Test Your Understanding

Transcription

1 074 Part 2 Analog Electronics EXEISE POBLEM Ex 5.3: For the switched-capacitor circuit in Figure 5.3b), the parameters are: = 30 pf, 2 = 5pF, and F = 2 pf. The clock frequency is 00 khz. Determine the low-frequency gain and the cutoff frequency. Ans. / 2 = 6, f 3dB = 6.63 khz) This discussion of switched-capacitor filters is a short introduction to the topic and is intended only to show another application of operational amplifiers. Switchedcapacitor filters are sampled-data systems ; that is, the analog input signal is not transmitted through the circuit as a continuous signal but passes through the system as a series of pulses. The equivalent resistance given by Equation 5.29) is valid only for clock frequencies much greater than the analog input signal frequency. Switched-capacitor systems can be analyzed and designed by z-transform techniques. Test Your Understanding TYU 5. a) Design a three-pole high-pass Butterworth active filter with a cutoff frequency of 200 Hz and a unity gain at high frequency. b) Using the results of part a), determine the magnitude of the voltage transfer function at i) f = 00 Hz and ii) f = 300 Hz. Ans. a) Let = 0.0 μf, then = k, 2 = 57.7 k, 3 = k ; b) i) T = db, ii) T = db) TYU 5.2 a) Design a four-pole low-pass Butterworth active filter with a 3 db frequency of 30 khz. b) Determine the frequency at which the voltage transfer function magnitude is 99 percent of its maximum value. Ans. a) Let = 00 k, then = 57.4 pf, 2 = pf, 3 = 38.6 pf, 4 = pf; b) f = 8.43 khz) TYU 5.3 One-, two-, three-, and four-pole low-pass Butterworth active filters are all designed with a cutoff frequency of 0 khz and unity gain at low frequency. Determine the voltage transfer function magnitude, in db, at 2 khz for each filter. Ans db, 4.88 db, 6.0 db, and 7.24 db) TYU 5.4 Simulate a 25 M resistance using the circuit in Figure 5.2a). What capacitor value and clock frequency are required? Ans. For example, for f = 50 khz, then = 0.8 pf) 5.2 OSILLATOS Objective: Analyze and design oscillators that provide sinusoidal signals at specified frequencies. In this section, we will look at the basic principles of sine-wave oscillators. In our study of feedback in hapter 2, we emphasized the need for negative feedback to provide a stable circuit. Oscillators, however, use positive feedback and, therefore, are actually nonlinear circuits in some cases. The analysis and design of oscillator circuits are divided into two parts. In the first part, the condition and frequency for oscillation are determined; in the second part, means for amplitude control is addressed. We consider only the first step in this section to gain insight into the basic operation of oscillators.

2 hapter 5 Applications and Design of Integrated ircuits Basic Principles for Oscillation The basic oscillator consists of an amplifier and a frequency-selective network connected in a feedback loop. Figure 5.4 shows a block diagram of the fundamental feedback circuit, in which we are implicitly assuming that negative feedback is employed. Although actual oscillator circuits do not have an input signal, we initially include one here to help in the analysis. In previous feedback circuits, we assumed the feedback transfer function β was independent of frequency. In oscillator circuits, however, β is the principal portion of the loop gain that is dependent on frequency. For the circuit shown, the ideal closed-loop transfer function is given by A f s) = As) As)βs) 5.33) v e v s A v o v fb Frequency selective network, b Figure 5.4 Block diagram of the fundamental feedback circuit and the loop gain of the feedback circuit is T s) = As)βs) 5.34) From our discussion of feedback in hapter 2, we know that the loop gain Ts) is positive for negative feedback, which means that the feedback signal v fb subtracts from the input signal v s. If the loop gain Ts) becomes negative, then the feedback signal phase causes v fb to add to the input signal, increasing the error signal v ε. If T s) =, the closed-loop transfer function goes to infinity, which means that the circuit can have a finite output for a zero input signal. As Ts) approaches, an actual circuit becomes nonlinear, which means that the gain does not go to infinity. Assume that T s) so that positive feedback exists over a particular frequency range. If a spontaneous signal due to noise) is created at v s in this frequency range, the resulting feedback signal v fb is in phase with v s, and the error signal v ε is reinforced and increased. This reinforcement process continues at only those frequencies for which the total phase shift around the feedback loop is zero. Therefore, the condition for oscillation is that, at a specific frequency, we have T jω o ) = A jω o )β jω o ) = 5.35) The condition that T jω o ) = is called the Barkhausen criterion. Equation 5.35) shows that two conditions must be satisfied to sustain oscillation:. The total phase shift through the amplifier and feedback network must be N 360, where N = 0,,2, The magnitude of the loop gain must be unity.

3 076 Part 2 Analog Electronics In the feedback circuit block diagram in Figure 5.4, we implicitly assume negative feedback. For an oscillator, the feedback transfer function, or the frequencyselective network, must introduce an additional 80 degree phase shift such that the net phase around the entire loop is zero. For the circuit to oscillate at a single frequency ω o, the condition for oscillation, from Equation 5.35), should be satisfied at only that one frequency Phase-Shift Oscillator An example of an op-amp oscillator is the phase-shift oscillator. One configuration of this oscillator circuit is shown in Figure 5.5. The basic amplifier of the circuit is the op-amp A 3, which is connected as an inverting amplifier with its output connected to a three-stage filter. The voltage followers in the circuit eliminate loading effects between each filter stage. 2 v I ) v A 2 2 v2 v v A 3 A 3 v v O Figure 5.5 Phase-shift oscillator circuit with voltage-follower buffer stages The inverting amplifier introduces a 80 degree phase shift, which means that each network must provide 60 degrees of phase shift to produce the 80 degrees required of the frequency-sensitive feedback network in order to produce positive feedback. Note that the inverting terminal of op-amp A 3 is at virtual ground; therefore, the network between op-amps A 2 and A 3 functions exactly as the other two networks. We assume that the frequency effects of the op-amps themselves occur at much higher frequencies than the response due to the networks. Also, to aid in the analysis, we assume an input signal v I ) exists at one node as shown in the figure. The transfer function of the first network is ) s v = v I ) 5.36) s Since the networks are assumed to be identical, and since there is no loading effect of one stage on another, we have ) v 3 s 3 v I ) = = βs) 5.37) s where βs) is the feedback transfer function. The amplifier gain As) in Equation 5.33) and 5.34) is actually the magnitude of the gain, or As) = v O = ) v 3

4 hapter 5 Applications and Design of Integrated ircuits 077 The loop gain is then ) ) 2 s 3 T s) = As)βs) = 5.39) s From Equation 5.35), the condition for oscillation is that T jω o ) =and the phase of T jω o ) must be 80 degrees. When these requirements are satisfied, then v O will equal v I ) and a separate input signal will not be required. If we set s = jω, Equation 5.39) becomes ) 2 jω) 3 T jω) = jω) 3 = 2 ) jω)ω) 2 [ 3ω ] jω[3 ω ] 5.40) To satisfy the condition T jω o ) =, the imaginary component of Equation 5.40) must equal zero. Since the numerator is purely imaginary, the denominator must become purely imaginary, or [ 3ω 2 o 2 2] = 0 which yields ω o = 3 5.4) where ω o is the oscillation frequency. At this frequency, Equation 5.40) becomes T jω o ) = 2 ) j/ 3)/3) 0 j/ 3)[3 /3)] = 2 onsequently, the condition T jω o ) = is satisfied when ) ) ) 2 = ) Equation 5.43) implies that if the magnitude of the inverting amplifier gain is greater than 8, the circuit will spontaneously begin oscillating and will sustain oscillation. EXAMPLE 5.4 Objective: Determine the oscillation frequency and required amplifier gain for a phase-shift oscillator. onsider the phase-shift oscillator in Figure 5.5 with parameters = 0. μf and = k. Solution: From Equation 5.4), the oscillation frequency is f o = 2π 3 = 2π = 99 Hz 30 3 ) ) The minimum amplifier gain magnitude is 8 from Equation 5.43; therefore, the minimum value of 2 is 8 k. omment: Higher oscillation frequencies can easily be obtained by using smaller capacitor values.

5 078 Part 2 Analog Electronics EXEISE POBLEM Ex 5.4: Design the phase-shift oscillator shown in Figure 5.5 to oscillate at f o = 22.5 khz. The minimum resistance to be used is 0 k. Ans. Set = 0 k, = 408 pf, 2 = 80 k ) Using Equation 5.36), we can determine the effect of each network in the phase-shift oscillator. At the oscillation frequency ω o, the transfer function of each network stage is jω o jω o = j/ 3) j/ 3) = j 5.44) 3 j which can be written in terms of the magnitude and phase, as follows: 3 90 tan / 3) = 2 [ 90 tan 0.577)] 5.45a)) or ) = b)) As required, each network introduces a 60 degree phase shift, but they each also introduce an attenuation factor of ) 2 for which the amplifier must compensate. The two voltage followers in the circuit in Figure 5.5 need not be included in a practical phase-shift oscillator. Figure 5.6 shows a phase-shift oscillator without the voltage-follower buffer stages. The three network stages and the inverting amplifier are still included. The loading effect of each successive network complicates the analysis, but the same principle of operation applies. The analysis shows that the oscillation frequency is ω o = ) 2 v O Figure 5.6 Phase-shift oscillator circuit and the amplifier resistor ratio must be 2 = ) in order to sustain oscillation.

6 hapter 5 Applications and Design of Integrated ircuits Wien-Bridge Oscillator Another basic oscillator is the Wien-bridge circuit, shown in Figure 5.7. The circuit consists of an op-amp connected in a noninverting configuration and two networks connected as the frequency-selecting feedback circuit. Again, we initially assume that an input signal exists at the noninverting terminals of the op-amp. Since the noninverting amplifier introduces zero phase shift, the frequency-selective feedback circuit must also introduce zero phase shift to create the positive feedback condition. 2 v y v x v O v I ) Z s Z p Figure 5.7 Wien-bridge oscillator The loop gain is the product of the amplifier gain and the feedback transfer function, or T s) = ) ) 2 Z p 5.48) Z p Z s where Z p and Z s are the parallel and series network impedances, respectively. These impedances are and Z p = s 5.49a)) Z s = s 5.49b)) s ombining Equations 5.49a)), 5.49b)), and 5.48), we get an expression for the loop gain function, T s) = 2 )[ 3 s /s) ] 5.50) Since this circuit has no explicit negative feedback, as was assumed in the general network shown in Figure 5.4, the condition for oscillation is given by T jω o ) = = )[ ] 2 5.5) 3 jω o /jω o )

7 080 Part 2 Analog Electronics Since T jω o ) must be real, the imaginary component of Equation 5.5) must be zero; therefore, jω o jω o = a)) which gives the frequency of oscillation as ω o = 5.52b)) The magnitude condition is then = ) ) a)) 3 or 2 = b)) Equation 5.53b)) states that to ensure the startup of oscillation, we must have 2 / )>2. DESIGN EXAMPLE 5.5 Objective: Design a Wien-bridge circuit to oscillate at a specified frequency. Specifications: Design the Wien-bridge oscillator shown in Figure 5.7 to oscillate at f o = 20 khz. hoices: An ideal op-amp is available and standard-valued resistors and capacitors are to be used. Solution: The oscillation frequency given by Equation 5.52b)) yields = = = π f o 2π ) A 0 k resistor and 796 pf capacitor satisfy this requirement. Since the amplifier resistor ratio must be 2 / = 2, we could, for example, have 2 = 20 k and = 0 k, which would satisfy the requirement. Trade-offs: Standard-valued resistors = 0 k and 2 = 20 k. In place of the ideal 796 pf capacitor, a standard-valued capacitor = 800 pf can be used. The oscillation frequency would then be f o = 9.9 khz. Element tolerance values should also be considered. omment: As usual in any electronic circuit design, there is no unique solution. easonably sized component values should be chosen whenever possible. omputer Simulation Verification: A omputer simulation was performed using the circuit in Figure 5.8a). Figure 5.8b) shows the output voltage versus time. Since the ratio of resistances is 2 / = 22/0 = 2.2, the overall gain is greater than unity so the output increases as a function of time. This increase shows the oscillation nature of the circuit. Another characteristic of the circuit is shown in Figure 5.8c). A mv sinusoidal signal was applied to the input of and the output voltage measured as the frequency was swept from 0 khz to 30 khz. The resonant nature of the circuit is observed. The oscillation frequency and the resonant frequency are both at approximately 8.2 khz, which is below the design value of 20 khz.

8 hapter 5 Applications and Design of Integrated ircuits kω 0 0 kω 2 3 v v U 4 A LM324 V v v 2 5 V 5 V 0 P P S 796 pf S 0 kω 796 pf 0 kω 0 a) V UA out) mv) Time ms) b) V UA out) mv) Frequency khz) Figure 5.8 a) ircuit used in the computer simulation for Example 5.5, b) output voltage versus time, and c) output voltage versus input frequency c) If the capacitor in the circuit is reduced from 796 pf to 720 pf, the resonant frequency is exactly 20 khz. This example is one case, then, when the design parameters need to be changed slightly in order to meet the design specifications.

9 082 Part 2 Analog Electronics EXEISE POBLEM Ex 5.5: Design the Wien-bridge circuit in Figure 5.7 to oscillate at f o = 800 Hz. Assume = = 0 k. Ans. = 0.02 μf, 2 = 20 k ) Additional Oscillator onfigurations Oscillators that use transistors and L tuned circuits or crystals in their feedback networks can be used in the hundreds of khz to hundreds of MHz frequency range. Although these oscillators do not typically contain an op-amp, we include a brief discussion of such circuits for completeness. We will examine the olpitts, Hartley, and crystal oscillators. olpitts Oscillator The ac equivalent circuit of the olpitts oscillator with an FET is shown in Figure 5.9. A circuit with a BJT can also be designed. A parallel L resonant circuit is used to establish the oscillator frequency, and feedback is provided by a voltage divider between capacitors and 2. esistor in conjunction with the transistor provides the necessary gain at resonance. We assume that the transistor frequency response occurs at a high enough frequency that the oscillation frequency is determined by the external elements only. Figure 5.20 shows the small-signal equivalent circuit of the olpitts oscillator. The transistor output resistance r o can be included in. A KL equation at the output node yields V o V o g V o mv gs sl = ) s s 2 and a voltage divider produces V gs = s 2 s 2 sl V o 5.55) Substituting Equation 5.55) into Equation 5.54), we find that )] V o [g m s 2 s 2 L 2 ) s = ) L 2 L Figure 5.9 The ac equivalent circuit, MOSFET olpitts oscillator 2 V gs g m V gs Figure 5.20 Small-signal equivalent circuit, MOSFET olpitts oscillator V o

10 hapter 5 Applications and Design of Integrated ircuits 083 If we assume that oscillation has started, then V o 0 and can be eliminated from Equation 5.56). We then have s 3 L 2 s2 L 2 s 2 ) g m ) = ) Letting s = jω, we obtain g m ) ω2 L 2 jω[ 2 ) ω 2 L 2 ] = ) The condition for oscillation implies that both the real and imaginary components of Equation 5.58) must be zero. From the imaginary component, the oscillation frequency is ω o = ) 5.59) 2 L 2 which is the resonant frequency of the L circuit. From the real part of Equation 5.58), the condition for oscillation is ωo 2L 2 = g m ombining Equations 5.59) and 5.60) yields 5.60) 2 = g m 5.6) where g m is the magnitude of the gain. Equation 5.6) states that to initiate oscillations spontaneously, we must have g m > 2 / ). Hartley Oscillator Figure 5.2 shows the ac equivalent circuit of the Hartley oscillator with a BJT. An FET can also be used. Again, a parallel L resonant circuit establishes the oscillator frequency, and feedback is provided by a voltage divider between inductors L and L 2. The analysis of the Hartley oscillator is essentially identical to that of the olpitts oscillator. The frequency of oscillation, neglecting transistor frequency effects, is ω o = 5.62) L L 2 ) Equation 5.62) also assumes that r π /ω 2 ). L 2 L Figure 5.2 The ac equivalent, BJT Hartley oscillator L s p rystal Oscillator A piezoelectric crystal, such as quartz, exhibits electromechanical resonance characteristics in response to a voltage applied across the crystal. The oscillations are very stable over time and temperature, with temperature coefficients on the order of ppm per. The oscillation frequency is determined by the crystal dimensions. This means that crystal oscillators are fixed-frequency devices. The circuit symbol for the piezoelectric crystal is shown in Figure 5.22a), and the equivalent circuit is shown in Figure 5.22b). The inductance L can be as high as a few hundred henrys, the capacitance s can be on the order of 0.00 pf, and the a) r b) Figure 5.22 a) Piezoelectric crystal circuit symbol and b) piezoelectric crystal equivalent circuit

11 084 Part 2 Analog Electronics capacitance p can be on the order of a few pf. Also, the Q-factor can be on the order of 0 4, which means that the series resistance r can be neglected. The impedance of the equivalent circuit in Figure 5.22b) is Zs) = s 2 /L s ) s p s 2 [ p s )/L s p )] 5.63) 2 Figure 5.23 Pierce oscillator in which the inductor in a olpitts oscillator is replaced by a crystal Equation 5.63) indicates that the crystal has two resonant frequencies, which are very close together. At the series-resonant frequency f s, the reactance of the series branch is zero; at the parallel-resonant frequency f p, the reactance of the crystal approaches infinity. Between the resonant frequencies f s and f p, the crystal reactance is inductive, so the crystal can be substituted for an inductance, such as that in a olpitts oscillator. Figure 5.23 shows the ac equivalent circuit of a Pierce oscillator, which is similar to the olpitts oscillator in Figure 5.9 but with the inductor replaced by the crystal. Since the crystal reactance is inductive over a very narrow frequency range, the frequency of oscillation is also confined to this narrow range and is quite constant relative to changes in bias current or temperature. rystal oscillator frequencies are usually in the range of tens of khz to tens of MHz. Test Your Understanding TYU 5.5 onsider the phase-shift oscillator in Figure 5.6. The value of is 5 k and the frequency of oscillation is f o = 20 khz. Determine the values of and 2. Ans. = 27 pf, 2 = 435 k ) *TYU 5.6 For the olpitts oscillator in Figure 5.9, assume parameters of L = μh, and 2 = nf, and = 4k. Determine the oscillator frequency and the required value of g m. Is this value of g m reasonable for a MOSFET? Why? Ans. f o = 7.2 MHz, g m = 0.25 ma/v) 5.3 SHMITT TIGGE IUITS Objective: Analyze and design various Schmitt trigger circuits. In this section, we will analyze another class of circuits that utilize positive feedback. The basic circuit is commonly called a Schmitt trigger, which can be used in the class of waveform generators called multivibrators. The three general types of multivibrators are: bistable, monostable, and astable. In this section, we will examine the bistable multivibrator, which has a comparator with positive feedback and has two stable states. We will discuss the comparator first, and will then describe various applications of the Schmitt trigger omparator The comparator is essentially an op-amp operated in an open-loop configuration, as shown in Figure 5.24a). As the name implies, a comparator compares two voltages