Applied Electronics II

Transcription

1 Applied Electronics II Chapter 3: Operational Amplifier Part 1- Op Amp Basics School of Electrical and Computer Engineering Addis Ababa Institute of Technology Addis Ababa University Daniel D./Getachew T./Abel G. April 2017 April / 46

2 Overview I 1 Introduction 2 The Ideal Op Amp The Op Amp Terminals Function and Characteristics of the Ideal Op Amp 3 The Inverting Configuration Closed-Loop Gain Effect of Finite Open-Loop Gain Input and Output Resistances An Important Application - The Weighted Summer 4 The Noninverting Configuration The Closed-Loop Gain Effect of Finite Open-Loop Gain Application - The Voltage Follower 5 Difference Amplifiers A Single-Op-Amp Difference Amplifier April / 46

3 Overview II The Instrumentation Amplifier 6 Integrators and Differentiators The Inverting Integrator The Op-Amp Differentiator 7 DC Imperfections Offset Voltage Input Bias and Offset Currents 8 Frequency Response Frequency Dependence of the Open-Loop Gain Frequency Response of Closed-Loop Amplifiers 9 Large-Signal Operation of Op Amps April / 46

4 Introduction Introduction The operational amplifier (Op amps) have been in use for a long time, their initial applications being primarily in the areas of analog computation and sophisticated instrumentation. Early op amps were constructed from discrete components (vacuum tubes and then transistors, and resistors). The introduction of integrated circuit (IC) reduced the cost and improved the performance. One of the reasons for the popularity of the op amp is its versatility. IC op amp has characteristics that closely approach the assumed ideal. April / 46

5 The Ideal Op Amp The Op Amp Terminals The Op Amp Terminals From a signal point of view the op amp has three terminals: two input terminals (1 and 2) and one output terminal (3). Most IC op amps require two dc power supplies, as shown in two terminals, 4 and 5, are brought out of the op-amp package and connected to a positive voltage V CC and a negative voltage V EE, respectively. Some times other terminals can include terminals for frequency compensation and terminals for offset nulling. April / 46

6 The Ideal Op Amp Function and Characteristics of the Ideal Op Amp Function and Characteristics of the Ideal Op Amp Op amp is designed to sense the difference between the voltage signals applied at its two input terminals and multiply this by a number A. v 3 = A(v 2 v 1 ) Characteristics of the Ideal Op Amp Infinite input impedance Zero output impedance Zero common-mode gain or, equivalently, infinite common-mode rejection Infinite open-loop gain A Infinite bandwidth Question: But, is an amplifier with infinite gain of any use? April / 46

7 The Ideal Op Amp Function and Characteristics of the Ideal Op Amp Function and Characteristics of the Ideal Op Amp An amplifiers input is composed of two components differential input (v Id ) - is difference between inputs at inverting and non-inverting terminals v Id = v 2 v 1 common-mode input (v Icm ) - is input present at both terminals v Icm = 1 2 (v 2 + v 1 ) The input signals v 1 and v 2 v 1 = v Icm v Id /2 and v 2 = v Icm + v Id /2 Similarly, two components of gain exist differential gain(a) - gain applied to differential input ONLY common-mode gain(a cm ) - gain applied to common-mode input ONLY April / 46

8 The Ideal Op Amp Function and Characteristics of the Ideal Op Amp Function and Characteristics of the Ideal Op Amp Figure: Equivalent circuit of the ideal op amp. Figure: Input signals in terms of differential and common-mode components. April / 46

9 The Inverting Configuration The Inverting Configuration Op amps are not used alone; rather, the op amp is connected to passive components in a feedback circuit. There are two such basic circuit configurations employing an op amp and two resistors: the inverting configuration and the noninverting configuration Figure: The inverting closed-loop configuration. April / 46

10 The Inverting Configuration Closed-Loop Gain Closed-Loop Gain Assuming an ideal op amp. How to analyze closed-loop gain for inverting configuration of an ideal op-amp? Step 1 Begin at the output terminal Step 2 If v o is finite, then the voltage between the op-amp input terminals should be negligibly small and ideally zero. v 2 v 1 = v o A = 0, because A is A virtual short circuit between v 1 and v 2. A virtual ground exist at v 1. Step 3 Define current in to inverting input (i 1 ). i 1 = v I v 1 = v I 0 = v I R 1 R 1 R 1 Step 4 Determine where this current flows? April / 46

11 The Inverting Configuration Closed-Loop Gain Closed-Loop Gain It cannot go into the op amp, since infinite input impedance draws zero current. i 1 will have to flow through R 2 to the low-impedance terminal 3. Step 5 Define v o in terms of current flowing across R 2. v o = v 1 i 1 R 2 = 0 v I R 1 R 2 = R 2 R 1 v I G = R 2 R 1 April / 46

12 The Inverting Configuration Effect of Finite Open-Loop Gain Effect of Finite Open-Loop Gain How does the gain expression change if open loop gain (A) is not assumed to be infinite? One must employ analysis similar to the previous. The voltage at v 1 becomes The current i 1 becomes v 2 v 1 = v o A v 1 = v o A i 1 = v I v 1 R 1 The output voltage v o becomes = v I + vo A R 1 v o = v 1 i 1 R 2 = v o A v I + vo A R 1 R 2 ; v o ( R 2 R 1 A ) = v 1 R 2 R 1 April / 46

13 The Inverting Configuration Effect of Finite Open-Loop Gain Effect of Finite Open-Loop Gain The Gain will be G A< = v o v i = R 2 /R ( 1 ) 1 + R2 /R A Figure: Analysis of the inverting configuration taking into account the finite open-loop gain of the op amp. April / 46

14 The Inverting Configuration Input and Output Resistances Input and Output Resistances The Input Resistance is R i = v i i i = v i (v i v 1 )/R 1 = v i v i /R 1 = R 1 For a Voltage amplification R i must be large. Then the gain would be reduced, so such configuration suffers from low R i. Consider the following circuit and find the expression of the closed loop gain April / 46

15 The Inverting Configuration Input and Output Resistances Input and Output Resistances The closed loop gain v o = R ( R 4 + R ) 4 v i R 1 R 2 R 3 It can be seen a higher R i can be achieved without compromising the closed loop gain. Since the output of the inverting configuration is taken at the terminals of the ideal voltage source A(v 2 v 1 ), it follows that the output resistance of the closed-loop amplifier is zero. R o = 0 April / 46

16 The Inverting Configuration An Important Application - The Weighted Summer An Important Application - The Weighted Summer Weighted Summer - is a closed-loop amplifier configuration which provides an output voltage which is weighted sum of the inputs. April / 46

17 The Noninverting Configuration The Noninverting Configuration The input signal v I is applied directly to the positive input terminal of the op amp while one terminal of R 1 is connected to ground. Then the polarity / phase of the output is same as input. Figure: The noninverting configuration. April / 46

18 The Noninverting Configuration The Closed-Loop Gain The Closed-Loop Gain For an ideal case the closed-loop gain by using the previous methods. v o v i = 1 + R 2 R 1 Figure: The noninverting configuration. April / 46

19 The Noninverting Configuration Effect of Finite Open-Loop Gain Effect of Finite Open-Loop Gain Assuming the op amp to be ideal except for having a finite open-loop gain A. The closed-loop gain G A< = v o v i = 1 + R 2 /R ( 1 ) 1 + R2 /R A For A 1 + R 2 R 1 the closed-loop gain can be approximated by the ideal value. The percentage error in G resulting from the finite op-amp gain A as Percentage gain error = 1 + R 2/R 1 A (R 2 /R 1 ) The input impedance R i of this closed-loop amplifier is ideally infinite, since no current flows into the positive input terminal of the op amp. The output is taken at the terminals of the ideal voltage source thus the output impedance R o of the noninverting configuration is zero. April / 46

20 The Noninverting Configuration Application - The Voltage Follower The Voltage Follower The property of high input impedance is a very desirable feature of the noninverting configuration. It enables using this circuit as a buffer amplifier to connect a source with a high impedance to a low-impedance load. Buffer amplifier is not required to provide any voltage gain This circuit is commonly referred to as a voltage follower, since the output follows the input. Figure: a) The unity-gain buffer or follower amplifier. (b) Its equivalent circuit model. April / 46

21 Difference Amplifiers Difference Amplifiers A difference amplifier is one that responds to the difference between the two signals applied at its input and ideally rejects signals that are common to the two inputs. Ideally, the amp will amplify only the differential signal (v Id ) and reject completely the common-mode input signal (v Icm ). However, a practical circuit will behave as below v o = A d v Id + A cm v Icm The efficacy of a differential amplifier is measured by the degree of its rejection of common-mode signals in preference to differential signals. CMRR = 20 log A d A cm Question: The op amp is itself a difference amplifier; why not just use an op amp? April / 46

22 Difference Amplifiers Difference Amplifiers A difference amplifier is one that responds to the difference between the two signals applied at its input and ideally rejects signals that are common to the two inputs. Ideally, the amp will amplify only the differential signal (v Id ) and reject completely the common-mode input signal (v Icm ). However, a practical circuit will behave as below v o = A d v Id + A cm v Icm The efficacy of a differential amplifier is measured by the degree of its rejection of common-mode signals in preference to differential signals. CMRR = 20 log A d A cm Question: The op amp is itself a difference amplifier; why not just use an op amp? very high (ideally infinite) gain of the op amp April / 46

23 Difference Amplifiers A Single-Op-Amp Difference Amplifier A Single-Op-Amp Difference Amplifier Analyzing the difference amplifier below using superposition. v o1 = R 2 R 1 v I1 v o2 = R 4 R 3 + R 4 ( 1 + R ) 2 v I2 R 1 We have to make the two gain magnitudes equal in order to reject common-mode signals. April / 46

24 Difference Amplifiers A Single-Op-Amp Difference Amplifier A Single-Op-Amp Difference Amplifier R 2 R 1 = R 2 /R R 2 /R 1 = The condition is obtained when R 4 R 3 + R 4 ( 1 + R ) 2 R 1 R 4 R 3 + R 4 = R 4/R R 4 /R 3 R 2 R 1 = R 4 R 3 Assuming the condition is satisfied, the output voltage v o = R 2 R 1 (v I2 v I1 ) In addition to rejecting common-mode signals, a difference amplifier is usually required to have a high input resistance. Assuming R 4 = R 2 and R 3 = R 1 and applying a differential input. April / 46

25 Difference Amplifiers A Single-Op-Amp Difference Amplifier A Single-Op-Amp Difference Amplifier v Id = R 1 i I R 1 i I Thus R Id = v Id = 2R 1 i I Note that if the amplifier is required to have a large differential gain (R 2 /R 1 ), then R 1 of necessity will be relatively small and the input resistance will be correspondingly low, a drawback of this circuit. Another drawback of the circuit is that it is not easy to vary the differential gain of the amplifier. April / 46

26 Difference Amplifiers The Instrumentation Amplifier The Instrumentation Amplifier The low-input-resistance problem can be solved by using voltage followers to buffer the two input terminals. But why not get some voltage gain. Solution: using a Noninverting Op Amp. April / 46

27 Difference Amplifiers The Instrumentation Amplifier The Instrumentation Amplifier The output v o v o = R ( R ) 2 (v I2 v I1 ) R 3 R 1 The Advantages are very high input resistance high differential gain symmetric gain (assuming that A 1 and A 2 are matched) The Disadvantage A d and A cm are equal in first stage - meaning that the common-mode and differential inputs are amplified with equal gain need for matching - if two op amps which comprise stage 1 are not perfectly matched, one will see unintended effects The Solution is to disconnect the two resistors (R 1 ) connected to node X from ground and connecting them together. April / 46

28 Difference Amplifiers The Instrumentation Amplifier The Instrumentation Amplifier For a differential input applied the gain would remain the same. For a common mode input voltage v Icm an equal voltage appears at the negative input terminals of A 1 and A 2, causing the current through 2R 1 to be zero. Thus v o1 and v o2 will be equal to the input. Thus the first stage no longer April / 46

29 Integrators and Differentiators The Inverting Integrator The Inverting Integrator By placing a capacitor in the feedback path and a resistor at the input, we obtain the circuit of below. We shall now show that this circuit realizes the mathematical operation of integration. Let the input be a time-varying function v I (t). The transient description v O (t) = 1 t CR v I (t)dt v O (t 0 ) 0 April / 46

30 Integrators and Differentiators The Inverting Integrator The Inverting Integrator The steady-state description V o (s) V i (s) = 1 scr Thus the integrator transfer function has magnitude of 1/ωCR and phase φ = +90 This configuration also known as a Miller integrator has a disadvantage. At ω = 0, the magnitude of the integrator transfer function is infinite. This indicates that at dc the op amp is operating with an open loop. Solution: By placing a very large resistor in parallel with the capacitor, negative feedback is employed to make dc gain finite. April / 46

31 Integrators and Differentiators The Inverting Integrator The Inverting Integrator The integrator transfer function becomes V o (s) V i (s) = R F /R 1 + scr F The lower the value we select for R F, the higher the corner frequency will be and the more nonideal the integrator becomes. Thus selecting a value for R F presents the designer with a trade-off between dc performance and signal performance. April / 46

32 Integrators and Differentiators The Op-Amp Differentiator The Op-Amp Differentiator Interchanging the location of the capacitor and the resistor of the integrator circuit results in the circuit which performs the mathematical function of differentiation. The transient description v O (t) = CR dv I(t) dt April / 46

33 Integrators and Differentiators The Op-Amp Differentiator The Op-Amp Differentiator The steady-state description V o (s) V i (s) = scr Thus the integrator transfer function has magnitude of ωcr and phase φ = 90 This configuration as a differentiator has a disadvantage. Differentiator acts as noise amplifier, exhibiting large changes in output from small (but fast) changes in input. As such, it is rarely used in practice. When the circuit is used, it is usually necessary to connect a small-valued resistor in series with the capacitor. This modification, unfortunately, turns the circuit into a nonideal differentiator. April / 46

34 DC Imperfections Offset Voltage Offset Voltage Now we consider some of the important nonideal properties of the op amp. What happens If the two input terminals of the op amp are tied together and connected to ground. Ideally since v id = 0, we expect v O = 0 In practice a finite dc voltage exists at the output. April / 46

35 DC Imperfections Offset Voltage Offset Voltage The causes of V OS is unavoidable mismatches in the differential stage of the op amp. It is impossible to perfectly match all transistors. General-purpose op amps exhibit V OS in the range of 1 mv to 5 mv. Also, the value of V OS depends on temperature. Analysis to determine the effect of the op-amp V OS on their performance is the same for both inverting and the noninverting amplifier configurations. [ V O = V OS 1 + R ] 2 R 1 April / 46

36 DC Imperfections Offset Voltage Offset Voltage How to reduced Offset Voltage offset nulling terminals A variable resistor (if properly set) may be used to reduce the asymmetry present and, in turn, reduce offset. capacitive coupling A series capacitor placed between the source and op amp may be used to reduce offset, although it will also filter out dc signals. April / 46

37 DC Imperfections Input Bias and Offset Currents Input Bias and Offset Currents In order for the op amp to operate, its two input terminals have to be supplied with dc currents, termed the input bias currents, I B. I B = I B1 + I B2 2 I OS = I B1 I B2 input offset currents, I OS - the difference between bias current at both terminals. The resulting output voltage V O = I B1 R 2 I B R 2 April / 46

38 DC Imperfections Input Bias and Offset Currents Input Bias and Offset Currents To reduce the value of the output dc voltage due to the input bias currents, logically it is R 2 but higher R 2 needed for gain. The solution is introducing a resistance R 3 in series with the noninverting input. The output voltage when calculated Assuming I B1 = I B2 = I B V O = I B2 R 3 + R 2 (I B1 I B2 R 3 /R 1 ) V O = I B [R 2 R 3 (1 + R 2 /R 1 )] April / 46

39 DC Imperfections Input Bias and Offset Currents Input Bias and Offset Currents Thus we can reduce V O to zero by selecting R 3 such that R 2 R 3 = = R 1R R 2 /R 1 R 1 + R 2 We conclude that to minimize the effect of the input bias currents, one should place in the positive lead a resistance equal to the equivalent dc resistance seen by the inverting terminal. This is the case for op amps constructed using bipolar junction transistors (BJTs). Those using MOSFETs in the first (input) stage do not draw an appreciable input bias current; nevertheless, the input terminals should have continuous dc paths to ground. April / 46

40 Frequency Response Frequency Dependence of the Open-Loop Gain Frequency Dependence of the Open-Loop Gain The differential open-loop gain A of an op amp is not infinite; rather, it is finite and decreases with frequency. It is high at dc, but falls off at a rather low frequency. Internal compensation - is the presence of internal passive components (caps) which cause op-amp to demonstrate STC low-pass response. Frequency compensation - is the process of modifying the open-loop gain to increase stability. Figure: Open-loop gain of a general-purpose internally compensated op amp. April / 46

41 Frequency Response Frequency Dependence of the Open-Loop Gain Frequency Dependence of the Open-Loop Gain The gain of an internally compensated op-amp may be expressed as shown below The transfer function in Laplace domain: A(s) = A s/ω b The transfer function in Frequency domain: A(jω) = A jω/ω b The transfer function for high frequnecy: A(jω) A 0ω b jω Magnitude gain for high frequnecy: Unity gain occurs at ω t A(jω) A 0ω b ω ω t = A 0 ω b = ω t ω April / 46

42 Frequency Response Frequency Response of Closed-Loop Amplifiers Frequency Response of Closed-Loop Amplifiers The effect of limited op-amp gain and bandwidth on the closed-loop transfer functions of the inverting configurations. Step 1 Define closed-loop gain of an inverting amplifier with finite open-loop gain (A) V o V i = R 2 /R (1 + R 2 /R 1 )/A Step 2 Insert frequency-dependent description of A V o V i = R 2 /R R 2/R 1 ) ( A0 1 + s/ω b = R 2 /R ( ) 1 ( ) R2 /R 1 A 0 + s 1+R2 /R 1 ω b A 0 Step 3 Assume A R 2 /R 1 V o V i = R 2 /R s(1+r 2/R 1 ) ω t April / 46

43 Frequency Response Frequency Response of Closed-Loop Amplifiers Frequency Response of Closed-Loop Amplifiers By using the same methods the effect of limited op-amp gain and bandwidth on the closed-loop transfer functions of the noninverting configurations. V o V i = 1 + R 2/R s(1+r 2/R 1 ) ω t 3dB frequency - is the frequency at which the amplifier gain is attenuated 3dB from maximum (aka. dc value). ω 3dB = ω t 1 + R 2 /R 1 April / 46

44 Large-Signal Operation of Op Amps Large-Signal Operation of Op Amps The following are limitations on the performance of op-amp circuits when large output signals are present. 1 Output Voltage Saturation Op amps operate linearly over a limited range of output voltages. If supply voltage +/- 15V is v O will saturate around +/- 13V. 2 Output Current Limits Another limitation on the operation of op amps is that their output current is limited to a specified maximum. If the circuit requires a larger current, the op-amp output voltage will saturate at the level corresponding to the maximum allowed output current. 3 Slew Rate Slew Rate is the maximum rate of change possible at the output of a real op amp. SR = dv o dt max April / 46

45 Large-Signal Operation of Op Amps Large-Signal Operation of Op Amps If slew rate is less than rate of change of input it becomes problematic. Slewing occurs because the bandwidth of the op-amp is limited, so the output at very high frequencies is attenuated. April / 46

46 Large-Signal Operation of Op Amps Large-Signal Operation of Op Amps 4 Full-Power Bandwidth Op-amp slew-rate limiting can cause nonlinear distortion in sinusoidal waveforms. Assume a unity-gain follower with a sine-wave input v I = V i sin ωt The rate of change dv I dt = ωv i cos ωt Now if ωv i exceeds the slew rate of the op amp, the output waveform will be distorted in the manner shown. April / 46

47 Large-Signal Operation of Op Amps Large-Signal Operation of Op Amps Full-power bandwidth (f M ) is the frequency at which an output sinusoid with amplitude equal to the rated output voltage of the op amp begins to show distortion due to slew-rate limiting. SR = ω M V omax f M = SR 2πV omax Maximum output voltage (V omax ) - is equal to (Av I ). Output sinusoids of amplitudes smaller than V omax will show slewrate distortion at frequencies higher than f M At a frequency ω higher than f M, the maximum amplitude of the undistorted output sinusoid is ( ωm ) V o = V omax ω April / 46