CHAPTER 01: Signals and Amplifiers Contents 1–1 Signals 1–2 Frequency Spectrum of Signals 1–3 Analog and Digital Signals 1–4 Amplifiers 1–5 Circuit Models for Amplifiers 1–6 Frequency Response of Amplifiers 2 Introduction The heart of the course is the study of the three basic semiconductor devices: the diode (Chapter 3); the bipolar transistor (Chapter 4); and the MOS transistor (Chapter 5). In each case, we study the device operation, its characterization, and its basic circuit applications. Chapter 2 provides an overview of semiconductor concepts at a level sufficient for the study of electronic circuits. Since the purpose of electronic circuits is the processing of signals, an understanding is essential of signals, their characterization in the time and frequency domains, and their analog and digital representations. This is provided in Chapter 1, which also introduces the most common signal-processing function, amplification, and the characterization and types of amplifiers. Besides diodes and transistors, the basic electronic devices, the op amp is also studied. This circuit element can be used in the design of powerful circuits, as we do in Chapter 6, without any knowledge of its internal construction. 3 IN THIS CHAPTER YOU WILL LEARN 1. That electronic circuits process signals, and thus understanding electrical signals is essential to appreciating the material in this course. 2. The Thévenin and Norton representations of signal sources. 3. The representation of a signal as the sum of sine waves. 4. The analog and digital representations of a signal. 5. The most basic and pervasive signal-processing function: signal amplification, and correspondingly, the signal amplifier. 6. How amplifiers are characterized (modeled) as circuit building blocks independent of their internal circuitry. 7. How the frequency response of an amplifier is measured, and how it is calculated, especially in the simple but common case of a single-time-constant (STC) type response. 4 1.1. Signals Signals contain information about a variety of things and activities in our physical world. Examples abound: Information about the weather is contained in signals that represent the air temperature, pressure, wind speed, etc. The voice of a radio announcer reading the news into a microphone provides an acoustic signal that contains information about world affairs. To monitor the status of a nuclear reactor, instruments are used to measure a multitude of relevant parameters, each instrument producing a signal. 5 To extract required information from a set of signals, the signal must first be converted into an electrical signal, that is, a voltage or a current. This process is accomplished by devices known as transducers. For instance, the sound waves generated by a human can be converted into electrical signals by using a microphone, which is in effect a pressure transducer. It is not our purpose here to study transducers; rather, we shall assume that the signals of interest already exist in the electrical domain and represent them by one of the two equivalent forms shown in Figure 1.1. 6 7 FIGURE 1.1 Two alternative representations of a signal source: (a) the Thévenin form; (b) the Norton form. For the two representations in this Figure to be equivalent, their parameters are related by: Example 1.1 The output resistance of a signal source, although inevitable, is an imperfection that limits the ability of the source to deliver its full signal strength to a load. To see this point more clearly, consider the signal source when connected to a load resistance RL as shown in Figure 1.2. For the case in which the source is represented by its Thévenin equivalent form, find the voltage vo that appears across RL, and hence the condition that Rs must satisfy for vo to be close to the value of vs. Repeat for the Norton-represented source; in this case finding the current io that flows through RL and hence the condition that Rs must satisfy for io to be close to the value of is. 8 FIGURE 1.2: Circuit for Example 1.1 9 Solution For the Thévenin-represented signal source shown in Figure 1.2(a): Solution (continued) From this equation we see that for: the source resistance Rs must be much lower than the load resistance RL: Thus, for a source represented by its Thévenin equivalent, ideally Rs = 0, and as Rs is increased, relative to the load resistance RL with which this source is intended to operate, the voltage vo that appears across the load becomes smaller, not a desirable outcome. 10 Solution (continued) Next, we consider the Norton-represented signal source in Figure 11 1.2(b). To obtain the current io that flows through the load resistance RL, we utilize the ratio of the current divider formed by Rs and RL : From this relationship we see that for the source resistance Rs must be much larger than the load resistance RL: Solution (continued) Thus for a signal source represented by its Norton equivalent, ideally Rs = ∞, and as Rs is reduced, relative to the load resistance RL with which this source is intended to operate, the current io that flows through the load becomes smaller, not a desirable outcome. Finally, we note that although circuit designers cannot usually do much about the value of Rs, they may have to devise a circuit solution that minimizes or eliminates the loss of signal strength that results when the source is connected to the load. 12 1.2. Frequency Spectrum of Signals An extremely useful characterization of a signal, and for that matter of any arbitrary function of time, is in terms of its frequency spectrum. Such a description of signals is obtained through the mathematical tools of Fourier series and Fourier transform. They provide the means for representing a voltage signal 𝒗𝒔 (𝒕) or a current signal 𝒊𝒔 (𝒕) as the sum of sine-wave signals of different frequencies and amplitudes. 13 Brief review of sinusoid’s properties Figure 1.3 shows a sine-wave voltage signal 𝒗𝒂 (𝒕), 𝑽𝒂 denotes the peak value or amplitude in volts (V) denotes the angular frequency in radians per second; that is, (rad/s) 𝝎 𝒇 is the frequency in hertz, 𝑻 is the period in seconds (s) (Hz) 14 FIGURE 1.3 The sine-wave signal is completely characterized by its peak value 𝑽𝒂 , its frequency 𝝎, and its phase with respect to an arbitrary reference time. In the case depicted in Fig. 1.3, the time origin has been chosen so that the phase angle is 0. 15 It is common to express the amplitude of a sine-wave signal in terms of its root-meansquare (rms) value, which is equal to the peak value divided by Thus the rms value of the sinusoid 𝒗𝒂 (𝒕) of Figure 1.3 is: For instance, when we speak of the wall power supply in our homes as being 120 V, we mean that it has a sine waveform of volts peak value. 16 Representation of signals as the sum of sinusoids The Fourier series allows us to express a given periodic function of time as the sum of an infinite number of sinusoids whose frequencies are harmonically related. For instance, the symmetrical Figure 1.4 can be expressed as. FIGURE 1.4 A symmetrical square-wave of amplitude V. square-wave signal in 17 The sinusoidal components in the series of Equation constitute the frequency spectrum of the square-wave signal. Such a spectrum can be graphically represented as in Figure 1.5. FIGURE 1.5 The frequency spectrum of the periodic square wave of Figure 1.4 𝑽 is the amplitude, and 𝝎𝟎 = 𝟐𝝅Τ𝑻 (T is the period of the square wave) is called the fundamental frequency. Note that in counting harmonics, the fundamental at 𝝎𝟎 is the first, the one at 𝟐𝝎𝟎 is the second, etc. 18 Representation of signals as the sum of sinusoids The Fourier transform can be applied to a non-periodic function of time, such as that depicted in Fig.1.5, and provides its frequency spectrum as a continuous function of frequency, as indicated in Fig. 1.6. 19 FIGURE 1.5 FIGURE 1.6 An arbitrary voltage signal 𝒗𝒔 (𝒕). The frequency spectrum of an arbitrary wave form such as that in Fig. 1.5. Unlike the case of periodic signals, where the spectrum consists of discrete frequencies (at 𝝎𝟎 and its harmonics), the spectrum of a non-periodic signal contains in general all possible frequencies. For instance, the spectrum of audible sounds such as speech and music extends from about 20 Hz to about 20 kHz—a frequency range known as the audio band. As another example, analog video signals have their spectra in the range of 0 MHz to 4.5 MHz. 20 1.3. Analog and Digital Signals A signal can be represented in his analog form or digital form. Figure 1.7 shows how signals can be converted from analog to digital form. FIGURE 1.7 Sampling the continuous-time analog signal in (a) results in the discrete-time signal in (b). 21 Now if we represent the magnitude of each of the signal samples in Fig. 1.7(b) by a number having a finite number of digits, then the signal amplitude will no longer be continuous; rather, it is said to be quantized, discretized, or digitized. Figure 1.8 shows the time variation of such a digital signal. the digital signals in binary systems need have only two voltage levels, which can be labeled low and high. 22 FIGURE 1.8 23 Variation of a particular binary digital signal with time. If we use N binary digits (bits) to represent each sample of the analog signal, then the digitized sample value can be expressed as LSB MSB Conventionally, this binary number is written as such a representation quantizes the analog sample into one of 2N levels. Figure 1.9 shows the block diagram form of the analog-to-digital converter (A/D or ADC). The dual circuit of the ADC is the digital-to-analog converter (D/A or DAC). It converts an N-bit digital input to an analog output voltage. 24 1.4. Amplifiers In this section, we shall introduce the most fundamental signalprocessing function, one that is employed in some form in almost every electronic system, namely, signal amplification. Signal Amplification The need for amplification arises because transducers provide signals that are said to be “weak,” that is, in the microvolt (μV) or millivolt (mV) range and possessing little energy. An amplifier that preserves the details of the signal waveform is characterized by the relationship: where 𝒗𝒊 and 𝒗𝒐 are the input and output signals, respectively, and A is a constant representing the magnitude of amplification, known as amplifier gain. 25 Amplifier Circuit Symbol FIGURE 1.10 (a) Circuit symbol for amplifier. (b) An amplifier with a common terminal (ground) between the input and output ports. 26 Voltage Gain Figure 1.11(a) shows a linear amplifier which accepts an input signal 𝒗𝑰 (𝒕) and provides at the output, across a load resistance 𝑹𝑳 , an output signal 𝒗𝑶 that is a magnified replica of 𝒗𝑰 (𝒕). FIGURE 1.11 (a) A voltage amplifier fed with a signal 𝒗𝑰 (𝒕) and connected to a load resistance 𝑹𝑳 . (b) Transfer characteristic of a linear voltage amplifier with voltage gain 𝑨𝒗 . 27 Power Gain and Current Gain The power gain of the amplifier in Figure 1.11(a) is defined as: where 𝒊𝑶 is the current that the amplifier delivers to the load (𝑹𝑳 ), 𝒊𝑶 = 𝒗𝑶 Τ𝑹𝑳 , and 𝒊𝑰 is the current the amplifier draws from the signal source. The current gain of the amplifier is defined as: From Equations of voltage gain and current gain, we note that: 28 Expressing Gain in Decibels Electronics engineers logarithmic measure. express amplifier 29 gain with a The Amplifier Power Supplies Amplifiers need dc power supplies for their operation. Figure 1.12(a) shows an amplifier that requires two dc sources: one positive of value VCC and one negative of value VEE. FIGURE 1.12 (a)An amplifier that requires two dc supplies (shown as batteries) for operation. (b) Simplified circuit. 30 The Amplifier Power Supplies Now, if the current drawn from the positive supply is denoted ICC and that from the negative supply is IEE (see Fig. 1.12a), then the dc power delivered to the amplifier is: If the power dissipated in the amplifier circuit is denoted Pdissipated, the power-balance equation for the amplifier can be written as where PI is the power drawn from the signal source and PL is the power delivered to the load. Since the power drawn from the signal source is usually small, the amplifier power efficiency is defined as 31 Example 1.2 Consider an amplifier operating from ±10-V power supplies. It is fed with a sinusoidal voltage having 1 V peak and delivers a sinusoidal voltage output of 9 V peak to a 1-kΩ load. The amplifier draws a current of 9.5 mA from each of its two power supplies. The input current of the amplifier is found to be sinusoidal with 0.1 mA peak. Find the voltage gain, the current gain, the power gain, the power drawn from the dc supplies, the power dissipated in the amplifier, and the amplifier efficiency. Solution Voltage gain 32 Current gain Power gain 33 Power drawn from the dc supplies Power dissipated in the amplifier Amplifier efficiency 34 Amplifier Saturation Practically speaking, the amplifier transfer characteristic remains linear over only a limited range of input and output voltages. For an amplifier operated from two power supplies the output voltage cannot exceed a specified positive limit and cannot decrease below a specified negative limit. The resulting transfer characteristic is shown in Figure 1.13, with the positive and negative saturation levels denoted 𝑳+ and 𝑳− , respectively. Each of the two saturation levels is usually within a fraction of a volt of the voltage of the corresponding power supply. Obviously, in order to avoid distorting the output signal waveform, the input signal swing must be kept within the linear range of operation, 35 36 FIGURE 1.13: An amplifier transfer characteristic that is linear except for output saturation. Symbol Convention Terminology: Figure 1.14 shows the waveform of a current 𝒊𝑪 (𝒕) that is flowing through a branch in a particular circuit. At a time 𝒕, the total instantaneous current 𝒊𝑪 (𝒕) is the sum of the dc current 𝑰𝑪 and the signal current 𝒊𝒄 (𝒕), where the signal current is given by 37 38 FIGURE 1.14: Symbol convention employed throughout the course. Some conventions Total instantaneous quantities are denoted by a lowercase symbol with uppercase subscript(s), for example, Direct-current (dc) quantities are denoted by an uppercase symbol with uppercase subscript(s), for example, 39 Incremental signal quantities are denoted by a lowercase symbol with lowercase subscript(s), for example, If the signal is a sine wave, then its amplitude is denoted by an uppercase symbol with lowercase subscript(s), for example, 40 Finally, although not shown in Figure 1.14, dc power supplies are denoted by an uppercase letter with a double-letter uppercase subscript, for example, A similar notation is used for the dc current drawn from the power supply, for example, 41 1.5. Circuit Models for Amplifiers In this section, we study simple but effective amplifier models. These models apply irrespective of the complexity of the internal circuit of the amplifier. The values of the model parameters can be found either by analyzing the amplifier circuit or by performing measurements at the amplifier. 42 Voltage Amplifiers Figure 1.15(a) shows a circuit model for the voltage amplifier. That is a voltage-controlled voltage source having a gain factor 𝑨𝒗𝒐 . FIGURE 1.15: (a) Circuit model for the voltage amplifier. 43 To be specific, we show in Figure 1.15(b) the amplifier model fed with a signal voltage source 𝒗𝒔 having a resistance 𝑹𝒔 and connected at the output to a load resistance 𝑹𝑳 . FIGURE 1.15: (b) The voltage amplifier with input signal source and load. 44 Using the voltage-divider rule we obtain: Thus the voltage gain is given by: It follows that in order not to lose gain in coupling the amplifier output to a load, the output resistance 𝑹𝒐 should be much smaller than the load resistance 𝑹𝑳 . 45 Ideal voltage amplifier An ideal voltage amplifier is one with : Also, the following equation Indicates that for Thus 𝑨𝒗𝒐 is the voltage gain of the unloaded amplifier, or the open-circuit voltage gain. 46 Input resistance The finite input resistance 𝑹𝒊 introduces another voltage-divider action at the input, with the result that only a fraction of the source signal 𝒗𝒔 actually reaches the input terminals of the amplifier; that is, in order not to lose a significant portion of the input signal, the amplifier must be designed to have: An ideal voltage amplifier is one with: 47 Overall voltage gain The overall voltage gain 𝒗𝒐 Τ𝒗𝒔 can be found by combining two of the previous Equations : 48 Cascaded Amplifiers To meet given amplifier specifications, we often need to design the amplifier as a cascade of two or more stages. To illustrate the analysis and design of cascaded amplifiers, we consider a practical example. 49 Example 1.3 Figure 1.16 depicts an amplifier composed of a cascade of three stages. The amplifier is fed by a signal source with a source resistance of 100 kΩ and delivers its output into a load resistance of 100 Ω. The first stage has a relatively high input resistance and a modest gain factor of 10. The second stage has a higher gain factor but lower input resistance. Finally, the last, or output, stage has unity gain but a low output resistance. We wish to evaluate the overall voltage gain, that is, 𝒗𝑳 Τ𝒗𝒔 , the current gain, and the power gain. 50 FIGURE 1.16: Three-stage amplifier for Example 1.3. 51 Solution The fraction of source signal appearing at the input terminals of the amplifier is obtained using the voltage-divider rule at the input, as follows: The voltage gain of the first stage is obtained by considering the input resistance of the second stage to be the load of the first stage; that is : Similarly, the voltage gain of the second stage is obtained by considering the input resistance of the third stage to be the load of the second stage : 52 Finally, the voltage gain of the output stage is as follows : The total gain of the three stages in cascade can be now found from : 53 To find the voltage gain from source to load, we multiply 𝑨𝒗 by the factor representing the loss of gain at the input; that is: 54 The current gain is found as follows : The power gain is found from: Note that : 55 Other Amplifier Types In the design of an electronic system, the signal of interest—whether at the system input, at an intermediate stage, or at the output—can be either a voltage or a current. For instance, some transducers have very high output resistances and can be more appropriately modeled as current sources. Similarly, there are applications in which the output current rather than the voltage is of interest. Thus, although it is the most popular, the voltage amplifier considered above is just one of four possible amplifier types. The other three are the current amplifier, the transconductance amplifier, and the transresistance amplifier. Table 1.1 shows the four amplifier types, their circuit models, the definition of their gain parameters, and the ideal values of their input and output resistances. 56 57 Relationships between the Four Amplifier Models Although for a given amplifier a particular one of the four models in Table 1.1 is most preferable, any of the four can be used to model any amplifier. For instance, the open-circuit voltage gain 𝑨𝒗𝒐 can be related to the short-circuit current gain 𝑨𝒊𝒔 as follows: The open-circuit output voltage given by the voltage amplifier model of Table 1.1 is : 𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊 The current amplifier model in the same table gives an open-circuit output voltage of : 𝒗𝒐 = 𝑨𝒊𝒔 𝒊𝒊 𝑹𝒐 noting that 𝒊𝒊 = 𝒗𝒊 Τ𝑹𝒊 , gives : 58 Similarly, And The we can show that : : expressions in the previous Equations can be used to relate any two of the gain parameters 𝑨𝒗𝒐 , 𝑨𝒊𝒔 , 𝑮𝒎 , and 𝑹𝒎 . 59 Determining Ri and Ro From the amplifier circuit models given in Table 1.1, we observe that : The input resistance 𝑹𝒊 of the amplifier can be determined by applying an input voltage 𝒗𝒊 and measuring (or calculating) the input current 𝒊𝒊 ; that is: The output resistance is found as the ratio of the open-circuit output voltage to the short-circuit output current. 60 Alternatively, the output resistance can be found by eliminating the input signal source (then 𝒊𝒊 and 𝒗𝒊 will both be zero) and applying a voltage signal 𝒗𝒙 to the output of the amplifier, as shown in Figure 1.17. FIGURE 1.17: Determining the output resistance. 61 Example 1.4 The bipolar junction transistor (BJT), which will be studied in Chapter 3, is a three-terminal device that when powered-up by a dc source (battery) and operated with small signals can be modeled by the linear circuit shown in Figure 1.18(a). The three terminals are the base (B), the emitter (E), and the collector (C). The heart of the model is a transconductance amplifier represented by an input resistance between B and E (denoted 𝒓𝝅 ), a short-circuit transconductance 𝒈𝒎 , and an output resistance 𝒓𝒐 . 62 63 FIGURE 1.18: (a) Small-signal circuit model for a bipolar junction transistor (BJT). (b) The BJT connected as an amplifier with the emitter as a common terminal between input and output (called a common-emitter amplifier). (c) An alternative small-signal circuit model for the BJT. Example 1.4 (continued) (a) Figure 1.18(b) shows a transistor amplifier known as a common-emitter or grounded-emitter circuit. Derive an expression for the voltage gain 𝒗𝒐 Τ𝒗𝒔 , and evaluate its magnitude for the case Rs = 5 kΩ, rπ = 2.5 kΩ, gm = 40 mA/V, ro = 100 kΩ, and RL = 5 kΩ. What would the gain value be if the effect of ro were neglected ? (b) An alternative model for the transistor in which a current amplifier rather than a transconductance amplifier is utilized is shown in Figure 1.18(c). What must the short-circuit current gain β be? Give both an expression and a value. 64 Solution (a) Refer to Figure 1.18(b). We use the voltage-divider rule to determine the fraction of input signal that appears at the amplifier input as : 65 66 Next we determine the output voltage 𝒗𝒐 by multiplying the current 𝒈𝒎 𝒗𝒃𝒆 by the resistance 𝑹𝑳 ∕∕ 𝒓𝒐 : Substituting for 𝒗𝒃𝒆 from previous Equation yields the voltage-gain expression : Observe that the gain is negative, indicating that this amplifier is inverting. For the given component values : 67 Neglecting the effect of 𝒓𝒐 , we obtain: which is quite close to the value obtained including 𝒓𝒐 . This is not surprising, since: 68 (b) For the model in Figure 1.18(c) to be equivalent to that in Figure 1.18(a), For the values given: 69 1.6 Frequency Response of Amplifiers From Section 1.2 we know that the input signal to an amplifier can always be expressed as the sum of sinusoidal signals. It follows that an important characterization of an amplifier is in terms of its response to input sinusoids of different frequencies. Such a characterization performance is known as frequency response. of the amplifier amplifier 70 Measuring the Amplifier Frequency Response Figure 1.19 depicts a linear voltage amplifier fed at its input with a sine-wave signal of amplitude 𝑽𝒊 and frequency 𝝎. As the figure indicates, the signal measured at the amplifier output also is sinusoidal with exactly the same frequency, 𝝎. 71 Measuring the Amplifier Frequency Response Figure 1.19: Measuring the frequency response of a linear amplifier: At the test frequency 𝝎, the amplifier gain is characterized by its magnitude 𝑽𝒐 Τ𝑽𝒊 and phase ∅. 72 If we denote the amplifier transmission, or transfer function as it is more commonly known, by 𝑻(𝝎) , then : These two plots together constitute the frequency response of the amplifier. It is a common practice to express the magnitude of transmission in decibels and thus plot 20 log |T(ω)| versus frequency. 73 Amplifier Bandwidth Figure 1.20 shows the magnitude response of an amplifier. The gain is almost constant over a wide frequency range, roughly between 𝝎𝟏 and 𝝎𝟐 . The band of frequencies over which the gain of the amplifier is almost constant, to within a certain number of decibels (usually 3 dB), is called the amplifier bandwidth. 74 75 Figure 1.20: Typical magnitude response of an amplifier: 𝑻(𝝎) is the magnitude of the amplifier transfer function—that is, the ratio of the output 𝑽𝒐 (𝝎) to the input 𝑽𝒊 (𝝎). Evaluating the Frequency Response of Amplifiers Above, we described the method used to measure the frequency response of an amplifier. We now briefly discuss the method for analytically obtaining an expression for the frequency response. Here is just a preview of this important subject, whose detailed study is in Chapter 9. To evaluate the frequency response of an amplifier, one has to analyze the amplifier equivalent circuit model, taking into account all reactive components. An inductance 𝑳 has a reactance or impedance 𝒋𝝎𝑳, and a capacitance 𝑪 has a reactance or impedance 𝟏Τ𝒋𝝎𝑪 or, equivalently, a susceptance or admittance 𝒋𝝎𝑪. 76 Thus in a frequency-domain analysis we deal with impedances and/or admittances. The result of the analysis is the amplifier transfer function, 𝑻(𝝎) : 𝑻(𝝎) is generally a complex function. to determine frequency response, the algebraic manipulations can be considerably simplified by using the complex frequency variable, s : Subsequently, we replace 𝒔 by 𝒋𝝎 to determine the transfer function for physical frequencies, 𝑻(𝒋𝝎) . 77 Single-Time-Constant Networks An STC network is one that is composed of, or can be reduced to, one reactive component (inductance or capacitance) and one resistance. Examples are shown in Figure 1.21. An STC network formed of an inductance, L and a resistance, R has a time constant : The time constant of an STC network composed of a capacitance, C and a resistance, R is given by : 78 79 FIGURE 1.21 Two examples of STC networks: (a) a low-pass network and (b) a high-pass network. Evaluating the Time Constant The first step in the analysis of an STC circuit is to evaluate its time constant 𝝉. Example E.1 : Reduce the circuit in Figure E.1(a) to an STC circuit, and find its time constant. FIGURE E.1 Circuit to be reduced. 80 Solution : The reduction process is illustrated in Figure E.1 and consists of repeated applications of Thévenin’s theorem. 81 From the final circuit (Figure E.1c), we obtain the time constant as : 82 Rapid Evaluation of 𝝉 In many instances, it will be important to be able to evaluate rapidly the time constant, 𝝉 of a given STC circuit. A simple method for accomplishing this goal consists first of reducing the excitation to zero: if the excitation is by a voltage source, short it, if by a current source, open it. 83 Then, if the circuit has one reactive component and a number of resistances, “grab hold” of the two terminals of the reactive component (capacitance or inductance) and find the equivalent resistance 𝑹𝒆𝒒 seen by the component. The time constant is then either : 84 As an example, in the circuit of Fig. E.1(a), we find that the capacitor C “sees” a resistance R4 in parallel with the series combination of R3 and R2 in parallel with R1. Thus : 85 In some cases it may be found that the circuit has one resistance and a number of capacitances or inductances. In such a case, the procedure should be inverted; that is, “grab hold” of the resistance terminals and find the equivalent capacitance 𝑪𝒆𝒒 , or equivalent inductance 𝑳𝒆𝒒 , seen by this resistance. The time constant is then found as : 86 Example E.2 Find the time constant of the circuit in Figure E.2. Solution After reducing the excitation to zero by shortcircuiting the voltage source, we see that the resistance R “sees” an equivalent capacitance 𝑪𝟏 + 𝑪𝟐 . Thus, the time constant τ is given by : 87 Finally, in some cases an STC circuit has more than one resistance and more than one capacitance (or more than one inductance). Such cases require some initial work to simplify the circuit, as illustrated by Example E.3. Example E.3 : Find the time constant of the circuit in Figure E.3 (a). 88 Solution The analysis steps are illustrated in Fig. E.3. In Fig. E.3(b) we show the circuit excited by two separate but equal voltage sources. There is an equivalence of the circuits in Fig. E.3(a) and E.3(b). 89 Application of Thévenin’s theorem to the circuit to the left of the line XX’ and then to the circuit to the right of that line results in the circuit of Figure E.3(c). 90 Using the principle of superposition, the output voltage, 𝒗𝑶 = 𝒗𝑶𝟏 + 𝒗𝑶𝟐 . The first component, 𝒗𝑶𝟏 , is the output due to the lefthand-side voltage source with the other voltage source reduced to zero. The circuit for calculating 𝒗𝑶𝟏 is shown in Figure E.3(d). It is an STC circuit with a time constant : 91 Similarly, the second component 𝒗𝑶𝟐 is the output obtained with the left-hand-side voltage source 92 reduced to zero. It can be calculated from the circuit of Figure E.3(e), Which is an STC circuit with The same time constant : Finally, it should be observed that the fact that the circuit is an STC one can also be ascertained by setting the independent source 𝒗𝑰 in Fig. E.3(a) to zero. Also, the time constant is then immediately obvious. Classification of STC Circuits STC circuits can be classified into two categories, low-pass (LP) and high-pass (HP) types, with each category displaying distinctly different signal responses. The simplest way of finding whether an STC circuit is of LP or HP type may be accomplished by using the frequency domain response. In Table E.1, which provides a summary of these results, s.c. stands for short circuit and o.c. for open circuit. 93 94 Figure E.4 shows examples of low-pass STC circuits, and Figure E.5 shows examples of high-pass STC circuits. Note that a given circuit can be of either category, depending on the input and output variables. Figure E.4 : STC circuits of the low-pass type. 95 Figure E.5 : STC circuits of the high-pass type. 96 EXERCISES 97 EXERCISES 98 Frequency Response of STC Circuits Low-Pass Circuits The transfer function, T(s) of an STC low-pass circuit can always be written in the form : which for physical frequencies, 𝒔 = 𝒋𝝎 becomes : Where K is the magnitude of the transfer function at 𝝎 = 𝟎 (dc) and 𝝎𝟎 is defined by : with 𝝉 being the time constant. 99 The magnitude response is given by : The phase response is given by : Figure E.6 sketches the magnitude and phase responses for an STC low-pass circuit. 100 101 Figure E.6 (a) Magnitude and (b) Phase response of STC circuits of the low-pass type. Straight-line asymptotes As shown, the magnitude curve is closely defined by two straight-line asymptotes. Low-frequency asymptote For 𝝎=𝟎 𝟐𝟎𝒍𝒐𝒈 𝑻 𝒋𝝎 Τ𝑲 = 𝟎 𝒅𝑩 is a horizontal straight line at 0 dB 102 High-frequency asymptote To find the slope, consider Equation of the magnitude : and let : resulting in : 103 104 It follows that if 𝝎 doubles in value, the magnitude is halved. On a logarithmic frequency axis, doublings of 𝝎 represent equally spaced points, with each interval called an octave. Halving the magnitude function corresponds to a 6 dB reduction in transmission 𝟐𝟎 𝐥𝐨𝐠 𝟎. 𝟓 = −𝟔 𝒅𝑩 . Thus the slope is –6 dB/octave. of the high-frequency asymptote This can be equivalently expressed as –20 dB/decade, where “decade” indicates an increase in frequency by a factor of 10. Break frequency The two straight-line asymptotes of the magnitude– response curve meet at the “corner frequency” or “break frequency” 𝝎𝟎 . The difference between the actual magnitude–response curve and the asymptotic response is largest at the corner frequency, where its value is 3 dB. To verify that this value is correct, simply substitute 𝝎 = 𝝎𝟎 in 𝑻(𝒋𝝎) to obtain : 𝑻(𝒋𝝎𝟎 ) = 𝑲Τ 𝟐 Thus at 𝝎 = 𝝎𝟎 , the gain drops by a factor of 𝟐 relative to the dc gain, which corresponds to a 3-dB reduction in gain. The corner frequency 𝝎𝟎 is appropriately referred to as the 3-dB frequency. 105 Figure E.6 (a) 106 Magnitude response of STC circuits of the low-pass type. Phase response Similar to the magnitude response, the phase– response curve, shown in Figure E.6(b), is closely defined by straight-line asymptotes. Note that at the corner frequency : 𝝎 = 𝝎𝟎 the phase is –45°, and for 𝝎 ≫ 𝝎𝟎 the phase approaches –90°. Also note that the –45°/decade straight line approximates the phase function, with a maximum error of 5.7°, over the frequency range 𝟎. 𝟏𝝎𝟎 0 to 𝟏𝟎𝝎𝟎 . 107 Figure E.6 (b) Phase response of STC circuits of the low-pass type. 108 High-Pass Circuits The transfer function, T(s) of an STC high-pass circuit can always be written in the form : which for physical frequencies 𝒔 = 𝒋𝝎 becomes : where K denotes the gain as 𝒔 ⟶ ∞ 𝒐𝒓 𝝎 → ∞ and 𝝎𝟎 is the inverse of the time constant 𝝉 : 109 The magnitude response : and the phase response : are sketched in Figure E.7. As in the low-pass case, the magnitude and phase curves are well defined by straight-line asymptotes. Because of the similarity (or, more appropriately, duality) with the low-pass case, no further explanation will be given. 110 111 Figure E.7 (a) Magnitude and (b) Phase response of STC circuits of the high-pass type. Table 1.2 provides a summary of the frequency-response results for STC networks of both types. 112 Example E.3 Figure E.8 shows a voltage amplifier having an input resistance 𝑹𝒊 , an input capacitance 𝑪𝒊 , a gain factor 𝝁, and an output resistance 𝑹𝒐 . The amplifier is fed with a voltage source 𝑽𝒔 having a source resistance 𝑹𝒔 , and a load of resistance 𝑹𝑳 is connected to the output. Figure E.8 Circuit for Example E.3. 113 (a) Derive an expression for the amplifier voltage gain 𝑽𝒐 Τ𝑽𝒔 as a function of frequency. From this find expressions for the dc gain and the 3-dB frequency. (b) Calculate the values of the dc gain, the 3-dB frequency, and the frequency at which the gain becomes 0 dB (i.e., unity) for the case Rs = 20 kΩ, Ri = 100 kΩ, Ci = 60 pF, 𝝁 = 𝟏𝟒𝟒 𝑽Τ𝑽 , Ro = 200 Ω, and RL = 1 kΩ. (c) Find 𝒗𝒐 (𝒕) for each of the following inputs : (c1) 𝑣𝑖 = 0.1 sin 102 𝑡 , V (c2) 𝑣𝑖 = 0.1 sin 105 𝑡 , V (c3) 𝑣𝑖 = 0.1 sin 106 𝑡 , V (c4) 𝑣𝑖 = 0.1 sin 108 𝑡 , V 114 Solution (a) Utilizing the voltage-divider rule, we can express 𝑽𝒊 in terms of 𝑽𝒔 as follows : where 𝒁𝒊 is the amplifier input impedance. It is easier to work in terms of 𝒀𝒊 = 𝟏Τ𝒁𝒊 thus obtaining Thus 115 This previous expression can be put in the standard form for a low-pass STC network (see the top line of Table 1.2) by extracting 𝟏 + 𝑹𝒔 Τ𝑹𝒊 from the denominator; thus we have : At the output side of the amplifier we can use the voltage-divider rule to write : These two equations can be combined to obtain the amplifier transfer function as: 116 From the expression of the amplifier transfer function, The time constant is given by : The transfer function is of the form 𝑲Τ 𝟏 + 𝒔Τ𝝎𝟎 , which corresponds to a low-pass STC network. The dc gain is found as: The 3-dB frequency, 𝝎𝟎 can be found from: 117 (b) Substituting the numerical values results in : 118 dc gain : 3-dB frequency : Since the gain falls off at the rate of –20 dB/decade, starting at 𝝎𝟎 the gain will reach 0 dB in two decades (a factor of 100); thus we haven : Unity-gain frequency (c) To find 𝒗𝒐 (𝒕) we need to determine the gain magnitude and phase at 𝟏𝟎𝟐 , 𝟏𝟎𝟓 , 𝟏𝟎𝟔 , and 𝟏𝟎𝟖 rad/s. This can be done either approximately utilizing the Bode plots or exactly utilizing the expression for the amplifier transfer function : 119 (c1) For 𝝎 = 𝟏𝟎𝟐 rad/s, which is 𝜔0 Τ104 , the Bode plots of Figure E.6 suggest that 𝑇 = 𝐾 = 100 and ∅ = 0° . The transfer function expression gives 𝑇 ≃ 100 and 𝜙 = −𝑡𝑎𝑛−1 10−4 ≃ 0°. Thus, (c2) For 𝝎 = 𝟏𝟎𝟓 rad/s, which is 𝜔0 Τ10 , the Bode plots of Figure E.6 suggest that 𝑇 = 𝐾 = 100 and ∅ = −5.7° . The transfer function expression gives 𝑇 ≃ 99.5 and 𝜙 = −𝑡𝑎𝑛−1 0.1 ≃ −5.7°. Thus, 120 𝟔 (c3) For or 𝝎 = 𝟏𝟎 rad/s = 𝝎𝟎 , 𝑇 = 100Τ 2 = 70.7 VΤV 37 dB and ∅ = −45°. Thus, (c4) For plots 𝝎 = 𝟏𝟎𝟖 suggest rad/s, which that 𝑇 =1 is 𝟏𝟎𝟎𝝎𝟎 , the Bode and ∅ = −90° . gives 𝑇 ≃ 1 and The transfer function expression 𝜙 = −𝑡𝑎𝑛−1 100 ≃ −89.4°. Thus, 121 Classification of Amplifiers Based on Frequency Response Amplifiers can be classified based on the shape of their magnitude-response curve. Figure E.9 shows typical frequency-response curves for various amplifier types. Figure E.9 Frequency response for (a) a capacitively coupled amplifier, (b) a direct-coupled amplifier, and (c) a tuned or bandpass amplifier. 122 In Figure E.9(a) the gain remains constant over a 123 wide frequency range, but falls off at low and high frequencies. This type of frequency response is common in audio amplifiers. Internal capacitances in the device (a transistor) cause the falloff of gain at high frequencies, just as Ci did in the circuit of Example 1.5. On the other hand, the falloff of gain at low frequencies is usually caused by coupling capacitors used to connect one amplifier stage to another, as indicated in Figure E.10. Figure E.10 Use of a capacitor to couple amplifier stages. 124 Figure E.9(b) shows the frequency response of a dc 125 amplifier. Such a frequency response characterizes what is referred to as a low-pass amplifier. IC amplifiers are usually designed as directly coupled or dc amplifiers (as opposed to capacitively coupled, or ac amplifiers). Figure E.9(c) shows the frequency response of Amplifiers called tuned amplifiers, bandpass amplifiers, or bandpass filters. The signal of a particular channel can be received while those of other channels are attenuated or filtered out.
