5.2 Operational Amplifiers

← Back to Fundamentals of Electric Circuits Overview

5.2 Operational Amplifiers

An operational amplifier is designed so that it performs some mathematical operations when e xternal components, such as resistors and capaci tors, are connected to its terminals. Thus,

An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration.

The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. A full discussion of what is inside the op amp is beyond the scope of this book. It will suffice

The term operational amplifier was introduced in 1947 by John Ragazzini and his colleagues, in their work on analog computers for the National Defense Research Council after World War II. The first op amps used vacuum tubes rather than transistors.

An op amp may also be regarded as a voltage amplifier with very high gain.

to treat the op amp as a circuit b uilding block and simply study what takes place at its terminals.

Op amps are commercially a vailable in integrated circuit packages in several forms. Figure 5.1 sho ws a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are:

• 1. The inverting input, pin 2.
• 2. The noninverting input, pin 3.
• 3. The output, pin 6.
• 4. The positive power supply V⁺, pin 7.
• 5. The negative power supply V⁻, pin 4.

The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown, the op amp has tw o inputs and one output. The inputs are mark ed with minus (−) and plus (+) to specify inverting and noninverting inputs, respectively. An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the in verting terminal will appear inverted at the output.

As an active element, the op amp must be powered by a voltage supply as typically shown in Fig. 5.3. Although the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked. By KCL,

$$i_o = i_1 + i_2 + i_+ + i_- \tag{5.1}$$

The equivalent circuit model of an op amp is shown in Fig. 5.4. The output section consists of a v oltage-controlled source in series with the

Figure 5.2 A typical op amp: (a) pin configuration, (b) circuit symbol.

Powering the op amp.

Figure 5.4 The equivalent circuit of the nonideal op amp.

Figure 5.1

A typical operational amplifier. © McGraw-Hill Education/Mark Dierker, photographer

The pin diagram in Fig. 5.2(a) corresponds to the 741 generalpurpose op amp made by Fairchild Semiconductor.

output resistance Ro. It is evident from Fig. 5.4 that the input resis tance Ri is the Thevenin equivalent resistance seen at the input terminals, while the output resistance Ro is the Thevenin equivalent resistance seen at the output. The differential input voltage vd is given by

$$v_d = v_2 - v_1 \tag{5.2}$$

where v1 is the voltage between the inverting terminal and ground and v² is the voltage between the noninverting terminal and ground. The op amp senses the difference between the two inputs, multiplies it by the gain A, and causes the resulting voltage to appear at the output. Thus, the output vo is given by

$$v_o = Av_d = A(v_2 - v_1)$$

(5.3)

Sometimes, voltage gain is expressed in decibels (dB), as discussed in Chapter 14.

A dB = 20 log10 A

A is called the open-loop voltage gain because it is the g ain of the op amp without any external feedback from output to input. Table 5.1 shows typical values of voltage gain A, input resistance Ri, output resistance Ro, and supply voltage VCC.

The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp. As Example 5.1 shows, when there is a feedback path from output to input, the ratio of the output voltage to the input voltage is called the closed-loop gain. As a result of the ne gative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths.

A practical limitation of the op amp is that the magnitude of its output voltage cannot e xceed |VCC|. In other w ords, the output v oltage is dependent on and is limited by the po wer supply voltage. Figure 5.5 illustrates that the op amp can operate in three modes, depending on the differential input voltage vd:

• 1. Positive saturation, v^o = VCC.
• 2. Linear region, −VCC ≤ v^o = Av^d ≤ VCC.
• 3. Negative saturation, v^o = −VCC.

If we attempt to increase vd beyond the linear range, the op amp becomes saturated and yields v^o = VCC or v^o = −VCC. Throughout this book, we will assume that our op amps operate in the linear mode. This means that the output voltage is restricted by

$$-V_{CC} \le v_o \le V_{CC} \tag{5.4}$$

TABLE 5.1

Typical ranges for op amp parameters.

Parameter	Typical range	Ideal values
Open-loop gain, A	105 to 108	`
	105	`Ω
Output resistance, Ro Supply voltage, VCC	10 to 100 Ω 5 to 24 V	0Ω
	Input resistance, Ri	to 1013 Ω

Op amp output voltage vo as a function of the differential input voltage vd.

, input resis tance

Although we shall al ways operate the op amp in the linear re gion, the possibility of saturation must be borne in mind when one designs with op amps, to a void designing op amp circuits that will not w ork in the laboratory.

of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closed-loop gain vo∕vs. Determine current i when

Throughout this book, we assume that an op amp operates in the linear range. Keep in mind the voltage constraint on the op amp in this mode.

Example 5.1

Figure 5.6

For Example 5.1: (a) original circuit, (b) the equivalent circuit.

A 741 op amp has an open-loop voltage gain of 2 × 105

Solution:

Using the op amp model in Fig. 5.4, we obtain the equivalent circuit of Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. 5.6(b) by using nodal analysis. At node 1, KCL gives

$$\frac{v_s - v_1}{10 \times 10^3} = \frac{v_1}{2000 \times 10^3} + \frac{v_1 - v_o}{20 \times 10^3}$$

Multiplying through by 2000 × 103 , we obtain

$$200v_s = 301v_1 - 100v_o$$

or

$$2v_s \simeq 3v_1 - v_o \quad \Rightarrow \quad v_1 = \frac{2v_s + v_o}{3} \tag{5.1.1}$$

At node O,

$$\frac{v_1 - v_o}{20 \times 10^3} = \frac{v_o - Av_d}{50}$$

But v^d = −v1 and A = 200,000. Then

$$v_1 - v_o = 400(v_o + 200,000v_1)$$

\n(5.1.2)

Substituting v1 from Eq. (5.1.1) into Eq. (5.1.2) gives

$$0 \simeq 26,667,067v_o + 53,333,333v_s \quad \Rightarrow \quad \frac{v_o}{v_s} = -1.9999699$$

This is closed -loop gain, because the 20 -kΩ feedback resistor closes the loop between the output and input terminals. When v^s = 2 V, v^o = −3.9999398 V. From Eq. (5.1.1), we obtain v¹ = 20.066667 μV. Thus,

$$i = \frac{v_1 - v_o}{20 \times 10^3} = 0.19999 \text{ mA}$$

It is evident that working with a nonideal op amp is tedious, as we are dealing with very large numbers.

Practice Problem 5.1

Figure 5.7 For Practice Prob. 5.1.

If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7, calculate the closed-loop gain vo∕vs. Find io when v^s = 1 V.

Answer: 9.00041, 657 μA.