The Operational Amplifier (Op-Amp)

Transcription

1 The Operational Amplifier (Op-Amp)

2 Chapter Goals Develop understanding of linear amplification concepts such as: Voltage gain, current gain, and power gain Gain conversion to decibel representation Input and output resistances Biasing for linear amplification Distortion in amplifiers Two-port representations of amplifiers Understand behavior and characteristics of ideal differential and op amps. Demonstrate circuit analysis techniques for ideal op amps. Characterize inverting, non-inverting amplifiers.

3 Amplification Introduction A complex periodic signal can be represented as the sum of many individual sine waves. We consider only one component with amplitude V s = 1 mv and frequency ω s with 0 phase (signal is used as reference): Amplifier output is sinusoidal with same frequency but different amplitude V o and phase θ:

4 Amplification Introduction (cont.) Amplifier output power is: Here, we desire P O = 100 W with R L = 8 Ω and V s = 1 mv Output power also requires output current which is: Input current is given by Output phase is zero because circuit is purely resistive.

5 Amplification Voltage Gain & Current Gain Voltage Gain: Magnitude and phase of voltage gain are given by and For our example, Current Gain: Magnitude of current gain is given by

6 Amplification Power Gain Power Gain: For our example,

7 Amplification Expressing Gain in Decibels (db) The logarithmic decibel or db scale compresses the huge numeric range of gains encountered in real systems.

8 Amplification Expressing Gain in db - Example For our example:

9 Mismatched Source and Load Resistances In introductory circuit theory, the maximum power transfer theorem is usually discussed. Maximum power transfer occurs when the source and load resistances are matched (equal in value). In most amplifier applications, however, the opposite situation is desired. A completely mismatched condition is used at both the input and output ports of the amplifier.

10 Mismatched Source and Load Resistances For the voltage amplifier shown If R in >> R s and R out << R L, then In an ideal voltage amplifier, and R out = 0

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12 Distortion in Amplifiers In this graph, different gains for positive and negative values of the input cause distortion in the output. Total Harmonic Distortion (THD) is a measure of signal distortion that compares undesired harmonic content of a signal to the desired component.

13 Total Harmonic Distortion dc desired output 2nd harmonic distortion 3rd harmonic distortion Numerator = rms amplitude of distortion terms Denominator = desired component

14 Differential Amplifier Basic Model An ideal differential amplifier produces an output that depends on the voltage difference between its two input terminals. Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 180 o out of phase with that applied at - input (inverting) terminal. v o = A v id A = open-circuit voltage gain v id = (v + - v - ) = differential input signal voltage R id = amplifier input resistance R o = amplifier output resistance

15 Differential Amplifier Model Impact of Source and Load R L = load resistance R S = Thevenin equivalent resistance of signal source v s = Thevenin equivalent voltage of signal source Op amp circuits are mostly dc-coupled amplifiers. Signals v o and v s may have a dc component representing a dc shift of the input away from the Q-point. Op-amp amplifies both dc and ac components.

16 Differential Amplifier Model Example including Source and Load Resistances Problem: Calculate voltage gain for an amplifier Given Data: A = 100, R id = 100kΩ, R o = 100Ω, R S = 10kΩ, R L = 1000Ω

17 Differential Amplifier Model Example including Source and Load Resistances Analysis: An ideal amplifier s output depends only on the input voltage difference and not on the source and load resistances. This can be achieved by using a fully mismatched resistance condition (R id >> R S or infinite R id, and R o << R L or zero R o ). Then: A = open-loop gain (maximum voltage gain available from the device)

18 Operational Amplifers Op-amp is an electronic device that amplify the difference of voltage at its two inputs. Most op-amps operate from plus and minus supply voltages, which may or may not be shown on the schematic symbol. Very high gain dc coupled amplifiers with differential inputs. One of the inputs is called the inverting input ( ); the other is called the non-inverting input. Usually there is a single output. +V 8 1 DIP 8 1 DIP 20 SMT SMT + V

19 The Ideal Op-Amp Ideally, op-amps have characteristics (used in circuit analysis): Infinite voltage gain Infinite input impedance (does not load the driving sources) Zero output impedance (drive any load) Infinite bandwidth (flat magnitude response, zero phase shift) Zero input offset voltage. The ideal op-amp has characteristics that simplify analysis of op-amp circuits. The concept of infinite input impedance is particularly a valuable analysis tool for several op-amp configurations. V in + Z in = A v V in A v = Z out = 0 V out

20 The Practical Op-Amp Real op-amps differ from the ideal model in various respects. In addition to finite gain, bandwidth, and input impedance, they have other limitations. Finite open loop gain. Finite input impedance. Non-zero output impedance. Input current. Input offset voltage. Temperature effects. V in + Z in A v V in Z out V out

21 Internal Block Diagram of an Op-Amp Internally, the typical op-amp has a differential input, a voltage amplifier, and a push-pull output. Recall from the discussion in Section 6-7 of the text that the differential amplifier amplifies the difference in the two inputs. V in + Differential amplifier input stage Voltage amplifier(s) gain stage Push-pull amplifier output stage V out

22 Input Signal modes As the input stage of an op-amp is a differential amplifier, there are two input modes possible: differential mode and common mode. In differential mode any one of the two scenarios can occur. Either one input is applied to one input while the other input is grounded (single-ended). Or opposite polarity signals are applied to the inputs (double-ended).

23 Differential Mode Operation V in + V out Single-ended differential amplifier V in + V out Double-ended differential amplifier

24 Common Mode Operation In common mode, two signals voltages of the same amplitude, frequency and phase are applied to the two inputs. The same input tend to cancel each other and the output is zero. This is called common-mode rejection. V in V in + V out This is useful to reject unwanted signal that appears to both inputs. It is cancelled and does not appear at the output

25 Common-Mode Rejection Ratio The ability of an amplifier to amplify differential signals and reject common-mode signals is called the common-mode rejection ratio (CMRR). CMRR is defined as Op-Amp parametrs CMRR = A A A cm is zero in ideal op-amp and much less than 1 is practical op-amps. A ol ranges up to 200,000 (106dB) ol cm where A ol is the open-loop differential-gain and A cm is the common-mode gain. CMRR = 100,000 means that desired signal is amplified 100,000 times more than un wanted noise signal. A ol CMRR can also be expressed in decibels as CMRR = 20log Acm

26 Example Common-Mode Rejection Ratio What is CMRR in decibels for a typical 741C op-amp? The typical open-loop differential gain for the 741C is 200,000 and the typical common-mode gain is 6.3. A ol CMRR = 20 log Acm 200, 000 = 20log = db (The minimum specified CMRR is 70 db.)

27 Maximum Output Voltage Swing V O(p-p) : The maximum output voltage swing is determined by the op-amp and the power supply voltages With no input signal, the output of an op-amp is ideally 0 V. This is called the quiescent output voltage. When an input signal is applied, the ideal limits of the peakto-peak output signal are ± V CC. In practice, however, this ideal can be approached but never reached (varies with load resistance).

28 Input Offset Voltage Ideal op-amp produces zero output voltage if the differential input is zero But practical op-amp produces a non-zero output voltage when there is no differential input applied. This output voltage is termed V OUT(error). It is due to unavoidable mismatches in the differential stage of the op amp. The amount of differential input voltage required between the inputs to force the output to zero volts is the input offset voltage V OS. Typical value of V OS is about 2mV.

29 Input Bias Current (I BIAS ) The input bias current is the average of the two dc currents required to bias the differential amplifier Ideally, input bias current is zero. I1+ I2 IBIAS = 2

30 Input Impedance The input impedance of an op-amp is specified in two ways: Differential input impedance and common-mode input impedance. Differential input impedance, Z IN(d), is the total resistance between inverting and noninverting input. Common-mode input impedance, Z IN(c), is the resistance between each input and ground. + + Z IN(d) Z IN(cm)

31 Input offset Current I OS Ideally, the two input bias currents are equal, and thus their difference is zero The input offset current is the difference of the input bias currents II oooo = II 1 II 2 The offset voltage developed by the input offset current is VV OOOO = II OOOO RR iiii The output error volt is VV OOUUUU(eeeeeeeeee) = AA vv II OOOO RR iiii

32 Output Impedance Z out : The output impedance is the resistance viewed from the output of the circuit. Z out +

33 Slew Rate The slew rate is the maximum rate of change of the output voltage in response to a step input voltage (V/µs) Slew rate is measured with an opamp connected as shown and is given as The slew rate is dependent upon the high-frequency response of the amplifier stages within the op-amp.

34 Example Determine the slew rate for the output response to a step input shown. Since this response is not ideal, the limits are taken at the 90% points.

35 Negative Feedback Negative feedback is the process of returning a portion of the output signal of an amplifier to the input with a phase angle that is opposite to the input signal. The advantage of negative feedback is that precise values of amplifier gain can be set. In addition, bandwidth and input and output impedances can be controlled. V in V f + Internal inversion makes V f 180 out of phase with V in. Negative feedback circuit V out

36 Importance of Negative Feedback The inherent open-loop voltage gain of a typical op-amp is very high. Therefore, an extremely small input voltage (even the input offset voltage) drives the op-amp into saturation. The usefulness of an opamp operated without negative feedback is generally limited to comparator applications With negative feedback the gain of op-amp (called closeloop gain A cl ) can be reduced and controlled so that an op-amp can function as a linear amplifier.

37 Op-Amps with Negative Feedback An op-amp can be connected using negative feedback to stabilize the gain and increase frequency response. This close-loop gain (A cl ) is usually much less than the openloop gain (A ol ). The close-loop voltage gain is the voltage of an op-amp with external feedback. The amplifier circuit consists of an op-amp and an external negative feedback circuit. The feedback from the output is connected to the inverting input of the op-amp. The negative feedback is determined and controlled by external components.

38 Noninverting Amplifier A noninverting amplifier is a configuration in which the signal is on the noninverting input and a portion of the output is returned to the inverting input. The feedback circuit is formed by input resistance R i and feedback resistance R f. This feedback creates a voltage divider circuit which reduces V out and connects the reduced voltage V f to the inverting input and can be expressed as: V in V f + R f R i V out Feedback circuit

39 Noninverting Amplifier The differential input is amplified by the openloop gain and produces the output voltage as: The attenuation, B, of the feedback circuit is

40 Noninverting Amplifier Substituting BV out for V f, we get The overall voltage gain of the amplifier can be expressed as The product A ol B is typically much greater than 1, so the equation simplifies to Which means A = + cl (NI) 1 R R f i

41 Noninverting Amplifier Determine the gain of the noninverting amplifier shown. A cl (NI) 1 R f = + Ri 82 kω = k Ω = 25.8 V in + R f 82 kω R i 3.3 kω V out

42 Voltage Follower A special case of the inverting amplifier is when R f =0 and R i =. This forms a voltage follower or unity gain buffer with a gain of 1. This configuration offers very high input impedance and its very low output impedance. These features make it a nearly ideal buffer amplifier for interfacing high-impedance sources and low-impedance loads V in + R f 82 kω R i 3.3 kω It produces an excellent circuit for isolating one circuit stage from another, which avoids "loading" effects. V out

43 Inverting Amplifier We have: and Since I in = I f, then The overall gain is

44 Example Inverting Amplifier Determine the gain of the inverting amplifier shown. R f A cl (I) R f = Ri 82 kω = 3.3 kω V in R i 3.3 kω + 82 kω V out = 24.8 The minus sign indicates inversion.

45 Closed-Loop Gain Analysis of the inverting configuration. The circled numbers indicate the order of the analysis steps. question: how will we step #4: define v Out in terms of current flowing across R 2 step #5: substitute v in / R 1 for 1 i. closed-loop gain G = -R 2 /R 1 Oxford University Publishing Electronic Microelectronic Devices, Circuits 9th edition by Adel S. Sedra and Kenneth C. Smith ( )

46 Effect of Finite Open-Loop Gain Q: How does the gain expression change if open loop gain (A) is not assumed to be infinite? A: One must employ analysis similar to the previous, result is presented below

47 Effect of Finite Open-Loop Gain Collecting terms, the closed-loop gain G is found as G A< non-ideal gain G A = vout R2/ R1 R2 = = v 1 ( R2 / R1) In + R A if A= then the previous gain expression is yielded ideal gain

48 Effect of Finite Open-Loop Gain Q: Under what condition can G = -R 2 / R 1 be employed over the more complex expression? A: If 1 + (R 2 /R 1 ) << A, then simpler expression may be used. R2 R2 R2/ R1 if 1 + << A then GA= = else GA< = R 1 ( R 1 R1 + 2/ R1) 1 + A ideal gain non-ideal gain

49 2012 Oxford Pearson University Education. Publishing Upper Saddle River, NJ, All rights Microelectronic reserved. Circuits by Adel S. Sedra and

50 Impedances of the Noninverting Amplifier Input Impedance From the Figure we get Substituting I in Z in = V d, where Z in is open loop input impedance Z ( ) in(ni) = 1+ ABZ ol in The input impedance of the noninverting amplifier with negative feedback is much greater than the internal open-loop input impedance of the op-amp.

51 Impedances of the Noninverting Amplifier Output Impedance Referring to the shown Figure, after some mathematical manipulations (Floyd 620) it can be shown that the output impedance of a noninverting (NI) amplifier can be given as Z out(ni) = ( 1+ AB) The output impedance of the noninverting amplifier with negative feedback is much less than the internal open-loop input impedance of the op-amp. Z out ol

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54 Impedances of the Voltage-Follower Since voltage-follower is a special case on noninverting amplifier, the same formulas are used with B = 1, therfore The input impedance of voltage-follower is very high, it can be seen that it is extremely high as compared to the noninverting amplifier as the feedback attenuation B = 1. The same is true for the output impedance where the factor of is removed so it becomes even less than the output impedance of the noninverting amplifier.

55 Input Impedance Impedances of the Inverting Amplifier The input impedance of the inverting (I) amplifier is This is because the input is in series with R i and that is connected to virtual ground so that is the only resistance seen by input.

56 Impedances of the Inverting Amplifier Output Impedance As with the noninverting amplifier, the output impedance of an inverting (I) amplifier is decreased by the negative feedback. In fact the the expression is the same as for the noninverting case. The output impedance of the both noninverting and inverting amplifier is low. In fact, practically it can be considered zero..

57 Impedances Noninverting amplifier: Z Z Z ( ) = + ABZ Zout = 1+ AB in(ni) 1 ol in out(ni) in(i) out(i) R ( ) Inverting amplifier: Z = i Z out ol ( 1+ AB) ol Generally, assumed to be Generally, assumed to be 0 Generally, assumed to be R i Generally, assumed to be 0 Note that the output impedance has the same form for both amplifiers.

58 Example

59 Solution

60 Example

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62 Determine the closed-loop gain of each amplifier in Figure. (a) 11 (b) 101 (c) 47.8 (d) 23

63 If a signal voltage of 10 mv rms is applied to each amplifier in Figure, what are the output voltages and what is there phase relationship with inputs?. (a) V out V in = 10 mv, in phase (b) V out = A cl V in = 10 mv, 180º out of phase (c) V out = 233 mv, in phase (d) V out = 100 mv, 180º out of phase

64 Effect of Input Bias Current - Inverting Amplifier An inverting amplifier with zero input voltage is shown. Ideally, the current through R i is zero because the input voltage is zero and the voltage at the inverting terminal is zero. The small input bias current, I 1, is through R f from the output terminal. This produces an output error voltage I 1 R f when it should be zero

65 Effect of Input Bias Current Voltage Follower A voltage-follower with zero input voltage and a source resistance, R s is shown. In this case, an input bias current, I 1, produces a drop across R s and creates an output voltage error -I 1 R s as shown.

66 Effect of Input Bias Current - Noninverting Amplifier Consider a noninverting amplifier with zero input voltage. The input bias current, I 1, produces a voltage drop across R f and thus creates an output error voltage of I 1 R f

67 Bias Current Compensation Voltage Follower The output error voltage due to bias currents in a voltagefollower can be sufficiently reduced by adding a resistor, R f, equal to the source resistance, R s, in the feedback path. The voltage drop created by I 1 across the added resistor subtracts from the output error voltage (if I 1 = I 2 ).

68 Bias Current Compensation Inverting and Noninverting To compensate for the effect of bias current in Inverting and Noninverting Amplifiers, a resistor R c is added at the noninverting terminal. The compensating resistor value equals the parallel combination of R i and R f. The input current creates a voltage drop across R c that offsets the voltage across the combination of R i and R f, thus sufficiently reducing the output error voltage.

69 Bias Current Compensation Inverting and Noninverting R f R f R i V out V out R i + V in + R c = R i R f R c = R i R f V in Noninverting amplifier Inverting amplifier

70 Effect of Input Offset Voltage The output of an op-amp should be zero when the differential input is zero. Practically this is not the case. There is always an output error voltage present whose range is typically in microvolts to millivolts. This is due to the imbalances within the internal op-amp transistors This output error voltage is aside from the one produced by the input bias

71 Input Offset Voltage Compensation Most ICs provide a mean of compensating for offset voltage. An external potentiometer to the offset null pins of IC package

72 Bandwidth Limitations Frequency response of amplifiers is shown in a plot called Bode Plot. In Bode plot, the frequency is on the horizontal axis and is in logarithmic scale. It means that the frequency change is not linear but ten-times. This ten-time change in frequency is called a decade. The vertical axis shows the voltage gain in decibel (db). The maximum gain on the plot is called the midrange gain. The point in the frequency response of amplifiers where the gain is 3dB less than the midrange gain is called the critical frequency.

73 Bandwidth Limitations A ol (db) Midrange db/decade roll-off Critical frequency Unity-gain frequency (f T ) k 10k 100k 1M f (Hz)

74 Bandwidth Limitations An open-loop response curve (Bode plot) for a certain op-amp is shown. The differential open-loop gain A ol of an op amp is not infinite; rather, it is finite and decreases with frequency. Note that although the gain is quite high at dc and low frequencies, it starts to fall off at a rather low frequency. The 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). These are units that have a network (usually a single capacitor) included within the same IC chip whose function is to cause the op-amp gain to have the single-time-constant low-pass response shown.

75 Gain-Versus-Frequency Analysis The RC lag (low-pass) circuits within an op-amp are responsible for the roll-off in gain as the frequency increases, just as for the discrete amplifiers. The attenuation of an RC lag circuit shown is expressed as: The critical frequency of an RC circuit is:

76 Gain-Versus-Frequency Analysis If an op-amp is represented by a voltage gain element with a gain of A ol(mid) plus a single RC lag circuit, as shown, it is known as a compensated op-amp. The total open-loop gain of the op-amp is the product of the midrange open-loop gain, A ol(mid), and the attenuation of the RC circuit as:

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78 Phase Shift An RC circuit causes a propagation delay from input to output, thus creating a phase shift between the input signal and the output signal. An RC lag circuit such as found in an op-amp stage causes the output signal voltage to lag the input. The phase shift, θ, is given by:

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80 Overall Frequency Response Most op-amps have a constant roll-off of -20 db/decade above its critical frequency. The more complex IC operational amplifier may consist of two or more cascaded amplifier stages. The gain of each stage is frequency dependent and rolls off at above its critical frequency. Therefore, the total response of an op-amp is a composite of the individual responses of the internal stages. db gains are added and phase lags of the stages are added as shown in next slide for a three stage op-amp.

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83 closed-loop frequency response Op-amps are usually used in a closed-loop configuration with negative feedback in order to achieve precise control of gain and bandwidth. Midrange gain of an op-amp is reduced by negative feedback as we have already seen in previous sections. Now we will see its effects on bandwidth.

84 Effect of Negative Feedback on Bandwidth The closed-loop critical frequency of an op-amp is given by: ff cc(cccc) = ff cc ooll (1 + BBAA oooo mmmmmm ) Where B is the feedback attenuation of the closed-loop op-amp. The above expression shows that the closed-loop critical frequency, ff cc(cccc), is higher than the open-loop critical frequency ff cc oooo by a factor of (1 + BBAA oooo mmmmmm ) Since ff cc(cccc) equals bandwidth therefore the closed-loop bandwidth, BBBB cccc, is also increased: BBBB cccc = BBBB ooll (1 + BBAA oooo mmmmmm )

85 Summary Bandwidth Limitations The Figure shows the concept of closed-loop response. When the openloop gain is reduced due to negative feedback, the bandwidth is increased. A v Open-loop gain This means that you can achieve a higher BW by accepting less gain. A ol(mid) A cl(mid) Closed-loop gain 0 f c(ol) f c(cl ) f

86 Gain-Bandwidth Product An increase in the closed-loop gain causes a decrease in the bandwidth and vice versa, such that the product of gain and bandwidth is constant. If AA cccc is the gain of an op-amp with ff cc(cccc) bandwidth then: AA cccc ff cc(cccc) = AA ooll ff cc(ooll) The equation, A cl f c(cl) = A ol f c(ol) shows that the product of the gain and bandwidth are constant. The gain-bandwidth product is always equal to the frequency at which the op-amp s open-loop gain is unity or 0 db (unity gain bandwidth, ff TT ) ff TT = AA cccc ff cc(cccc)

87 Example Determine the bandwidth of each of the amplifiers shown. Both op-amps have an open-loop gain of 100 db and a unitygain bandwidth (f T ) of 3 MHz.

88 Solution

89 Selected Key Terms Operational A type of amplifier that has very high voltage amplifier gain, very high input impedance, very low output impedance and good rejection of common-mode signals. Differential A mode of op-amp operation in which two mode opposite-polarity signals voltages are applied to the two inputs (double-ended) or in which a signal is applied to one input and ground to the other input (single-ended). Common mode A condition characterized by the presence of the same signal on both inputs

90 Selected Key Terms Open-loop The voltage gain of an op-amp without external voltage gain feedback. Negative The process of returning a portion of the output feedback signal to the input of an amplifier such that it is out of phase with the input. Closed-loop The voltage gain of an op-amp with external voltage gain feedback. Gainbandwidth the frequency at which the op-amp s open-loop A constant parameter which is always equal to product gain is unity (1).

91 Quiz 1. The ideal op-amp has a. zero input impedance and zero output impedance b. zero input impedance and infinite output impedance c. infinite input impedance and zero output impedance d. infinite input impedance and infinite output impedance

92 Quiz 2. The type of signal represented in the figure is a a. single-ended common-mode signal b. single-ended differential signal c. double-ended common-mode signal d. double-ended differential signal V in + V out

93 Quiz 3. CMRR can be expressed in a. amps b. volts c. ohms d. none of the above

94 Quiz 4. The difference in the two dc currents required to bias the differential amplifier in an op-amp is called the a. differential bias current b. input offset current c. input bias current d. none of the above

95 Quiz 5. To measure the slew rate of an op-amp, the input signal is a a. pulse b. triangle wave c. sine wave d. none of the above

96 Quiz 6. The input impedance of a noninverting amplifier is a. nearly 0 ohms b. approximately equal to R i c. approximately equal to R f d. extremely large

97 Quiz 7. The noninverting amplifier has a gain of 11. Assume that V in = 1.0 V. The approximate value of V f is a. 0 V b. 100 mv c. 1.0 V V in V f + R f 10 kω V out d. 11 V R i 1.0 kω

98 Quiz 8. The inverting amplifier has a gain of 10. Assume that V in = 1.0 V. The approximate value of the voltage at the inverting terminal of the op-amp is a. 0 V R f b. 100 mv c. 1.0 V d. 10 V V in R i 1.0 kω + 10 kω V out

99 Quiz 9. To compensate for bias current, the value of R c should be equal to a. R i R f b. R f c. R i R f R i V out d. R i + R f V in + R c

100 Quiz 10. Given a noninverting amplifier with a gain of 10 and a gain-bandwidth product of 1.0 MHz, the expected high critical frequency is a. 100 Hz b. 1.0 khz c. 10 khz d. 100 khz

101 Quiz Answers: 1. c 6. d 2. d 7. c 3. d 8. a 4. b 9. c 5. a 10. d