Operational Amplifiers (OpAmps)

Transcription

1 Chapter 1 Operational Amplifiers (OpAmps) 1.1 Introduction [1] An opamp is an electronic unit that behaves like a voltage-controlled voltage source.an opamp may also regarded as a voltage amplifier with very high gain. An opamp is an integrated circuit (IC) that amplifies the difference between two input voltages and produces a single output. An op amp can sum signals, amplify a signal, integrate it, or differentiate it. The ability of the op amp to perform these mathematical operations is the reason it is called an operational amplifier. It is also the reason for the widespread use of op amps in analog design. The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Op amps are commercially available in integrated circuit packages in several forms. Figure 1.1 shows a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 1.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to this course. The five important terminals are: 1 P a g e

2 1. The inverting input, pin The noninverting input, pin The output, pin The positive power supply V +, pin The negative power supply V, pin 4. The circuit symbol for the op amp is the triangle in Fig. 1.2(b); as shown, the op amp has two inputs and one output. The inputs are marked 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 inverting terminal will appear inverted at the output. Figure 1.1 A typical operational amplifier. [1] Figure 1.2 A typical op amp: (a) pin configuration, (b) circuit symbol. [1] 2 P a g e

3 As an active element, the op amp must be powered by a voltage supply as shown in Fig By KCL, i o = i 1 + i 2 + i + + i (1.1) Figure 1.3 Powering the op amp. [1] 1.2 OpAmp Equivalent Circuit [1] The equivalent circuit model of an op amp is shown in Fig The output section consists of a voltage-controlled source in series with the output resistance R o. It is evident from Fig. 1.3 that the input resistance R i is the Thevenin equivalent resistance seen at the input terminals, while the output resistance R o is the Thevenin equivalent resistance seen at the output. The differential input voltage v d is given by v d = v 2 v 1 (1.2) where v 1 is the voltage between the inverting terminal and ground and v 2 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 v o is given by v o = A od (v 2 v 1 ) (1.3) 3 P a g e

4 A is called the open-loop voltage gain because it is the gain of the opamp without any external feedback from output to input. Figure 1.4 The equivalent circuit of the nonideal op amp. [1] Table 1.1 shows typical values of voltage gain A, input resistance R i, output resistance R o, and supply voltage V CC. Table 1.1 Typical ranges for op amp parameters. [1] A practical limitation of the op amp is that the magnitude of its output voltage cannot exceed V CC. In other words, the output voltage is dependent on and is limited by the power supply voltage. Figure 1.4 illustrates that the opamp can operate in three modes, depending on the differential input voltage v d : 4 P a g e

5 1. Positive saturation, v o = V cc. 2. Linear region, V cc v o = AV d V cc. 3. Negative saturation, v o = V cc. Figure 1.5 Opamp output voltage v o as a function of the input voltage v d. [1] If we attempt to increase v d beyond the linear range, the op amp becomes saturated and yields v o = V cc or v o = V cc. Here, we will assume that our op amps operate in the linear mode. This means that the output voltage is restricted by V cc v o V cc (1.4) 5 P a g e

6 Example 1: A 741 op amp has an open-loop voltage gain of , input resistance of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of the figure below. Find the closed-loop gain V 0 /V S. Determine current i when V S = 2 V. Figure 1.6 For Example 1 [1] Solution: From the question, we obtain the equivalent circuit shown in Figure 1.6. Figure 1.7 For Example 1 [1] 6 P a g e

7 At node 1, KCL gives VS V1 V V S V0 or V1 3 At node 0, 3 V1 V V1 V But V Then V 3 1 d V0 AVd 50 V and A 200,000 V ( V V0 We obtain V S 0 200,000V ) When V S = 2 V, V 0 = V, and then we obtain V 1 = µv 1 Thus, i V V ma Practice problem 1: If the same 741 op amp is used in the circuit of this figure, calculate the closed-loop gain V o /V s. Find i 0 when V s = 1 V. Answer: , ma. Figure 1.8 For Problem 1 [1] 7 P a g e

8 1.3 Ideal OpAmp [1] Opamp is ideal if it has the following characteristics: Infinite open-loop gain, A od. Infinite input resistance, R i. Zero output resistance, R o 0. For circuit analysis, the ideal op amp is illustrated in Fig Two important characteristics of the ideal op amp are: 1. The currents into both input terminals are zero: i 1 = i 2 = 0 (1.5) 2. The voltage across the input terminals is negligibly small; i.e., v d v 0 v v (1.6) 2 v1 or 1 2 Example 2: Rework Practice Prob. 5.1 using the ideal op amp model. Figure 1.9 For Example 2 [1] 8 P a g e

9 Solution: We may replace the op amp in Fig. 1.8 by its equivalent model in Fig. 1.9 as we did in problem 1. We just need to keep Eqs. (1.5) and (1.6) in mind as we analyze the circuit in Fig Notice that v 2 = v s Since i 1 = 0, the 40-kΩ and 5-kΩ resistors are in series; the same current flows through them. v 1 is the voltage across the 5-kΩ resistor. Hence, using the voltage division principle, According to Eq. (1.6), v 1 = 5 v = v 0 9 v 2 = v 1 Therefore, v s = v 0 9 v 0 v s = 9 which is very close to the value of obtained with the nonideal model in Practice Prob. 1. This shows that negligibly small error results from assuming ideal op amp characteristics. At node O, i 0 = v v 0 20 ma when v s = 1 V, v o = 9 V. Substituting for v o = 9 V produces i 0 = = 0.65 ma This, again, is close to the value of ma obtained in Practice Prob. 1 with the nonideal model. 9 P a g e

10 1.4 Inverting Amplifiers [1] In this and the following sections, we consider some useful op amp circuits that often serve as modules for designing more complex circuits. The first of such op amp circuits is the inverting amplifier shown in Fig In this circuit, the noninverting input is grounded, v i is connected to the inverting input through R 1, and the feedback resistor R f is connected between the inverting input and output. Our goal is to obtain the relationship between the input voltage v i and the output voltage v o. Figure 1.10 The inverting amplifier. [1] Applying KCL at node 1, (1.7) But v 1 = v 2 = 0 for an ideal op amp, since the noninverting terminal is grounded. Hence, 10 P a g e

11 or (1.8) The voltage gain is A v (closed loop gain) = v 0 v i = R f R i. The designation of the circuit in Fig as an inverter arises from the negative sign. Thus, an inverting amplifier reverses the polarity of the input signal while amplifying it. Example 3: Consider the op amp in Fig If v i = 0.5 V, calculate: (a) the output voltage v o, and (b) the current in the 10-kΩ resistor. Solution: (a) Using Eq. (1.8), Figure 1.11 For Example 3 [1] 11 P a g e

12 (b) The current through the 10-kΩ resistor is Practice problem 2: Find the output of the op amp circuit shown in Fig Calculate the current through the feedback resistor. Figure 1.12 For practice problem 2. [1] Answer: 1.2 V, 10 A. 12 P a g e

13 1.5 Summing Amplifiers [1] Besides amplification, the op amp can perform addition and subtraction. The addition is performed by the summing amplifier covered in this section. A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. Figure 1.13 The summing amplifier. [1] The summing amplifier, shown in Fig. 1.13, is a variation of the inverting amplifier. It takes advantage of the fact that the inverting configuration can handle many inputs at the same time. Here, the current entering each op amp input is zero. Applying KCL at node a gives, But (1.9) 13 P a g e

14 (1.10) We note that v a = 0 and substitute Eq. (1.10) into Eq. (1.9). We get (1.11) indicating that the output voltage is a weighted sum of the inputs. For this reason, the circuit in Fig is called a summer. Example 4: Calculate v o and i o in the op amp circuit in Fig Figure 1.14 For Example 4 [1] Solution: Using Eq. (1.11), 14 P a g e

15 The current i o is the sum of the currents through the 10-kΩ and 2-kΩ resistors. Both resistors have voltage v o = 8 V across them, since v a = v b = 0. Hence, Practice problem 3: Find v o and i o in the op amp circuit shown in Fig Figure 1.15 For Practice problem 3 [1] Answer: 3.8 V, ma. 15 P a g e

16 1.6 Non-Inverting Amplifiers [1] Figure 1.16 The noninverting amplifier. [1] Another important application of the op amp is the noninverting amplifier shown in Fig In this case, the input voltage v i is applied directly at the noninverting input terminal, and resistor R 1 is connected between the ground and the inverting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the inverting terminal gives But v 1 = v 2 = v i. Equation (1.12) becomes (1.12) 16 P a g e

17 Or (1.13) The voltage gain is A v = v o /v i = 1+ R f / R 1, which does not have a negative sign. Thus, the output has the same polarity as the input. Therefore, a noninverting amplifier is an op amp circuit designed to provide a positive voltage gain. Notice that if feedback resistor R f = 0 (short circuit) or R 1 = (open circuit) or both, the gain becomes 1. Under these conditions (R f = 0 and R 1 = ), the circuit in Fig becomes that shown in Fig. 1.17, which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus, for a voltage follower v O = v i (1.14) Figure 1.17 The voltage follower. [1] Such a circuit has a very high input impedance and is therefore useful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as shown in Fig The voltage follower minimizes interaction between the two stages and eliminates interstage loading. 17 P a g e

18 Figure 1.18 A voltage follower used to isolate two cascaded stages of a circuit. [1] Example 5: Calculate v o in the op amp circuit in Fig Solution: Using Eq. (1.13), Figure 1.19 For Example 5 [1] 18 P a g e

19 Practice problem 4: Calculate v o in the circuit of Fig Figure 1.20 For Practice problem 4 [1] Answer: 7 V 1.7 Difference Amplifier [1] Difference (or differential) amplifiers are used in various applications where there is need to amplify the difference between two input signals. Consider the op amp circuit shown in Fig Keep in mind that zero currents enter the op amp terminals. Applying KCL to node a, or (1.15) 19 P a g e

20 Applying KCL to node b, Figure 1.21 Difference amplifier. [1] or (1.16) But v a = v b. Substituting Eq. (1.16) into Eq. (1.15) yields Or (1.17) 20 P a g e

21 Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that v o = 0 when v 1 = v 2. This property exists when Thus, when the op amp circuit is a difference amplifier, Eq. (1.17) becomes (1.18) (1.19) If R 2 = R 1 and R 3 = R 4, the difference amplifier becomes a subtractor, with the output (1.20) Example 6: Design an op amp circuit with inputs v 1 and v 2 such that v o = 5v 1 + 3v 2. Solution: If we desire to use only one op amp, we can use the op amp circuit of Fig Comparing this requirement with Eq. (1.17), we see Also, 21 P a g e

22 or >> If we choose R 1 = 10 kω and R 3 = 20 kω, then R 2 = 50 k Ω and R 4 = 20 k Ω. 1.8 Cascaded Op Amp Circuits [1] It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achieve a large overall gain. In general, two circuits are cascaded when they are connected in tandem, one behind another in a single file. Therefore, a cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next. Figure 1.22 A three-stage cascaded connection. [1] When op amp circuits are cascaded, each circuit in the string is called a stage; the original input signal is increased by the gain of the individual stage. Figure 1.22 displays a block diagram representation of three op amp circuits in cascade. Since the output of one stage is the input to the next stage, the overall gain of the cascade connection is the product of the gains of the individual op amp circuits, or A = A 1 A 2 A 3 (1.21) 22 P a g e

23 Example 7: Find v o and i o in the circuit in Fig Figure 1.23 For Example 7 [1] Solution: This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp, At the output of the second op amp, The current i o is the current through the 10-k resistor. 23 P a g e

24 But, v a = v b = 100 mv, therefore Practice problem 5: Determine v o and i o in the circuit in Fig Answer: 24 V, 2 ma. Figure 1.24 For Practice problem 5 [1] 24 P a g e

25 1.9 Homework 1 1. Find I X, I Y, and V OUT /V IN in this circuit. Figure 1.25 For Prob (a) Show that the output voltage V OUT of the circuit is V OUT R R 3 R 4 R 2V1 R1V2 R R (b) If V 1 = 5 V, V 2 = 10 V and R 1 = R 2 = R 3 = R 4 = 1 k, Find I X Figure 1.26 For Prob P a g e

26 3. Obtain I X and I Y in the Op-amp circuit, using the ideal op-amp model. Figure 1.27 For Prob Calculate i x and v 0 in the circuit. Find the power dissipated by the 60 k. Figure 1.28 For Prob. 4 [1] 26 P a g e

27 5. Use the ideal op-amp model to find I O, I 12, V X and V O. Figure 1.29 For Prob Use the ideal op-amp model, determine I 20, I 5, I O and V O in the circuit. Figure 1.30 For Prob P a g e

28 7. For the circuit in Figure 1.31, use the ideal op-amp model to find I X, V A and V O. Figure 1.31 For Prob Use the ideal op-amp model, determine I X, I Y, I Z and V O in the circuit shown in Figure Figure 1.32 For Prob P a g e

29 9. For the circuit in Figure 1.33, use the ideal op-amp model to find I X, I Y and V O. Figure 1.33 For Prob Use the ideal op-amp model, determine I X, I Y and V O in the circuit. Figure 1.34 For Prob P a g e