Applications of Operational Amplifiers: Inverting and Non-inverting Configuarations

Inverting summer:

The configuration is shown in fig. 2. With three input voltages v_a, v_b & v_c. Depending upon the value of R_f and the input resistors R_a, R_b, R_c the circuit can be used as a summing amplifier, scaling amplifier, or averaging amplifier.

Again, for an ideal OPAMP, v1 = v2. The current drawn by OPAMP is zero. Thus, applying KCL at v2 node This means that the output voltage is equal to the negative sum of all the inputs times the gain of the circuit Rf/ R; hence the circuit is called a summing amplifier. When Rf= R then the output voltage is equal to the negative sum of all inputs. vo= -(va+ vb+ vc)

Fig. 2

If each input voltage is amplified by a different factor in other words weighted differently at the output, the circuit is called then scaling amplifier.

The circuit can be used as an averaging circuit, in which the output voltage is equal to the average of all the input voltages.

In this case, R_a= R_b= R_c = R and R_f / R = 1 / n where n is the number of inputs. Here R_f / R = 1 / 3.

v_o = -(v_a+ v_b + v_c) / 3

In all these applications input could be either ac or dc.

Non-inverting configuration:

If the input voltages are connected to non-inverting input through resistors, then the circuit can be used as a summing or averaging amplifier through proper selection of R₁, R₂, R₃ and R_f. as shown in fig. 3.

To find the output voltage expression, v1 is required. Applying superposition theorem, the voltage v1 at the non-inverting terminal is given by Hence the output voltage is

Fig. 3

This shows that the output is equal to the average of all input voltages times the gain of the circuit (1+ R_f / R₁), hence the name averaging amplifier.

If (1+R_f/ R₁) is made equal to 3 then the output voltage becomes sum of all three input voltages.

v_o = v _a + v_b+ v_c

Hence, the circuit is called summing amplifier. 

Example - 1

Find the gain of V_O / V_i of the circuit of fig. 4.

Solution:

Current entering at the inveting terminal . Applying KCL to node 1, Applying KCL to node 2, Thus the gain A = -8 V / VO

Fig. 4

Example - 2

Find a relationship between V_O and V₁ through V₆ in the circuit of fig. 5.

Fig. 5

Solution:
Let's consider of V₁ (singly) by shorting the others i.e. the circuit then looks like as shown in fig. 6.

The current flowing through the resistor R into the i/e. The current when passes through R, output an operational value of Let as now consider the case of V2 with other inputs shorted, circuit looks like as shown in fig. 7.	Fig. 6
Now VO is given by same thing to V4 and V6 net output V" = V2 + V4 + V6     (3) From (2) & (3) V' + V" = (V2 + V4 + V6 ) - (V1 + V3 + V5) So VO = V2 + V4 + V6 - V1 - V3 - V5.	Fig. 7

Example - 3

1. Show that the circuit of fig. 8 has A = V_O / V_i = - K (R₂ / R₁) with K = 1 + R₄ / R₂ + R₄ / R₃, and R_i = R₁.
2. Specify resistance not larger than 100 K to achieve A = -200 V / V and R_i = 100 K.

Fig. 8

Solution: