INTODUTION In this lesson, we investigate some forms of wave-form generation using op amps. Of course, we could use basic transistor circuits, but it makes sense to simplify the analysis by considering the ideal op amp. We also see in this lesson how the op amp can be used in switching applications. YOU AIMS At the end of the lesson, you should be able to: analyse and design a range of sine-wave oscillators use an op amp as a switch understand the design of multivibrators. STUDY SKILLS There is a fair amount of mathematical manipulation in this unit. Make sure you follow the steps. When you come to do the Self-Assessment Questions, remember that there are many ways of using electrical theory to analyse a circuit. Use that with which you are most comfortable. Teesside University 0
OSILLATOS Oscillators are waveform generators. Usually, it is their frequency of oscillation and wave shape that is important. Oscillators can be classified as: sinusoidal square wave others, e.g. triangular, staircase, trapezoidal, etc. We have found the general condition for oscillation in Lesson of this topic. emember the condition for oscillation is a gain of at least when the phase shift round a feedback loop is 0 at some frequency. We shall first investigate some oscillators giving sinusoidal output. SINE WAE OSILLATOS wien bridge oscillator Z Z 0 FIG. Teesside University 0
3 onsider the network of FIGUE. Z jω Series and / jω and Z ( jω ) Parallel and / jω /jω jω Hence, ( jω ) Z jω ( jω ) Z ( jω) Z ( jω) ( jω) ω / j jω / jω jω jω ( ω) 3 jω ()... Now, when does this expression have either 0 or 80 phase shift? When the above expression is purely real! The only way this can apply in this equation is if the real term in the denominator is 0; then the 'j' factors in the numerator and denominator cancel. Teesside University 0
4... ω 0 i.e. ( jω ) jω jω 0 3jω 3... 3 So the circuit will oscillate at a frequency given by Equation (): ω / or f ω / π / π Hz provided that is fed back to with a gain of 3 (to make the loop gain or more). Let us look at how we design an amplifier with a gain of 3. See FIGUE. 0 FIG. What should / be? Suggest suitable values, bearing in mind that resistor tolerances could bring the ratio below the critical value..... Teesside University 0
5 The gain of this amplifier is /, so / must be greater than. kω and 0 kω would be suitable, unless the resistor tolerance is very high. What determines the amplitude of the oscillation?...... Only the output range of the op amp limits it. Since the characteristic of the amplifier has a hard saturation, the circuit above would generate a distorted sine wave as illustrated in FIGUE 3. This is because there is a sudden gain reduction to almost 0. To minimize distortion, we should build in a soft saturation. FIGUE 4 shows a possible modification, and shows the complete oscillator. ve saturation 0 ve saturation FIG. 3 Teesside University 0
6 3 0 0 FIG. 4 We arrange for / > (say.) and ( )/ < (say.5). Then, 3 when the Zener diodes conduct, the gain falls below the crucial level required to maintain oscillation. This damps the oscillation, preventing it from building up. Suppose we want a ±0 peak-peak output. If 0, 0/3 (since the gain is nominally 3). Hence, there is about (0-0/3) 6.7 across and this is the value of the Zener diode we should choose (say 6.3). Another way of controlling the amplitude of oscillation is by using a voltage dependent resistor instead of. Teesside University 0
7 OLPITTS OSILLATO i x o L Z 0 FIG. 5 We want to find o / i in FIGUE 5. o x ( jω ) jω / jω jωl / jω ω L... 4 Also, Z (jω) is the complex impedance obtained from j ω L j ω j ω Z ( jω ) jωl / jω / jω jωl / jω / jω jω 3 ω L L jω... ( 5) Teesside University 0
8 We are now able to derive an expression for o / i. o i o x x i ω L Z Z ω L ω L j 3 jω... ( 6) Make sure you can derive this equation from the previous line. This is real when the imaginary terms of the denominator sum to 0. jω3l jω 0 ω L... ( 7) At this frequency, o i... 8 The negative sign in this equation indicates that there is a 80 phase shift. So, to oscillate at the frequency defined by Equation (7), we require an amplifier with a gain of / as shown in FIGUE 6. Note that the resistor plays no direct part in the circuit; it merely provides a convenient impedance between the amplifier and the frequency dependent elements. The above analysis is only one way of finding o / i ; you could use Thevenin or nodal analysis. It comes to the same thing. FIGUE 6 shows the complete oscillator. Teesside University 0
9 3 L 0 FIG. 6 SUMMAY OF SINE-WAE OSILLATOS The analysis of all sine-wave oscillators follows more or less the same pattern. The analysis can be fairly complex, and there are plenty of opportunities for making algebraic errors. The process steps are as follows.. Analyse a complex network.. Find at what ω the gain of the network is purely real. 3. Find the magnitude of the gain at this ω. 4. Design an amplifier to make the overall gain of. 5. Include a soft saturation characteristic. The ordinary 74 op amp is only useful over a low frequency range, say up to about 3 to 4 khz. Above these frequencies, slew rate limitations start to show up. So you need a higher frequency op amp, or you can use a simple transistor amplifier! Teesside University 0
0 THE OPEATIONAL AMPLIFIE AS A SWITH o o 0 o o o ve Saturation 0 ve Saturation FIG. 7 With a high gain op amp, any small difference between the inputs will send the output into saturation. If, in the circuit above, is positive then, if >, the output will go to negative saturation; on the other hand, a signal less than will cause the amplifier to go to positive saturation. emember that o A ( ). Hence, the circuit acts as a switch; o changes when. If we want to avoid the uncertain values of amplifier saturation, we can add the Zener diode network as shown above. The output is then clamped at voltages of Z D as shown above. The value of the resistor is not critical, but it has to supply the load current and the Zener diode current. o( sat) ( Z ) D > Load current Zener current Teesside University 0
LEEL DETETO WITH HYSTEESIS in ( Z D ) 3 0 FIG. 8 o ( Z D ) 0 in in in ( Z D ) h h FIG. 9 In this circuit, we have positive feedback from the output to the non-inverting input. So the output can only be stable in the two saturation limits. There are therefore only two stable levels for. Switching of the output occurs when and so there are two different input voltages at which the output will switch. This is similar in concept to a thermostat which is used to switch on at a low temperature and off at a higher temperature to control an oven. Teesside University 0
Since there are two stable output levels, another name for this circuit is a bistable, or sometimes bistable multivibrator. It also performs the same function as a Schmitt trigger circuit. You will find all these names in textbooks. Switching occurs when. Z D But... 9 3 Note that can be either positive or negative, depending on the amplifier output. Also, Hence, in... 0 in So we can write down the equations for the two values of in at which the circuit output switches. in Z D h... 3 ( ) in Z D 3 h... ( ) where h is half the hysteresis voltage as shown in FIGUE 9.... 3 ± o Z D Teesside University 0
3 NOTES. Note that is the mid-voltage between the two input switching levels. (e.g. if we want to design a level detector that switches at 4 and 8, then ).. The voltage difference between the two input switching levels is the hysteresis h. 3. Many other forms of level detector are possible. They can all be analysed by considering the conditions when. 4. In practice, the slew rate of the common op amps limits the switching speed of the circuit. It is therefore better to use a comparator, an integrated circuit which much resembles the op amp, but is specifically designed for high slew rates. WOKED EXAMPLE Design a level detector which will switch at ± 5. Take ( Z D ) as 7. It is clear that, by adding equations () and (), we can eliminate h. in in So, if in 5 and in 5 then 0. Teesside University 0
4 Now substitute the figures into equation (). 5 0 7 3 5 5 4 3 5 9 3 /.8. 3 We should choose 3 8 kω and 0 kω. The value of is not critical; choose 0 kω. In designing level detectors, the significant errors will clearly be: accuracy and stability of the reference voltage stability and matching of the Zener diodes resistor tolerance errors due to op amp bias current and offset voltage. Teesside University 0
5 ASTABLE MULTIIBATO 3 ( Z D ) 0 FIG. 0 This circuit is a modification of the level detector and provides a continuous square-wave output. Typical waveforms are shown in FIGUE. A capacitor is charged and discharged via a resistor. Switching takes place when. Immediately after the output switches, the polarity of also changes, and the voltage across the capacitor charges towards this new level exponentially. Teesside University 0
6 ( Z D ) 0 t ( Z D ) T t 0 T FIG. Typical Waveforms As in the level detector, Z 3 D... ( 4) The equation for can be written by considering the initial and final voltages of an exponential with a time constant of. ( ) exp t /... 5 Z D You may find this a little hard to see. ompare this equation with that for the simple circuit shown below. We have to consider an initial condition; the exponential starts from and would continue until the asymptotic value [ Z D ], if switching did not intervene. Teesside University 0
7 B Final voltage ( B 0 ) t 0 t 0 Switch closes Initial capacitor voltage B As a check, put t 0 in equation (5) and remember e 0. Then put t infinity. Z D Z D Switching occurs when, and t T half the period of the cycle. Z 3 D Z D Z 3 D ( exp( T / ) ) Z 3 D Teesside University 0
8 Simplify by dividing by [ Z D ] and multiplying by ( 3 ): exp T / 3 3 exp( T / ) 3 3 exp( T / ) So exp( T / ) / 3 The full cycle period is T, since the circuit is symmetrical. It is interesting that the frequency does not depend on the Zener voltages. What errors would you expect between predicted and actual performance?....... The errors would be the same as for the level detector, plus the limitation of the amplifier slew rate, which means that the waveform will not be square, but a trapezoid. Teesside University 0
9 WOKED EXAMPLE What is the oscillation frequency of the astable multivibrator if 3 0 kω and 0 nf? Substitute the values into equation (6) to obtain the half period. T 04 0 0 9 ln 09. 86μs Frequency, f T 455 Hz. As with the level detector, there are many similar square-wave oscillator circuits. A very popular integrated circuit which can be used for oscillators or timers (as in the next section) is the type 555. MONOSTABLE MULTIIBATO This is the third circuit in this family of oscillators. The level detector had two stable states, while the astable had none; the monostable has one stable state. In other words, the monostable stays in a stable condition until it is triggered externally. It then produces a single pulse of defined length before recovering to its initial condition. Teesside University 0
0 T T Z D 3 0 FIG. ( Z D ) D 0 t ( Z D ) t 0 T m T r FIG. 3 Illustrative Waveforms The diode in FIGUE clamps the capacitor voltage at about 0.7, thus preventing the voltage rising to ; the circuit is in a stable condition. A negative trigger pulse, T, pulls down to the level of the capacitor voltage and initiates a negative output pulse of defined width. Immediately after the trigger pulse, the waveform is defined by the following equation, obtained by the same sort of considerations that generated the astable exponential equation. Teesside University 0
The capacitor starts to charge from D to ( Z D ). ( ) D Z D exp t / D... ( 7) Also ± Z D... 8 3 eset occurs when, and t T m the monostable period. Z 3 D ( exp( )) D D Z D Tm / e-arrange this equation in the same way as we did for the astable circuit. T m ( ) D Z / 3 ln... 9 ( Z D) Prove for yourself that this is true. NOTES. To ensure triggering, T must be greater than ( D ).. To reverse the output and trigger sense, change the polarity of the diode connection. 3. An expression for the recovery time can also be derived using a similar analysis. The recovery time is less than the monostable time, but can still be too long for convenience. If we demand another pulse before recovery Teesside University 0
is complete, the resultant period T m will be shorter. FIGUE 4 shows a modification to reduce the recovery time whilst leaving T m unaffected. 4. The monostable suffers from the same errors as the astable, and you will note that in this case T m depends on [ Z D ]. 4 << 4 ( Z D ) FIG. 4 WOKED EXAMPLE 3 Design a monostable with a T m of 0 seconds. As in all design problems there are choices to be made. The student finds it much easier to calculate T m if all the parameters of equation (9) are known. It is much harder to choose sensible values to give a desired answer! Let us choose Z 5.6, a readily available device, and assume that D 0.7. It also seems sensible to try making 3 (say 00 kω). Hence the 'ln' term yields the following value. Teesside University 0
3 ln ( ) ( ) D Z / 3 07 ln ( Z D). 56. ( 56. 07. ) 080. Hence, T m / 0. 8. 5. It is best to use a high value of to keep the value of low and cheap. A reasonable solution would be to make 0 μf and. MΩ. Teesside University 0
4 SELF-ASSESSMENT QUESTIONS. Find the formulae for oscillation frequency and gain required for a Wien bridge oscillator where the two resistors are not equal as shown below. Z Z 0. Derive the formula for the frequency of oscillation of the Hartley oscillator based on the circuit below. What sort of amplifier is needed and what gain will be required if L mh and L 0.5 mh? Z L L 0 3. A phase-shift oscillator has the circuit shown opposite. Prove that j ω 6 ω 5jω j ω 3 3 Teesside University 0
5 emember that the into the amplifier loads the point. Hence find the ratio F / to cause oscillation. hoose components to give an oscillation frequency of khz. F x y 0 0 4. Design a level detector to operate at input signal levels of 6 and. Assume [ Z D ] 7.5. 5. Design a square-wave generator to oscillate at a frequency of khz. 6. Show that the recovery time for the monostable circuit of FIGUE is given by the equation below. Tr ln D / Z { } 3 Teesside University 0
6 7. Find the input voltages at which the circuit below switches. in 0 kω 0 kω 0 z 8. 8. In the level detector circuit below, the two resistors no longer have the same value. BUT the is defined to be 5 or 5 (depending on the design requirement). Derive new formulae for the input switching levels and design a circuit to switch at 6 and. Assume ( Z D ) 0. in a b 3 Z D 0 Teesside University 0
7 NOTES Teesside University 0
8 ANSWES TO SELF-ASSESSMENT QUESTIONS. Put Z / jω jω jω and Z / jω / jω /jω jω Now Z Z Z jω jω jω jω Multiply above and below by jω (jω ). jω ( ) jω jω jω jω ( [ ] ) ω jω This is real when [ ω ( )] 0 i.e. ω At this frequency, jω jω( ). Teesside University 0
9 Therefore, to ensure oscillation, the gain of an accompanying amplifier must be greater than / /. Note: If we make, then ω / and the gain requirement is 3 as in the lesson.. We will follow the same steps as in the olpitts oscillator. jω L ω L / jω jωl ω L Z ( jω L ) ωl ω ωl ( / jω ωl ) ( j ) / j j j jωl jωl / jω ( jωl )( L ) ω ω L L ω ω L L Z Z ω L ω L ( j ω L )( ω L ) ω L L ( jωl ) ω L ω ( L L ). This simplifies to the expression below. jω 3 L L ω L L L jω ω L Teesside University 0
30 To make this real, the real part of the denominator must be zero. ω L L 0 ω L L i.e. At this frequency, jω 3 LL jωl ( ω L ) ω L ω L Now, using the frequency condition, substitute / L L for ω. L L To obtain an overall gain of, we need an amplifier gain of L /L. If L mh and L 0.5 mh, the required gain is. 3. Algebraically, this is probably one of the hardest problems which has been set. onsider the currents at the nodes x, y and. Note that, at, the current flows to the virtual earth via the right-hand resistance. x jω x y j ω x /... A x y jω y j ω y /... B y j ω /... Multiply each equation by and put jω K. Teesside University 0
3... A K K x x y x K K... B x y y y ( ) K y... Arrange each of these three equations so that terms are gathered.... ( A) K K x Ky 0 Kx K y K 0... B Ky K 0... We must now eliminate x and y from these simultaneous equations. (A) (): ( )... ( D) K K K 0 x K (B) (K ) (): K K K K 0... D x ( ) This eliminates y. Now we need to remove x using equations (D) and (D). Teesside University 0
3 K (D) (K ) (D) gives equation (E) below. ( ) K3 K K K K K K 0 Gather all the terms in, multiplying out the brackets. An equation of the following form results. ( E) K3 F 0 where K 3 K K 3 K 4K 3 4K K 4K 4K F The coefficient F can be considerably simplified. K3 K3 6K ( 5K ) 0 The voltage gain may then be obtained by rearrangement. K 3 K3 6K 5K Now replace K with jω. j( ω) 6 ω 5jω j ω 3 3. Teesside University 0
33 This has been hard going; you need a clear head and a logical procedure to be successful. If you refer back to Lesson of this topic, you will find that the required gain is 9, and the frequency is defined as below. ω 6. f ω / π 0 3 From this, we can derive the required value of. π 6 03 3. 5 0 6 If we choose nf, then 3.5 kω. Select 33 kω as the nearest value in the E series. 4. Use equations () and () of the lesson text and substitute in the values given. 6 7. 5 3 7. 5 3 Add these equations to obtain the value of. 4 Teesside University 0
34 Substitute this value of into the first equation. 6 5 3 earrange to give a relationship bewtween and 3. 4 4 5 3 75. 3 hoosing 0 kω and 3 7 kω would be close. hoose 0 kω (or any other convenient value). 5. f /T where T half the period of oscillation. So T / f 05. 0 3 Use equation (6) and substitute this value of T. 05. 0 3 ln / 3 hoose 3 (this is almost always a sensible choice). 0. 5 0 / ln3 0. 455 0 3 3 Some trial and error with standard component values gives 56 nf and 8. kω as a close fit. Teesside University 0
35 6. The exponential equation for the recovery time, T r, is as follows. Z D exp t / When t T r, D. Substitute in these values. D Z D exp Tr / Substitute for. D Z D Z 3 D exp Tr / Z 3 D earrange this equation to isolate the exponential term. Z D D 3 ( Z D) 3 r exp( T / ) exp ( T / ) r Z ( Z ) / D 3 Finally, Tr ln D / Z / 3 A few intermediate steps have been missed out here. Make sure you can fill them in. Teesside University 0
36 7. With 7.5 Zener diodes, the output voltages are ( ) / Z D / 75. 07. / 8.. Since 0, switching occurs when 0. Input current must flow via the 0 kω and kω resistors, giving the following relationship. in 0 0 8. / 0 3 3 Hence, the voltage at which the circuit switches is determined as follows. 0 8. in / / 373. This is an example of a different comparator, one that gives a positive output transition for a rising input voltage. 8. In this level detector, switching still takes place when, and is still defined in the same way; the only change is the relationship between the input voltage, and. in a in b a a b / / b in a b a b Teesside University 0
37 So we now have the two equations for the two input switching levels. in / Z D a b 3 /... A a b in ( ) / Z D a b 3 /... B a b Substitute the values in 6 and in and add equations (A) and (B). 6 / a b is either 5 or 5, but we must choose 5 in this case. b / a 30 / 8 3. 75 hoose b 56 kω and a 5 kω as appropriate E resistances to yield this ratio. Now substitute in equation (A), remembering that ( Z D ) 0. 6 ( 0) ( / 3. 75) ( ) 3 5 / 3. 75. 67 3 / 533. 3 It is not possible to get very close to this ratio with just two resistors. 3 0 kω and kω is about the best. Teesside University 0
38 SUMMAY In this lesson, we have learnt how to design sine-wave oscillators, by using the Barkhausen criterion of seeking the condition when the loop gain is when the phase shift round a feedback loop is 0. We have also investigated a family of circuits defined as follows: Bistable multivibrator or Level detector or Schmitt trigger or comparator Astable multivibrator or square-wave oscillator Monostable multivibrator or timer. All these circuits have been treated as op amp applications, but there are many other ways of obtaining the same result. Teesside University 0