hill - 2013.

The circuit shown in the Figura 47-04.1 represents an instrumentation amplifier with two operational amplifiers. Deduct an expression for V_o in terms of V₁ and V₂. Explain how this amplifier can be used as a subtractor.

Solution of the Problem 47-4

This circuit is composed of an operational amplifier ( OP 1) connected in the non-inverting configuration and another ( OP 2) in the configuration inverter. Whereas the current flowing through the R₄ is the same that circulates through R₃, we can write:

( V_o - V₃ ) / R₄ = ( V₃ - V_x ) / R₃

Of the circuit, virtually, V₂ = V₃ and thus rearranging the terms:

V_o = V₂ (( R₃ + R₄ ) / R₃ ) - V_x ( R₄ / R₃ )

On the other hand, the following relation is obtained from OP 1:

V_x = V₁ (( R₁ + R₂ ) / R₁ )

Substituting this value into the previous equation finds the desired relation.

V_o = V₂ (( 1 + ( R₄ / R₃ )) - V₁ ( R₄ / R₃ )( 1 + ( R₂ / R₁ ))

For this circuit to act as a subtractor, it is sufficient to do R₄ = R₁ and also R₃ = R₂. Thus, the end result is a subtractor circuit with gain equal to K = 1 + ( R₄ / R₃). Then, V_o will be given by:

V_o = ( V₂ - V₁ ) ( 1 + ( R₄ / R₃))

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