# Part I: Operational Amplifiers & Their Applications

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1 Part I: Operational Amplifiers & Their Applications

2 Contents Opamps fundamentals Opamp Circuits Inverting & Non-inverting Amplifiers Summing & Difference Amplifiers Integrators & Differentiators Opamp Applications: ectifiers Comparators & elaxation Oscillators Feedback (sine-wave) Oscillators

3 ecommended textbook: Administrative matters A.S. Sedra and K.C. Smith: Microelectronic Circuits, Oxford University Press, New York, 998.

4 Operational Amplifiers ecommended Text: A.S. Sedra and K.C. Smith: Microelectronic Circuits, Oxford University Press, New York, 998. pp. 6-84

5 Contents Introduction Basics Ideal opamps How to use opamps? Inverting configuration: Closed-loop gain Input and output resistances Integrator Differentiator Weighted summer Noninverting configuration Closed-loop gain Voltage follower

6 Introduction-Basics

7 Introduction-Basics Basic Configuration of Op amps Three terminals (2 inputs and output) Bipolar power supply eference ground is the power supply ground

8 Introduction-Basics

9 Introduction-Basics Characteristics of Opamps Output voltage is proportional to the difference between the two inputs. V = V 3 2 A( V )

10 Introduction-Ideal Op amps The open-loop gain A is very very large, and can be considered as infinite The two input terminals are virtually open. That is, the input impedance of the terminals are very very big (infinite). The output impedance is very very small and can be considered as zero.

11 We would like opamps to Introduction- How to use OpAmps amplify the amplitude of input signal by a factor of any arbitrary value; However, the open-loop gain is fixed and too large, therefore some external circuits should be used to make the system close-loop; There are two configurations in terms of using the external circuits: Non-inverting and Inverting.

12 Inverting configuration

13 Inverting configuration

14 Inverting close-loop configuration The gain A is very large (ideally infinite), If we assume that the circuit is "working" and producing a finite output voltage at terminal 3, then the voltage between the op amp input terminals should be negligibly small. Specifically, if we call the output voltage v o, then, by definition. v 2 v = voltage at the inverting input terminal (v ) is given by v =v 2. That is, because the gain A approaches infinity, the voltage v approaches v 2. We speak of this as the two input terminals "tracking each other in potential." We also speak of a "virtual short circuit" that exists between the two input terminals. Here, the word virtual should be emphasized, and one should not make the mistake of physically shorting terminals and 2 together while analyzing a circuit vo A

15 Inverting close-loop configuration A virtual short circuit means that whatever voltage is at 2 will automatically appear at because of the infinite gain A. But terminal 2 happens to be connected to ground; thus v = 0 and v 2 = 0. We speak of terminal I as being a virtual ground that is, having zero voltage but not physically connected to ground Now that we have determined v we are in a position lo apply Ohm's law and find the current i through as follows: i = v I v vi Where will this current go? It cannot go into the op amp, since the ideal op amp has an infinite input impedance and hence draws zero current. It follows that it will have to flow through 2 to the terminal 3. We can then apply Ohm's law to 2 and determine v o that is, v O = v i = 0 vi 2 thus v v O I = 2

16 Inverting close-loop configuration

17 Inverting close-loop configuration (Effect of finite open-loop gain) If we denote the output voltage v O, then the voltage between two input terminals of the opamp will be v O /A. Since the positive input terminal is grounded, the voltage at the negative input terminal must be - v O /A. the current i I and I can now be found as: i I v = I ( v O / A) = v I + v O / A

18 Inverting close-loop configuration The infinite input impedance of the opamp forces the current i I to flow entirely through 2. The output voltage v O can thus be determined from: v O = v O v O vi + vo / A / A i2 A = Collecting terms, the closed loop gain G is found as: vo 2 / G = v + I ( + / )/ A This is virtual ground assumption we used in our earlier analysis when the opamp was assumed to be ideal. Finally, note that this equation in fact indicates that to minimize the dependence of the closed-loop gain G on the value of open-loop gain A, we should make: / << A

19 Input and output resistances

20 Input and output resistances Assuming an ideal op amp with infinite open-loop gain, the input resistance of the closed-loop inverting amplifier is simply equal. v v O I i = = ii vi /

21 Input and output resistances

22 General impedances Consider first the generalized inverting configuration where impedances Z (s) and Z 2 (s) replace resistors and 2, respectively. The resulting circuit is shown below and has the closed-loop gain or, more appropriately, the closed-loop transfer function VO ( s) Z2( s) = V ( s) Z ( s) I

23 Inverting integrator (Miller integrator) Let the input he a time-varying function v I (t). The virtual ground at the inverting op-amp input causes v I (t) to appear in effect across, and thus the current i (t) will be v I (t)/. This current flows through the capacitor C, causing charge lo accumulate on C. If we assume that the circuit begins operation at time t = 0, then at an arbitrary time t the current i (t) will have deposited on C a charge equal to.

24 Inverting integrator (Miller integrator)

25 Inverting integrator (Miller integrator) Thus the circuit provides an output voltage that is proportional to the timeintegral of the input, with Vc being the initial condition of integration and C is the integrator time-constant. Note that as expected there is a negative sign attached to the output voltage, and thus this integrator circuit is said to be an inverting integrator. It is also known as a Miller integrator after an early worker in this area.

26 Inverting integrator (Miller integrator) For physical frequencies, s = jω and V V O I ( jω) = ( jω) jωc Thus the integrator transfer function has magnitude and phase V V O I = ωc o ϕ = +90

27 The Bode plot for the integrator magnitude response can be obtained by noting that as ω doubles (increases by an octave) the magnitude is halved (decreased by 6 db). Thus the Bode plot is a straight line of slope -6 db/octave (or equivalently, -20 db/ decade). This line intercepts the 0-dB line at the frequency that makes Vo/Vi =, which) is ω int = Inverting integrator (Miller integrator) C The frequency ω int is known as the integrator frequency and is simply the inverse of the integrator rime constant.

28 Inverting integrator (Miller integrator)

29 Inverting integrator (Miller integrator)

30 Op Amp Differentiator The differentiator can be thought of as that of an STC high pass filter with a corner frequency at infinity.

31 Op Amp Differentiator V V O I ( s) ( s) = Z Z 2 ( s) ( s) = / sc = sc V V O I ( jω) ( jω) Z2 ( jω) = jωc = ωc ω = / C when A = Z ( jω) = db 0 There is stability problem. In practice a small resistance in series with the capacitor is connected to guarantee stability.

32 Weighted Summer i i i 2 n v = v = o v = 2 2 n n

33 Weighted Summer i = i + i i n v = i = i + i +... i ) o f ( 2 + n f f f v o = i f = ( v + v f n v n )

34 Non-inverting Configuration

35 Noninverting Configuration v v = = + v i i v = = v i v v i 0 = i2 + v = 2 + vi = ( + 2 ) v i

36 Noninverting Configuration When A is not infinity, we have G = v v / = + 2 / + A i Therefore the condition for an op amp to be IDEAL is: + / 2 << A

37 Voltage Follower Since the noninverting configuration has a gain greater than or equal to unity, depending on choice of 2 /, If 2 / =0, than the closed-loop gain is exactly, and the circuit is called voltage follower

38 A Difference Amplifier Now, we can combine the non-inverting amplifier and inverting amplifier configurations to be able to take a difference between two inputs. You can use superposition or brute force it

39 A Difference Amplifier V o = 2

40 A Difference Amplifier

41 A Difference Amplifier when 2 4 =, 3 then

42 A Difference Amplifier - Input esistance = 4 = 2, 3 v = 2 v = i + i 2i therefore in = 2

43 A Difference Amplifier: Why Useful? Difference amplifiers find application in many areas, most notably an the design instrumental ion systems. As an example, consider the case of a transducer that produces between its two output terminals a relatively small signal, say mv. However, between each of the two wires (leading from the transducer to the instrumentation system) and ground there may be large picked-up interference, say V. The required amplifier, known as instrumentation amplifier, must reject this large- interference signal, which is common the two wires (a common-mode signal), and amplify the small difference (or differential) signal. This situation is illustrated in next Fig. where v cm denotes the common-mode signal and v d denotes the differential signal.

44 A Difference Amplifier

45 An Instrumentation Amplifier Problem with the circuit as an instrumentation amplifier is the low resistance. Therefore the following circuit was proposed:

46 An Instrumentation Amplifier The difference amplifier studied in the previous example is not entirely satisfactory as art instrumentation amplifier its major drawbacks are its low input resistance and that its gain cannot be easily varied. A much superior instrumentation amplifier circuit. Analyze the circuit to determine v o as a function of v and v 2, and determine the differential gain. Suggest a way for making the gain variable. Also find the input resistance. Design the circuit to provide a gain that can be varied over the range 2 to 000 utilizing a 00-kΩ variable resistance (a potentiometer, or "pot" for short)

47 An Instrumentation Amplifier The circuit consist of two stages; The first stage is formed by op amps A and A 2 : and their associated resistors, and the second stage is formed by op amp A 3. Together with its four associated resistors. We recognize the second stage as that of the difference amplifier studied above.

48 An Instrumentation Amplifier

49 An Instrumentation Amplifier

50 An Instrumentation Amplifier From the differential-gain expression in Eq. (2-6) we observe that the gain value can be varied by varying the single resistor any other arrangement involves varying two resistors simultaneously. Since both of the input-stage op amps are connected in the noninverting configuration. The input impedance seen by each of v and v 2 is (ideally) infinite. This is a major advantage of this instrumentation amplifier configuration.

51 Effect of Non-ideal Open-Loop Gain The differential open-loop gain of an op amp decreases with frequency

52 Effect of Non-ideal Open-Loop Gain Note that although the gain is quite high at dc and low frequencies, it starts to fall off at a rather low frequency (0 Hz in our example). The uniform -20-dB/decade gain roll-off shown is typical of internally compensated op amps. These are units that have a network (usually a single capacitor) included on the same IC chip whose function is to cause the op-amp gain to have the single-time-constant low-pass response shown. This process of modifying the open-loop gain is termed frequency compensation, and its purpose is to ensure that opamp circuits will be stable (as opposed to oscillatory).

53 Effect of Non-ideal Open-Loop Gain

54 3-dB frequency ( break frequency) : Effect of Non-ideal Open-Loop Gain A( ω b ) = A 2 Unit-Gain frequency (or unit-gain bandwidth) the frequency at which the gain is : for ω >> ω b A( ω) A0 ωb ω 0 Hence Unit-Gain frequency is ωt A0ω b The open-loop gain of an op amp can be estimated using ωb ωt A( ω) A0 = ω ω The unit-gain bandwidth is usually given by data sheet

55 Frequency esponse of Closed-Loop Amplifiers b s A A ω + = 0 ) / /( / ) ( ) ( s A s V s V t i o = ω The closed-loop gain for inverting amplifier, assuming a finite opamp open-loop gain A, was already derived as: + + = 2 2 / ) ( ) ( A s V s V i o Substituting A gives:

56 For A >> + which is usually the case 0 2 / Vo ( s) V ( s) i 2 / = s + ω /( + t 2 / ) which is of the same form as that for a low-pass single-lime-constant network), ω t Thus the inverting amplifier has an STC low-pass response with a dc gain of magnitude equal to 2/. The closed-loop gain rolls off at a uniform 20 db/decade slope with a comer frequency (3-dB frequency) given by ω 3dB = ωt + / 2

57 Frequency esponse of Closed-Loop Amplifiers

58 Large-Signal Operation of Op-Amps input step waveform Slew-rate limited Exponential rising output

59

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