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# CHAPTER 01 SIGNALS and AMPLIFIERS

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```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 Th&eacute;venin 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.

microphone provides an acoustic signal that contains

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 Th&eacute;venin
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 Th&eacute;venin 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 Th&eacute;venin-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 Th&eacute;venin 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,
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
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 &plusmn;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
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
 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
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
Th&eacute;venin’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 Th&eacute;venin’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&deg;,
and for 𝝎 ≫ 𝝎𝟎 the phase approaches –90&deg;.
 Also
note that the –45&deg;/decade straight line
approximates the phase function, with a
maximum error of 5.7&deg;, 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&deg; .
The transfer function expression gives 𝑇 ≃ 100
and
𝜙 = −𝑡𝑎𝑛−1 10−4 ≃ 0&deg;. 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&deg; .
The transfer function expression gives 𝑇 ≃ 99.5 and
𝜙 = −𝑡𝑎𝑛−1 0.1 ≃ −5.7&deg;. Thus,
120
𝟔
(c3) For
or
𝝎 = 𝟏𝟎 rad/s = 𝝎𝟎 , 𝑇 = 100Τ 2 = 70.7 VΤV
37 dB and ∅ = −45&deg;. Thus,
(c4) For
plots
𝝎 = 𝟏𝟎𝟖
suggest
that
𝑇 =1
is
𝟏𝟎𝟎𝝎𝟎 , the Bode
and ∅ = −90&deg; .
gives 𝑇 ≃ 1
and
The transfer function expression
𝜙 = −𝑡𝑎𝑛−1 100 ≃ −89.4&deg;. 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.
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