How Analog Amplifier works

analog amplifiers classification and generalization. what is the difference between an analog amplifier and a digital repeater pdf free download
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J. C. G. Lesurf ˜ 1 ˜ Electronics from http://www.audiomisc.co.uk These notes are for the alf of the Electronics course given by Dr Lesurf. There are just nine lectures in this half of the course, but there are also a set of laboratories. These notes should be combined with the labwork booklet to cover the analog portion of the course. There is also additional information, colour versions of the diagrams, etc, on the ”Scots Guide to Electronics• website. To view this, direct a web-browser to the URL http://www.st-and.ac.uk/www_pa/Scots_Guide/intro/electron.htm and then click the link for the ”analog and audio• section of the guide. Items in other sections of the Scots Guide may also be of interest. The examples given here and on the website tend to focus on applications like audio (hi-fi) as I think this helps to make the topic more understandable and interesting. Hence the website contains extra information on audio-related topics which you can understand on the basis of the explanations provided in this course. The lectures in this booklet are: 1) Analog and amplifiers pg 2 2) Limits and class prejudice pg 10 3) Frequencies and filters pg 18 4) Feedback pg 27 5) Power supplies pg 35 6) Wires and cables pg 44 7) Transmission and loss pg 51 8) Op-Amps and their uses pg 58 9) Semiconductor devices. pg 66 There are also some tutorial questions at the end of this set of notes. Please read through the notes for each lecture before the lecture itself is given. Then see of you can answer the relevant tutorial questions after you have attended the lecture. The notes should mean you can then concentrate upon the explanations given in the lecture. If you are unclear as to any point you can ask about this during the lectures, or during one of the whole-class tutorials. J. C. G. Lesurf October 2007 J. C. G. Lesurf ˜ 2 ˜ Electronics from http://www.audiomisc.co.uk Lecture 1 – Analog and Amplifiers This course is intended to cover a range of electronic techniques which are likely to be of some use or interest to physical scientists and engineers. The lectures covered in these notes focus on the Analog‘ side of things. In general, electronic systems tend to be made up by combining a range of simpler elements, circuits, and components of various kinds. We will examine a number of these in detail in later lectures, but for now we‘ll start with a simple over-view of amplifiers and their uses. 1.1 The uses of amplifiers. At the most basic level, a signal amplifier does exactly what you expect — it makes a signal bigger However the way in which this is done does vary with the design of the actual amplifier, the type of signal, and the reason why we‘re wanting to enlarge the signal. We can illustrate this by considering the common example of a Hi-Fi‘ audio system. Loudspeakers CD low-level signals Power Amplifier high-level preamp signals vol Figure 1·1 — Simple electronic system In a typical modern hi-fi system, the signals will come from a unit like a CD Player, FM Tuner, or a Tape/MiniDisc unit. The signals these produce have typical levels of the order of 100mV or so when the music is moderately loud. This is a reasonably large voltage, easy to detect with something like an oscilloscope or a voltmeter. However the actual power levels of these signals is quite modest. Typically, these sources can only provide currents of a few milliamps, which by P = V I means powers of just a few milliwatts. A typical loudspeaker will require between a few Watts and perhaps over 100 Watts to produce loud sounds. Hence we will require some form of Power Amplifier to boost‘ the signal power level from the source and make it big enough to play the music. We usually want to be able to control the actual volume — e.g. turn it up to annoy neighbours, or down to chat to someone. This means we have to have some way of adjusting the overall Gain of the system. We also want other functions — the most obvious being to select which signal source J. C. G. Lesurf ˜ 3 ˜ Electronics from http://www.audiomisc.co.uk we wish to listen to. Sometimes all these functions are built into the same piece of equipment, however is often better to use a separate box which acts as a Pre-Amp to select inputs, adjust the volume, etc. In fact, even when built into the same box, most amplifier systems use a series of stages‘ which share the task of boosting and controlling the signals. Microphones inside microphone low-noise preamp sound sensor very low EQ/Mix level signal output to power amps. Figure 1·2 — Simple PA system The system shown in figure 1.2 is similar to the previous arrangement, but in this case is used for taking signals from microphones and mixing‘ them together to produce a combined output for power amplification. This system also include an EQ‘ unit (Equalising) which is used to adjust and control the frequency response. Microphones tend to produce very small signal levels. typically of the order of 1 mVrms or less, with currents of a few tens of microamps or less — i.e. signal powers of microwatts to nanowatts. This is similar to many other forms of sensor employed by scientists and engineers to measure various quantities. It is usual for sensors to produce low signal power levels. So low, in fact, that detection and measurement can be difficult due to the presence of Noise which arises in all electronic system. For that reason it is a good idea to amplify these weak, sensor created, signals, as soon as possible to overcome noise problems and give the signal enough energy to be able to travel along the cables from the source to its destination. So, as shown in figure 1.2, in most studios, the microphone sensor will actually include (or be followed immediately) by a small Low- Noise Amplifier (LNA). Due to the range of tasks, amplifiers tend to be divided into various types and classes. Frequently a system will contain a combination of these to achieve the overall result. Most practical systems make use of combinations of standard building block‘ elements which we can think of as being the words‘ of the language of electronics. Here we can summarise the main types and their use. Most types come in a variety of forms so here we‘ll just look at a few simple examples. J. C. G. Lesurf ˜ 4 ˜ Electronics from http://www.audiomisc.co.uk 1.2 The Voltage Amplifier vacuum op-amp bipolar J-FET (triode) (inverting) (NPN) (N-channel) V + V + V + R 1 R 1 v ou t R v 1 ou t v R ou t 1 v i n v i n v R i n 2 v i n v ou t – + R 2 R 2 R 2 Figure 1·3 — Examples of voltage amplifiers Figure 1·3 shows four examples of simple voltage amplifier stages using various types of device. In each case the a.c. voltage gain will usually be approximated by R 1 A ® - ... (1.1) v R 2 provided that the actual device has an inherent gain large enough to be controlled by the resistor values chosen. Note the negative sign in expression 1.1 which indicates that the examples all invert the signal pattern when amplifying. In practice, gains of the order of up to hundred are possible from simple circuits like this, although for reasons we will discuss in later lectures (on feedback and distortion) it is usually a good idea to keep the voltage gain below this. Note also that vacuum state devices tend to be called ”valves• in the UK and ”tubes• in the USA. Many practical amplifiers chain together a series of voltage amplifier stages to obtain a high overall voltage gain. For example a PA system might start with voltages of the order on 0·1mV from microphones, and boost this to perhaps 10 — 100 V to drive loudspeakers. This requires an overall 9 voltage gain of × 10 , so a number of voltage gain stages will be required. 1.2 Buffers and Current Amplifiers In many cases we wish to amplify the signal current level as well as the voltage. The example we can consider here is the signal required to drive the loudspeakers in a Hi-Fi system. These will tend to have a typical input impedance of the order of 8 Ohms. So to drive, say, 100 watts into such a loudspeaker load we have to simultaneously provide a voltage of 28 V and 3·5 A . rms rms Taking the example of a microphone as an initial source again a typical source impedance will be around 100 Ohms. Hence the microphone will provide just 1 nA when producing 0·1mV. This means that to take this and drive 100 W into a loudspeaker the amplifier system must amplify the 9 signal current by a factor of over ×10 at the same time as boosting the voltage by a similar 18 amount. This means that the overall power gain required is ×10 — i.e. 180 dB J. C. G. Lesurf ˜ 5 ˜ Electronics from http://www.audiomisc.co.uk This high overall power gain is one reason it is common to spread the amplifying function into separately boxed pre- and power-amplifiers. The signal levels inside power amplifiers are so much larger than these weak inputs that even the slightest leakage‘ from the output back to the input may cause problems. By putting the high-power (high current) and low power sections in different boxes we can help protect the input signals from harm. In practice, many devices which require high currents and powers tend to work on the basis that it is the signal voltage which determines the level of response, and they then draw the current they need in order to work. For example, it is the convention with loudspeakers that the volume of the sound should be set by the voltage applied to the speaker. Despite this, most loudspeakers have an efficiency (the effectiveness with which electrical power is converted into acoustical power) which is highly frequency dependent. To a large extent this arises as a natural consequence of the physical properties of loudspeakers. We won‘t worry about the details here, but as a result a loudspeaker‘s input impedance usually varies in quite a complicated manner with the frequency. (Sometimes also with the input level.) 16 12 Z 8 Ohms 4 0 2 kHz 20 kHz 200 Hz 20 Hz Frequency + 90 + 45 f Z 0 Deg. − 45 − 90 Figure 1·4 ˜ Impedance properties of a typical 8 Ohm‘ loudspeaker. Figure 1·4 shows a typical example. In this case the loudspeaker has an impedance of around 12 Ohms at 150 Hz and 5 Ohms at 1 kHz. So over twice the current will be required to play the same output level at 1 kHz than is required at 150 Hz. The power amplifier has no way to ”know in advance• what kind of loudspeaker you will use so simply adopts the convention of asserting a voltage level to indicate the required signal level at each frequency in the signal and supplying whatever current the loudspeaker then requires. This kind of behaviour is quite common in electronic systems. It means that, in information terms, the signal pattern is determined by the way the voltage varies with time, and — ideally — the current required is then drawn. Although the above is based on a high-power example, a similar situation can arise when a sensor is able to generate a voltage in response to an input stimulus but can only supply a very limited current. In these situations we require either a current amplifier or a J. C. G. Lesurf ˜ 6 ˜ Electronics from http://www.audiomisc.co.uk buffer. These devices are quite similar, in each case we are using some form of gain device and circuit to increase the signal current level. However a current amplifier always tries to multiply the current by a set amount. Hence is similar in action to a voltage amplifier which always tries to multiply the signal current by a set amount. The buffer differs from the current amplifier as it sets out to provide whatever current level is demanded from it in order to maintain the signal voltage it is told to assert. Hence it will have a higher current gain when connected to a more demanding load. op-amp bipolar V + v i v n ou t – v + i n v ou t R 1 Figure 1·5 — Examples simple buffers Figure 1·5 shows two simple examples of buffers. In each case the gain device (an NPN transistor or an op-amp in these examples) is used to lighten the current load on the initial signal source that supplies v . An alternative way to view the buffer is to see it as making the load impedance i n seem larger as it now only has to supply a small current to ensure that a given voltage is output. V + Tr2 b i 1 d I = b b i i o u t 2 1 d d Tr1 Figure 1·6 — photodiode and 2-stage current amplifier Figure 1·6 shows an example of a 2-stage current amplifier in a typical application. The photodiodes often used to detect light tend to work by absorbing photons and releasing free charge carriers. We can then apply a potential difference across the photodiode‘s junction and sweep out‘ these charges to obtain a current proportional to the number of photons per second J. C. G. Lesurf ˜ 7 ˜ Electronics from http://www.audiomisc.co.uk 16 striking the photodiode. Since we would require around 10 electrons per second to obtain an amp of current, and the efficiency of the a typical photodiode is much less than one electron per photon‘ this means the output current from such a photodector is often quite small. By using a current amplifier we can boost this output to a more convenient level. In the example, two bipolar transistors are used, one an NPN-type, the other PNP-type, with current gains of b and 1 b . For typical small-signal transistors b ® 50 - 500 so a pair of stages like this can be expected 2 to amplify the current by between × 250 and × 250,000 depending upon the transistors chosen. In fact, if we wish we can turn this into a voltage by applying the resulting output current to a resistor. The result would be to make the circuit behave as a high-gain voltage amplifier. 1.3 The diff-amp There are many situations where we want to amplify a small difference between two signal levels and ignore any common‘ level both inputs may share. This can be a achieved by some form of Differential Amplifier. Figure 1·7 shows an example of a differential amplifier stage that uses a pair of bipolar transistors. V + I I 1 2 R R 1 2 v ou t v v 1 2 Tr2 Tr1 I E Figure 1·7 — ‘Long-tailed pair’ differential amp This particular example uses a Long-Tailed Pair connected to a Current Source. We can understand how the circuit works simply from an argument based upon symmetry. For simplicity in this diagram we represent the Current Source by the standard symbol of a pair of linked circles. We will look at how a current source actually works later on. For now we can just assume it is an arrangement that insists upon always drawing a fixed current which in this case has the value I . E Assume that R = R and that the two transistors are identical. Assume that we start with the 1 2 two input voltages also being identical. The circuit is arranged so that I = I + I . By E 1 2 symmetry, when v = v it follows that I = I , hence I = I = I / 2. 1 2 1 2 1 2 E Now assume we increase, say, v by a small amount. This will mean that the transistor, Tr1, will 1 try to increase its current level and hence lift the voltage present at its emitter. However as it does this the base-emitter voltage of Tr2 will fall. Since the current in a bipolar transistor depends J. C. G. Lesurf ˜ 8 ˜ Electronics from http://www.audiomisc.co.uk upon its base-emitter voltage the result is that the current, I , rises, and I falls. The sum of these 1 2 currents, I , does not alter very much, but the balance between the two transistor currents/ E voltages changes. The result is that the rise in v , keeping v fixed, causes more current to flow 1 2 through R and less through R . The reduction in I means the voltage drop across R will reduce. 1 2 2 2 hence the voltage at its lower end moves up towards V . + To evaluate this more precisely, we can assume that for each transistor in the pair the collector- emitter current is related to the base-emitter voltage via an expression H I = V ... (1.2) C E BE i.e. we can say that H I = (V - V ) ; H I = (V - V ) ... (1.3 & 1.4) 1 B1 E 2 B2 E where V is the emitter voltage which the two transistors have in common and V and V are E B1 B2 the base voltages. The value H represents the voltage-current gain of each transistor. The sum of these two currents always has a fixed value imposed by the current source, but any difference between them will be such that H (I - I ) = v - v ... (1.5) 1 2 1 2 where V vanishes as it is common to both of the initial expressions. E Since I + I = I we can say that the above is equivalent to saying that 1 2 E ( ) H I - 2I = v - v ... (1.6) E 2 1 2 When only concerned about a.c. signals we can ignore the constant I and say that this becomes E 1 ( ) i = - v - v ... (1.7) 2 1 2 2h i b where h is one of the small-signal h-parameter values for a bipolar transistor, i represents the i b 2 change in the current in R produced when we change the input voltage so that their imbalance 2 from being equal is (v - v ). Since this change in current appears across R , and the potential at 1 2 2 the top of this is held fixed at V it follows that the lower end of R which we are using as the + 2 output will change in voltage by R 2 v = (v - v ) ... (1.8) ou t 1 2 2h i b i.e. the stage has an effective voltage gain of R / 2h . ( ) 2 i b Differential amplifiers are particularly useful in three applications: • When we have an input which has come from some distance and may have had some added interference. Using a pair of wires to send the signal we can then take the difference in potential between them as the signal and reject any common mode‘ voltages on both wires as being induced by interference. • In feedback arrangements we can use the second input to control the behaviour of the amplifier. • When we wish to combine two signals we can feed one into one transistor, and the second signal into the other. Most Operational Amplifier integrated circuits have differential amplifier input stages and hence amplify the difference between two given input levels. Many use bipolar pairs of the kind shown in figure 1·7, but similar arrangements using Field Effect Transistors are also often used. J. C. G. Lesurf ˜ 9 ˜ Electronics from http://www.audiomisc.co.uk The Current Source Having made use of one, we should finish with an explanation of the Current Source. Figure 1·8 shows a simple example. I C V + R 1 3 diodes R 2 Figure 1·8 — Simple Current Source A positive voltage is applied to the base of a bipolar transistor, but this base is also connected to a lower potential via a series of three diodes. The potential across such a diode when conducting current will tend to be around 0·6 V. Since there are three of these, the base of the transistor will sit at a potential about 1·8 V above the lower end of R . The base-emitter junction of a bipolar 2 transistor is also a diode, hence it will also drop about 0·6 V when conducting. As a result, there will be 1·8 - 0·6 = 1·2 V across R . This means the current through this resistor will be 1·2 / R . 2 2 since the transistor will probably have a current gain of over × 100 this means that more than 99% of this current will be drawn down from the transistor‘s collector. Hence approximately, we can say that 1·2 Volts I = ... (1.9) c R 2 in practice the current will vary slightly with the chosen values of R and V . However provided 1 + the current through the diodes is a few mA the above result is likely to be reasonably accurate. Provided that V is much larger than 1·8 V the exact value does not matter a great deal. The + circuit therefore tends to draw down much the same current whatever voltage appears at the transistor‘s collector — assuming that it, too, is more than a couple of volts. So the circuits acts to give a fixed‘ current and behaves as a steady current source. Although this example uses a bipolar transistor, as with many other forms of circuit we can make similar arrangements using other types of device. Summary You should now know the difference between some of the basic types of amplifier and the kinds of electronic building blocks‘ that can be used as amplifier stages. It should also be clear that the range of tasks and signal details/levels varies enormously over a wide range of applications. You should now also know how the basic amplifier arrangements such as Long-Tailed Pairs, etc, work. J. C. G. Lesurf ˜ 10 ˜ Electronics from http://www.audiomisc.co.uk Lecture 2 – Limits and Class Prejudice Every real amplifier has some unavoidable limitations on its performance. The main ones that we tend to have to worry about when choosing and using them are: • Limited bandwidth. In particular, for each amplifier there will be an upper frequency beyond which it finds it difficult/impossible to amplify signals. • Noise. All electronic devices tend to add some random noise to the signals passing through them, hence degrading the SNR (signal to noise ratio). This, in turn, limits the accuracy of any measurement or communication. • Limited output voltage, current, and power levels. This will mean that a given amplifier can‘t output signals above a particular level. So there is always a finite limit to the output signal size. • Distortion. The actual signal pattern will be altered due non-linearities in the amplifier. This also reduces the accuracy of measurements and communications. • Finite gain. A given amplifier may have a high gain, but this gain can‘t normally be infinite so may not be large enough for a given purpose. This is why we often use multiple amplifiers or stages to achieve a desired overall gain. Let‘s start by looking at the limits to signal size. 2.1 Power Limitations Figure 2·1 shows a simple amplifier being used to drive output signals into a resistive load. V + v R ou t C v i n R L R E V - Figure 2·1 — Amplifier and resistive load The amplifier is supplied via power lines at V and V so we can immediately expect that this + - will limit the possible range of the output to V v V as the circuit has no way to + ou t - provide voltages outside this range. In fact, the output may well be rather more restricted than this as we can see using the following argument. In effect, we can regard the transistor as a device which passes a current whose level is primarily determined by its base-emitter voltage, and hence by the input, v . However any current it i n draws from its collector must pass through R or R , and any current emerging through its C L J. C. G. Lesurf ˜ 11 ˜ Electronics from http://www.audiomisc.co.uk emitter must then flow through R . In practice it really acts a sort of variable resistor‘ placed in E between these other resistors. We can therefore represent it as being equivalent to the circuit shown in Figure 2·2a where the transistor is replaced by a variable resistance, VR. V V V + + + v v v R ou t R ou t R ou t C C C = – VR R VR R VR = 0 R L L L R R R E E E V V V - - - 2·2a 2·2b 2·2c Figure 2·2 — Transistor as variable resistor When we lower the input voltage, v , we reduce the transistor‘s willingness to conduct, which is i n equivalent to increasing its effective collector-emitter resistance. Taking this to the most extreme case we would reduce the current it passes to zero, which is equivalent to giving it an effective collector-emitter resistance of infinity. This situation is represented in figure 2·2b. Alternatively, when we increase v we encourage the transistor to pass more current from collector to emitter. i n This is equivalent to reducing its effective collector-emitter resistance. Again, taking this to the extreme we would reduce its resistance to zero as shown in 2·2c. We can now use the extreme cases shown in 2·2b and 2·2c to assess the maximum output the circuit can provide, and the implications for the design. From 2·2b we can see that the highest (i.e. most positive) current and voltage we can provide into a load, R , will be L V R V + L + v = V = : I = ... (2.1 & 2.2) ou t max max R + R R + R L C L C as the collector resistor acts a a Potential Divider with the load to limit the fraction of the available positive power line voltage we can apply to the output. Looking at expressions 2.1 and 2.2 we can see that we‘d like R Ç R to enable the maximum output to be able to be as big as C L possible — i.e. to approach V . For the sake of example, lets assume that we have therefore chosen + a value for R equal to R / 10. C L Now consider what happens when we lower the transistor‘s resistance (i.e. increase the current it will conduct) to the maximum case shown in 2·2c. We now find that the minimum (most negative) output voltage and current will be R V V L a a v = V = : I = ... (2.3 & 2.4) ou t mi n mi n R + R R + R a L a L where we can define R V + R V R R E + C - C E V ¬ : R = ... (2.5 & 2.6) a a R + R R + R E C E C Either by considering the diagram or examining these equations we can say that in order for J. C. G. Lesurf ˜ 12 ˜ Electronics from http://www.audiomisc.co.uk V to be able to approach V we require R Ç R / / R (where / / represents the value of the mi n - E C L parallel combination). Since we have already chosen to set R Ç R this is equivalent to choosing C L a value of R Ç R . Again, we can assume for the sake of example that R is chosen to be E C E (R / / R ) / 10 which means it will also apprimate to R / 100. C L L To consider the implications of the above lets take a practical example based on a power amplifier for use as part of an audio system. This means we can expect R to be of the order of 8 Ohms. For L simplicity, lets assume R = 10W. Following the assumptions above this means we would choose L R = 1W and R = 0·1W. Again, for the sake of example, lets assume that the voltage rails C E available are ±25 V. This would mean that for an ideal amplifier we could expect to output peak levels of about 90% of ±25 Vwhich for a sinewave would correspond to about 30 Watts rms maximum power delivered into the 10W load. Now consider what happens when the amplifier is on, but isn‘t trying to deliver a signal to the load — i.e. when v = 0. This means that there will be 25 Volts across R . Since this has a ou t C resistance of 1W it follows that I = 25 Amps when v = 0. i.e. the amplifier will demand C ou t 25 Amps from its positive power supply. Since none of this is being sent to the load this entire current will be passed through the transistor and its emitter resistor and away into the negative power rail supply. The result is that 25 × 50 = 1250 Watts will have to be dissipated in the combination of R , the transistor, and R in order to supply no actual output C E It would be fair to describe this kind of design as being ”quite inefficient• in power terms. In effect it will tend to require and dissipate over 1 kW per channel if we wanted to use it as a 30 Watt power amplifier. This is likely to require large resistors, a big transistor, and lots of heatsinks. The electricity bill is also likely to be quite high and in the summer we may have to keep the windows open to avoid the room becoming uncomfortably warm. Now we could increase the resistor values and this would reduce the ”standing current• or Quiescent Current the amplifier demands. However it would also reduce the maximum available output power, so this form of amplifier would never be very efficient in power terms. To avoid this problem we need to change the design. This leads us to the concept of amplifier Classes. 2.2 Class A The circuit illustrated in Figure 2·1 and discussed in the previous section is an example of a Class A amplifier stage. Class A amplifiers have the general property that the output device(s) always carry a significant current level, and hence have a large Quiescent Current. The Quiescent Current is defined as the current level in the amplifier when it is producing an output of zero. Class A amplifiers vary the large Quiescent Current in order to generate a varying current in the load, hence they are always inefficient in power terms. In fact, the example shown in Figure 2·1 is particularly inefficient as it is Single Ended. If you look at Figure 2·1 again you will see that the amplifier only has direct control over the current between one of the two power rails and the load. The other rail is connected to the output load through a plain resistor, so its current isn‘t really under control. We can make a more efficient amplifier by employing a Double Ended or Push-Pull arrangement. Figure 2·3 is an example of one type of output stage that works like this. You can see that this new arrangement employs a pair of transistors. One is an NPN bipolar transistor, the other is a PNP bipolar transistor. Ideally, these two transistors have equivalent properties — e.g. the same current gains, etc — except for the difference in the signs of their voltages and currents. (In effect, a PNP transistor is a sort of electronic ”mirror image• of an NPN one.) J. C. G. Lesurf ˜ 13 ˜ Electronics from http://www.audiomisc.co.uk The circuit shown on the left now V + has two transistors which we can control using a pair of input I 1 voltages, V and V . We can R 1 2 E therefore alter the two currents, I 1 and I independently if we wish. 2 V 1 v ou t In practice, the easiest way to use the circuit is to set the Quiescent I Current, I to half the maximum L Q V R 2 L level we expect to require for the load. Then adjust the two transistor I 2 currents in opposition‘. R E It is the imbalance between the two transistor currents that will pass V - through the load so this means the transistors share‘ the burden of Figure 2·3 — Push-Pull Output driving the output load. When supplying a current, I to the load this means that we will have L I I L L I = I + : I = I - ... (2.7 & 2.8) 1 Q 2 Q 2 2 In linear operation this limits us to a current range 0 Î 2I in each transistor, and load Q currents in the range -2I Æ I Æ 2I . Hence this arrangement is limited in a similar way to Q L Q before. However the important difference to before becomes clear when we take the same case as was used for the previous example, with ±25 V power lines and a 10W load. Since we wish to apply up to 25 V to a 10W load we require a peak output load current of 2·5 Amps. This means we can choose I = 1·25 A. As a result, each transistor now only has to Q dissipate 1·25 × 25 = 31 Watts — i.e. the output stage dissipates 62 Watts in total when providing zero output. Although this still isn‘t very efficient, it is rather better than the value of well above 1 kW required for our earlier, single ended, circuit 2·3 Class B Note that as is usual in electronics there are a number of arrangements which act as Class A amplifiers so the circuit shown in Figure 2·3 is far from being the only choice. In addition we can make other forms of amplifier simply by changing the Quiescent Current or Bias Level and operating the system slightly differently. The simplest alternative is the Class B arrangement. To illustrate how this works, consider the circuit shown in Figure 2·4a. Once again this shows a pair of devices. However their bases (inputs) are now linked by a pair of diodes. The current in the diodes is mainly set by a couple of constant current stages which run a bias current, i , through them. If we represent the forward voltage drop across each diode to be b i as V we can say that the input to the base of the upper transistor sits at a voltage of v + V , d i n d and the input to the base of the lower transistor sits at v - V . Now since the base-emitter i n d junction of a bipolar transistor is essentially a diode we can say that the drop between the base and emitter of the upper transistor will also be V . Similarly, we can say that the emitter of the d lower (PNP) transistor will be V above its base voltage. d J. C. G. Lesurf ˜ 14 ˜ Electronics from http://www.audiomisc.co.uk i i b i as b i as V V + + D 1 V V 1 1 D 1 R R v v E D E ou t ou t 2 v v i n i n D R 3 R R L R L E E D 2 V V 2 2 D 4 V V - - i i b i as b i as 2·4a Class B 2·4b Class AB Figure 2·4 — Class B and AB Power stages This leads to the very interesting result where the emitter voltages in 2·4a will be V = V = v ... (2.9) 1 2 i n This has two implications. Firstly, when v = 0 the output voltages will be zero. Since this i n means that the voltages above and below the emitter resistors (R ) will both be zero it follows that E there will be no current at all in the output transistors. The Quiescent Current level is zero and the power dissipated when there is no output is also zero Perfect efficiency The second implication is that as we adjust v to produce a signal, the emitter voltages will both i n tend to follow it around. When we have a load connected this will then draw current from one transistor or the other — but not both. In effect, when we wish to produce an positive voltage the upper transistor conducts and provided just enough current to supply the load. None is wasted in the lower transistor. When we wish to produce a negative voltage the lower transistor conducts and draws the required current through the load with none coming from the upper transistor. Again this means the system is highly efficient in power terms. This all seems to be too good to be true, there must be a snag... and yes, there are indeed some serious drawbacks. When power efficiency is the main requirement then Class B is very useful. However to work as advertised it requires the voltage drops across the diodes and the base-emitter junctions of the transistors to all be exactly the same. In practice this is impossible for various reasons. The main reasons are as follows: • No two physical devices are ever absolutely identical. There are always small differences between them. • The diodes and the transistors used will have differently doped and manufactured junctions, designed for different purposes. Hence although their general behaviours have similarities, the detailed IV curves of the transistor base-emitter junctions won‘t be the same as the IV curves of the diodes. • The current through the transistors is far higher than through the diodes, and comes mostly from the collector-base junctions which are intimately linked to the base-emitter. Hence the J. C. G. Lesurf ˜ 15 ˜ Electronics from http://www.audiomisc.co.uk behaviour is altered by the way carries cross the device. • The transistors will be hotter than the diodes due to the high powers they dissipate. Since a PN junction‘s IV curve is temperature sensitive this means the transistor junctions won‘t behave exactly like the diodes, and may change with time as the transistor temperatures change. • When we change the applied voltage it takes time for a PN junction to react‘ and for the current to change. In particular, transistors tend to store a small amount of charge in their base region which affects their behaviour. As a result, they tend to lag behind‘ any swift changes. This effect becomes particularly pronounced when the current is low or when we try to quickly turn the transistor ”off• and stop it conducting. The overall result of the above effects is that the Class B arrangement tends to have difficulty whenever the signal waveform changes polarity and we have to switch off one transistor and turn on the other. The result is what is called Crossover Distortion and this can have a very bad effect on small level or high speed waveforms. This problem is enhanced due to Non-Linearities in the transistors — i.e. that the output current and voltage don‘t vary linearly with the input level — which are worst at low currents. (We will be looking at signal distortions in a later lecture.) This places the amplifier designer/user in a dilemma. The Class A amplifier uses devices that always pass high currents, and small signals only modulate these by a modest amount, avoiding the above problems. Alas, Class A is very power inefficient. Class B is far more efficient, but can lead to signal distortions. The solution is to find a half-way house‘ that tries to take the good points of each and minimise the problems. The most common solution is Class AB amplification, and an example of this is shown in Figure 2·4b. 2·4 Class AB The Class AB arrangement can be seen to be very similar to the Class B circuit. In the example shown it just has an extra pair of diodes. The change these make is, however, quite marked. We now find that — when there is no output — we have a potential difference of about 2 × V d between the emitters of the two transistors. As a consequence there will be a Quiescent Current of V d I ® ... (2.10) Q R E flowing through both transistors when the output is zero. For small output signals that require output currents in the range -2I I 2I both transistors will conduct and act as a Q L Q double ended Class A arrangement. For larger signals, one transistor will become non-conducting and the other will supply the current required by the load. Hence for large signals the circuit behaves like a Class B amplifier. This mixed behaviour has caused this approach to be called Class AB. When driving sinewaves into a load, R , this means we have a quasi-Class A system up to an rms L voltage and output power of 2 2 V R 2V R d L d L V ' ® : P ' ® ... (2.11 & 2.12) r ms L 2 R R E E If we take the same audio example as before (R = 10W), assume a typical diode/base-emitter L drop of 0·5V, and chose small value emitter resistors of R = 0·5W this leads to a maximum E Class A‘ power level of 20 Watts. The Quiescent Current will be around 1 Amp, so with ±25V power rails this system would still dissipate around 50 Watts in total. In this case the system is more like Class A than Class B. We do, however have some control over the circuit and can adjust the bias current level. J. C. G. Lesurf ˜ 16 ˜ Electronics from http://www.audiomisc.co.uk The obvious way to change the bias current is to choose a different value for R . Choosing a E larger value will reduce I and P ' whilst still preserving a region of Class A behaviour which can Q L be large enough to avoid the Crossover and imperfection problems mentioned earlier. However if we make R too large they will get in the way‘ as all the current destined for the load must pass E though these emitter resistors. Most practical Class AB systems therefore use a different approach. The most common tends to be to replace the diodes with an arrangement often called a rubber diode‘ or rubber Zener‘. i b i as V + v i n V 1 R 1 R v E ou t V Z R R L E R 2 V 2 V - i b i as 2·5a Rubber diode 2·5b Rubber diode in use Figure 2·5 — Rubber diode used to set Quiescent Current Figure 2·5a shows the basic rubber zener‘ arrangement. Looking at this we can see that it consists of a pair of resistors and a transistor. The resistors are connected together in series to make a potential divider which feeds a voltage to the base of the transistor. However the transistor can shunt‘ away any current applied and control the potential difference appearing across the pair of resistors. The arrangement therefore settles into a state controlled by the fact that the base-emitter junction of the transistor is like a diode and the transistor conducts when the base-emitter voltage, V , is approximately 0·5 V. Since V is also the voltage across R we can expect that d d 2 V R Z 2 V ® ... (2.13) d R + R 1 2 provided that the base current is small enough compared to the currents in the two resistors to be ignored. The above just comes from the normal action of a potential divider. However since in this case the transistor is involved it will act to adjust V to keep V at about 0·5V. The result is Z d therefore that we can set the voltage V (R + R ) d 1 2 V ® ... (2.14) Z R 2 This means we can choose whatever voltage value, V , we require by selecting an appropriate pair Z of resistors. In a real power amplifier the pair of fixed resistors is often replaced by a potentiometer and the circuit in use looks as shown in Figure 2·5b. The Quiescent current can now be adjusted and set to whatever value gives the best result since we can now alter V and Z hence the potential difference between the emitters of the output transistors. In practice the optimum value will depend upon the amplifier and its use. Values below 1mA are common in op-amp IC‘s, whereas high-power audio amps may have I values up to 100mA or more. Q J. C. G. Lesurf ˜ 17 ˜ Electronics from http://www.audiomisc.co.uk Class AB arrangements are probably the most commonly employed system for power amplifier output sections, although Pure‘ Class A is often used for low current/voltage applications where the poor power efficiency isn‘t a problem. Although we won‘t consider there here, it is worth noting that there are actually a number of other arrangements and Classes of amplifier which are used. For example, Class C‘ and Class E‘ which tend to use clipped or pulsed versions of the signal. In linear audio amplifiers the most well known alternative is the Current Dumping‘ arrangement developed by QUAD (Acoustical Manufacturing) during the 1970‘s. This combines a small Class A amplifier and a pair of Dumping‘ devices. The Dumping devices just supply wadges of current and power when required, but carry no Quiescent Current so they are highly power efficient, but prone to distortion. The accompanying Class A amplifier is arranged to ‘correct‘ the errors and distortions produced by the Dumping devices. In effect this means it just has to fill in the gaps‘ so only has to provide a relatively small current. Hence the Quiescent level required by the Class A fill in‘ amplifier can be small, thus avoiding a poor overall efficiency. Summary You should now be aware of of the basic building blocks that appear in Power Amplifiers. That Class A amplifiers employ a high Quiescent Current (sometimes called bias current or standing current) large enough that the transistor currents are large even when the output signal level is small. It should be clear that the power efficiency of Class A amplifiers is therefore poor, but they can offer good signal performance due to avoiding problems with effects due to low-current level nonlinearities causing Distortion. You should now also see why a Double Ended output is more efficient than a Single Ended design. You should also know that Class B has a very low (perhaps zero) Quiescent Current, and hence low standing power dissipation and optimum power efficiency. However it should be clear that in practice Class B may suffer from problems when handling low-level signals and hence Class AB, with its small but controlled Quiescent Current is often the preferred solution in practice. It should also be clear how we can employ something like a Rubber Zener to set the optimum Quiescent Current and power dissipation for a given task. J. C. G. Lesurf ˜ 18 ˜ Electronics from http://www.audiomisc.co.uk Lecture 3 – Frequencies and filters One of the limitations which affect all real amplifiers is that they all have a finite Signal Bandwidth. This means that whatever amplifier we build or use, we find that there is always an upper limit to the range of sinewave frequencies it can amplify. This usually means that high frequencies signals won‘t be boosted, and may attenuated instead. It also often means that high frequency signals may be Distorted — i.e. their waveshape may be changed. Some amplifiers also have a low-frequency limitation as well, and are unable to amplify frequencies below some lower limit. We therefore need to be aware of, and be able to assess, these effects to decide if a specific amplifier is suitable for a given task. In addition to the above, there are many situations where we deliberately want to alter the signal frequencies a system will pass on or amplify. There are many situations where deliberate filtering is needed. Some common examples are: • Low Pass Filters used to limit the bandwidth of a signal before Analog to Digital Conversion (digital sampling) to obey the Sampling Theorem • Bandpass filters to select a chosen narrowband‘ signal and reject unwanted noise or interference at other frequencies. • Bandreject filters used to block‘ interference at specific frequencies. e.g. to remove 50/100 Hz hum‘ (or 60/120 Hz if you happen to be in the USA) • Tone Controls‘ or EQ‘ (Equalisation) controls. These tend to be used to alter the overall tonal balance of sound signals, or correct for other unwanted frequency-dependent effects. • A.C.‘ amplifiers that don‘t amplify unwanted low frequencies or d.c. levels. In this lecture we will focus on some examples of filters, but bear in mind that similar effects can often arise in amplifiers. 3.1 Ways to characterise filters The most common and most obvious way to describe the effect of a filter is in terms of is Frequency Response. This defines the filter‘s behaviour in terms of its Gain as a function of frequency. figure 3·1 shows a typical example. 0.0 -5.0 Measured Toko 208BLR3315N filter gain normalised to 0 dB at 1 kHz -10.0 -15.0 -20.0 -25.0 -30.0 -35.0 -40.0 -45.0 -50.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 Frequency (kHz) Figure 3·1 — Example of Low-Pass Filter Power (dB)+ + J. C. G. Lesurf ˜ 19 ˜ Electronics from http://www.audiomisc.co.uk In this case Figure 3·1 shows the measured power gain (or loss) as a function of frequency for a specific low-pass filter. This example is based upon a commercial audio filter sold under the Toko‘ brand name for use in hi-fi equipment. This filter is Passive — i.e. includes no gain devices, just passive components like resistors, capacitors, or inductors. As a result we need to counteract any unwanted losses in the filter. The overall circuit measured in this case is shown in figure 3·2. Powered by a pair of PP3s this gives up to ± 6V out. Response approx 1 Hz to 16kHz (about -3dB points). Mid-band voltage gain around × 50 2K7 27K 2K7 27K 1K2 1m F 4k7 TL071 1m F TL071 1K Toko LPF 100K 1500pF 4k7 100K 100K Figure 3·2 — Amplifiers and Low-Pass Filter. Here the filter actually sits in between two amplifiers which raise the signal level. The filter‘s job in the system this example is taken from is to stop high frequencies from reaching the output which is to be sampled by an ADC. The overall voltage gain is around ×50 at 1 kHz. The Toko filter passes signals from d.c. up to around 18 kHz but attenuates higher frequencies. Note also the pair of 1mF capacitors in the signal path. These, taken in conjunction with the following 100kW resistors, act as High Pass filters to remove any d.c. as it this was unwanted in this case. The combination of the 1kW resistor and 1500pF capacitor at the input also act as an additional low-pass filter. 0.0 180.0 150.0 -5.0 Phase 120.0 -10.0 90.0 -15.0 60.0 -20.0 30.0 -25.0 0.0 -30.0 -30.0 -60.0 -35.0 -90.0 -40.0 -120.0 -45.0 -150.0 -50.0 -180.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 Frequency (kHz) Figure 3·3 — As fig 3·1 but including phase plot Although a frequency response of the kind shown in figure 3·1 is the most common way to display the behaviour of a filter, it doesn‘t tell the whole story. When sinewave components pass through the filter they may also be delayed by an amount that depends upon their frequency. Power (dB) Phase (deg) J. C. G. Lesurf ˜ 20 ˜ Electronics from http://www.audiomisc.co.uk This effect is often represented in terms of a plot of the Phase Delay as a function of signal frequency. For the same filter as before the measured Phase Delay as a function of frequency is shown in figure 3·3. In figure 3·3 the phase seems to jump around‘ in quite a complex way. However the underlying behaviour is simpler than it appears. There are two reasons for this. Firstly, the phase measurements are always in the range -180° Î 180°, but the actual signal delay may be more than a half-cycle. When making phase measurements we can‘t normally distinguish a delay of, say, 50 degrees, from 360 + 50 = 410 degrees, or any other multiple number of cycles plus 50 degrees. If we look at the phase plot shown in figure 3·3 the phase seems to jump‘ from -180° to about +160° at around 16 kHz. In fact, what has happened is that the real signal delay has increased from -180° to around -200°. However, since the measurement system only gives results in the limited range it assumes that this is the same as +160°. Despite the apparent hopping around, the actual signal phase delay tends to increase fairly steadily with frequency. The second reason for the complex plot is that at some points the signal level becomes very small. e.g. in figure 3·3 at around 25kHz. When we try to make a phase measurement on a tiny signal the result becomes unreliable due to measurement errors and noise. Hence the value at such a frequency can‘t be taken as being as reliable as at other frequencies. Measured plots of phase delay versus frequency therefore have to be interpreted with some care. That said, the phase behaviour is important as we require it along with the amplitude behaviour to determine how complex signals may be altered by passing through the filter. In practice, though, it is often more convenient to measure or specify the filter‘s time-effects in terms of a time delay rather than a phase. We can say that a phase delay of f f (radians) at a frequency f is equivalent to a time delay of f f Dt = ... (3.1) 2pf Figure 3·4 shows the time delay as a function of frequency for the same filter as used for the earlier figures. 30 25 20 15 10 5 0 4 6 8 10 12 14 16 18 Frequency (kHz) Figure 3·4 — Time delay versus frequency Looking at figure 3·4, we can see that when expressed in terms of a time, the delay seems rather more uniform and does not vary as quickly with frequency as does the phase. This behaviour is Time Delay (microseconds)

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