5.4 Inverting Amplifier

← Back to Fundamentals of Electric Circuits Overview

5.4 Inverting Amplifier

In this and the following sections, we consider some useful op amp cir cuits that often serv e as modules for designing more comple x circuits. The first of such op amp circuits is the inverting amplifier shown in Fig. 5.10. In this circuit, the nonin verting input is grounded, vi is con nected to the in verting input through R1, and the feedback resistor Rf is connected between the inverting input and output. Our goal is to obtain

• – ^v2 0 V + +

– 1

i 2

Figure 5.10 The inverting amplifier.

vi

R1

i 1

v1 0 A

For Example 5.2.

Rf

–

vo +

–

the relationship between the input v oltage vi and the output v oltage vo. Applying KCL at node 1,

$$i_1 = i_2 \Rightarrow \frac{v_i - v_1}{R_1} = \frac{v_1 - v_o}{R_f}$$

(5.8)

But v¹ = v² = 0 for an ideal op amp, since the nonin verting terminal is grounded. Hence,

or

$$\frac{v_i}{R_1} = -\frac{v_o}{R_f}$$ $$v_o = -\frac{R_f}{R_1}v_i$$

(5.9)

The voltage gain is A^v = vo∕vⁱ = −Rf∕R1. The designation of the circuit in Fig. 5.10 as an inverter arises from the negative sign. Thus,

An inverting amplifier reverses the polarity of the input signal while amplifying it.

Notice that the gain is the feedback resistance di vided by the in put resistance which means that the gain depends only on the external elements connected to the op amp. In view of Eq. (5.9), an equivalent circuit for the inverting amplifier is shown in Fig. 5.11. The inverting amplifier is used, for example, in a current-to-voltage converter.

Example 5.3 Refer to the op amp in Fig. 5.12. If v^{i =} 0.5 V, calculate: (a) the output voltage vo, and (b) the current in the 10-kΩ resistor.

Solution:

(a) Using Eq. (5.9),

$$\frac{v_o}{v_i} = -\frac{R_f}{R_1} = -\frac{25}{10} = -2.5$$ $$v_o = -2.5v_i = -2.5(0.5) = -1.25$$

V

(b) The current through the 10-kΩ resistor is

$$i = \frac{v_i - 0}{R_1} = \frac{0.5 - 0}{10 \times 10^3} = 50 \,\mu\text{A}$$

Practice Problem 5.3 Find the output of the op amp circuit shown in Fig. 5.13. Calculate the current through the feedback resistor.

Answer: −3.15 V, 11.25 μA.

A key feature of the inverting amplifier is that both the input signal and the feedback are applied at the inverting terminal of the op amp.

Note there are two types of gains: The one here is the closed-loop voltage gain Av, while the op amp itself has an

open-loop voltage gain A.

Figure 5.11

An equivalent circuit for the inverter in Fig. 5.10.

Figure 5.13 For Practice Prob. 5.3. Determine vo in the op amp circuit shown in Fig. 5.14.

Solution:

Applying KCL at node a,

$$\frac{v_a - v_o}{40 \text{ k}\Omega} = \frac{6 - v_a}{20 \text{ k}\Omega}$$ $$v_a - v_o = 12 - 2v_a \implies v_o = 3v_a - 12$$

But v^a = v^b = 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence,

$$v_o = 6 - 12 = -6$$

V

Notice that if v^b = 0 = va, then v^o = −12, as expected from Eq. (5.9).

For Example 5.4.

Two kinds of current -to-voltage converters (also known as transresis- Practice Problem 5.4 tance amplifiers) are shown in Fig. 5.15.

(a) Show that for the converter in Fig. 5.15(a),

$$\frac{v_o}{i_s} = -R$$

(b) Show that for the converter in Fig. 5.15(b),

$$\frac{v_o}{i_s} = -R_1 \left( 1 + \frac{R_3}{R_1} + \frac{R_3}{R_2} \right)$$

Answer: Proof.

Another important application of the op amp is the nonin verting amplifier shown in Fig. 5.16. In this case, the input v oltage vi is applied directly at the nonin verting input terminal, and resistor R1 is connected

Figure 5.16 The noninverting amplifier.

Example 5.4