Lecture 10 - 1 Lecture 10 Introduction to electronic...

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Unformatted text preview: 1 Lecture 10: Introduction to electronic analog circuits 361-1-3661 7. Positive-Feedback Oscillators © Eugene Paperno, 2008 7.1. Introduction Our aim is to develop an electronic circuit that is able to produce sustained sinusoidal oscillations at a single frequency. No external ac source will drive the circuit; it should start from an initial state, and its zero-input response should approach a sinewave. We assume that in steady state, the circuit that we develop can be described by a linear, lumped, time-invariant (LLTI) small-signal equivalent model. We shall call such a circuit linear sine-wave oscillator. Terminology Before we start the development of an oscillator, let us pay some attention to the terminology accepted in basic circuit theory [1]. We recall that the zero-input response (ZIR) of an LLTI circuit is also called the natural response. It is the response of the circuit to an initial state only, with all the independent sources suppressed. The response of the circuit to an external excitation only, with the initial state equals zero, is called zero-state response (ZSR) or forced response. The complete response of a circuit is a superposition of its natural and the forced responses. The complete response of a stable circuit can also be divided into the transient response and the steady-state response. The circuit transient response is contributed by both the initial state and the external excitation. For a stable circuit, the transient response is decaying and the circuit steady state depends on the external excitation only. Note, that the oscillator that we develop should behave in a different manner compared to a stable circuit: its forced response is zero, and its steady state should be defined by its natural response. Hence, the oscillator natural response should not decay with time. Instead, it should approach a sinewave at a desired frequency and amplitude. Applications In terms of applications, oscillators, either sinusoidal or not, single or multi-frequency, linear or nonlinear, are generally used to synchronize a system (computers), to carry and detect information (communication), and to test a system (instrumentation). Principles of operation Depending on the principle of operation, oscillators can be classified as negative resistance, parametric, relaxation, and positive-feedback. We shall treat here the latter type only. Positive-feedback oscillators Positive-feedback oscillators employ an LLTI passive frequency-selective electric network (feedback network) β(jω) and an active, electronic amplifier AOL interconnected in a positive-feedback loop. We shall assume here that AOL is frequency-independent. Passive feedback network Recall that a passive electric network is that combined of passive elements: resistors, capacitors, and inductors. A passive resistor never dissipates and a passive capacitor and inductor never store negative energy [1]. A circuit or its elements are active if they are not passive. It is quite clear that being always stable, a passive network alone is not able to produce sustained oscillations: it has no poles in the right half of the Laplace plane, and, therefore, its natural response approaches zero. To illustrate this, we recall that the transfer function of a network in terms of the Laplace transform [1], ∞ L[ f (t )] ≡ ∫ f (t ) e − st dt , (1) 0− is defined as follows [1]: m H (s) ≡ L[zero − state response] =K L[input ] ∏ (s − z j ) j =1 n , (2) ∏ ( s − pi ) i =1 where s is the complex frequency, s=σ+jω, K is a constant, the zj's are called zeros, since H ( z j ) = 0 , and the pi's are called poles, since H ( pi ) = ∞ , provided that z j ≠ p i . The partial-fraction extension of (2) can be written as n H ( s) = ∑ K i i =1 1 , s − pi (3) where the Ki's are called the residues of the particular poles [1]. Applying the inverse Laplace transform L-1[H(s)] to (3)— this is usually done with the help of a look up table—one can obtain the natural response of the network, n natural response = ∑K e i pi t , (4) i =1 where the poles pi's are also called natural frequencies (or natural modes [2]) of the network. The pi's in (4) depend on 2 Lecture 10: Introduction to electronic analog circuits 361-1-3661 β(jω) vo(t) iin(t) R L C Im[β(s)]=0 (a) jω Q= ∞ jω0 jωd s plane s =σ+jω [email protected]|sLD β(s)| 4 (d) 3 2 1 0 −2 1 0 jω −1 σ −1 0 Q=0.5 1 −α σ 0 −jωd −jω0 1.5 1 (b) 0.5 Im[β(s)]=0 vo(t) Ke−α t jω 0 2π/ωd −0.5 0 (c) −1 (e) −1.5 −1 −0.5 0 0.5 1 σ Fig. 1. (a) A parallel RLC circuit, (b) locus of its natural frequencies, and (c) its natural time response. The characteristic equation: s2+2αs+ω02= s2+(ω0/Q)s+ω02. The resonant frequency ϖ 0 = 1 / LC , L, and C are kept constant, and R varies. Q=R/(ω0L). Since R is never equal infinity, the time response is always decaying. β(jω0) never equals infinity. (d) Network function | β(s)| of the RLC circuit. The red line is the locus of purely real values of β (s). (e) Contour plot of the network function | β (s)|. the network topology, and the Ki's (possibly complex) depend, in addition, on the initial state. Recall that a network function (a particular transfer function from an input variable to an output variable) and the natural frequencies of a network can be determined by applying an excitation that does not change the natural structure of the circuit [1], [2]. This can be done either with an independent current source connected in parallel to a network branch, or with an independent voltage source connected in series with a network branch. As far as the natural response is concerned, the above independent sources turn into an open or closed circuit, correspondingly; thus, not altering the natural structure of the network. Consider for example a passive RLC network of Fig. 1(a). Its network function, β ( s) = L[vo (t )] 1 s , ≡ Z ( s) = C ( s − p )( s − p*) L[iin (t )] p = −α + jϖ d , p* = −α − jϖ d α= 1 , ϖ0 = 2 RC , (5) 1 2 , ϖd ≡ ϖ0 −α2 LC can be found by connecting an independent current source iin(t), as shown in Fig. 1(a). Locus of the natural frequencies for the RLC network is shown in Fig. 1(b). This figure and (5) clearly demonstrate that the poles of a practical network (with a finite and positive R and, hence, α) are always located to the left of the jω axis. Or in other words, the denominator of (5) is never zero for any s with a nonnegative σ. As a result, a practical RLC network cannot generate a sustained sinewave. Its output is always decaying, as shown in Fig. 1(c), and its natural response approaches zero. 3 Lecture 10: Introduction to electronic analog circuits 361-1-3661 One way to reach non-decaying oscillations is to compensate for the network positive (passive) resistance R (read energy losses) with an element having negative (active) resistance –R (read energy pumping into the circuit), as shown in Fig. 2(a). Negative-resistance oscillators are based just on this idea. They employ nonlinear electronic elements, such as the tunnel diode, having negative slope in a part of their voltage-current characteristic, to obtain—for relatively small signals—negative resistance. Negative-resistance approach to oscillators is generally used at radio (RF) and especially at microwave frequencies, where it becomes difficult to construct a feedback circuit without introducing excess phase shift. At relatively low frequencies, from kilohertz up to hundreds of megahertz, positive-feedback approach is generally used. β (jω) iin(t) vo(t) −R C jω jω0 0 s plane s =σ+jω σ −jω0 Another way to make a passive network to oscillate continuously is to integrate it in a feedback loop, as shown in Fig. 3(a), such that the total network function becomes So AOL P( s) . = =K Q( s ) S in 1 − AOL β ( s ) L (a) Feedback loop H ( s) = R vo(t) 0 Ke−0 t (b) 2π/ω0 t (6) Note that contrary to (5), the denominator of (6) can reach zero for an s1 with a non-negative σ, namely, it happens when the loop gain AOLβ(s1)=1. Barkhausen criterion To reach sustained oscillations in steady state, the loop gain AOLβ(s1) should be equal to unity at a physical, ω —not complex, s —frequency, namely, ⎧ ⎪ AOL 1 β ( jω1 ) = 1 , amplitude criterion . (7) AOL 1 β ( jω1 ) = 1 ⇒ ⎨ ⎪∠AOL 1 β ( jω1 ) = 0, phase criterion ⎩ where AOL 1 β ( jϖ 1 ) and ω1 are the oscillator's steady-state loop gain and frequency; index 1 emphasizes that in steady state the loop gain AOL 1β ( jω1 ) is unity. Since for a physical frequency, σ =0, the oscillator's steadystate poles (natural frequencies) are located directly on the jω axis of the s plane, as shown in Fig. 3(b), where according to (6) and (7), H ( jω1 ) = ∞ . Equations (7) determine the steadystate oscillation conditions and are called the Barkhausen criterion. Initial pole location We now have to decide on the pole location in the initial state. Let us first suppose that there should be no difference between the initial and final pole locations; in both the cases the poles can be located directly on the jω axis. (c) Fig. 2. A parallel RLC network employing negative-resistance (a), its natural frequencies (b), and its natural time response (c). Since RII(−R)=∞, H(jω0) does equal infinity. Magnitude of the natural response (4) in this case [see Fig. 3(b)] is a function of the initial state. It is an undesirable situation for the three following reasons. First, we do not intend to supply the circuit with additional means to control the initial state. Second, any circuit is subject to disturbances and noise. In the case of an oscillator, the uncontrolled initial state, disturbances, and noise will make its steady state unpredictable. Third, it is impossible to provide the initial pole locations exactly on the jω axis because of the non-zero tolerances of the circuit components. Since it is undesirable to locate the oscillator's initial-state poles directly on the jω axis, they should be located in the right half of the s plane at a distance from the jω axis. This distance [see Fig. 3(b)] should be large enough to ensure the desirable pole location for any deviations of the circuit element values within their tolerances and temperature ranges. The oscillator in this case will generate an increasing sinewave. When the sinewave magnitude will become sufficiently large, a negative feedback should stabilize it by shifting the poles towards the jω axis, as shown in Figs. 3(b) and (c). It is undesirable to locate the poles in the left half of the s plane because, in this case, the oscillator steady state will be zero. To summarize the above: we want the oscillator to be unstable in the initial state and conditionally stable in the 4 Lecture 10: Introduction to electronic analog circuits 361-1-3661 Natural response at ω1 vo Sε Σ So AOL t 0 Ss (a) Sf β(jω) ϕ=0 Forced response at ω vo ω1 H(jω) t 0 Initial pole location Undesirable Desirable jω jω1 Forced response at ω vo s plane ω1 ϕ=π AOLβ(s) =1 AOLβ(jω) =1 t 0 AOLβ(s) =1 ϕ=0 σ (b) 0 Natural and forced responses at ω vo ω1 σ >0 −jω1 vo t 0 ϕ=0 0 ϕ=0 (c) t Natural and forced responses at ω vo ϕ=π +VCC Sε 0 So AOL ϕ=0 Fig. 4. The effect of disturbances and noise on the oscillator output. AOL max So=Sε AOL max +VCC So< Sε AOL max So ~ Sε AOL ~ (1/3)VCC 0 t (d) −VCC So(ω1) ω1 max Sε (ω1) −VCC Fig. 3. Positive-feedback oscillator (note that suppressing Ss returns the network to its original topology; would a "-" used at the summing point, we would add an inverter to the original topology) (a), locus of its natural frequencies as a function of the lop gain AOLβ(jω) (b), its natural time response (c), and nonlinear properties of the amplifier AOL (d). steady state. We never want it to be stable. The simplest—but not always the best (see the concluding part of this Section)—way to determine the initial poles location in the right half of the s plane and to provide their movement to the jω axis, when the oscillations approach their steady-state magnitude, is to use the nonlinear properties of the amplifier AOL, as shown in Fig. 3(d). In the initial state, the small-signal value of AOL is simply set to be greater than it is needed to satisfy the Barkhausen criterion (7), AOL > AOL 1 . As a result, the initial pole location will be either to the right or to the left of the jω axis [see Fig. 3(b)], depending on the specific behavior of the feedback transmission β(s). For the reasons discussed above, β(s) for which the initial poles are in the left half of the s plane should not be employed in our design. Below (see Section III), we will show how one can easily decide whether AOL > AOL 1 shifts poles to the right or to the left of the jω axis. 5 Lecture 10: Introduction to electronic analog circuits 361-1-3661 Negative feedback Meanwhile, let us suppose that AOL > AOL 1 shifts the poles as desired, to the right of the jω axis. This forces unstable operation of the oscillator. It starts from an initial state and builds up its oscillations, as shown in Fig. 3(c). When the oscillations became large enough and reach nonlinear part of the amplifier transfer characteristic AOL=So/Sε in Fig. 3(d), the amplifier gain AOL decreases (negative feedback) and approaches AOL 1 . Accordingly, the poles move towards their steady-state location on the jω axis, as shown in Fig. 3(c). The steady state is reached when the amplifier gain at frequency ω1 equals AOL 1 , and the Barkhausen criterion (7) is satisfied. The effect of disturbances and noise A disturbance and noise affecting the circuit can be translated to the input of the oscillator and represented by an equivalent source Sin, as shown in Fig. 3(a). The source Sin will contribute to the oscillator output: it will either decrease or increase the oscillation caused by the initial state (the natural response). Recall that a circuit having a single couple of purely imaginary poles is conditionally stable. (A circuit having more than a couple of equal imaginary poles has linearly increasing natural response and, hence, is unstable.) A conditionally stable circuit has a bounded steady state only if it is not excited at the frequency of the poles (see Fig. 4). If the circuit is excited at the frequency of the poles then its forced response linearly increases with time. Let us suppose now that the spectrum of the disturbance and/or noise is continuous and always includes a sinusoidal excitation at the frequency of the oscillator poles. This excitation will contribute a linearly increasing part (forced response) to the oscillator natural response as shown in Fig. 4. Depending on the phase difference between the natural and forced responses, the oscillator output will either increase immediately after adding the external excitation or first decrease and after that increase (see Fig. 4). The increased oscillations will decrease the amplifier gain AOL below its steady state value AOL 1 , and the decreased gain will shift the oscillator poles in the left half of the s plane. The oscillator natural response will decay, and the oscillator steady state will be defined by its steady state response to the input signal Sin in accordance with the network function (6). Since in the new steady state, caused by a disturbance or noise, the loop gain AOL β ( jω1 ) will be smaller than unity, the network function (6) will be finite, and so will be the oscillator output. The specific value of the amplifier gain AOL in the new steady state and thus the specific value of the network function and the oscillator output depend on the magnitude of the disturbance or noise and on the nonlinear properties of the amplifier. Note that under influence of a disturbances or noise, the oscillator operates as a frequencyselective amplifier with the total gain described by the network function (6). At the frequency of oscillation, the total gain is at maximum and, at the other frequencies, it decreases in accordance with the frequency behavior of the feedback transmission β ( jω ) . The disadvantage of the use of the nonlinear properties of the amplifier is that its output signal becomes distorted. To avoid distortions, it is better to control with the help of a negative feedback the value of β(jω1) in such a way that the amplifier output is always within one third of its full scale [see Fig. 3(d)]. This can be done, for example, by using in the feedback network photo- or thermo-resistors, which values depend on average, not instant, magnitude of oscillations. The Nyquist stability criterion We shall finally illustrate how one can easily decide whether AOL > AOL 1 shifts the oscillator poles from their steady-state position on the jω axis to the right or to the left of this axis. Let us first assume that the denominator in (6) has only two zeros: Q ( s ) = 1 − AOL β ( s ) = ( s − z )( s − z*), Q ( s ) = ( s − z )( s − z*) . (8) ∠Q ( s ) = ∠( s − z ) + ∠( s − z*) Let us also assume that the two zeros of Q(s) lie in the right half of the s plane, as shown in Fig. 5. We consider as well that for a passive β(jω), Q(s) has no poles in the right half of the s plane. If we now encircle the entire right half of the s plane with a closed contour (so called Nyquist D-contour), as shown in Fig. 5(a), another closed contour, Q(jω), is created in the complex Q(s) plane around its origin, Q(s)=(0, j0), as shown in Fig. 5(b). The function Q(s) is said to map the right half of the s plane inside the contour Q(jω) in the Q(s) plane. [Note also that contour Q(jω) in Fig. 5(b) encircles the origin of the Q(s) plane in the same direction as the right half of the s plane is encircled in Fig. 5(a).] The above result is known as the Principle of the Argument. This principle can be easily understood if one considers an opposite case. Consider, for example, that Q(jω) does not encircle the origin of the Q(s) plane, as sown in Fig. 5(c). In this case, the vector Q(jω) will not rotate by 720o for −∞ < jϖ < ∞ as Fig. 5(a) and (8) state. We are now ready to formulate the Nyquist stability criterion for a positive-feedback oscillator combined of a stable feedback network β(jω) and a stable amplifier AOL: such an oscillator is unstable if its loop gain AOLβ(jω) does encircle the point (1, j0), as shown in Fig. 6. 6 Lecture 10: Introduction to electronic analog circuits 361-1-3661 jω s Im s plane s 360 R→∞ σ 0 z* = Q(s) = (s−z) (s−z*) Q(s)=(s−z)(s−z*) s−z z Q(s) plane Q(s)= (s−z) + (s−z*) Q( jω) σ >0 σ >0 1−AOLβ(s) plane Im s plane Q(s)=1−AOLβ(s) Q(jω) 0 720 jω H(s)= ∞ 1−AOLβ(s)= 0 AOLβ(s)=1 R→∞ Re σ 0 −1 0 Re Nyquist D-contour (b) (a) H(s) = AOL P(s) =K 1−AOLβ (s) Q(s) Im Im Q(s) plane −AOLβ(s) plane AOLβ(s) plane Im AOLβ(jω) Q ( s) < 720 0 σ >0 −1 0 Re 0 Re 1 Re Q(jω) (c) Fig. 5. Mapping of the right half of the s plane: the Principle of the Argument. Fig. 6 illustrates that AOLβ(jω1)>1 can cause both unstable and stable operations of a circuit. Or in other words, AOLβ(jω1)>1 can shift the circuit's poles either to the right or to the left of the jω axis, depending on the specific behavior of the feedback transmission β(jω). Fig. 6. The Nyquist stability criterion. H(s) = AOL 1−AOLβ(s) A β(jω )>1 OL 1 AOLβ (jω) H(s) is unstable 7.2. Oscillators for low frequencies: RC oscillators 0 Our current aim is to develop oscillators for low frequencies. Since the value L=1/(ω02C) and, therefore, the size and cost of inductances should be high at a low frequency of oscillation ω0, we will not use them. Capacitors are much smaller and less expensive; therefore, our feedback networks and oscillators at low frequencies will be of the RC type. 7.2.1. Phase-shift oscillator According to the above, our aim is to design a circuit that is able to produce sustained oscillations as a response to the initial conditions only. We would also like this circuit to oscillate at a single frequency. As a result we would like the Barkhausen criterion to hold at this frequency only. We start the design from choosing a CS configuration for the amplifier (see. Fig. 7). We suppose that the CS amplifier has a real negative gain in the entire frequency band of interest. As a result, the feedback network has to have a real negative transmission at a single finite frequency to satisfy the phase conditions of the Barkhausen criterion. One of the simplest networks satisfying this request is a three-stage RC high-pass filter shown in Figs. 7 and 8. We know that an elementary RC stage is able to rotate the phase by an angle AOLβ (s) plane Im 1 Re (a) AOLβ(jω1) =1 AOLβ(jω1) >1 Im AOLβ (jω) H(s) is stable 0 1 Re (b) Fig. 6. Examples for an unstable (a) and stable (b) networks. Note that in the both cases the loop gain AOLβ(jω1) can be greater than unity. less than 90○. Therefore, to obtain a real negative transmission, reed a 180○ phase shift, we need at least three such stages. 7 Lecture 10: Introduction to electronic analog circuits 361-1-3661 VDD β(jω) RD 0.1C C vo 0.01C 10R R CC RS 10R R vo AOL 0.01C 0.1C C 100R 100R R>>RD Im[β(s)]=0 100 β(jω) AOL G D 0.1C C 0.01C vo gmvgs 10R R RD||ro [email protected]|sLD β(s)| 1 0 −2 0 jω 100R σ 0 −1 (a) iD |AOL|=5 iD VDS V'DS<VDS |AOL|=8 1 3 −1/RD 0.2 0.125 0.1 0.08 0.5 Q Im[β(s)]=0 0.06 0.25 Vt vGS VGS V'DS VDS VDD vDS 0.04 0.02 jω 0 − 0.25 Linear gain − 0.5 Nonlinear gain t |AOL|=25 0.3 1 0.75 gm Q 2 t Fig. 7. Phase-shift oscillator. Note that the transistor transfer characteristic, iD−vGS, drops off with decreasing VDS. To find the solution for AOLβ(jω1)=1 in a simple way, we decouple the stages of the feedback network from each other and also from the amplifier. To do this, we increase the input impedance of each following stage compared to the output impedance of the amplifier and the output impedance of the preceding stages (see Fig. 7). This helps us to find the oscillator return ratio as: − 0.75 −2 − 1.5 −1 − 0.5 0 0.5 σ (b) Fig. 8. (a) Network function |β(s)| of the phase-shift oscillator feedback network. The red line is the locus of purely real values of β(s). (b) Contour plot of the network function. The pole of β(s) is shown in blue, and the poles of the network function H(s) of the oscillator are shown in red for |AOL|>8, green for |AOL|=8 [see also Fig. 8(c)], and brawn for |AOL|<8. 8 Lecture 10: Introduction to electronic analog circuits 361-1-3661 AOL β ( jϖ ) ≡ + v ′′ Sε′′ = + gs Sε v gs ′′ = v gs vgs =1 ⎛ ⎞ 1 ≈ − g m ( RD || ro ) ⎜ 1 + 1/ jϖ RC ⎟ ⎝ ⎠ ≈ − g m RD 3 , (9) Abs@ AfHs(s)| |H LD 1 3 3 1 − − 1+ jϖ RC (ϖ RC )2 j (ϖ RC )3 Note the "+" in the definition of the return ratio AOLβ(jω) in (9): we write "+" and not "−" because the feedback in the present case is positive and not negative. Solving (9) for AOLβ(jω) zero-phase condition, we find the frequency of oscillations, ω1: ∠AOL β ( jϖ 1 ) = 0 ⇒ 3 1 − =0 jϖ RC j (ϖ RC )3 , ⇒ 1 (ϖ 1RC ) 2 (10) =3 1 3 RC ⇒ ϖ1 = Solving (9) for the |AOLβ(jω1)|=1 condition, we find the transconductance gain of the transistor, gm: | AOL β ( jϖ 1 ) |= 1 ⇒ − g m RD 1 1− = − g m RD 3 (ϖ 1RC ) 1 1 − 3⋅ 3 2 , = − g m RD ⇒ gm = (11) 1 1 = g m RD = 1 1 − 3⋅ 3 8 8 RD Fig. 8 shows the locus of real values of β(s) in the Laplace domain. One can see that for some s=jω1, |β(s)|=0.125. This means that for AOLβ(s)=1 or |AOL|=1/|β(s)|=8 the poles of the oscillator transfer function Η(s) [see also Fig. 8(c) on this 50 1 0 −2 0 σ jω 0 −1 Fig. 8. (c) Network function of the phase-shift oscillator for |AOL|=8. page] will be located at the jω axis. One can also see that |β(s)|<0.125 on the right of the jω axis and |β(s)|>0.125 on the left of the jω axis. This means that for |AOL|>8, the oscillator poles will be in the right half of the Laplace plane, and for |AOL|<8, the oscillator poles will be in the left half of the Laplace plane. We choose therefore the static state of the CD amplifier in such a way that it provides a |AOL|=|gmRD|>8. As a result, the oscillator will generate an increasing sinewave at a real frequency approaching ω1. Due to the nonlinearity of the transistor transfer characteristic (see Fig. 7) and its dependence on VDS, the amplifier gain |AOL|=|gMRD| will decrease with increasing the amplitude of the oscillations, and at an |AOL|=|gMRD|>8, the steady state will be reached. (Note that gM is the large-signal transconductance gain.) 9 Lecture 10: Introduction to electronic analog circuits 361-1-3661 ϕ Frequency stability ϕβ 270 Let us now use the phase response (see Fig. 9) of the feedback network to analyze the frequency stability of the phase-shift oscillator. According to the Barkhausen criterion, at the oscillation frequency the phase of the return ratio ∠ AOLβ(jω1)=ϕA1+ϕβ1 =0 or 360○, were ϕA1 is the phase of the small-signal open-loop gain, and ϕβ1 is the phase of the feedback transmission. In the above example, we assumed that both ϕA=180○ and ϕβ=180○. Let us assume now that ϕA1 in Fig. 9 increases, for example, due to a temperature change, by ΔϕA1, ϕ'A1=ϕA1+ΔϕA1. ϕβ1 ϕA1 180 such that ϕA Δϕβ1 (a) Δω1 0 ω ω1 ω'1 HRLC(jω) vo (12) (13) ϕ'β1 dϕβ /dω ΔϕA1 90 iin R The phase condition of the Barkhausen criterion will now hold true for a different frequency of oscillations, ω'1=ω'1+Δω1, at which ϕ'β1=ϕβ1−ΔϕA1, ϕ'A1 L C (b) |HRLC(jω)| normalized 1 0.707 ϕ'A1+ϕ'β1=ϕA1+ΔϕA1+ϕβ1−ΔϕA1=360○. (14) Note in Fig. 8, that the translation of ΔϕA1 into Δω1 depends on the slope of the feedback-network phase response at ω1: the steeper the slope, the less sensitive the oscillation frequency is to the amplifier phase fluctuations. To describe the slope dϕβ/dω at ω1 quantitatively it is a common practice to substitute the phase response of an RLC resonant circuit with the same dϕRLC/dω at ω0 for the feedback-network phase response and to use the quality factor of the resonant circuit QRLC, which is proportional to dϕ/dω (see Appendix) QRLC = − ϖ 0 dϕ RLC 2 dϖ , ϖ 0 = ϖ1 , (15) R1 (c) R2>R1 0 ω ω0 ϕRLC Δω ω0 90 Q= Δω 45 Q= ω0L 0 R ω0 (d) ω −45 ϖ =ϖ 0 as a measure for dϕβ/dω at ω1 (see Fig. 9). For example, an RLC resonant circuit with QRLC=0.65 has dϕ/dω at ω0, which equals dϕβ/dω at ω1 of the phase response of the phase-shift oscillator feedback network. We say, therefore, that the phaseshift oscillator feedback network has the equivalent quality factor Qequiv=0.65. For such a small Qequiv the frequency stability is very low a 1% relative fluctuation of the amplifier phase corresponds to a 1.3% fluctuation of the frequency of oscillations [see (A2) in Appendix]. For example, the frequency fluctuations are as large as 13 Hz for a frequency of oscillations of 1 kHz. −90 dϕ RLC/dω ∝ QRLC ϕβ 270 dϕ RLC /dω = dϕβ /dω 180 ϕRLC 90 0 −90 90 Qequiv= QRLC 0 ω0= ω1 Fig. 9. Equivalent quality factor as a measure for the slope of the phase response of the oscillator feedback-network at ω1. 10 Lecture 10: Introduction to electronic analog circuits 361-1-3661 va Im[H(jω)] AOL vb vo 2R Zs βb HHPF (jω) ω vR HPF βa(jω) Zp R* vin=1 vR R C 0 vin=1 C R vC ϕHPF vC vC LPF 1/3 R*=R ω βb R*<R ω1 0 HPF ϕβ (b) 0 βa(jω) ω1 ϕβ RC Re[β(jω)] ∞ ∞ 1/3 ω ω |βa(jω)| |β (jω)| ω R*=R ω β (jω) 0 −1/3 1/3−β R vR HLPF (jω) βb βb Re[H(jω)] 0 Im[β(jω)] Wien bridge |β (jω)| ∞ vin ϕLPF ∞ (a) ω Im[β (jω)] LPF R*=R ω ω=ω1 R ϕβ 0 ∞ ZC Zp | β RC (|jϖ(jω|)| = βa 1 ) Zp Z p + Zs = Zp Z p + 2Z p = 1 3 −1/3 (c) R*<R β (jω) ω1 Re[β (jω)] ω1 ω β ( jϖ 1 ) = 1 − βb R 3 = 1 R* − 3 2 R + R* Zs Fig. 11. Finding quantitatively the Wien-bridge phase response. (a) Graphical solution (hodograph of Nyquist) for the phase response of elementary R-C high- and low-filters. (b) Graphical solution for the phase response of βRC(jω) and (b) its translation to the phase response of β(jω)=βRC(jω)−βR. ϕβ 180 R*=R 90 0 R*<R ω1 ω −90 −180 Fig. 10. Wien-bridge oscillator. For R*=0.913, β(jω1) =1/3−R*/(2R*+ R).=0.02. 7.2.2. Wien-bridge oscillator Our current aim is to develop an oscillator with much better frequency stability than that of the treated above phase shift oscillator. To achieve this aim, we will utilize a very interesting property of the Wien bridge (developed by Max Wien in 1891), to provide the approaching infinity slope of its phase response (see Fig. 10). Note that this is obtained due to the differential output of the bridge. The Wien bridge comprises two voltage dividers, one of them is purely resistive, with the transmission βb, and the other is reactive, with the transmission βa(jω). The total transmission of the bridge equals the difference β(jω)=βa(jω)−βb. The bridge is said to be in equilibrium when R*= R (we will see below why). When the bridge is in equilibrium, we can easily find its amplitude response for ω→0, ω1=1/RC (ωC=R at this frequency), and ω→∞ (see Fig. 10). To find quantitatively the amplitude response for 0<ω<ω1 and ω1<ω<∞, we note that in the first frequency range, where ωC>R, the reactive voltage divider behaves like the elementary R-C high-pass filter, and in the other frequency range, where ωC<R, it behaves like the elementary R-C low-pass filter. To find quantitatively the bridge phase response, we first refer to Fig. 11. One can see in Fig. 11(a) that in the complex plane the loci of the tips of the output phasors of the elementary R-C high- and low-filters rotates making a circle. It is reasonable to expect a similar behavior from βa(jω) [see Fig. 11(b)]. To translate βa(jω) to β(jω), we simply subtract 11 Lecture 10: Introduction to electronic analog circuits 361-1-3661 from the phasor βa(jω) the phasor βb. The result is shown in Fig. 11(c) for R*= R and for R*< R. According to Fig. 11(c), we plot in Fig. 10 the phase response ϕβ(jω) of the Wien bridge. One can see from Fig. 10 that the slope of ϕβ(jω) approaches infinity at ω1 when R* approaches R. Note, however, that simultaneously the bridge amplitude response β(jω), measured at ω1, approaches zero. It becomes clear, therefore, that the frequency stability of an oscillator based on the Wien bridge depends on the available voltage gain of the amplifier: the higher the amplifier gain, the lower β(jω1), which provides AOLβ(jω1)=1, the higher slope of ϕ(jω) at ω1 is, and the higher frequency stability is of the oscillator. In a home exercise you will see that having an amplifier with a voltage gain of 2000 provides a Qequiv=223, 300 times as much compared to Qequiv=0.65 of the phase-shift oscillator. Now a 1% relative fluctuation of the amplifier phase corresponds to a 0.004% fluctuation of the frequency of oscillations, or to 0.04 Hz, instead of 13 Hz in the above case, at a f1=1 kHz. The first Wien-bridge oscillator was developed by William Hewlett (the cofounder of the Hewlett-Packard company) in 1939. To stabilize the amplitude of the oscillation before the amplifier approaches nonlinear region, Hewlett replaced the resistor R* in Fig. 10 with a small lamp having at room temperature a resistance Rinitial* that is small enough to provide the oscillator return ratio AOLβ(jω1)>1. This causes the oscillator instability (see Appendix) and its natural response is a growing sinewave. Due to heating, the lamp resistance increases with increasing the amplitude of oscillation. When the lamp resistance reaches the value R*>Rinitial*, for which AOLβ(jω1)=1, the oscillation amplitude stops to grow. For a given AOL, the resistance R* can be found as follows Rjϖ L 1 R + jϖ L jϖ C β RLC ( jϖ ) = R Z L Z C = Rjϖ L 1 + R + jϖ L jϖ C = * ϕ RLC ( jϖ ) = arctan R⎞ ⎛ 1 + j ⎜ Rϖ C − ϖL⎟ ⎝ ⎠ − Rϖ C + R ϖL 1 ϕ RLC ( jϖ ) → 0 = R ϖL − Rϖ C and then we obtain dϕRLC/dω as a function of QRLC ⎛R ⎞ d⎜ − Rϖ C ⎟ dϕ RLC ( jϖ ) ⎝ϖ L ⎠ = − R − RC = 2 dϖ dϖ ϖ 2L ϖ 2 =ϖ 0 =− =− R ϖ 02 L R − RC LC LC ϖ 2 = 0 R − ϖ 02 L ϖ 02 L =− . 1 LC 2R ϖ 02 L Q =− R RLC ≡ ϖ 0L (A2) 2QRLC ϖ0 Finally we get ϖ0 . (16) * ⇒ AOL R − AOL R* = 1.5R * +3R AOL − 3 1.5 + AOL =− dϕ RLC dϕ o RLC π dϕ o RLC . (A2) 100% ≈ −0.87 =− 2QRLC 2QRLC 180 2QRLC c= dϖ R * +2 R − 3 R * 2R − 2R * = AOL =1 3R * + 6 R 3R * +6 R ⇒ R* = R R (A1) R* ⎞ ⎛1 AOL β ( jϖ 1 ) = AOL ⎜ − ⎟ =1 ⎝ 3 R * +2 R ⎠ ⇒ AOL Rjϖ L = − Rϖ LC + R + jϖ L 2 1 1 a − jb a b ; = = −j 2 + b2 2 + b2 a + jb a + jb a − jb a a c= 1 a + b2 2 , APPENDIX ∠c = arctan Relationship between dϕRLC/dω and QRLC To find the dependence of the slope of the phase response dϕRLC/dω of the parallel resonant circuit at ω0 (see Fig. 9) on its quality factor QRLC, we first find the phase response −b a REFERENCES [1] [2] C. A. Desoer, E. S. Kuh, Basic circuit theory, McGraw-Hill, 1969. S. Sedra, K. C. Smith, Microelectronic Circuits, 4th ed. New York: Oxford University Press, 1998. 12 Lecture 10: Introduction to electronic analog circuits 361-1-3661 Wien-bridge oscillator in the Laplace domain 1 0.8 [email protected] sLD 0.6 0.4 0.2 0 −2 1 0 jω 100 80 [email protected] sLD 60 40 20 0 −2 −1 1 0 σ −1 0 σ −1 0 1 1 Fig. A1. Abs[β(s)]. Fig. A2. Abs[H(s)] for AOLβ( jω1) =1. 1.5 1.5 1 1 0.5 jω 0.5 jω 0 0 − 0.5 − 0.5 −1 −1 − 1.5 −1 − 0.5 0 0.5 − 1.5 1 −1 σ − 0.5 0 0.5 1 σ Fig. A3. Abs[β(s)]. Fig. A4. Abs[H(s)] for AOLβ( jω1) =1. 1.5 1.5 1 1 0.5 jω jω −1 0.5 jω 0 0 − 0.5 − 0.5 −1 −1 − 1.5 −1 − 0.5 0 0.5 σ Fig. A5. Abs[H(s)] for AOLβ( jω1) =0.5. 1 − 1.5 −1 − 0.5 0 0.5 σ Fig. A6. Abs[H(s)] for AOLβ( jω1) =2. 1 13 Lecture 10: Introduction to electronic analog circuits 361-1-3661 Illustration Note that the instant shift of the oscillator poles to the imaginary axis of the Laplace plane by reducing the gain from -8.1 to 8.0 results in the immediate stabilization of the amplitude of the oscillations. There is no transient. The signals at the poles shifting time, 1 ms, are the initial conditions for the new time response. From 1 ms on, the poles are located on the imaginary axis, and the envelope of the oscillations becomes horizontal. 1.0V 0.5V 0V -0.5V -1.0V 0s 0.5ms 1.0ms 1.5ms 2.0ms V(GAIN12:OUT) Time 1.0V 0.5V 0V -0.5V -1.0V 0.990ms 0.995ms V(GAIN12:OUT) 1.000ms Time 1.005ms 1.010ms ...
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This note was uploaded on 01/14/2012 for the course EE 361-1-3711 taught by Professor Prof.eugenepaperno during the Fall '11 term at Ben-Gurion University.

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