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FEEDBACK AMPLIFIERS AND OSCILLATORS for Electronic Circuits by Prof. Michael Tse September 2004 Contents Feedback Basic feedback conﬁguration Advantages The price to pay Feedback Ampliﬁer Conﬁgurations Series-shunt, shunt-series, series-series, shunt-shunt Input and output impedances Practical Circuits with loading effects Compensation Op-amp internal compensation Oscillation Oscillation criteria Sustained oscillation Wein bridge, phase shift, Colpitts, Hartley, etc. C.K. Tse: Feedback ampliﬁers and 2 oscillators Basic feedback conﬁguration The basic feedback ampliﬁer consists of a basic ampliﬁer and a feedback network. + e si A so input – output basic ampliﬁer Careful!! sf f feedback network A = basic ampliﬁer gain f = feedback gain C.K. Tse: Feedback ampliﬁers and 3 oscillators Characteristics The input is subtracted by a + feedback signal which is part of the si e A so output, before it is ampliﬁed by the input – output basic ampliﬁer. basic ampliﬁer sf so = Ae = A(si - s f ) f But, since sf = f so, we get feedback network so = A(si - fso ) Hence, the overall gain is so A si Ao so Ao = = si 1+ Af 1 If Af >> 1, Ao ª f C.K. Tse: Feedback ampliﬁers and 4 oscillators Simple viewpoint + e si A so input – output basic ampliﬁer sf f feedback network If A is large, then e must be very small in order to give a ﬁnite output. So, the input si must be very close to the feedback signal sf . That means sf ≈ si . But, sf is simply a scaled-down copy of the output so. so 1 Hence, f so = si or ª si f C.K. Tse: Feedback ampliﬁers and 5 oscillators Obvious advantage If the feedback network is constructed from passive elements having stable characteristics, the overall gain becomes very steady and unaffected by variation of the basic ampliﬁer gain. Quantitatively, we wish to know how much the overall gain Ao changes if there is a small change in A. Let assume A becomes A + dA. From the formula of Ao, we have dAo Ê dA ˆÊ 1 ˆ = Á ˜Á ˜ Ao Ë A ¯Ë1+ Af ¯ Obviously, if Af is large, then dAo/Ao will be reduced drastically. 1 Feedback reduces gain sensitivity! In fact, the gain is just Ao ª . f C.K. Tse: Feedback ampliﬁers and 6 oscillators so Another advantage Suppose the basic ampliﬁer is distortive. So, the output does output A not give a sine wave for a sine e wave input. But, with feedback, we see that the gain is about 1/f anyway, input regardless of what A is (or as long as Af is large enough). so This gives a very good property 1/f of feedback ampliﬁer in terms si of eliminating distortion. C.K. Tse: Feedback ampliﬁers and 7 oscillators Other advantages • Improve input and output resistances (to be discussed later). • Widening of bandwidth of ampliﬁer (to be discussed later). • Enhance noise rejection capability. ni ni si A so + + si e A’ A so input + – output basic ampliﬁer Signal-to-noise ratio: sf È so ˘ È si ˘ f Í ˙=Í ˙ feedback network Î no ˚ Î ni ˚ È so ˘ È si ˘ Signal-to-noise ratio improves! Í ˙ = A¢Í ˙ Î no ˚ Î ni ˚ C.K. Tse: Feedback ampliﬁers and 8 oscillators The price to pay Of course, nothing is free! Feedback comes with reduced gain, and hence you may need to add a pre- ampliﬁer to boost the gain. Also, wherever you have a loop, there is hazard of oscillation, if you don’t want it. Later, we will also see how we can use feedback to create oscillation deliberately. C.K. Tse: Feedback ampliﬁers and 9 oscillators Terminologies Basic ampliﬁer gain = A Feedback gain = f A 1 Overall gain (closed-loop gain) = ª 1+ Af f Loop gain (roundtrip gain) = Af Some books use T to denote Af. C.K. Tse: Feedback ampliﬁers and 10 oscillators Feedback ampliﬁers What is an ampliﬁer? si A so Signals can be voltage or current. General model for voltage ampliﬁer: Ro + + + vin Rin – Avin vo – – voltage ampliﬁer C.K. Tse: Feedback ampliﬁers and 11 oscillators Models of ampliﬁers Ro iin io + + Aiin + vin Rin – Avin vo Rin Ro – – voltage ampliﬁer current ampliﬁer iin Ro io + + Aiin + vin Rin Ro Rin – Aiin vo – – transresistance ampliﬁer transconductance ampliﬁer C.K. Tse: Feedback ampliﬁers and 12 oscillators Feedback ampliﬁer conﬁgurations Voltage ampliﬁer + e si A so input – output basic ampliﬁer sf f voltage voltage feedback network To subtract voltage from To copy voltage, we should use voltage, we should use series parallel (shunt) connection connection + • + vi – A vo – • – vf + Hence, series-shunt feedback C.K. Tse: Feedback ampliﬁers and 13 oscillators Series-shunt feedback (for voltage ampliﬁer) Ro + + + + vo vi ve Ri – Ave – – – + – fvo vo A Overall gain (closed-loop gain) : Ao = = v i 1+ Af C.K. Tse: Feedback ampliﬁers and 14 oscillators Series-shunt feedback (for voltage ampliﬁer) To ﬁnd the input resistance, we consider the ratio of vi and ii, with output opened. ii Ro + + + + vo vi ve Ri – Ave – – – vi vi RIN RIN = = ii v e /Ri v e + fv o + = Ri – fvo ve = Ri (1+ Af ) The input resistance has been enlarged by (1+Af). This is a desirable feature for voltage ampliﬁer as a large input resistance minimizes loading effect to the previous stage. C.K. Tse: Feedback ampliﬁers and 15 oscillators Series-shunt feedback (for voltage ampliﬁer) To ﬁnd the output resistance, we consider shorting the input source and calculate the ratio of vo and io. Ro io First, we have ve = – fvo. + + Also, + ve Ri + vo v o - Av e v o + Afv o vi – Ave io = = – – – Ro Ro Hence, ROUT vo Ro ROUT = = + fvo io 1+ Af – The output resistance has been reduced by (1+Af). This is a desirable feature for voltage ampliﬁer as a small output resistance emulates a better voltage source for the load. C.K. Tse: Feedback ampliﬁers and 16 oscillators Series-shunt feedback (for voltage ampliﬁer) Summary of features Equivalent model A 1 Ro Closed-loop gain = ª 1+ Af 1+ Af f + + Av i + Input resistance = Ri ( 1 + Af ) vi Ri ( 1 + Af ) – 1+ Af vo – – Ro Output resistance = 1+ Af NOTE: We did not consider loading effect of the feedback network, i.e., we assume that the feedback network is an ideal ampliﬁer which feeds a scaled-down copy of the output to the input. + – ∞ feedback network C.K. Tse: Feedback ampliﬁers and 17 oscillators Feedback ampliﬁer conﬁgurations Transresistance ampliﬁer + e si A so input – output basic ampliﬁer sf f current voltage feedback network To subtract current from To copy voltage, we should use current, we should use shunt parallel (shunt) connection (connection) connection ii • + A vo – • Hence, shunt-shunt feedback C.K. Tse: Feedback ampliﬁers and 18 oscillators Shunt-shunt feedback (for transresistance ampliﬁer) ii ie Ro + + vo Ri – Aie – fvo vo A Overall gain (closed-loop gain) : Ao = = ii 1+ Af C.K. Tse: Feedback ampliﬁers and 19 oscillators Shunt-shunt feedback (for transresistance ampliﬁer) To ﬁnd the input resistance, we consider the ratio of vi and ii, with output opened. ii ie Ro + + + vo vi Ri – Aie – – v i Riie RIN = = RIN ii ii ie = Ri ie + fv o fvo Ri = 1+ Af The input resistance has been reduced by (1+Af). This is a desirable feature for transresistance ampliﬁer as a small input resistance ensures better current sensing from the previous stage. C.K. Tse: Feedback ampliﬁers and 20 oscillators Shunt-shunt feedback (for transresistance ampliﬁer) To ﬁnd the output resistance, we consider opening the input source (putting ii = 0) and calculate the ratio of vo and io. First, we have ie = – fvo. ii = 0 ie Ro io + Also, + v o - Aie v o + Afv o Ri – Aie vo io = = – Ro Ro Hence, ROUT vo Ro ROUT = = io 1+ Af fvo The output resistance has been reduced by (1+Af). This is a desirable feature for transresistance ampliﬁer as a large small resistance emulates a better voltage source for the load. C.K. Tse: Feedback ampliﬁers and 21 oscillators Shunt-shunt feedback (for transresistance ampliﬁer) Summary of features Equivalent model A 1 Ro Closed-loop gain = ª 1+ Af 1+ Af f ii Aii + Ri Ri + Input resistance = – 1+ Af vo 1+ Af 1+ Af – Ro Output resistance = 1+ Af Similar, we can develop the feedback conﬁgurations for transconductance ampliﬁer and current ampliﬁer. Transconductance ampliﬁer: series-series feedback Current ampliﬁer: shunt-series feedback C.K. Tse: Feedback ampliﬁers and 22 oscillators Series-series feedback (for transconductance ampliﬁer) io + + Ro vo ve Ri – Ave – io + – f io io A Overall gain (closed-loop gain) : Ao = = v i 1+ Af Input resistance: RIN = Ri (1+ Af ) Desirable! Output resistance: ROUT = Ro (1+ Af ) Desirable! C.K. Tse: Feedback ampliﬁers and 23 oscillators Shunt-series feedback (for current ampliﬁer) ii ie io Ro Ri Aie io f io Overall gain (closed-loop gain) : io A Ao = = ii 1+ Af Input resistance: Ri RIN = 1+ Af Desirable! Output resistance: Ro ROUT = Desirable! 1+ Af C.K. Tse: Feedback ampliﬁers and 24 oscillators Practical feedback circuits (with loading effects) In practice, the input source has resistance and the feedback network has resistance. Example: shunt-shunt feedback ie Ro + ii + vo Ri – Aie – fvo What are the effects on the gain, input and output resistances? C.K. Tse: Feedback ampliﬁers and 25 oscillators Systematic analysis using 2-port networks The best way to analyze feedback circuits with loading effects is to use two-port models. For shunt-shunt feedback, input and output sides are both parallel connected. Thus, the loading can be combined by summing the conductances. Also, voltage is common at both sides. So, y-parameter is best. The ﬁrst step is to put everything in y-parameter: + + ii yi vi y11 y22 vo – y21vi – 1 2 y22f y11f y21fvo C.K. Tse: Feedback ampliﬁers and 26 oscillators Systematic analysis of shunt-shunt feedback using y-parameter + + ii yi vi y11 y22 vo – y21vi – 1 2 y22f y11f y12fvo In order to use the standard results, we have to convert this model to the standard form (slide 19). C.K. Tse: Feedback ampliﬁers and 27 oscillators Systematic analysis of shunt-shunt feedback using y-parameter y11f y22f + + ii yi vi y11 y22 vo – y21vi – 1 2 y12fvo One step closer… C.K. Tse: Feedback ampliﬁers and 28 oscillators Systematic analysis of shunt-shunt feedback using y-parameter y11f y22f + + ii vi yi y11 y22 vo – y21vi – 1 2 y12fvo One more step closer… C.K. Tse: Feedback ampliﬁers and 29 oscillators Systematic analysis of shunt-shunt feedback using y-parameter y11+y11f+yi y22f +y22 + + ii vi vo – y21vi – 1 2 y12fvo Yet another step closer… C.K. Tse: Feedback ampliﬁers and 30 oscillators Systematic analysis of shunt-shunt feedback using y-parameter in conductance (S) in resistance (Ω) y11+y11f+yi 1/(y22f +y22) + + ii + vo vi – – -y 21v i – y 22f + y 22 1 2 Use Thevenin y12fvo Yet another step closer… C.K. Tse: Feedback ampliﬁers and 31 oscillators Systematic analysis of shunt-shunt feedback using y-parameter in conductance (S) in resistance (Ω) ie y11+y11f+yi 1/(y22f +y22) + + ii + vo vi – – – 1 2 -y 21ie ( y 22f + y 22 )( y11 + y11f + y i ) y12fvo Finally, we get the same standard form. C.K. Tse: Feedback ampliﬁers and 32 oscillators Systematic analysis of shunt-shunt feedback using y-parameter We can simply apply the standard results: -y 21 -y 21 Basic ampliﬁer gain A= = ( y 22f + y 22 )( y11 + y11f + y i ) y oT y iT Feedback gain f = y21f A 1 1 Overall (closed-loop) gain Ao = ª = 1+ Af f y12f 1 Input resistance RIN = (y11 + y11f + y i )(1+ Af ) 1 Output resistance ROUT = (y 22f + y 22 )(1+ Af ) C.K. Tse: Feedback ampliﬁers and 33 oscillators Appropriate 2-port networks for analyzing feedback circuits For shunt-shunt feedback, use y-parameter. For shunt-series feedback, use g-parameter. For series-series feedback, use z-parameter. WHY? For series-shunt feedback, use h-parameter. The procedure is essentially the same as in the previous shunt-shunt case. C.K. Tse: Feedback ampliﬁers and 34 oscillators General procedure of analysis 1. Identify the type of feedback. 2. Use appropriate 2-port representation. 3. Lump all loading effects in the basic ampliﬁer, giving a modiﬁed basic ampliﬁer. 4. Apply Thevenin or Norton to cast the model back to the standard form (without loading). 5. Apply standard formulae to ﬁnd A, f, RIN and ROUT. C.K. Tse: Feedback ampliﬁers and 35 oscillators Example Rf – a is + + vo RL – Type of feedback: shunt-shunt Appropriate 2-port type: y-parameter So, the ﬁrst step is to represent the circuit in y-parameter networks. C.K. Tse: Feedback ampliﬁers and 36 oscillators Example – a is + + vo RL – Rf C.K. Tse: Feedback ampliﬁers and 37 oscillators Example Converting to y-parameter – Ro + + is vi Ri – vo RL + avi – Note: this goes to the Rf –ve input of A. y11f y22f y12fvo 1 -1 1 y11f = y12f = y 22f = Rf Rf Rf C.K. Tse: Feedback ampliﬁers and 38 oscillators Example Converting to y-parameter – Ro + + is vi Ri – vo RL + avi – y11f y22f y12fvo 1 -1 1 REMEMBER: y11f = y12f = y 22f = Rf Rf Rf y11f and y22f are conductance! C.K. Tse: Feedback ampliﬁers and 39 oscillators Example Casting it to standard form – Ro + + is vi Ri || R f – Rf||RL vo + avi – y12fvo 1 -1 1 y11f = y12f = y 22f = Rf Rf Rf C.K. Tse: Feedback ampliﬁers and 40 oscillators Example Casting it to standard form – Ro || R f || RL + + is vi Ri || R f – a(R f || RL ) vo + vi – Ro + R f || RL -v o Rf C.K. Tse: Feedback ampliﬁers and 41 oscillators Example Finally, we get the standard form ie – Ro || R f || RL + + is vi Ri || R f – vo + Aie – Using Thévenin theorem, R f || RL Aie = av i -v o (R f || RL ) + Ro Rf A = -a (R || R )(R || R ) i f f L (R || R ) + R f L o Ri R 2 RL f 1 A = -a (Ri + R f ) ( R f RL + Ro R f + Ro RL ) C.K. Tse: Feedback ampliﬁers and 42 oscillators Example Apply standard results: Ri R 2 RL f 1 Basic ampliﬁer gain (transresistance) A = -a (Ri + R f ) ( R f RL + Ro R f + Ro RL ) Feedback gain: -1 f = Rf A 1 Overall (closed-loop) gain Ao = ª = -R f if Af >> 1 1+ Af f Ri || R f Input resistance RIN = 1+ Af Ro || R f || RL Output resistance ROUT = 1+ Af C.K. Tse: Feedback ampliﬁers and 43 oscillators Frequency response Gain and bandwidth Suppose the basic ampliﬁer has a pole at p1, i.e., + e ALF si A(jw) so A( jw ) = jw input – output 1+ basic ampliﬁer p1 sf 20log10|A| (dB) f feedback network ALF slope = –20dB/dec w p1 C.K. Tse: Feedback ampliﬁers and 44 oscillators Frequency response Gain and bandwidth Hence, we see that the overall gain has a pole at The overall (closed-loop) gain is pc = p1(1 + fALF) and the low-frequency gain is lowered A( jw ) to ALF Ao ( jw ) = Ao,LF = 1+ A( jw ) f 1+ fALF ALF = Ê jw ˆ Á1+ ˜ + fALF Ë p1 ¯ 20log10|A| (dB) È ˘ ALF Í 1 ˙ ALF basic ampliﬁer = Í ˙ 1+ fALF Í1+ jw ˙ feedback ampliﬁer Í ˙ Î p1 (1+ fALF ) ˚ Ao,LF w p1 pc C.K. Tse: Feedback ampliﬁers and 45 oscillators Stability of feedback ampliﬁer Deﬁnition: A feedback system is said to be stable if it does not oscillate by itself at any frequency under a given circuit condition. Note that this is a very restrictive deﬁnition of stability, but is appropriate for our purpose. Therefore, the issue of stability can be investigated in terms of the possibility of sustained oscillation. feedback circuit sustained oscillation at certain frequency C.K. Tse: Feedback ampliﬁers and 46 oscillators Why and how does it oscillate? The feedback system oscillates because of the simple fact that it has a closed loop in which signals can combine constructively. Let us break the loop at an arbitrary point along the loop. + si A so input – output f B B’ Signal at B, as it goes around the loop, will be multiplied by f and A, and also –1. SB’ = – A f SB C.K. Tse: Feedback ampliﬁers and 47 oscillators Why and how does it oscillate? Clearly, if SB’ and SB are same in magnitude and have a 360o phase difference, then the closed loop will oscillate by itself. Oscillation criteria: 1. Af = 1 2. Af = ±180o This is known as the Barkhausen criteria. The idea is If the signal, after making a round trip through A and f, has a gain of 1 and a phase shift of exactly 360o, then it oscillates. But, in the negative feedback system, there is already a 180o phase shift. Therefore, the phase shift caused by A and f together will only need to be 180o to cause oscillation. C.K. Tse: Feedback ampliﬁers and 48 oscillators The loop gain T An important parameter to test stability is the loop gain, usually denoted by T. T = Af |T| (dB) crossover frequency (where the gain is 1) wo 0dB w f w fT If fT = –180o, OSCILLATES! C.K. Tse: Feedback ampliﬁers and 49 oscillators Phase margin Phase margin is an important parameter to evaluate how stable the system is. Phase margin fPM = –180o – fT |T| (dB) crossover frequency (where the gain is 1) wo 0dB w f w fT –180o phase margin fPM (the larger the better) C.K. Tse: Feedback ampliﬁers and 50 oscillators Compensation Compensation is to make the ampliﬁer more stable, i.e., to increase fPM. REMEMBER: We should always look at T, not A or Ao. |T| (dB) crossover frequency (where the gain is 1) wo 0dB w p1 p2 f w fT –180o phase margin fPM (how to increase it?) C.K. Tse: Feedback ampliﬁers and 51 oscillators Method 1: Lag compensation Add a pole at a low frequency point. The aim is to make the crossover point appear at a much lower frequency. The drawback is the reduced bandwidth. Compensation |T| (dB) crossover frequency after compensation function Gc is crossover frequency 1 Gc ( jw ) = before compensation jw 1+ pa 0dB w pa p1 p2 f w before compensation after compensation –180o phase margin fPM phase margin fPM before compensation after compensation C.K. Tse: Feedback ampliﬁers and 52 oscillators Method 2: Lead compensation Add a zero near the ﬁrst pole. The aim is to reduce the phase shift and hence increase the phase margin and keep a wide bandwidth. But the drawback is the more difﬁcult design. Compensation |T| (dB) function Gc is crossover frequency crossover frequency jw before compensation 1+ after compensation za Gc ( jw ) = jw 0dB za 1+ p1 p2 w pa f w before compensation after compensation –180o phase margin fPM phase margin fPM after before compensation compensation C.K. Tse: Feedback ampliﬁers and 53 oscillators Op-amp stability problem The op-amp has a high DC gain, and hence at crossover it is likely that the phase shift is signiﬁcant. The worst-case scenario is when the feedback gain is 1 (maximum for passive feedback). We call this unity feedback condition, and use this to test the stability of an op-amp. Under unity-gain feedback condition, the loop gain T = Af = A, because f = 1. |A| (dB) op-amp frequency response p1 p2 –90o phase margin too small –180o C.K. Tse: Feedback ampliﬁers and 54 oscillators Op-amp internal compensation Usually, op-amps are internally compensated. The technique is by lag compensation, i.e., adding a pole at low frequency such that the phase margin can reach at least 45o. Suppose we add a low-frequency dominant pole at pa. If we can put pa such that p1 (original dominant pole) is at crossover, then the phase margin is about 45o. |A| (dB) op-amp frequency response before compensation op-amp frequency response after compensation pa p1 p2 –90o phase margin too small –180o phase margin ≈ 45o C.K. Tse: Feedback ampliﬁers and 55 oscillators Op-amp internal compensation Typically, pa is about a few Hz, say 5 Hz. Then, we have to create a pole at such a low frequency. First, consider the input differential stage of an op-amp. One way to add the pole is to put a capacitor between the two collectors of the differential stage. Equivalent model: RL RL next stage to next stage rπ ro//RL 2C RIN C The dominant pole is 1 pa = = 2p (5) 2(ro || RL || RIN )C We can ﬁnd C from this equation. C.K. Tse: Feedback ampliﬁers and 56 oscillators Op-amp internal compensation If we use the previous method of inserting a C between collectors of the differential stage, the size of C required is very large, as can be found from 1 Using this method, C can be as large as hundreds pa = = 2p (5) of pF, which is too large to be implemented on 2(ro || RL || RIN )C chip. NOT practical! Better solution: Use to active load Miller effect. C RL RL output stage Miller effect can expand capacitor size by a factor of the gain magnitude. So, we may put the capacitor across the input and output of the main main gain stage gain stage in order to use CE stage Miller effect. In this way, C can be much smaller, say a few pF. C.K. Tse: Feedback ampliﬁers and 57 oscillators Example: Op-amp 741 internal compensation +Vcc Data: Q1 differential Q2 Q13B Q13A input stage DC gain = 70 dB + – Poles: Q14 30 kHz Q3 Q4 500 kHz +Vcc output 10 MHz Q20 –VEE Q16 Q23 Q17 Q5 Q6 main gain stage CE stage –VEE C.K. Tse: Feedback ampliﬁers and 58 oscillators Example: Op-amp 741 internal compensation Unity-gain feedback (worst case stability problem): T = A p1 = 30 kHz Bad stability because of the substantial phase shift! C.K. Tse: Feedback ampliﬁers and 59 oscillators Example: Op-amp 741 internal compensation Compensation trick (based on lag compensation approach): • Introduce a low-frequency pole at pa such that p1 is at crossover. • This ensures the phase angle at crossover = –135º. Hence, PM = 45º. |A| (dB) op-amp frequency response after compensation pa p1 –90o –180o phase margin ≈ 45o C.K. Tse: Feedback ampliﬁers and 60 oscillators Example: Op-amp 741 internal compensation Graphical construction method p1 = 30 kHz pa Bad stability because of the substantial phase shift! C.K. Tse: Feedback ampliﬁers and 61 oscillators Example: Op-amp 741 internal compensation Exact calculation of pa : 0 - 70 -70 slope = –20 dB/dec = = 70 dB log p1 - log pa 4.477 - log pa Hence, pa = 9.5 Hz pa p1 = 30 kHz C.K. Tse: Feedback ampliﬁers and 62 oscillators Example: Op-amp 741 internal compensation After compensation, the phase margin is 45º. C.K. Tse: Feedback ampliﬁers and 63 oscillators Example: Op-amp 741 internal compensation Question: How to create the 9.5 Hz pole with a reasonably small C ? Solution: Take advantage of Miller effect to boost capacitance. +Vcc Q1 differential Q2 Q13B Q13A + input stage – Cc Q14 Q3 Q4 +Vcc output Q20 –VEE Q16 Q23 Q17 Q5 Q6 main gain stage CE stage –VEE C.K. Tse: Feedback ampliﬁers and 64 oscillators Example: Op-amp 741 internal compensation +Vcc Given: Q1 differential Q2 Ro17 = 5 MΩ Q13B Q13A + input stage – Ro13 = 720 kΩ Cc Ri23 = 100 kΩ Q14 Q3 Q4 +Vcc output Q20 –VEE Q16 Q23 Q17 Q5 Q6 Gain of CE stage: main gain stage CE stage ACE = Gm[Ro17||Ro13||Ri23] –VEE Miller-effect capacitor Gm = 6 mA/V CM = Cc (ACE + 1) = 518.74 Cc C.K. Tse: Feedback ampliﬁers and 65 oscillators Example: Op-amp 741 internal compensation +Vcc Given: Q1 differential Q2 Ri16 = 2.9 MΩ Q13B Q13A + input stage – Ro4 = 10 MΩ Cc Ro6 = 20 MΩ Q3 Q4 Equivalent ckt: +Vcc Ro4||Ro6 –VEE Q16 Ri16 CM Q17 Q5 Q6 main gain stage CE stage pa = 1 / 2π CM [Ro4||Ro6|| Ri16] and CM = 518.74 Cc –VEE Hence, Cc = 15 pF C.K. Tse: Feedback ampliﬁers and 66 oscillators Oscillation In designing feedback ampliﬁers, we want to make sure that oscillation does not occur, that is, we want stable operation. However, oscillation is needed to make an oscillator. As shown before, the criteria for oscillation in a feedback ampliﬁer are 1. Loop gain magnitude | T | = 1 2. Roundtrip phase shift fT = ±180o Thus, the same feedback structure can be used to make an oscillator. In other words, we construct a feedback ampliﬁer, but try to make it satisfy the above two criteria. In practice, T is a function of frequency, and the above criteria are satisﬁed for one particular frequency. This frequency is the oscillation frequency. C.K. Tse: Feedback ampliﬁers and 67 oscillators Oscillator principle As T = Af, we can deliberately create phase shift in A or f. NOTE: Since this model is a + negative feedback, we need the total si A(jw) so phase shift of A(jw) and f(jw) to be input – output 180o at the frequency of oscillation. basic ampliﬁer If a positive feedback is used, we need the total phase shift to be 360o. f(jw) |A(jw) f(jw)| (dB) feedback network wo w w –180o C.K. Tse: Feedback ampliﬁers and 68 oscillators Sustained oscillation There are two problems! How does oscillation start? And how can oscillation be maintained? First, there is noise everywhere! So, signals of all frequencies exist and go around the loop. Most of them get reduced and do not show up as oscillation. But the one at the oscillation frequency starts to oscillation as it satisﬁes the Barkhausen criteria. If | T | is slightly bigger than 1, oscillation amplitude will grow and go to inﬁnity. But if | T | is slightly less than 1, oscillation subsides. The question is how to maintain oscillation with a constant magnitude. We need a control that changes | T | continuously. Typically, this is done by a nonlinear amplitude stabilizing circuit, for example, an ampliﬁer whose gain drops when its output increases, and rises when its output decreases. C.K. Tse: Feedback ampliﬁers and 69 oscillators The Wien bridge oscillator R2 Model: R1 – + A + Zp C R Zs C R Zp Zs R2 Basic ampliﬁer gain A = 1+ R1 Feedback gain -Z p R 1+ jwCR f = where Z p = and Z s = Zs + Z p 1+ jwCR jwC C.K. Tse: Feedback ampliﬁers and 70 oscillators The Wien bridge oscillator Oscillation frequency Note that we deﬁne the standard feedback structure with negative feedback. So, the loop gain is Ê ˆR -Á1+ 2 ˜ Ë R1 ¯ T( jw ) = Ê 1 ˆ 3 + jÁwCR - ˜ Ë wCR ¯ Applying the oscillation criteria, we can ﬁnd the oscillation frequency and the resistor values as follows: R T( jw ) = 1 ﬁ 1+ 2 = 3 ﬁ R2 = 2R1 R1 1 1 fT = ±180 o ﬁ w oCR = ﬁ wo = w oCR CR We can choose R2/R1 to be slightly larger than 2, say 2.03, to start oscillation. C.K. Tse: Feedback ampliﬁers and 71 oscillators The Wien bridge oscillator Frequency response viewpoint Suppose the ampliﬁer has a ﬁxed gain of A. The feedback network, however, has a bandpass frequency response. WHY OSCILLATE? A Clearly, the roundtrip gain will be 1 + for f = fo if the basic ampliﬁer has a gain of 3. |f| The world is noisy. Signals of all 1/3 frequencies exist everywhere! freq But signals at all frequencies except fo ff will be reduced after a round trip. fo Only signals at fo will have a + 90 roundtrip gain of 1. freq –90 Hence, the oscillation frequency is fo. From the ﬁlter structure, we can ﬁnd that fo is equal to 1/2πCR. C.K. Tse: Feedback ampliﬁers and 72 oscillators The Wien bridge oscillator +15V Amplitude control 3k If we choose R2/R1 = 2.03, then amplitude may D2 grow. We have to stabilize the amplitude. The following is an amplitude limiter circuit. 10k 20.3k 1k Diode D1 (D2) conducts when vo reaches its – A vo positive (negative) peak. + Just when D1 conducts, we have vA = vB. 1k 10k A 20.3k vo 16n 10k B 1k 16n 10k B 3k 3k D1 –15V –15V 3 1 -15 + (v o + 15) = v o ﬁ v o = 9 V. Similar procedure applies for the negative peak. 4 3 So, the amplitude is 9 V. C.K. Tse: Feedback ampliﬁers and 73 oscillators The phase shift oscillator This circuit matches exactly our negative feedback model. The basic ampliﬁer gain is R2/R1, and the feedback network is frequency dependent. For the feedback network, we want to ﬁnd the frequency at which the phase shift is R2 exactly 180o. At this frequency, if the R1 – roundtrip gain is 1, oscillation occurs. Note that the negative feedback already gives 180o + phase shift. |f| 1 C C C 29 freq R' R R fo ff R1 || R'= R f 180o freq C.K. Tse: Feedback ampliﬁers and 74 oscillators The phase shift oscillator C C C From the ﬁlter characteristic of the feedback network, we know that the phase shift is 180o at fo, where its gain R R R is 1/29. So, oscillation starts at fo if A ≥ 29. This means we need |f| to have R2 ≥ 29R1. 1 We can prove that 29 freq 1 fo = fo 2p 6CR ff 180o freq Note that the leftmost resistor in the feedback ﬁlter is R’ (not R). But R’//R1 is exactly R. This will adjust the loading effect of the basic ampliﬁer and make the overall ﬁlter circuit easier to analyze since it is then simply composed of three identical RC sections. C.K. Tse: Feedback ampliﬁers and 75 oscillators Resonant circuit oscillators A general class of oscillators can be constructed by a pure reactive π-feedback network. For a voltage ampliﬁer implementation, this structure can be modelled as a series-shunt feedback circuit: A + Ro jX1 jX2 + vi Ri – – Avi jX3 pure reactive π-feedback network jX3 – vi jX1 jX2 + C.K. Tse: Feedback ampliﬁers and 76 oscillators Resonant circuit oscillators Analysis: -A( jX1 )( jX 2 ) The loop gain is T( jw ) = j ( X1 + X 2 + X 3 ) Ro + jX 2 ( jX1 + jX 3 ) AX1 X 2 = j ( X1 + X 2 + X 3 ) Ro - X 2 ( X1 + X 3 ) For oscillation to start, we need T = –1. + Ro + Thus, the oscillation criteria become vi Ri – – Avi X1 + X 2 + X 3 = 0 AX1 =1 X1 + X 3 In practice, we may have jX3 (a) X1 and X2 are capacitors and X3 is inductor. – vi jX1 jX2 OR + (b) X1 and X2 are inductors and X3 is capacitor. C.K. Tse: Feedback ampliﬁers and 77 oscillators Colpitts oscillator When X1 and X2 are capacitors and X3 is inductor, we have the Colpitts oscillator. In this case, we have 1 1 jX1 = and jX 2 = jwC1 jwC2 jX 3 = jwL3 + Ro + From X1 + X 2 + X 3 = 0 vi Ri – – Avi the oscillation frequency can be found: 1 wo = Ê C1C2 ˆ L3 Á ˜ Ë C1 + C2 ¯ L3 – vi C1 C2 + C.K. Tse: Feedback ampliﬁers and 78 oscillators A practical form of Colpitts oscillator The basic ampliﬁer can be realized by a common-emitter ampliﬁer. The loop gain is 1/sC1 T(s) = Gm ZT sL + 1/sC1 where 1 1 1 ZT = sC2 + + ZT 1 ro || Rc sL + sC1 Putting s = jw, and applying the Barkhausen criterion: 1 wo = Ê CC ˆ L3 Á 1 2 ˜ Ë C1 + C2 ¯ Gm C2 (ro || Rc ) C1 Gm Rc ro >1 ﬁ > for oscillation to start. C1 C2 Rc + ro C.K. Tse: Feedback ampliﬁers and 79 oscillators Hartley oscillator When X1 and X2 are inductors and X3 is capacitor, we have the Hartley oscillator. In this case, we have jX1 = jwL1 and jX 2 = jwL2 1 jX 3 = jwC3 + Ro + From X1 + X 2 + X 3 = 0 vi Ri – – Avi the oscillation frequency can be found: 1 wo = C3 ( L1 + L2 ) For both the Colpitts and Hartley oscillators, the – C3 gain of the ampliﬁer has to be large enough to vi L1 L2 ensure that the loop gain magnitude is larger than + 1. C.K. Tse: Feedback ampliﬁers and 80 oscillators