1. Operational Amplifiers 1

2. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith2 Introduction (The first IC op amp was introduced in mid-1960s.)  We can do almost anything with op amp, such as, summation, subtraction, amplification, differentiation, integration, …. Operational Amplifier  Op amp has almost ideal characteristics. Why op amp is so popular ?  Op amp works almost same as it is expected. You should be able to design nontrivial circuit by the end of this chapter ! What’s inside of op amp ?Chap. 9, complex but useful for high level engineer. 2.1 The Ideal Op Amp 2.1.1 The Op-Amp Terminals No terminal of op-amp package is physically connected to ground.

3. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith3 TABLE 2.1 Characteristic of the ideal Op Amp 1. Infinite input impedance 2. Zero output impedance 3. Zero common-mode gain or, equivalently, infinite common-mode rejection 4. Infinite open-loop gain A 5. Infinite bandwidth 2.1.2 Function and Characteristics of the Ideal Op Amp A: open-loop gain, differential gain Op amp is a differential-input single ended-output amplifier. Op amp is a directly-coupled or dc amplifier. - useful, but can cause some serious practical problems (How are we going to use it?) ( Feedback, closed loop) 2.1.3 Differential and Common-mode Signals

4. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith4 Figure 2.5 The inverting closed-loop configuration. (Output is 180 o phase-shifted w.r.t the input.) 2.2 The Inverting Configuration Figure 2.12 The noninvertin g configuration. p.77 Sect.2.3 (Output is in phase with the input.) Negative feedback 2.2.1 The closed-Loop Gain Ideal op amp: open loop gain A = ∞ The terminal 1and 2 are virtually shorted. The terminal 1 is a virtual ground ! The virtual ground is not an actual ground. Do not short the inverting input to ground to simplify analysis !

5. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith5 ① Procedure to find output voltage υ o. υ o will be 180 o phase-shifted with respect to the input wave. Inverting amplifier ! We can make the closed-loop gain as accurate as we want by selecting passive components of appropriate accuracy. The closed-loop gain is (ideally) independent to the op-amp gain.

6. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith6 2.2.2 The Effect of Finite Open-Loop Gain (A≠∞) ① EXAMPLE 2.1 p.72 10 3 10 4 10 5 υ1υ1

7. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith7 The production spread in the value of open-loop gain A between op- amp units of the same type (part number) is very wide.

8. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith8 2.2.3 Input and output Resistance of Inverting Amplifier (not op-amp) ① Ideal input resistance of inverting amplifier, not op-amp. Therefore, input resistance R 1 should be much larger than output resistance of previous stage. Unfortunately, the internal resistance of most sensors are large. For high gain (- R 2 / R 1 ), we need small R 1, otherwise, R 2 would be impractically large. R What would happen if R 1 is too small ? Contradiction ! Solution: example 2.2

9. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith9 EXAMPLE 2.2 p73 (a) Find(b) Design the inverting amplifier with gain of 100 and an input resistance 1 MΩ. (a) (b) input resistance 1 MΩR 1 =1 MΩ maximum resistance in practical circuits: 1 MΩ R 2, R 4 =1 MΩ gain of -100R 3 =10.2 k Ω (If we adopt a typical inverting amplifier and R 1 =1 MΩ, R 2 =100 MΩ, impractically large !)

10. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith10 Figure 2.9 A current amplifier based on the circuit of Fig. 2.8. The amplifier delivers its output current to R 4. It has a current gain of (1 + R 2 /R 3 ), a zero input resistance, and an infinite output resistance. The load (R 4 ), however, must be floating (i.e., neither of its two terminals can be connected to ground). Transresistance (transimpedance) amplifier and current amplifier Photodiode (photodetector) generates electron- hole pairs proportional to the incident light power. Therefore, photodiode is a current source (I). υ o = - IR 2 means current input, voltage output ! Transresistance amplifier ! Input resistance is R 2 /A, very small. (Try this!) Since the photodiode has capacitance inside, if a large resistor for high output voltage is used instead of op-amp, time constant would be large, which means it can not be used in high frequency circuit. Transresistance amplifier is currently used as a part of receiver in high speed fiber-optic communication systems.

11. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith11 Figure 2.10 A weighted summer. 2.2.3 An Important Application-The Weighted Summer. Mom, virtual grounds are extremely handy ! Figure 2.11 A weighted summer capable of implementing summing coefficients of both signs.

12. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith12 Figure 2.12 The noninverting configuration. 2.3 The Non-inverting Configuration The inverting configuration. 2.3.1 The close-Loop Gain  Another derivation υ o is divided to R 1 and R 2.  the Role of Negative feedback 1. Let υ I increase. 2. υ Id is increased. 3. υ o is increased. 4. Fraction of υ o will be fed back to (-) terminal. 5. This feedback will be counteract the increase in υ Id, driving back to zero. – degenerative feedback Let’s derive the expression of the voltage gain.

13. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith13 2.3.2 Characteristics of the noninverting Configuration - Input and output signals have same phase.-noninverting. - Input impedance is infinite. - output impedance is zero. 2.3.3 Effect of Finite Open-Loop Gain - Follow the procedure used for inverting amplifier. - non-inverting amplifier - inverting amplifier - denominators are identical ! How come? identical negative feedback (think about 0 V input !) - numerators are different ! How come? different amount of feedback (closed loop gain) In order to minimize the effect of the finite open-loop gain,

14. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith14 2.3.4 The Voltage Follower - Input impedance is infinite. - very desirable feature. - Buffer Amplifier ! to connect a source with a high impedance to a low-impedance load Usually, gain = 1 (Sect. 1.5) equivalent circuit model. The unity-gain buffer or voltage follower. Why do we need this? - 100% feedback ! - Used as impedance transformer or power amplifier. - Elegant in simplicity !

15. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith15 2.4 Difference (differential) Amplifiers (very important application) differential gaindifferential inputCommon mode gain≈0 Common mode input The efficacy of a differential amp is measured by the degree of its rejection of common-mode signals in preference to differential input signals. Transducer (sensor) Noise (Electromagnetic Interference, EMI) from motor, spark, lightning… We want to amplify the sensor output only. (we want to reject common-mode noise signal.) 2.4.1 A single Op-Amp Difference Amplifier Gain of the inverting amplifier = -R 2 /R 1 Gain of the non-inverting amplifier = 1+R 2 /R 1 Combine these two amp.Independently adjustable or matched gain

16. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith16 Figure 2.16 A difference amplifier. Superposition ! To make gain R 2 /R 1, For easier matching, R 3 =R 1, R 4 =R 2

17. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith17 Differential input Resistance R id Drawback 1: for large gain, very small R 1 is required. Drawback 2: to vary the gain, two resistances should be changed. 2.4.2 A Superior Circuit-The Instrumentation Amplifier The Goal:High differential gain with high input resistance and easy gain control HOW?Voltage follower with gain + Difference amplifier Disadvantages 1: in the first stage, common-mode gain = differential gain may causes saturation of op-amp. 2: in the first stage, two amplifiers have to be perfectly matched. 3: to vary the gain, two resistances (two R 1 ) should be changed. There is a simple solution !

18. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith18 Instrumentation Amplifier 1 1 2 3 4 5 6 7 8 9

19. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith19 EXAMPLE 2.3 Design The instrumentation amplifier circuit to provide a gain that can be varied over the range of 2 to 1000 utilizing a 100 kΩ variable resistance ( a potentiometer or “pot” for short). (Sol.) - It is usually preferable to obtain all the required gain in the first stage. (Low Noise) - The second stage (difference amp) is usually designed for a gain of 1. - We select all the second-stage resistors to be 10 kΩ, practically convenient value. R 1f =100.2 Ω, R 2 = 50.050 kΩ, R 1f =100 Ω, 1% R 2 = 49.9 kΩ, 1%

20. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith20 2.5 Effect of Finite Open-Loop Gain and Bandwidth on Circuit Performance A circuit designer has to be thoroughly familiar with the characteristics of practical op amps and the effects of such characteristics on the performance of op-amp circuits. 2.5.1 Frequency Dependence of the Open-Loop Gain Figure 2.22 Open-loop gain of a typical general- purpose internally compensated op amp. - Internally compensated op amps are unit that have a network (usually a single capacitor) within the IC chip. - Capacitor’s function is to cause the op-amp gain to have the STC low-pass response. - This process of modifying the open-loop gain is termed to frequency compensation. - The purpose of frequency compensation is to ensure that the op-amp will be stable (as oppose to oscillation: Ch. 8). - f t = ω t /2π is specified on the data sheet as unity-gain band width.

21. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith21 -If f t is known, one can easily determine the magnitude of the op-amp gain at a given frequency when f >> f b. The production spread in the value of f t between op- amp units of the same type (part number) is much smaller than that of A 0 and f b. f t is preferred as a specification parameter. - An op amp having this -6 dB/octave (= - 20 dB/decade) gain roll off is said to have a single-pole model. - Since this single pole dominates the amplifier frequency response, it is called a dominant pole. (for more on poles and zeros, refer to Appendix E.) 2.5.2 Frequency Dependence of the Closed-Loop Gain The effect of limited op-amp gain and bandwidth on the closed-loop transfer functions.

22. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith22 Low-Pass STC Network ! EXAMPLE 2.4 f t = 1 MHz, find 3-dB frequency of closed-loop amp with gain of 1000, 100, 10, 1, -1, -10, -100, -1000 Gain-bandwidth product = 1000 V/V-kHz = 60dB-kHz Non-inverting amplifier

23. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith23 2.6 Large-signal Operation of Op Amps 2.6.1 Output Voltage Saturation Rated output voltage (output saturation voltage) = ± L (L=V DD – α V ) 2.6.2 Output Current Limits For example, maximum output current of 741 is ± 20 mA. EXAMPLE 2.5 Output saturation voltage = ± 13 V, output current limits = ± 20 mA (a) Output voltage for V p =1 V, R L = 1 kΩ υ o / υ I = (1+R 2 /R 1 ): V op = 10 V, I op = 10 mA (b) Output voltage for V p =1.5 V, R L = 1 kΩ V op = 15 V, I op = 14.3 mA (c) for R L = 1 kΩ, V pmax = ? for undistorted output. 13/10 = 1.3 V ( 14.3 mA) (d) for V p =1.5 V, R Lmin = ? for undistorted output.

24. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith24 2.6.3 Slew Rate - Another nonlinear distortion due to the large output signal Slew Rate : maximum rate of change possible at the output of a real op amp Why this happens ? : Chap. 9 SR is distinct from the finite op-amp bandwidth. - The limited bandwidth : linear, no distortion, but reduced gain at higher frequency. - The limited Slew-rate : nonlinear distortion, even at low frequency when output is too large. When V is sufficiently small, the output can be the exponentially rising ramp. This is a low-pass STC response. Output from capacitor of RC network !

25. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith25 Figure 2.27 Effect of slew-rate limiting on output sinusoidal waveforms. 2.6.4 Full-Power Bandwidth 2.7 DC Imperfection 2.7.1 Offset Voltage Op amps are direct-coupled devices.They are prone to dc problems. With inputs being zero, the amplifier output rests at some dc voltage level instead of zero. The equivalent dc input offset voltage is 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.

26. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith26 Figure 2.28 Circuit model for an op amp with input offset voltage V OS. To include effect of offset voltage, If υ id =0, Thus, CMRR is a measure of how total offset voltage v OS changes from its dc value V OS when common-mode voltage is applied. (another interpretation of CMMR) R1 =1.2 kΩ, R2 = 1 MΩ, υ O = ? When an input signal is applied to the amplifier, the corresponding signal output will be superimposed on the 2.5 V dc. Then the allowable signal swing at the output will be reduced. If signal is dc, we would not know where the output is due to VOS or the signal. * To overcome dc offset problem * The gain will fall off at the low- frequency.

27. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith27 2.7.2 Input Bias and Offset Current Bias currents ( base currents in BJTs or gate currents in MOSFETs or JFETs) are similar in value with directions depending on internal amplifier circuit type. In order for the op amp to operate, its two input terminals have to be supplied with current, termed the input bias current. Why? * To find the dc output voltage of the closed-loop amp due to the input bias current How to reduce the dc output voltage due to the input bias current

28. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith28 Conclusion: To minimize the effect of the input bias current, Place in the positive lead a resistance equal to the dc resistance seen by the inverting terminal. There should be a continuous dc path between each input terminal and ground. This circuit will not work.This circuit will work.

29. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith29 2.8 Integrators and Differentiators 2.8.1 The Inverting Configuration with General Impedances Example 2.6, p106 * Is this STC circuit? * dc gain=?, 3 dB frequency=? * Design a circuit of dc gain=40 dB, 3 dB frequency=1 kHz, input resistance = 1 kΩ * Frequency of gain=1 and phase at this frequency Sol) 40 dB = 100 V/V

30. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith30 (b) Frequency response of the integrator. 2.8.2 The inverting Integrator Figure 2.39 (a) The Miller or inverting integrator. CR : integrator time-constant. * An integrator behaves as a low-pass filter with a corner frequency of zero. * At f=0, the magnitude of the integrator transfer function is infinite and the op amp is operating with an open loop. * Any tiny dc component input signal will theoretically produce an infinite output ! Serious problem !!! * In practice, the output saturates at power supply voltage.

31. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith31 * Deleterious effect from the presence of the op-amp input dc offset voltage Op amp will saturate very soon !!! The dc offset current produces a similar problem !! * Deleterious effect from the presence of the op-amp input dc offset current ( R is added in (+) to reduce the effect of bias current.) Solution Place in the positive lead a resistance equal to the dc resistance seen by the inverting terminal. Let’s provide a dc current (V OS and I OS ) path ! Low R F is better for dc of High R F is better for ideal integrator. Selecting a value for R F present the designer with a trade-off between dc performance and signal performance. (example 2.7, p110)

32. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith32 EXAMPLE 2.7 Find the output by the Miller integrator due to Sol) Constant current (1 V/10 kΩ= 0.1 mA) to capacitor. Important Applications: square-wave input to triangular output Exercise 2.27, (Function Generator !!), : Active Filter !! (Ch. 12)

33. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith33 2.8.3 The Op-Amp Differentiator * Differentiator is an STC highpass filter with a corner frequency at infinity. * Differentiator is a noise magnifier. Input Output * Spikes in output could cause EMI problems. * Due to the EMI and stability (Ch.8) problem, differentiator is seldom used in practice. * A small resistor in series with the capacitor might solve this problem, but the circuit becomes non-ideal differentiator.

34. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith34 Figure 2.45 A linear macromodel used to model the finite gain and bandwidth of an internally compensated op amp. 2.9 The SPICE Op-Amp Model and Simulation Examples The differential gain : A0d of the voltage-controlled voltage source Ed Frequency response : Rb-Cb low-pass filter with corner frequency Eb with gain of 1 is used to isolate the low-pass filter from any load at output. Figure 2.46 A comprehensive linear macromodel of an internally compensated op amp.

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39. Copyright  2004 by Oxford University Press, Inc. Microelectronic Circuits - Fifth Edition Sedra/Smith39 Small Out-line PackageDual In-line Package