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4 AMPLIFIERS By the end of this unit you should be able to: OBJECTIVES 1. Define or explain the meaning of the following terms: . . . . . . . . . . . . . . . . . . . . . . . . . . Unit 4: Amplifiers ac amplifier bandwidth buffer amplifier closed-loop gain comparator current gain dc amplifier decibel equivalent circuit gain-bandwidth product hysterisis in a comparator ideal op-amp input resistance inverting amplifier negative feedback non-inverting amplifier open-loop gain operational amplifier output resistance power gain saturation summing amplifier tuned amplifier virtual ground/earth voltage follower voltage gain 2. Understand the basic operation of amplifier circuits and perform simple calculations on voltage gain, current gain and power gain in an amplifier, given the appropriate data. 3. Understand the operation of operational amplifier circuits for the amplification, summation, and comparison of voltage signals. 4. Calculate the voltage gain and input resistance of simple inverting, non-inverting and voltage - follower operational amplifier circuits, given the appropriate data. 5. Calculate the output of simple operational amplifier summing circuits, given the appropriate data. 6. Sketch the output of operational amplifier comparator circuits for given input signals. 4-1 7. INTRODUCTION Design simple inverting, non-inverting and summing operational amplifier circuits given the appropriate data. An amplifier is a very signifcant components in a telecommunications system. Its basic function is to take an electronic signal of small amplitude and create a similar version of it with a much greater amplitude, so that it is less corruptible by noise that naturally occurs within any communications system. In the last Unit we studied the bipolar junction transistor (BJT) and the metal-oxide-semiconductor field-effect transistor (MOSFET) and described how these devices could provide signal amplification. We also considered simple, complete, single-stage amplifier circuits based on both devices. It is possible to extend such amplifier circuits to have several stages, and consequently greater signal gain, as well as other desirable amplifier characteristics such as high input resistance and low output resistance. Such amplifiers can be manufactured on a tiny piece of silicon and supplied as single integrated circuit (IC) package. AMPLIFIER EQUIVALENT CIRCUITS To understand or use amplifiers, we do not need to have a detailed knowledge of the complete circuitry inside the amplifier, but rather we can consider the amplifier as a black box, which can be represented by the equivalent network shown in Figure 4.1 Ro Io Ii + + Figure 4.1 Equivalent Circuit of a Voltage Amplifier Vi - + Ri - Gvo Vi Vo - When the voltage signal to be amplified, Vi is connected to the input terminals of the amplifier, a current Ii flows. From Ohm’s law, the ratio Ri = Vi / Ii is referred to as the input resistance of the amplifier. Thus the input side of the amplifier can be modelled as a resistance Ri comnnected between the two input terminals. The ability of the amplifier to produce an amplified signal requires a source of controlled energy and this is represented by the voltage generator GvoVi , where Gvo denotes the open-circuit voltage gain of the amplifier, ie the gain when there is no load connected to the amplifier output terminals. In practice, when the amplifier is connected to a load (other electronic circuitry), the voltage gain will be reduced slightly from its open-circuit value. This is modelled by the inclusion of the component Ro which represents the output resistance of the amplifier. In the open-circuit condition the cuurent Io = 0 and the output voltage Vo = GvoVi . When the amplifier output is connected to other electronic circuitry, the current Io is not zero and some of the amplified voltage will be lost across the output resistance Ro , just as we saw how some voltage is lost across the internal resistance of a battery in Unit 1. In most practical amplifiers the input resistance is relatively high and the output resistance is relatively low. Ideally Ri would be infinitely large, thereby reducing the input current and the input power to zero, and Ro would be zero minimising the output power loss in Ro. 4-2 Communications Technology 1 Figure 4.2 shows an amplifier, represented by its equivalent circuit with an input signal source Vs and internal resistance Rs driving a resistive load RL. The circuit may be easily analysed as follows. Figure 4.2 Equivalent Circuit of a Voltage Amplifierwith Input Signal Source and Output Load Rs + Ro Ii + + Vi Vs Io Ri - - Gvo Vi Vo RL - Ii = Vs Ri and Vi = Ii Ri = .Vs Rs + Ri Rs + Ri Similarly Vo = RL .GvoVi Ro + RL Therefore the overall voltage gain Gv is Vo Ri RL = .Gvo . Vs Rs + Ri Ro + RL Gv = The output signal current is Io = GvoVi G IR = vo i i Ro + RL Ro + RL Therefore the current gain Gi is Gi = Io G R = vo i Ii Ro + RL The power gain Gp is given by Gp = V2 / R I2R VI signal power into load = o2 L = o2 L = o o = Gv Gi signal power into amplifier Vi / Ri Vi Ii Ii Ri The voltage gain of an amplifier is often expressed in logarithmic units called decibels, or dBs ie GvdB = 20 log10 Gv = 20 log10 Vo Vs In practice a number of amplifiers may be connected in cascade as shown in Figure 4.3 for two amplifiers in which the output of the first amplifier is connected to the input of the second. Thus the second amplifier acts as a load for the first. The overall voltage gain Gvo is Gvo = Unit 4: Amplifiers Vo Vo1 Vo = . Vi Vi Vo1 4-3 Amplifier 1 Figure 4.3 Amplifiers in Cascade Vi Amplifier 2 Vo1 Vo Expressing this in decibels gives V V V V dB Gvo = 20 log10 o1 . o = 20 log10 o1 + 20 log10 o V V V V i o1 i o1 Thus the overall voltage gain in decibels is equal to the sum of the individual voltage gains in decibels. This is a very useful result and can be extended to any number of amplifiers in cascade. SAQ 1 FREQUENCY RESPONSE An integrated circuit amplifier has an open-circuit voltage gain of 60dB, an input resistance of 90kΩ and an output resistance of 50Ω. Its input is connected to a 1mV (rms) signal source with an internal resistance of 10kΩ and its output to a 450Ω resistive load. Determine the rms value of the output voltage across the load. So far we have been assuming that the gain of an amplifier is constant no matter what the frequency of the input signal is. Unfortunately practical amplifiers do not amplify all input signal frequencies to the same degree. This often results from a natural practical limitation of the amplifier but may also be due to the fact that the amplifier has been specifically designed to amplify a small range of frequencies only. The general shape of the frequency response of an ampifier which uses capacitors at the input and output of the amplifier is shown in Figure 4.4(a). Such amplifiers are referred to as capacitively coupled or ac amplifiers (see, for example, Figures 3.24 and 3.29). For low-frequency input signals, the presence of the input and output coupling capacitors causes the amplifier frequency response to fall off as shown. The frequency at which the voltage gain of the amplifier has fallen to 0.707Gv (-3 decibels) is referred to as the low-frequency cut-off of the amplifier fL. At high frequencies the amplifier gain decreases also due to internal transistor capacitances. At the high-frequency end, the freqency at which the gain is down 3dBs on its nominal value is referred to as the high-frequency cut-off fH. The frequency difference fH- fL is referrred to as the bandwith of the amplifier and defines the range of input signal frequencies over which the amplifier can provide the specified gain. It is possible to construct an amplifier which doesn’t use input and output coupling capacitors. Consequently such an amplifier doesn’t have a low-frequency cut-off and for low-frequency input signals the gain remains constant right down to 0 Hz, as shown in Figure 4.4(b). Another common type of amplifer is the so-called tuned amplifier which is common in radio circuits and is designed to amplify a very narrow range of frequencies only. A typical tuned amplifier response is shown in Figure 4.4(c). 4-4 Communications Technology 1 gain bandwidth of tuned amplifier (b) dc amplifier Gv 0.707Gv (a) ac amplifier 1 10 10 2 10 3 (c) tuned amplifier 10 4 10 5 Gain (db) db Gv Frequency (Hz) (log scale) 3db bandwidth of ac amplifier Figure 4.4 Frequency Response of an AC, DC and Tuned Amplifiers NEGATIVE FEEDBACK bandwidth of dc amplifier 1 10 10 2 10 3 10 4 10 5 Frequency (Hz) (log scale) Another important technique that is used in amplifier design is negative feedback . A detailed study of this topic is beyond the scope of this unit, but basically feedback is a process whereby a signal (voltage or current) is taken at the output side of th amplifier and a fraction of it is fed back and subtracted from a signal on the input side of the amplifier. This process is used mainly to alter the gain of the amplifer but also alters its input resistance, its output resistance and its frequency response. Figure 4.5 shows a basic amplifier with an open-loop (non-feedback) voltage gain Gv = A in which feedback is applied such that a fraction β of the output voltage across the load is fed back and subtracted from the input voltage. The circuit may be easily analysed as follows. Amplifier + Vin + Vf - + Vi - Figure 4.5 Amplifier with Negative Feedback Io + Gv =A RL - Vo β Feedback Network The feedback signal Vf is V f = βVo Vin = Vi + V f = Vi + βVo Vo = AVin = A(Vi + βVo ) Vo − AβVo = AVi ⇒ the voltage gain with feedback Gvf = Unit 4: Amplifiers Vo A = Vi 1 − Aβ 4-5 If the magnitude of (1 - Aβ) is greater than 1 then the magnitude of Gvf is less than that of A and the feedback is said to be negative, otherwise the feedback would be positive and the amplifier output could potentially oscillate, ie generate a sinusoid at a particular frequency . The easiest way to ensure the feedback is negative is to use an amplifier which provides a 180°phase shift between input and output, which means that A will be negative and the magnitude of (1 - Aβ) is guaranteed to be greater than unity. Thus in this case the gain with feedback can be written as Gvf = Vo A 1 = = 1 Vi 1 + Aβ A+β If the term 1/A is much smaller than the quantity β, as is generally the case, then the gain with feedback is given approximately by: Gvf ≈ 1 β This is an important result because it indictes that in a feedback amplifier the gain is determined essentially by the feedback network and not the amplifier itself. We will see examples of this in the next section. To conclude this section on negative feedback let’s consider its effect on the frequency response of an amplifier. Figure 4.6 shows the typical response of an ac amplifer without negative feedback. Its gain is Gv = A and its upper and lower frequencies are fH and fL respectively. When feedback is applied the gain falls to a value Gvf which is much smaller than Gv = A, and the characteristic becomes much flatter. It follows that the new lower and upper cut-off frequencies fLf and fHf are further apart and the bandwidth of the feedback amplifier is increased. In fact the product of gain and bandwidth ( the gain-bandwidth product ) in an amplifier is a constant. Thus if b denotes the bandwith of the original amplifier and bf denotes the bandwidth of the feedback amplifier then we can say that Gv × b = Gvf × b f Gain Gain without Feedback Gv = A b 0.707A bandwidth without feedback Figure 4.6 Effect of Negative Feedback on Amplifier Frequency Response 4-6 Gain with Feedback Gvf 0.707Gvf bf bandwidth with feedback f Lf fL f Hf Frequency (Hz) Communications Technology 1 OPERATIONAL AMPLIFIERS There are certain types of amplifier which have the general capability of amplifying the difference between two input signals. When such an amplifier is manufactured on a tiny piece of silicon and supplied as a single integrated circuit (IC) package, it is generally referred to as an operational amplifier or op-amp. The term ‘operational’ derives from the fact that an op-amp can electronically perform such mathematical operations as addition, multiplication and integration. Fortunately, it is not necessary to understand the detailed circuitry inside an op-amp in order to analyse and design circuits which contain op-amp devices. We may therefore treat the op-amp as a ‘black box’, as shown in Figure 4.7. + VS Inverting Input Figure 4.7 Circuit Symbol for an Operational Amplifier Positive Supply v2 - v1 + Non-Inverting Input vo Negative Supply - VS Firstly for an op-amp to work two DC power supplies, ±VS , must be connected to the device. These supplies are typically in the range ±5V to ±15V. To simplify circuit diagrams containing op-amps, the power supply connections are often omitted; but it is important to remember that they must be connected for the amplifier to function. There are two separate input terminals: the non-inverting input terminal (marked +) and the inverting input terminal (marked -). When a voltage V1 is applied to the non-inverting input, an amplified voltage Vo = +AV1 appears at the output terminal, where A is the voltage gain of the amplifier. The + sign indicates that the input and output voltages are in phase. If a voltage V2 is applied to the inverting input, an amplified voltage Vo = -AV2 appears at the output terminal. The - sign indicates that the input and output voltages are 180∞ out of phase (antiphase). If voltages V1 and V2 are applied simultaneously then the output voltage Vo is given by the expression: Vo = +AV1 - AV2 = A(V1 - V2 ) That is, the input signal V2 is subtracted from the input signal V1 and the result of the subtraction is amplified. This signal differencing and amplification function is very useful in many practical applications. For example, when an electrocardiogram (ecg), which is a signal indicatiing the electrical activity of the heart, is acquired by placing a pair of electrodes on the human body (chest), one electrode is the reference electrode and the ecg signal potential at the other electrode is with respect to this reference electrode. The ecg signal is very small, of the order only of a few microvolts, and requires a high level of amplification. Unfortunately both electrodes pick up high levels of other electrical Unit 4: Amplifiers 4-7 activity in the human body which appears as high amplitude-noise signals on both electrodes, thus masking the desired signal on the ecg electrode. This is where an op-amp can prove extremely useful. The reference electrode is fed to the inverting input of the op-amp and the ecg signal electrode is fed to the non-inverting input. Since the noise on both electrodes is virtually the same, the differencing action of the op-amp causes the noise to be cancelled and the small ecg signal is amplified by the high gain of the op-amp. We can draw a ‘black box’ equivalent circuit of an op-amp as shown in Figure 4.8 and summarise its main properties as follows: . A very high voltage gain, called the open-loop gain. This is typically in the range 105 - 106 at DC and low frequencies, but decreases with frequency as shown in Figure 4.9. . A very high input resistance Ri between the inverting and noninverting input terminals, which is typically of the order 1012Ω. . A very low output resistance Ro which is typically 100Ω. + V1 Ro Vd = V1 - V2 Figure 4.8 ‘Black Box’ Representation of an Op-Amp Vd Ri V2 Vo AVd _ Open-Loop Voltage Gain 10 6 10 5 10 4 10 3 10 2 10 Figure 4.9 Open-loop voltage gain as a function of frequency 4-8 1 0 10 10 2 10 3 10 4 10 5 10 6 10 7 10 8 Frequency (Hz) Communications Technology 1 THE IDEAL OP-AMP The above properties of an op-amp - high voltage gain, high input resistance and low output resistance - imply that it is close to being an ideal amplifying device. In fact, the analysis and design of op-amp circuits is much simplified if ideal devices are assumed. The properties of an ideal op-amp are: . Infinite open-loop gain (A = ∞). . Infinite input resistance (Ri = ∞). That is, the op-amp does not draw any current from any source connected to it. . Zero output resistance (Ro = 0 Ω). That is, the output acts as an ideal voltage source. The equivalent circuit of an ideal op-amp is shown in Figure 4.10. v0 = A (v1 - v 2 ) + Figure 4.10 Equivalent circuit of an ideal op-amp INVERTING AMPLIFIER v2 v1 The very high gain of an op-amp (typically 105 - 106) is far too high for many applications and some method of controlling the gain is required. This is achieved using negative feedback as described above, in which resistors are connected around the op-amp in such a way as to reduce the gain. Consider the circuit shown in Figure 4.11. i2 i1 v in R1 P v2 v1 + R2 v0 v0 -R 2 = v in R1 Figure 4.11 Op-Amp inverting amplifier With the resistors R1 and R2 connected, the gain G = Vo /Vin is called the closed-loop gain and will always be less than the open-loop gain, A, which is the gain without any resistors connected. V1 and V2 denote the values of the voltages at the non-inverting and inverting input terminals of the op-amp. Since the non-inverting input terminal is connected to 0V or ‘ground’, V1 = 0. Vo = - AV2 where A is the open-loop gain of the amplifier. If the DC supply voltages ±VS applied to the op-amp are ±15V, and the op-amp gain A = 105, then the maximum AC output voltage swing VO (max) from the amplifier is 30V and the maximum input voltage swing at point P, V2 (max) is given by: Unit 4: Amplifiers 4-9 V2 (max) = Vo (max) 30 = 5 = 300 µV ( ±150 µV ) A 10 This is a very small voltage. In fact, if we assume an ideal op-amp in which A = ∞, then V2 = 0 and the point P is virtually at ground or earth potential. Thus, the point P is often referred to as a virtual earth or virtual ground. The current i1 through R1 is given by: i1 = Vin − V2 Vin = , since V2 = 0 R1 R1 Since the input resistance of the op-amp is so high, a negligibly small current flows into the inverting input terminal and therefore all of the current i which flows through R1 also flows through R2. V o = V 2 - i2 R 2 = -i1R2, since V2 = 0 and i1 = i2 The gain G is given by: G= Vo −i1 R2 R = =− 2 Vin i1 R1 R1 The gain of the circuit is therefore simply the ratio of the two resistors o connected around the op-amp. The minus sign indicates that a 180 phase shift exists between the input voltage Vin and the output voltage, Vo. Hence the term ‘inverting amplifier’. The input resistance Rin of the amplifier is given by: Rin = SAQ 2 Vin Vin = = R1 i1 Vin / R1 An inverting operational amplifier circuit has R2 = 33 kΩ and R1 = 3.3 kΩ. Calculate the voltage gain of the amplifier. SAQ 3 Design an operational amplifier circuit which has a gain of 100 and an input resistance of 1 kΩ. 4-10 Communications Technology 1 NON-INVERTING AMPLIFIER Consider the circuit of Figure 4.12, which is similar to the previous circuit, except in that the input voltage source is now directly connected to the non-inverting input terminal and the input end of the resistor is connected to ground. i R2 P i R1 v in vin v2 v1 + v0 Figure 4.12 Op-amp non-inverting amplifier v0 v0 v in = (1 + R2 ) R1 An expression for the voltage gain of the circuit can again be calculated, by assuming that no current flows into the amplifier input terminals and that the voltage between the inverting and non-inverting input terminals is zero. For zero voltage between the inverting and non-inverting input terminals, the voltage at point P must equal the input voltage Vin. i= Vin R1 Vo = Vin + iR2 = Vin + ∴ G= Vin R R2 = Vin 1 + 2 R1 R1 Vo R = 1+ 2 Vin R1 Because the gain is positive, the output voltage is in phase with the input voltage and hence the name ‘non-inverting amplifier’. In this case, the input resistance Rin of the amplifier is equal to the very high input resistance of the op-amp, between the non-inverting and inverting terminals. SAQ 4 A non-inverting operational amplifier circuit has R1 = 1 kΩ and R2 = 10 kΩ. Calculate the voltage gain of the amplifier. SAQ 5 Design an operational amplifier circuit with a gain of +50, the smallest resistor used being 1 kΩ. VOLTAGE FOLLOWER (BUFFER) AMPLIFIER Consider the simple circuit of Figure 4.13. Again assuming an ideal opamp, with zero voltage difference between the inverting and noninverting input terminals, we can say that Vo = Vin or Unit 4: Amplifiers Vo =1 Vin 4-11 Figure 4.13 Voltage Follower or Buffer Amplifier v0 v in + Thus this circuit provides a gain of +1 and the output voltage simply ‘follows’ the input voltage - hence the term ‘voltage follower’. The input resistance of the circuit is the high differential input of the operational amplifier. This property of the circuit, together with the low output resistance of the op-amp, makes it useful for converting a high impedance source to a low - impedance one. When it is used in this way it is often referred to as a buffer amplifier. SUMMING AMPLIFIER It is often required to have a circuit which can add two or more voltages together. Consider the circuit of Figure 4.14 which is capable of adding two voltage signals V1 and V2 and amplifying the result. Assuming an ideal op-amp. i1 = v V1 v , i2 = 2 and i = o R R RF i1 i R P i2 v1 R RF + v0 v2 Figure 4.14 Summing Amplifier Circuit From Kirchoff’s Laws: i + i1 + i2 = 0, or i = −(i1 + i2 ) ∴ Vo V V = − 1 + 2 R R RF ⇒ Vo = − RF (V1 + V2 ) R The output signal, V0, is the sum of the input signal V1 and V2 amplified by the factor - RF/R. More than two signals can be summed by having additional input resistors, R. It is also possible to multiply the input signals by different gain factors by having unequal input resistors R and hence different gain factors RF/R. 4-12 Communications Technology 1 SAQ 6 In the circuit of Figure 4.14, R = 10 kΩ, RF = 10 kΩ. Derive an expression for the output voltage for the following input voltages. (a) V1 = 2 sin (400πt), V2 = 3 sin (400πt) (b) V1 = + 3V (dc), V2 = 4 sin (200πt) SAQ 7 OP-AMP VOLTAGE COMPARATOR Design an op-amp summing amplifier which adds three input signals V1, V2 and V3 such that V0 = -(V1+2V2+4V3 ). The smallest resistor to be used is 1 kΩ. Consider an operational amplifier connected in the open-loop configuration with input signals V1 and V2 connected to the inverting and non-inverting inputs as shown in Figure 4.15. From our previous study we know that the output voltage in this configuration is given by V0 = A(V1 - V2) where A is the open-loop gain of the amplifier which normally has a value in the range 105 - 106. The voltage difference between the inputs V1 and V2 is amplified and appears at the output. Since the value of A is so large a voltage difference of only a few hundred microvolts between V1 and V2 causes the amplifier output to saturate, i.e. have a value close to the supply voltage - either + VS or - VS. When V1 is slightly greater than V2, then V0 ≅ + VS and when V1 is slightly less than V2, V0 ≅ - VS. Thus the circuit can be used as a comparator to determine whether V1 is greater than or less than V2. Suppose in a voltage comparator V1 is a triangular waveform and V2 is a dc voltage as shown in Figure 4.16(a). The output voltage waveform is as shown in Figure 4.16(b). When the instantaneous value of the triangular signal, V1, is less than the d.c signal, V2, the output signal has value close to - VS and when V1 is greater than V2, the output signal has a value close to + VS. Thus V0 repeatedly switches between approximately -VS and +VS producing a pulse waveform. The mark - to - space ratio, ie the ratio of the high-level voltage to low-level voltage in each period, of the output waveform depends on the switching threshold which is determined by the value of the d.c voltage V2 . +Vs + Figure 4.15 Op-Amp Voltage Comparator Circuit Unit 4: Amplifiers v2 v1 v0 -Vs 4-13 v1 vin v2 +v s Figure 4.16 Comparator Input and Output Signals t (b) Comparator Output Signals v0 -vs SAQ 8 t (a) Comparator Input Signals In a simple voltage comparator circuit the power supply voltages are ± 15V. Input signal V1 is a triangular voltage waveform with an amplitude of 10V and a period of 4 seconds. Calculate the mark-to-space ratio when V1 is a dc voltage of: (i) 0V (ii) -5V (iii) +5V (iv) +12V A disadvantage of the simple comparator configurations above is that their behaviour is erratic in the presence of noise. If the input voltage has superimposed noise, and has an amplitude close to Vref the output may be forced to change states several times. This erratic behavior may be suppressed by introducing hysteresis into the comparator. Hysteresis is a phenomenon in which the transition point is different when switching from the high-to-low state as compared with switching from the low-tohigh state. Comparators with hysteresis are often referred to as Schmitt triggers. Figure 4.17 indicates the circuitry of an inverting Schmitt Trigger. Vo V+ Sat Vi _ Vo R1 Figure 4.17 (a) Circuitry of a Schmitt Trigger, (b) Output characteristic. 4-14 -Vt Vt Vi _ R2 (a) V Sat (b) Communications Technology 1 Resistors R1 and R2 form a voltage divider, establishing a voltage at the noninverting terminal Vt, proportional to Vo. Vt = R2 Vsat R1 + R2 when, Vo=+Vsat, the noninverting input = Vt Vo=-Vsat, the noninverting input = -Vt When Vi is negative, Vo = +Vsat and V+ = Vt. In order for the output to change signs, Vi must exceed the value of Vt. To return to the previous state the input voltage must reach a value of Vi < -Vt, causing the output to return to +Vsat. By choosing a sufficiently large value of Vt the effects of noise at the transition points may be minimised. Care however must be taken to ensure that Vt does not become too large, in which case the accuracy may be degraded for certain applications. The rectangle on the input-output curve is referred to as a hysteresis loop. Note, it is important to label the curve with arrows to indicate directions of change. It is also possible to develop a non-inverting Schmitt trigger, which is based on the same principles as above; however this circuit will have a slightly different hysteresis loop. SAQ 9 Design an inverting Schmitt Trigger having thresholds close to +/- 50mV based on supply voltages to the op-amp of +/15V. The exact value of Vt is not critical in this application. Assume op-amp saturation voltages of +/- 14V. 1. An amplifier is an electronic device for increasing the amplitude of a signal, and may be characterised by its voltage gain, its input resistance, its output resistance, and its frequency response. 2. Negative feedback is used to alter the gain of an amplifier but also affects its input resistance, its output resistance and its frequency response. 3. An operational amplifier is a high gain, BJT, FET or BJT/FET amplifier with a high input resistance and low output resistance. It is normally supplied as a single integrated (IC) package. 4. An ideal operational amplifier has infinite gain, infinite input resistance and zero output resistance. 5. The voltage gain G of the inverting amplifier circuit of Figure 4.11 is given by the expression: SUMMARY G=− R2 R1 The input resistance Rin = R1 . The input and output voltages are out o of phase by 180 . Unit 4: Amplifiers 4-15 6. The voltage gain G of the non-inverting amplifier circuit of Figure 4.12 is given by the expression: G = 1+ R2 R1 The input resistance Rin = ∞ . The input and output voltages are in phase. 4-16 7. A voltage follower or buffer amplifier has a high input impedance, low output impedance and a gain of +1. 8. A summing amplifier circuit may be used to add two or more voltages together - in different ratios if required. 9. A voltage comparator indicates whether the instantaneous value of one input signal is greater than or less than that of another. Communications Technology 1 ANSWERS TO SAQS SAQ 1 dB Gvo = 20 log10 Gv = Vo = 60 ⇒ Gvo = anti log10 3 = 1000 Vi Vo Ri 90 RL = .Gvo . = × 1000 × Vs Rs + Ri Ro + RL 10 + 90 450 = 0.9 × 1000 × 0.9 = 810 50 + 450 Vo = Gv Vs = 810 × 1mV = 0.81V (rms) SAQ 2 Voltage Gain G = − R2 33 × 10 3 =− = −10 R1 3.3 × 10 3 SAQ 3 Input Resistance Rin = R1 Rin = 1 kΩ => R1 = 1 kΩ Voltage Gain G = − R2 = −100 R1 or R2 = 100 x R1 = 100 x 1 kΩ = 100 kΩ. Therefore, the circuit is as shown in Figure 4.18. 100k Ω v in 1k Ω v0 + Figure 4.18 OP-AMP Circuit for SAQ 3 Unit 4: Amplifiers 4-17 SAQ 4 R 10 Voltage Gain G = 1 + 2 = 1 + = 11 R 1 1 SAQ 5 R Voltage Gain G = 1 + 2 = 50 R1 ⇒ R2 = 49 or R2 = 49 R1 R1 Let R1 = 1 kΩ R2 = 49 x R1 = 49 x 1 kΩ = 49 kΩ. The circuit is shown in Figure 4.19. 49k Ω 1k Ω Figure 4.19 OP-AMP Circuit for SAQ 5 v0 v in + SAQ 6 (a) Vo = − RF 10 (V1 + V2 ) = − 10 (2 sin 400πt + 3 sin 400πt ) = −5 sin 400πt R (b) Vo = −1(3 + 4 sin 200ωt ) = −3 − 4 sin 200ωt 4-18 Communications Technology 1 SAQ 7 The basic summing circuit is shown in Figure 4.20. Let R3 = 1 kΩ => RF = 4 kΩ. ∴ R2 = 2 kΩ and R1 = 4 kΩ. v1 R1 RF v2 R2 + v3 R3 v0 Figure 4.20 SAQ 8 Consider the sketch of Figure 4.21. The mark -to-space ratios are as follows: (i) 1 : 1 (ii) T 3T : = 1: 3 4 4 (iii) 3T T : = 3 :1 4 4 (iv) Output voltage always low (≅ -15V) Voltage (volts) V2 +15 +12V +10 V1 +5 +5V 0 -5 Figure 4.21 0V T/4 T/2 3T/4 T t -5V -10 -15 Unit 4: Amplifiers 4-19 SAQ 9 Vt = R2 Vsat R1 + R2 50 × 10 −3 = R2 .14 R1 + R2 0.05( R1 + R2 ) = 14 R2 ⇒ 0.05 R1 = 13.95 R2 ⇒ R1 = 279 R2 Let R2 = 100Ω and R1 = 27.9kΩ (28kΩ) 4-20 Communications Technology 1